SYLLABUS- ELECTRO MAGNETIC FIELD THEORY, Unit-I: Coordinate Systems and Transformation: Cartesian coordinates, circular

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*Table of contents : Electromagnetic Field TheoryDedicationPreface to the Revised EditionPreface to the First Edition 1Preface to the First Edition 2SyllabusBrief ContentsDetailed Contents 1Detailed Contents 2Detailed Contents 3Detailed Contents 4Detailed Contents 5Detailed Contents 6Detailed Contents 7Detailed Contents 8Detailed Contents 9Detailed Contents 10Detailed Contents 11Unit-1Unit-1: Fundamentals of Co-ordinate systems and Their Transformation with Vector CalculusUnit-2Section-A : ElectrostaticsSection-B : Electric Field in Material SpaceUnit-3Section-A : MagnetostaticsSection-B : Magnetic Forces, Materials and DevicesUnit-4Section-A : Waves and ApplicationsSection-B : Electromagnetic Wave PropagationUnit-5Unit-5: Transmission Lines*

Krishna's

Electromagnetic Field Theory

By Dr. S.K. Gupta Associate Professor & Head

Department of Physics D.N. (P.G.) College, Gulaothi C.C.S. University, Meerut (U.P.) and

Sankalp Gupta (B.Tech.)

KRISHNA Prakashan Media (P) Ltd. KRISHNA HOUSE, 11, Shivaji Road, Meerut-250 001 (U.P.), India

Jai Shri Radhey Shyam

Dedicated to

Lord

Krishna Author & Publishers

Preface to the Revised Edition

T

he wide range acceptibility, tremendous popularity, stunning success and over whelming appreciation gained by the previous editions of the book entitled ''Electromagnetic Field Theory'' provides the necessary inspiration to the authors for undertaking a complete revision of the book in accordance to the latest revised and existing syllabus of UPTU, Lucknow, UKTU, Dehradun and other technical Universities recently established in various states and cities. We feel immense pleasure in presenting thoroughly redesigned, revised new edition of Electromagnetic Field Theory for the students of B.Tech. Course of EC, EEC and TEC branches and for B.Sc. Students. The purpose of the book is to treat the fundamental principles in sufficient detail. The approach is fairly pragmatic throughout, aiming to provide a physical as well as mathematical understanding of the wide range of phenomena. In the light of meaningful comments and suggestions received from the teachers, colleagues, friends and students from various corners, some worthwhile additions, subtractions and modifications have been made in various articles of different sections of different units to provide extensive coverage of the field. The revised market leading text book maintained its classic strength, contemporary approach, flexible chapter construction, simplicity and lucidity. We trust this new edition will be helpful in creating great interest towards fundamental physics applicable in technologies to the engineers and keeping away any horror among students. We believe this new edition in present form would satisfy both average and brilliant user. We feel highly obliged to numerous authors and renowned scholors of the field whose invaluable work have been consulted in the course of preparations of the matter. We would like to extend our hearties gratitude to esteemed colleagues, professor's, friends and students who readily interacted with us for the enhancement of the ability and utility of the book. From the innermost core of the heart, we owe a special debt to Prof. D.S. Srivastava, Mangalayatan University, Aligarh, Dr. H.M. Agrawal, Professor and Head, Dept. of Physics, College of Basic Science and Humanities, G.B. Pant University of Agriculture & Technology, Pantnagar, Dr. M.K. Bansal, Head, Dept of physics, Bharat Institute of Technology, Meerut, Dr. S.K. Gupta, ABES College, Ghaziabad for giving their inspiration, constant encouragement and cooperation for successful complition of the book. We are grateful to Mr. S.K. Rastogi, (Managing Director), Mr. Sugam Rastogi (Executive Director and Chief Editor), Mrs. Kanupriya Rastogi (Director) Krishna Prakashan Media (P) Ltd., Meerut and their subordinate staff for publishing, composing and printing the present edition in the recorded short time. We would greatly welcome constructive criticism and suggestions from the reader for further improvement of the book. – Subodh Kumar Gupta – Er. Sankalp Gupta (v)

Preface to the First Edition

T

his book entitled “Electromagnetic Field Theory” has been written especially and strictly in accordance with the latest syllabus framed for one semester course (second year, third semester) by U.P. Technical University (UPTU), Lucknow for B.E./B.Tech./B. Arch. for

Engineering Students of all U.P. and Uttaranchal Engineering Colleges. In addition, it also covers the syllabus of our semester of Electrical Engineering Branch of all Engineering Colleges and B.Sc. (Honours and pass) course of all Universities in India and abroad. The contents of the book also covers the needs of those students who are appearing for various examinations such as Engineering Services (ES), GATE, AMIETE and other degree and Engineering Examinations. In the fast developing country, like India, for the quick and proper use of new revolutionized discoveries, tremendous and rapid advances and technological developments in the wireless technology, information technology, electrical engineering, communication engineering, electronics and instrumentation engineering, computer science, etc., and for their applications in many diverse areas especially in realistic human life the perfect and clear knowledge of electromagnetic field theory and their principles are of utmost importance. Keeping this fact in view this book is designed and styled in a very systematic and logical manner to give learners a sound base in fundamentals of electromagnetics and an introduction to its wide range of utility in engineering and technologies. Since undergraduates have insufficient acquaintance with the subject, the matter is presented in such a simple and lucid fashion that even students of ordinary calibre comprehend it easily without any strain and difficulty.The conversational style of language is intentionally accepted to make the reader feel that the teacher is always close to him and will be available for his assistance whenever needed. The present book gives a balance account of traditional topics and recent developments in the field of electromagnetics. A reasonably wide coverage in sufficient depth has been attempted. To clarify the concept and for making the matter interesting, enjoyable and pleasant a generous number of “in-text” solved examples of varying standards starting from elementary sort are incorporated.The entire subject matter has been divided in five units. Each unit begins with a clear statement of pertinent definitions, principles with illustrative, well supported by mathematical derivations and other descriptive material. Worked out numerical problems are arranged in a proper sequence starting from simple and fundamental to typical ones so that students may not encounter with any difficulty in solving unsolved numericals. The solved examples serve to illustrate and amplify the theory, bring into sharp focus those points without which the student continually feel himself on unsafe grounded and provide the repetition of basic principles so vital to effective teaching. In keeping with modern trends and to bring the confidence among readers multiple choice questions are arranged at the end of each unit. A set of theoretical question bank based on subject matter, which were asked in the examinations of various technical universities and engineering colleges have also been incorporated in the order in which the theory is developed. To assess the learning efficiency and for doing healthy (vi)

mental exercise a set of unsolved numerical problems is arranged in each unit. At the end of each unit a separate sheet of answers to the multiple choice questions and unsolved numerical problems is provided. To make the puzzled free mental exercise hints to some selected unsolved numerical problems have also been incorporated at the end of each unit. I hope that if a student goes through seriously with all these he or she would both appreciate and enjoy the subject. Author lays no claim to the original research in preparing the manuscript, but he has used his rich experiences gained in teaching the basic courses in electromagnetics in the presentation and the development of subject matter in the simplest possible manner keeping in view the stumbling blocks and the limitations of average students. The author wishes to acknowledge his indebtedness to numerous authors of those foreign standard books which were consulted in the course of preparation of the matter. The author takes the opportunity to express his deep sense of gratitude to Dr. B.K. Sharma, Reader in Chemistry, N.A.S. College, Meerut, Dr. D.K. Gupta, Retd. Reader and Head, Deptt. of Physics, S.V. College, Aligarh, Dr. H.M. Agrawal and Dr. B.R.K. Gupta, Professors (both) in Department of Physics, College of Basic Science and Humanities, G.B. Pant Univ. of Agri. and Tech., Pantnagar, Dr. Arun Kumar, Professor, Dept. of Electronics, M.M. Engineering College, Gorakhpur, Dr. D.K. Jain, Dr. O.D. Sharma, Deptt. of Physics, Meerut College, Meerut, Dr. Shalendra Agarwal, Deptt. of Physics, Saraswati College, Hathras, Dr. S.R. Verma, D.B.S. College, Dehradun, Dr. Bhagwat Swarup, Retd. PrincipaL, S.C. College, Bulandshaher, Dr. M.S. Tomar, Principal D.N. College, Gulaothi, colleagues, friends and well wishers whose constant encouragements and co-operation made me capable of completing this work. I would like to express my deep sense of appreciation and heart feet gratitude to my wife Dr. Madhu Bala, Reader, Dept. of Pol. Sc., N.A.S. College, Meerut, daughter Ms. Richa and son Mr. Sankalp whose dedication, encouragement and stimulation made this book possible and also for maintaining harmonious atmosphere, which is utmost necessity for being creative. The largest measure of my appreciation is reserved for dynamic publisher Mr. Satyendra Rastogi, Managing Director, Krishna Prakashan Media (P) Ltd., Meerut and Er. Anish Jain, Director, DEBUG CC (Computer Concern), Ghaziabad for publishing, composing, designing and printing the book excellently in such a fine format and for bringing out the book in a short time. In the last but not least all members of Krishna Prakashan Media (P) Ltd., and DEBUG (Computer Concern), Ghaziabad deserve all praise and congratulations for contributing directly or indirectly in this project. Errors might have crept in here and there in despite utmost care to avoid them. Author shall be very grateful for bringing them to his notice. The author will welcome helpful suggestions and comments for the improvement of the book.

August 2004

— Subodh Kumar Gupta

(vii)

S

Electromagnetic Field Theory

yllabus

UPTU, Lucknow B.Tech.-IV Sem. NEC-404

Unit-I: Coordinate Systems and Transformation: Cartesian coordinates, circular cylindrical coordinates, spherical coordinates. Vector calculus: Differential length, area and volume, line surface and volume integrals, del operator, gradient of a scalar, divergence of a vector and divergence theorem, curl of a vector and Stoke’s theorem, Laplacian of a scalar. Unit-II: Electrostatics: Electrostatic fields, Coulombs law and field intensity, Electric field due to charge distribution, Electric flux density, Gausses’s Law— Maxwell’s equation, Electric dipole and flux lines, energy density in electrostatic fields. Electric Field in Material Space: Properties of materials, convection and conduction currents, conductors, polarization in dielectrics, dielectric constants, continuity equation and relaxation time, boundary condition. Electrostatic boundary value problems: Poission’s and Laplace’s equations, general procedures for soling Poission’s or Laplace’s equations, resistance and capacitance, method of images. Unit-III: Magnetostatics: Magneto-static fields, Biot-Savart’s Law, Ampere’s circuit law, Maxwell’s equation, application of ampere’s law, magnetic flux density- Maxwell’s equation, Maxwell’s equation for static fields, magnetic scalar and vector potential. Magnetic Forces, Materials and Devices: Forces due to magnetic field, magnetic torque and moment, a magnetic dipole, magnetization in materials, magnetic boundary conditions, inductors and inductances, magnetic energy. Unit-IV: Waves and Applications: Maxwell’s equation, Faraday’s Law, transformer and motional electromotive forces, displacement current, Maxwell’s equation in final form. Electromagnetic Wave Propagation: Wave propagation in lossy dielectrics, plane waves in lossless dielectrics, plane wave in free space, plain waves in good conductors, power and the pointing vector, reflection of a plain wave in a normal incidence. Unit-V: Transmission lines: Transmission line parameters, Transmission line equations, input impedance, standing wave ratio and power, The Smith chart, Some applications of transmission lines. (viii)

B

rief

C ontents

Dedication......................................................................................................(iii) Preface to the Revised Edition ....................................................................(v) Preface to the First Edition.....................................................................(vi-vii) Syllabus......................................................................................................(viii) Brief Contents ..........................................................................................(ix) DetailedFundamentals Contents.....................................................................................(x-xx) Unit-1: of Co-ordinate systems and Their Transformation with

vector calculus ................................................................................................(01-92) Unit-2:

Electrostatics and Electric Field in Material Space ......................(93-288) Section-A : Electrostatics..................................................................................(95-194) Section-B : Magnetic Forces, Materials and Devices ...................................(195-288)

Unit-3:

Magnetostatistic & Magnetic Forces, Materials & Devices........(289-414) Section-A : Magnetostatics ...........................................................................(291-352) Section-B : Magnetic Forces, Materials and Devices ...................................(353-414)

Unit-4: Waves and Applications & Electromagnetic Wave Propagation .........................................................................................................................(415-564) Section-A : Waves and Applications ............................................................(417-456) Section-B : Electromagnetic Wave Propagation ..........................................(457-564)

Unit-5:

Transmission Lines .................................................................................(565-656)

(ix)

D

etailed

C ontents

Unit-1: Fundamentals of Co-ordinate systems and Their Transformation with vector calculus........................(01-92) Introduction

03

Fundamentals of Vector Analysis

04

Resolution of a Vector

04

Resolution of vector along three Directions: Direction Cosines

> Solved Examples

06

Product of Two Vectors (Dot and Cross Product)

09

Scalar Product in Terms of Components

09

Geometrical Interpretation

11

> Solved Examples

11

The Vector Product in Terms of Components Triple Product

05

12

15

Scalar Triple Product in Terms of the Components Geometrical Interpretation

15

16

Vector Triple Product in Terms of Components

> Solved Examples

17

17

Co-ordinate Systems and Transformations

19

Cartesian or Rectangular Co-ordinate System

19

Circular Cylindrical Co-ordinate Systems

20

Transformation of Cylindrical Co-ordinates into Rectangular Co-ordinates Spherical Coordinates Systems

21

23

Transformation of Spherical Coordinates (or Polar Coordinates) into Cartesian Coordinates (or Rectangular Coordinates) 24 > Solved Examples 29 Differential Length, Area and Volume

44

Line, Surface and Volume Integrals

47 (x)

Line Integral

47

Surface Integral

48

Importance of Surface Integral Volume Integral

49

> Solved Examples Del Operator

49

49 53

Gradient of a Scalar

54

Physical Interpretation of Gradient

54

Properties of Gradient of Scalar Field

> Solved Examples

57

57

Definition and Physical Interpretation of Divergence of a Vector Divergence of a Vector

61

62

Properties of Divergence of a Vector Field

> Solved Examples

64 64

Divergence Theorem or Gauss-Divergence Theorem Alternative Proof

67

68

> Solved Examples

69

Curl of a Vector

72

Properties of Curl

76

> Solved Examples

76

Stoke's Theorem

83

> Solved Examples

84

Laplacian of Scalar

87

Exercises $ $ $

Question Bank 89 Unsolved Numerical Problems 90 Answers to Unsolved Numerical Problems

92

Unit-2: Electrostatics and Electric Field in Material Space ............................................................................(93-288) Section-A : Electrostatics...................................................................(95-194) Introduction

95

Coulomb's Law of Force

96

Importance and Limitations of Coulomb's Law

> Solved Examples Electric Field

99

99 104 (xi)

Electric Field Intensity

105

Intensity of Electric Field Due to a Point Charge

106

Electric Field Intensity Due to A System of Charges

106

Electric Fields Due to Charge Distributions

107

Electric Field Due to Charge Distribution

108

Electric Field Intensity Due to a Line Charge (Infinitely Long Charged Wire) Intensity of Electric Field Due to a Finite Line of Charge

109 111

Electric Field Intensity Due to Infinite Sheet of Charge

113

Electric Field Intensity at an Axial Point of a Charged Ring

115

Electric Field Intensity at an Axial Point of a Solid Circular Charged Disc

116

Electric Field Intensity Due to a Uniformly Charged Annular Disc Electric Field Intensity Due to a volume Charge

> Solved Examples Electric Flux Density

120 126

Electric Flux Density Due to Charge Distributions Gauss's Law-Maxwell's Equation Proof of the Law

117

118

128

132 133

Applications of Gauss's Law: Some Symmetrical Charged Distributions

134

Electric Field Intensity Due to a Uniformly Charged Spherical Shell

135

Electric Field Intensity Due to a Uniformly Charged Sphere Hollow Spherical Conductor

136

138

Electric Field Intensity Due to a Co-axial Cable

138

Electric Field Intensity Due to Infinite Line Charge

140

Electric Field Intensity Due to Infinite Plane Sheet of Charge

> Solved Examples

142

Electrostatic Potential

148

141

Relation Between Electric Potential Difference and Electric Field Electric Potential Due to a Point Charge Electric Field as Negative Gradient of Potential Equipotential Surface

149

150 151

152

Potential Due to Charge Distributions

152

Potential Difference Due to Infinitely Long Line of Charge

153

Electric Potential Due to Surface Charge Distribution

153

Electric Potential Due to Volume Charge Distribution

154

Electric Potential Due to a Uniformly Charged Sphere Potential Due to a Solid Sphere

154

156

Electric Potential on the Axis of a Charged Ring

158

Electric Potential Between Two Conducting Concentric Spherical Shells Electric Potential Due to Two Concentric Spheres

158

160

Electric Potential at a Point Between Two Coaxial Cylinders (xii)

161

Electric Potential Due to Uniformly Charged Disc

> Solved Examples

161

164

An Electric Dipole and Flux Lines

173

Potential Due to An Electric Dipole

173

Flux Lines or Field Lines

176

> Solved Examples

177

Energy Density in Electrostatic Field

179

Expression for Energy Density or Electrostatic Energy Energy Stored in Concentric Spherical Shell

179 182

Electrostatic Energy Stored in a Charged Capacitor Energy Density in a Capacitor

> Solved Examples

182

183 183

Exercises $ Question Bank 190 $ Unsolved Numerical Problems $ Answers to Unsolved Numerical Problems

192 194

Section-B : Electric Field in Material Space.......................................(195-288) Introduction

195

Properties of Materials

196

Convection and Conduction Currents Conduction Current

197

197

Convection Current

198

Current Density

198

Conduction and Convection Current Densities Conduction Current Density

199

199

Convection Current Density

200

Drift Velocity of Charge Carriers in a Conductor

201

Relation between Drift Velocity of Electrons and Current Density Continuity Equation

> Solved Examples Conductors

202 203

207

Conductor in an Electric Field

208

Ohm's Law in Point Form and Resistance of a Conductor Resistance and Power Resistance

209 209

Power and Joule's Law

> Solved Examples

210 211 (xiii)

208

201

Boundary Conditions in Static Electric Field (or at Conductor Surface in Free Space)

> Solved Examples

212

214

Dielectrics

216

Dielectric An Atomic View

217

Polar and Non-Polar Dielectrics

217

Non-Polar Dielectric in An Electric Field

218

Polar Dielectric in An Electric Field

218

Polarization in Dielectrics

219

Field due to a Polarized Dielectric

221

Laws of Electrostatic Field in Presence of Dielectrics

223

Dielectric Constant (Relative Permittivity) of a Medium

225

Homogeneous, Linear and Isotropic Media Electrical Susceptibility

226 226

Relation between Electric Susceptibility and Dielectric Constant Dielectric Strength

227

227

Gauss's Law in Dielectrics

227

Three Electric Vectors

229 →

Electric Displacement D

230 →

Relation between Dielectric Polarization P and Dielectric constant K:

> Solved Examples

231

231

Electrostatic Boundary Conditions at Dielectric-Dielectric Surface

233

Refraction of Electrostatic Lines of Force at the Boundary between two Dielectric Media

> Solved Examples

236

Poisson's and Laplace's Equations

238

Solutions of Laplace’s and Poisson’s Equations in One Dimension

239

Electric field and Potential Distribution near a Semiconductor Junction: Using Poisson’s Equation Uniqueness Theorem for Electrostatics

> Solved Examples

244

246

Capacitance and Capacitor Capacitor

257

257

Capacitance of a Parallel Plate Capacitor with Compound Dielectric Special Cases

259

260

Capacitance of a Spherical Capacitor with Inner Sphere Earthed Capacitance between Two Parallel Wires Combinations of Capacitors

> Solved Examples

235

262

264 265

266

Method of Electrical Images Applied to Plane Boundaries (xiv)

274

241

Electric Potential V and Field Strength E

276

Electric Field on the Conductor

277

Surface Charge Density of Induced Charge on the Conducting Plane

277

Force of Attraction between the Point Charge (+Q) and the Earthed Conducting Plane

> Solved Examples

277

280

Exercises $ Question Bank 284 $ Unsolved Numerical Problems 286 $ Answers to Unsolved Numerical Problems

288

Unit-3: M a g n e t o -s t a t i c & M a g n e t i c F or c e s , M a t e r i a l s & Devices...............................................................(289-414) Section-A : Magnetostatic ...............................................................(291-352) Introduction

291

Biot-Savart Law

292

Applications of Biot-Savart's Law

294

Magnetic Field Intensity Due to Infinite Long Straight Conductor Alternate Method Magnetic Field Intensity Due to a Straight Conductor of Finite Length Magnetic Field Intensity on the Axis of a Current Carrying Circular Loop

> Solved Examples

311

Applications of Ampere's Law

312

Divergence of Magnetic Induction B

320

321

Magnetic Flux Density-Maxwell's Equation

327

Maxwell's Equation for Static Fields

> Solved Examples

298

300

Ampere's Circuital Law

> Solved Examples

295 296

329

329

Magnetic Scalar and Vector Potentials

330

Magnetic Scalar Potential (Vm )

331

Magnetic Scalar Potential For a Current Loop

332

Application of a Magnetic Scalar Potential: Equivalence of a Small Current Loop and a Magnetic Dipole 334 Magnetic Scalar Potential in the Region Between the Inner and Outer Conductors of a Coaxial Line 335 Magnetic Vector Potential

335

Derivation of Magnetic Vector Potential

(xv)

337

M agnetic Potential and Field Due to a Long Straight Current Carrying Wire B iot-Savart Law From Magnetic Vector Potential

341

Ampere Circuital Law from Magnetic Vector Potential

> Solved Examples

339

342

342

Exercises $ Question Bank 349 $ Unsolved Numerical Problems $ Answers to Unsolved Numerical Problems

350 352

Section-B : Magnetic Forces, Materials and Devices .........................(353-414) Introduction

353

Forces Due to Magnetic Field Lorentz Force

354

355

Force on a Differential Current Element

355

Force on a Current Carrying Conductor

356

Ampere's Law of (Magnetic) Force

357

Magnetic Interaction of Two Current Elements

358

Magnetic Force Between Two Long Parallel Wires

> Solved Examples

361

361

Magnetic Torque and Moment

368

Torque on a Differential Current Loop

369

Torque on a Current Carrying Loop Magnetic Dipole

371

372

> Solved Examples

374

Magnetisation in Materials

375

Magnetic Induction or Magnetic Flux Density (B)

377

→

Magnetic Field Intensity ( H)

377

Magnetic Susceptibility (χ m )

378

Magnetic Permeability (µ)

378

Relation Between Relative Permeability (µ r ) and Magnetic Susceptibility (χ m )

> Solved Examples

379

Classification of Magnetic Materials

383

Magnetic Boundary Conditions

386

> Solved Examples

378

389

Inductors and Inductance

391

Inductance or Self Inductance of a Long (or Infinitely Long) Solenoid Inductance (or Self Inductance) of a Toroid

393

394

Inductance (or Self Inductance) of Two Parallel Wires (xvi)

395

Inductance (or Self Inductance) Per Unit Length of a Co-axial Cable

396

Mutual Inductance Between two Coaxial Solenoids (or Coils)

396

Inductances in Series

397

Inductances in Parallel

398

Co-efficient of Coupling

399

> Solved Examples

399

Energy Stored in a Magnetic Field

403

Energy Density in a Magnetic Field

404

Energy Density in Terms of Magnetic Vector Potential

405 →

Inductance of a Coil (or Circuit) in Terms of Magnetic Vector Potential A

> Solved Examples

407

409

Exercises $ Question Bank 412 $ Unsolved Numerical Problems $ Answers to Unsolved Numerical Problems

413 414

Unit-4: Waves and Applications & Electromagnetic Wave Propagation.........................................................(415-564) Section-A: Waves and Applications .................................................(417-456) Introduction

417

Faraday's Law Second Law

418 419

Deduction of Faraday's Laws

419

Integral and Differential Forms (or Vector Forms) of Faraday's Law Transformer and Motional Electromotive Forces

422

Motional E.M.F. from Lorentz Force

> Solved Examples

421

425

426

Displacement Current

431

Modified Ampere's Law and Displacement Current

432

Modified Ampere's Law and Displacement Current Density Displacement Current from Displacement Current Density Maxwell's Equation Physical Significance

434 435

436 437

Derivations of Maxwell's Equations Maxwell's Equations in Integral Form

440 445

Maxwell's Equations for Harmonically Time Varying Fields (or Sinusoidally Varying Fields)

> Solved Examples

448 (xvii)

446

Isotropic and Anisotropic Medium

453

Exercises $ Question Bank 454 $ Unsolved Numerical Problems $ Answers to Unsolved Numerical Problems

455 456

Section-B: Electromagnetic Wave Propagation .................................(457-564) Introduction

457

Electromagnetic Wave Equations in Different Media & Their Solutions Electromagnetic Wave Equations in Free Space

458

461

Propagation of Uniform Plane Electromagnetic Waves in Vacuum

463

Plane Wave Equation and its Solution in Isotropic Dielectric (or Perfect Dielectric or Lossless) Medium

468

Plane Electromagnetic Wave and its Solution in Conducting Media (Lossy dielectric or Partially Conducting Media)

471

Differential & Integral forms of Maxwell's Equations in Different Media (In compiled form)

> Solved Examples Phasor Notation

477 482

Phase Velocity and Group Velocity Phase Velocity

484

485

Relative Phase Velocity

486

Group Velocity

487

Relation between Vp and Vg

490

> Solved Examples

491

Conductors and Dielectrics

492

Wave Propagation in Good Dielectrics (that is, Dielectric Having low or Partially Conducting Media or Lossy Dielectric)

495

Intrinsic or Characteristic Impedance of Good Dielectric or Lossy Dielectric: Wave Propagation in a Good Conductors or Conducting Media Characteristic Impedance of Conducting Medium Depth of Penetration-Skin Depth

> Solved Examples

501 507

509

Parameters for Various Media Polarisation

511

511

Superposition of Two Linearly Polarized Electromagnetic Waves Special Cases

498 500

512

514

Poynting Vector and Poynting Theorem

517

Poynting Vector and Flow of Power with an Electromagnetic Wave Average Power in Lossy Dielectric:

520

521

Poynting Vector for an Elliptically Polarised Electromagnetic Wave (xviii)

522

475

> Solved Examples

524

Reflection and Refraction of Plane Electromagnetic Waves at Plane Boundaries Properties of Reflection and Transmission Coefficients

530

534

Reflection of Uniform Plane Electromagnetic Wave by Perfect Conductor for Normal Incidence

> Solved Examples

537

Standing Wave Ratio

542

535

Reflection of Uniform Plane Electromagnetic Wave by a Perfect Dielectric (or Insulator) for Oblique Incidence

546

→

Case I: Electric Vector E perpendicular to the Plane of Incidence (Perpendicular Polarisation) →

Case II : E Parallel to the Plane of Incidence (Parallel Polarisation) Brewster Angle and Total Internal Reflection Total Internal Reflection

550

551

553

555

> Solved Examples

556

Exercises $ Question Bank 559 $ Unsolved Numerical Problems $ Answers to Unsolved Numerical Problems

562 564

Unit-5: Transmission Lines ............................................(565-656) Introduction

567

Transmission Lines

568

Types of Transmission Lines

569

Advantages and Disadvantages of two Wire Line and Co-axial Cables: Transmission Lines Equations Voltage Equation

570

571

571

Current Equation

572

Harmonic time Dependent Voltage and Current Equations Solution of Transmission Line Equation

572

573

Characteristic or Surge Impedance

574

Low Loss Transmission Lines

576

Low Frequency Transmission Lines

577

High Frequency Transmission Lines Infinite Transmission Line

578 579

Secondary Line Constants or Coefficients of Transmission Line in Terms of Primary Constants

> Solved Examples

582

Characteristic Impedance or Surge Impedance

587

Transmission Line Terminated in a Load Impedance Z L Input Impedance of a Loss-Less Transmission Line

588 589

(xix)

580

Short-Circuited and Open Circuited Loss-Less Lines

592

Standing Wave and Reflection Losses in Transmission Line Reflection Coefficients

596

597 600

Relation between input Impedance Zs and Reflection Coefficient Γ Standing Wave Ratio (SWR)

601

Relation between Standing Wave Ratio (SWR) S and Reflection Coefficient Γ

601

Relations for Reflection and Transmission Coefficients and Voltage Standing Wave Ratio of a Line

604

Power Transmitted along a Low-Loss Line

> Solved Examples

604

Impedance Matching

619

604

Impedance Matching with Quarter Wave Transformer Impedance Matching by Stubs

619

620

Disadvantage of Using Single Stub for Impedance Matching

625

The Position (ls ) and Length (lt ) of the Stub in Terms of Reflection Coefficient Merits of a Short Circuited Stub Over an Open Circuited Stub Impedance Matching by Double Stubs 633

Basis of Smith Chart

> Solved Examples

638 639

Losses and Distortions in Transmission Lines Copper Losses Dielectric Losses

644 644

Various Distortions in Lines

644

Attenuation Loss

644 645

Transmission Loss

645

Return Loss Insertion Loss

645 646

Application of Transmission Lines Waveguides

643

643

Radiation and Induction Losses

Reflection loss

630

630

Applications of Smith Chart

646

647

Waveguide Modes

649

Cut-off wavelength or Frequency

Exercises $ Question Bank 651 $ Unsolved Numerical Problems $ Answers to Unsolved Numerical Problems

650

653 654

(xx)

625

1 Fundamentals of Coordinate Systems and Their Transformation with Vector Calculus

FundamentalsTof Coordinate N Systems and Their Transformation with Vector C Solved Examples heory umerical Qalculus uestionnaire

3

U nit

1 F undamentals of C o-ordinate S ystems and T heir T ransformation With V ector C alculus

I

ntroduction

h e seeds for the growth of an entirely new crop of physics-electromagnetics were sowed with the discovery of interdependency of electricity and magnetism. Three dimensional understanding of electricity and magnetism, that is, electromagnetics is very important in real world. Many components and materials, such as inductors, electric motors, generators, microwave appliances, radio, television, computers, microprocessors, transmission lines, radars, bioelectromagnetic instruments, medical electronics, Satellite communication, automobiles, etc, as used in Electrical Engineering, Electronics and Communication Engineering, Electronics and Instrumentation Engineering, Microwave Engineering, and Radar Engineering, require a knowledge of electromagnetic fields. Many electromagnetic devices are inadvertently coupled to other systems. Video-display unit of computers and television sets may emit sufficient radiation that it can be picked up, decoded and the screen displays, reproduces at distances of a kilometer. Any device that radiates is coupled to the entire universe. To understand and enjoy the real world the study of electromagnetic field theory is essential.

T

Vector plays a very important role in the study of electromagnetic field theory and to understand the physical laws that mathematics describes. This is so because most of the fundamental and basic quantities involved in the study of electromagnetic theory are inherently vector in nature. Hence, the subject can be logically developed only by adopting vectors. This enables us to have a better understanding of the physical concept and to put the various mathematical formulae in compact form. In fact the basic concepts and principles of electromagnetic are based on electric and magnetic fields culminating in Maxwell’s equations. The laws of electricity and magnetism have been expressed in terms of electric and

E lectromagnetic F ield Theory

4

magnetic fields. A time varying electric field is accompanied by a magnetic field and vice-versa. Time varying electric and magnetic fields are coupled, producing electromagnetic fields. Like any other law, the laws of electricity and magnetism are most compactly expressed as differential equations. Since, electric →

→

field E and magnetic field B are vectors, the differential equations involve vector derivatives: divergence and curl. Hence, for better understanding of electromagnetic field theory the knowledge of vector analysis is very essential. In fact vector analysis is a concise language which greatly facilitates the analysis of electric and magnetic fields. In this section of the text the vector analysis is developed in a very concise manner. The brief introduction to vector analysis included here is for the benefit of those readers which are not already familiar with this tool.

F undamentals of Vector Analysis All physical quantities are divided into two main classes: Scalars and Vectors.

Scalars and Vectors The quantities which can be completely defined by a number and a unit, or which have magnitude only, and do not involve any direction are known as scalars or scalar quantities. The examples of scalar quantities are: mass, length, time, volume, speed, pressure, energy, work, temperature, charge, current, potential, frequency, specific heat etc. The quantities which have both magnitude and directions and combine according to certain rules of addition, are called vectors or vector quantities. The examples of vectors are : displacement, velocity, force, weight, stress, momentum, electric and magnetic field intensities, magnetic moment, temperature-gradient, current density, magnetic induction, impulse, gravitational intensities, etc. It should be remembered that, a physical quantity, which has both magnitude and direction, but does not add up according to the rules of vector addition will not be a vector quantity. For example, the current in a wire. Current has both magnitude and direction, but does not obey the rules of vector addition and subtraction. Hence, current is not a vector quantity. Another example of vector quantity is velocity. It has both magnitude and direction. It follows the laws of vector addition so long its magnitude is small compared with the speed of light. At enormous velocity it deviates from the laws of vector addition and is no longer a vector quantity, although it has both magnitude and direction.

R esolution of a Vector The splitting of a vector into its vector components is called resolution of a vector. The vectors into which a given vector splitted are called components of a vector or vector components. →

Let us consider that a given vector A, which makes an angle θ with the horizontal (Fig. 1) is to be resolved into two rectangular components. Taking the tail of →

Y

A Ay

vector A as origin, perpendicular axes OX and OY are drawn. From the head of →

vector A two perpendiculars are dropped on OX and OY , which intersect OX and

R

Q

θ O

Ax

Fig. 1

P

X

Fundamentals of Coordinate Systems and Their Transformation with Vector Calculus →

5

→

OY at points P and Q respectively. The vectors A x and A y drawn from O to P and O to Q are the →

respective rectangular components of vector A such that →

→

→

A = Ax + A y

In triangle ORP, we have

OP A x or A x = A cos θ = OR A PR OQ A y or A y = A sin θ = = sin θ = OR OR A

cos θ = and

→ →

...(1) ...(2)

→

where A, A x and A y are the magnitudes of vectors A, A x and A y respectively. From eqns. (1) and (2), we have

A x2 + A y2 = A2 cos2 θ + A2 sin2 θ = A2 ∴ A =

A x2 + A y2 and tan θ =

Ay Ax

→

Hence, we can determine both the magnitude of A in terms of its rectangular components and direction θ. Now, let aɵ x and aɵ y be the unit vectors along the X- and Y-axes, then →

→

A x = A x aɵ x and A y = A y aɵ y ∴

→

A = A x aɵ x + A y aɵ y

→

...(3) →

→

We can use the analytical method for vector addition. Let C be the sum of vectors A and B in X – Y plane, that is, →

→

→

C = A+ B

...(4)

Let us write these vectors in terms of their corresponding vector components, as

Cy

→

→

→

B

By

C

Ay A

A = A x aɵ x + A y aɵ y , B = B x aɵ x + B y aɵ y and C = C x aɵ x + C y aɵ y

From eqn. (4), we have C x aɵ x + C y aɵ y = ( A x + B x ) aɵ x + ( A y + B y ) aɵ y or C x = A x + B x and C y = A y + B y

Ax

Bx Cx

Thus, we see that the component of the resultant vector along any axis is the algebraic sum of the individual components in that direction. It is also evident from Fig 2.

Fig. 2

R esolution of vector along three Directions: Direction Cosines Y

The resolution of vector along three perpendicular directions is →

shown in Fig.3. Let OA = A be the given vector such that its one of the extremity lies at origin O of the coordinate axes X , Y and →

Ay

Z . Let the length of projections of A along X , Y and Z -axes be → A y,

Then,

and

→ Az ,

A=

A

Z

respectively. →

A

γ ax

A x , A y and Az and their corresponding component vectors are → Ax ,

ay az

Ax

→ Ax

+

→ Ay

+

→ Az

Az

Fig. 3

X

E lectromagnetic F ield Theory

6

If aɵ x , aɵ y and aɵz are the unit vectors along, X , Y and Z −axes then, →

→

→

→

→

A x = A x aɵ x , A y = A y aɵ y and Az = Az aɵz ∴ A = A x aɵ x + A y aɵ y + Az aɵz →

From the geometry of figure 3, the magnitude of A may be represented as

A2 = A2x + A2y + A2z or A = ( A2x + A2y + A2z )1 / 2 →

If α,β and γ are the angles which A makes with the directions of aɵ x , aɵ y and aɵz respectively, then

A x = A cos α , A y = A cos β and Az = A cos γ so that

cos α =

Ay Ay Ax Ax = , cos β = = 2 2 2 2 A A ( A x + A y + Az ) ( A x + A2y + A2z )

and

cos γ =

Az Az = 2 A ( A x + A2y + Az2 )

cos α, cos β and cos γ are known as ''direction cosines'' and have the property that

cos2α + cos2β + cos2 γ = 1 →

→

→

Example 1: Show that the vectors A = 3 aɵ x – 2 aɵ y + aɵ z , B = aɵ x – 3 aɵ y + 5 aɵ z and C = 2 aɵ x + aɵ y − 4 aɵ z form a right angled triangle. →

→

→

Solution: Here A = 3 aɵ x − 2 aɵ y + aɵz , B = aɵ x − 3 aɵ y + 5 aɵz and C = 2 aɵ x + aɵ y − 4 aɵz . The three vectors will form → → →

a triangle if one of the vectors A, B, C is the vector sum of the remaining two vectors. That is, →

→

→

B + C = (aɵ x – 3 aɵ y + 5 aɵz ) + (2 aɵ x + aɵ y – 4 aɵz ) = 3 aɵ x – 2 aɵ y + aɵz = A

→ →

→

Hence, vectors A , B and C form a right angled triangle.

Example 2: Deduce the angle made by the vector 4 aɵ x – 3 aɵ y + 5 aɵ z with Z-axis. →

Solution: The magnitude of a vector A = A x aɵ x + A y aɵ y + Az aɵz is A = ( A2x + A2y + A2z ) In the given problem, A x = 4, A y = – 3 and Az = 5 ∴ A = [(4)2 + (–3)2 + (5)2 ] = 5 2

5 A The angle made by the vector 4 aɵ x – 3 aɵ y + 5 aɵz with Z -axis is γ = cos –1 z = cos –1 = 45° A 5 2 →

→

Example 3: Given A = (2 x + 3 y ) aɵ x – (2 y + 3 z ) aɵ y + (3 x – y )aɵ z . Determine the unit vector parallel to A at point P (1,–1, 2 ).

[UPTU, B.Tech. IV Sem. (old) 2007]

Fundamentals of Coordinate Systems and Their Transformation with Vector Calculus →

→

7 →

Solution: Suppose B = B x aɵ x + B y aɵ y + Bz aɵz is a unit vector parallel to A. The condition for vectors A and →

B to be parallel is Ax B x A y B y = , = A B A B

A2x + A2y + Az2 =

A=

∴

and

Az Bz = A B

(2 x + 3 y)2 + (2 y + 3 z)2 + (3 x – y)2

→

Since B is a unit vector, B = 1 In the given vector, A x = (2 x + 3 y), A y = –(2 y + 3 z) and Az = (3 x – y) 2x + 3 y Ax B = = x ∴ 2 2 2 A 1 [(2 x + 3 y) + (2 y + 3 z) + (3 x – y) ] or

Bx =

Similarly

By =

and

Bz =

2x + 3 y 2

(2 x + 3 y) + (2 y + 3 z)2 + (3 x – y)2 ] – (2 y + 3 z) [(2 x + 3 y)2 + (2 y + 3 z)2 + (3 x – y)2 ] (3 x – y) 2

[(2 x + 3 y) + (2 y + 3 z)2 + (3 x – y)2 ]

at point (1,–1, 2), B x , By and Bz are

Bx =

2 × 1 + 3 × (–1) 2

[(2 × 1 + 3 × (–1)] + [2 × (–1) + 3 × 2]2 + [3 × 1 – 1 × (–1)]2 2 –3

=

2

[(2 – 3) + (–2 + 6)2 + (3 + 1)2 ] –1

or

Bx =

Similarly,

By =

or

By =–

and

Bz =

2

2

2

[1 + 4 + 4 ]

=–

–[2 × (–1) + 3 × 2] 2

2

2

[1 + 4 + 4 ]

=–

1 33 –2 + 6 33

4 33

[3 × 1 – 1 × (–1)] 2

2

2

[1 + 4 + 4 ]

=

3 +1 4 = 33 33

→

Hence unit vector parallel to A at point P (1,–1, 2) is →

B = B x aɵ x + B y aɵ y + Bz aɵz

or

or

→

B=– →

B=

4 aɵ y 4 aɵz aɵ x – + 33 33 33

– aɵ x – 4 aɵ y + 4 aɵ z 33

E lectromagnetic F ield Theory

8

Example 4: What vector should be added to the sum of 2 aɵ x + 3 aɵ y + 4 aɵ z and aɵ x – 2 aɵ y – 2 aɵ z such that the resultant vector is a unit vector along Z-axis? →

Solution: The given vectors are, 2 aɵ x + 3 aɵ y + 4 aɵz and aɵ x – 2 aɵ y + 4 aɵz . Let the vector to be added is A so →

that the sum of A with two given vectors must be equal to a unit vector along Z -axis, that is aɵz . →

→

A + (2 aɵ x + 3 aɵ y + 4 aɵz ) + (aɵ x – 2 aɵ y – 2 aɵz ) = aɵz ∴ A = –3 aɵ x – aɵ y – aɵ z

Thus,

Example 5: A unit vector is 0 .4 aɵ x + 0 .8 aɵ y + caɵ z . Calculate the value of c and also the cosine of the angle which this vector makes with X-axis. →

→

→

→

Solution: Let A = 0.4 aɵ x + 0.8 aɵ y + caɵz , where A is the unit vector, then the magnitude of A, i.e., | A| = 1. →

| A| = 1 = [(0.4)2 + (0.8)2 + (c)2 ] = ( A2x + A2y + Az2 )

or

(0.4)2 + (0.8)2 + c2 = 1 or c = (0 . 2 )

or

→

Let α be the angle made by A with X-axis, then by the definition of direction cosine

A (0.4) or α = cos –1(0.4) = 66. 4 ° cos α = →x = 1 | A| →

→

→

→

Example 6: Given A = 5 aɵ x – 2 aɵ y + aɵ z , find the expression of a unit vector B such that B ||A. →

[GBTU, B.Tech. III Sem. 2010]

→

Solution: For A and B to be parallel their components must make same angles with the coordinate axes →

→

or the direction cosines of A must be equal to the respective direction cosines of B . That is, Ax B x A y B y A B = , = and z = z A B A B A B →

→

→

Suppose the unit vector B is , B = B x aɵ x + B y aɵ y + Bz aɵz for a unit vector | B | = B = B2x + B2y + B2z = 1 →

→

Given vector A = 5 aɵ x – 2 aɵ y + aɵz ∴ | A| = A = (5)2 + (–2)2 + (1)2 = 30 and A x = 5, A y = –2 and Az = 1

5 B = x, 1 30

Hence,

→

B=

∴

By –2 = and 1 30

5 aɵ x – 30

2 aɵ y + 30

–2 5 1 B 1 , By = and Bz = = z or B x = 1 30 30 30 30 1 aɵ z 30

→

→

Example 7: Find a unit vector along r = aɵ x + 3 aɵ y + 4 aɵ z . Also calculate the direction cosines of vector r . →

Solution: Here r = aɵ x + 3 aɵ y + 4 aɵz . →

Let nɵ be the unit vector along r , then nɵ =

→

r

→

| r|

→

→

, where | r | is the modules of r

Fundamentals of Coordinate Systems and Their Transformation with Vector Calculus →

| r | = [(r x )2 + (r y )2 + (rz )2 ] or aɵ x + 3 aɵ y + 4 aɵ z nɵ = (26 )

∴ Hence,

9

→

| r | = [(1)2 + (3)2 + (4)2 ] = (26)

By definition, the direction cosines are ry 4 rz 1 3 r cos α = →x = = , cos β = → = and cos γ = → (26 ) (26 ) (26 ) |r | |r | |r |

P roduct of Two Vectors (Dot and Cross Product) Multiplication of a vector by a vector does not follow the laws of ordinary algebra. There are two type of product of two vectors. If the product of two vectors is a scalar, then it is called scalar product. If the →

→

product is a vector, then it is called vector product. If A and B are two vectors then their scalar product → →

→

→

is written as A . B and read as A dot B. Hence scalar product is also called dot product. →

→

→

→

The vector product is written as A × B and read as A cross B. Hence the vector product is also called cross product.

1. Scalar Product or Dot Product of Two Vectors The scalar product of two vectors is a scalar which is equal to the product of magnitudes of the two vectors and the cosine of the smaller angle between them. The scalar product or dot product of two →

→

vectors A and B may be expressed as → →

→

→

A . B =| A|| B |cos θ →

→

→

→

where | A| and | B | are the magnitudes of A and B respectively and θ is the →

B

→

smaller angle between A and B as in fig. 4.

θ →

A physical example in mechanics is that of work done by a force F in moving a body through a distance d along a straight line. Thus, Work done = the component of force F in the direction of displacement

A Fig. 4

× magnitude of displacement → →

W = F cos θ . d = F . d

Where θ is the angle which the force makes with d. Since the dot product of two vectors is a scalar the physical scalar quantities like work, gravitational potential energy, electric potential, electromagnetic energy density etc. can be describe as scalar product of two vectors.

Scalar Product in Terms of Components →

→

Let A and B are two vectors whose components along X, Y, Z-axes are A x , A y , Az and B x , B y , Bz respectively. If aɵ x , aɵ y , and aɵz are unit vectors along X , Y , and Z , axes then →

→

A = A x aɵ x + A y aɵ y + Az aɵz and B = B x aɵ x + B y aɵ y + Bz aɵz

E lectromagnetic F ield Theory

10 → →

A . B = ( A x aɵ x + A y aɵ y + Az aɵz ).(B x aɵ x + B y aɵ y + Bz aɵz )

so

= A x B x (^ a x . aɵ x ) + A x B y (aɵ x . aɵ y ) + A x Bz (aɵ x . aɵz ) + A y B x (aɵ y . aɵ x ) + A y B y (aɵ y . aɵ y ) A y Bz (aɵ y . aɵz ) + Az B x (aɵz . aɵ x ) + Az B y (aɵz . aɵ y ) + Az Bz (aɵz . aɵz ) ...(1) But we know that, aɵ x . aɵ x = aɵ y . aɵ y = aɵz . aɵz = 1 and aɵ x . aɵ y = aɵ y . aɵz = aɵz . aɵ x = 0 Hence, from eqn. (1), we have → →

A. B = A x Bx + A y B y + Az Bz Thus, the scalar product of two vectors is equal to the sum of the products of their corresponding

x, y, z components. → →

→ → A. B The angle between two vectors in terms of scalar products is, cos ( A, B ) = → → | A|| B| →

→

→

→

the magnitudes of A and B are ; | A| = ( A2x + A2y + A2z ) and | B | = (B2x + B2y + B2y )

A x Bx + A y B y + Az Bz

→ →

∴

cos ( A, B) =

( A2x

2 2 2 + A2y + A2 z ) ( Bx + B y + Bz )

2. Vectors Product or Cross Product of Two Vectors The cross product or vectors product of two vectors is a vector whose magnitude is equal to the product of their magnitudes and the sine of the smaller angle between them, and direction perpendicular to a plane containing the two vectors in accordance with right handed screw rule or right handed thumb rule. If θ is the smaller angle through →

→

→

(A×B)

θ

→

which A should be rotated to reach B , then the cross product of A and B is expressed as →

B

n

A B×A

→

A × B =| A || B|sin θ nɵ →

→

→

Fig. 5 →

where | A|, | B | be the magnitudes of A and B respectively and nɵ is the →

→

→

→

unit vector along the plane containing A and B (fig.5). The direction of A × B is decided either by right handed screw rule or by right handed thumb rule.

1. Right Handed Screw Rule →

→

According to this rule, if a right handed screw is rotated from A to the vector B through small angle →

→

between them, then the direction of advancement of screw gives the direction of A × B [Fig. 6(a)].

2. Right Hand Thumb Rule →

→

According to this rule if the right hand is held such that the curled fingers follow the rotation of A into B, →

→

→

→

the extended right thumb will point in the direction of A × B . If we consider the product B × A the

Fundamentals of Coordinate Systems and Their Transformation with Vector Calculus

11

magnitude AB sin θ remains unchanged, but since the sense of rotation has reversed the vector →

→

representing A × B will now point downward as shown in Fig. 6 (b). A×B

Direction of advancement of screw

Direction of advancement of screw

B

B θ θ A

Direction of rotation of screw

A

Direction of rotation of screw

–B×A

(a)

(b)

Fig. 6

The moment of force or torque about an axis of rotation is the physical example of vector multiplication. →

The product of magnitude of F and perpendicular distance of line of action (d sin θ) of the force, that is, →

→

F × d = F. d sin θ nɵ →

→

having a direction perpendicular to the plane containing F and d . The other example of vector multiplication is found in the lifting force of a screw jack. If friction is

→

assumed as neglected and a force f is applied at the end of lever arm of length l, then the lifting force F produced by the jack will be

p → → → F= f × l 2π where p is the pitch of the screw. Since the vector product of two vectors is a vector, the physical quantities like torque, angular moments, force on a moving charge in a magnetic field etc. can be described as vector product of two vectors.

Geometrical Interpretation →

→

→

→

A × B represents the vector area of the parallelogram whose adjacent sides are formed by A and B , or →

→

twice the area of the triangle with two sides A and B .

A×B

→

→

The vector product of A and B is

n

→

→

B

A × B = ( AB sin θ) nɵ →

B

→

where nɵ unit vector perpendicular to the plane of A and B . According to the fig. 7. →

→

A × B = AB sin θ nɵ = OA. OB sin θ nɵ = OA. BM nɵ (∵ BM = OB sin θ) = Base × Height nɵ = 2 × area of the triangle OAB × nɵ →

→

A × B = Vector area of parallelogram OACB

θ

O A

C

M A

Fig. 7

E lectromagnetic F ield Theory

12

The Vector Product in Terms of Components →

→

Let A and B are two vectors whose components along X, Y, Z-axes are A x , A y , Az and B x , B y , Bz respectively. If aɵ x , aɵ y and aɵz are unit vectors along perpendicular axes X, Y, and Z then →

→

A = A x aɵ x + A y aɵ y + Az aɵz and B = B x aɵ x + B y aɵ y + Bz aɵz →

→

A × B = ( A x aɵ x + A y aɵ y + Az aɵz ) × (B x aɵ x + B y aɵ y + Bz aɵz )

so that, →

→

A × B = A x B x (aɵ x × aɵ x ) + A x B y (aɵ x × aɵ y ) + A x Bz (aɵ x × aɵz ) + A y B x (aɵ y × aɵ x ) + A y B y (aɵ y × aɵ y )

or

+ A y Bz (aɵ y × aɵz ) + Az B x (aɵz × aɵ x ) + Az B y (aɵz × aɵ y ) + Az Bz (aɵz × aɵz ) But, we know that, aɵ x × aɵ x = aɵ y × aɵ y = aɵz × aɵz = 0 and aɵ x × aɵ y = aɵz , aɵ y × aɵz = aɵ x and aɵz × aɵ x = aɵ y →

→

→

→

A × B = 0 + A x B y (aɵz ) + A x Bz (– aɵ y ) + A y B x (– aɵz ) + 0

Thus, we have

+ A y Bz (aɵ x ) + Az B x (aɵ y ) + Az B y (– aɵ x ) + 0

A × B = aɵ x ( A y Bz – Az B y ) + aɵ y ( Az Bx – A x Bz ) + aɵ z ( A x B y – A y Bx )

or

which may also be written as

aɵ x aɵ y aɵ z →

→

A × B = A x A y Az Bx B y

Bz

→

→

Example 8: If A = αaɵ x + 2 aɵ y + 10 aɵ z and B = 4αaɵ x + 8 aɵ y – 2αaɵ z , find out the value of α for which the two vectors become perpendicular. [UPTU, B.Tech. IV Sem. 2007] → →

Solution: Two vectors are perpendicular to each other if A . B = 0 → →

A . B = ( αaɵ x + 2 aɵ y + 10 aɵz ) .(4αaɵ x + 8aɵ y – 2αaɵz ) = 0

That is, or

α × 4α + 2 × 8 + 10 × (–2α ) = 0 or 4α 2 + 16 – 20α = 0 or α 2 – 5α + 4 = 0

or

α 2 – 4α – α + 4 = 0 or α (α – 4) – 1(α – 4) = 0 or (α – 1)(α – 4) = 0 or α = 4 or α = 1 →

→

Example 9: Find the angle between the vectors A = 4 aɵ x – 2 aɵ y + 4 aɵ z and B = 3 aɵ x – 6 aɵ y – 2 aɵ z →

→

Solution: Here A = 4 aɵ x – 2 aɵ y + 4 aɵz and B = 3 aɵ x – 6 aɵ y – 2 aɵz →

→

→

→

Let θ be the angle between A and B . The scalar product of A and B is → →

→ →

→ → A. B A . B =| A|| B |cos θ or cos θ = → → | A|| B | →

→

→

→

where | A| and | B | are the magnitudes of A and B respectively.

Fundamentals of Coordinate Systems and Their Transformation with Vector Calculus → →

→

13

→

But we know that A . B = A x B x + A y B y + Az Bz and | A| = ( A2x + A2y + A2z ), and | B | = (B2x + B2y + Bz2 ) → →

A . B = (4 aɵ x – 2 aɵ y + 4 aɵz ).(3 aɵ x – 6 aɵ y – 2 aɵz ) = (4) (3) + (–2) (–6) + (4) (– 2) = 12 + 12 – 8 = 16

∴

→

→

| A| = [ (4)2 + (–2)2 + (4)2 ] = 6 and | B | = [ (3)2 + (–6)2 + (–2)2 ] = 7 → →

cos θ =

∴

A. B AB

=

16 8 = = 0. 38 Hence, θ = cos –1(0. 38) = 67.5° 6 × 7 21

Example 10: Compute the sin of the angle between the vectors (4 aɵ y – 7 aɵ z ) cm and (5 aɵ x + 3 aɵ y ) metre. →

→

Solution: Let A = 4 aɵ y – 7aɵz cm and B = 5 aɵ x + 3 aɵ y metre = 500 aɵ x + 300 aɵ y cm →

→

→

→

A × B =| A|| B| sin θ nɵ

We know that

→

→

where θ is the angle between the vectors and nɵ is the unit vector in a direction perpendicular to A and B both. So we may write →

→

→ → → → | A × B| | A × B | =| A|| B |sin θ or sin θ = → → | A|| B |

aɵ x aɵ y aɵz →

→

A × B = A x A y Az =

Further,

aɵ y

0

4

aɵz – 7 = 2100 aɵ x – 3500 aɵ y – 2000 aɵz

500 300 0

B x B y Bz →

aɵ x

→

| A × B| = [(2100)2 + (–3500)2 + (–2000)2 ] = 4545

Hence,

→

→

| A| = [(4)2 + (7)2 ] = (65) and | B | = [(500)2 + (300)2 ] = (340000)

Similarly, we have

sin θ =

Hence,

→

→

→

→

| A × B| | A|| B |

=

4545 = 0.967 (65). (340000) →

→

Example 11: Prove that the vector A = 3 aɵ x + 2 aɵ y – 6 aɵ z and B = 4 aɵ x − 3 aɵ y + aɵ z are perpendicular to each other. →

→

Solution: Here A = 3 aɵ x + 2 aɵ y – 6 aɵz and B = 4 aɵ x – 3 aɵ y + aɵz The two vectors are perpendicular to each other if their scalar product is zero, that is, → →

→ →

A . B = 0. We know that A . B = A x B x + A y B y + Az Bz

∴

→ →

A . B = (3 aɵ x + 2 aɵ y – 6 aɵz ) .(4 aɵ x – 3 aɵ y + aɵz ) = 3 × 4 + 2 × (–3) + (–6) × 1 = 12 – 6 – 6 = 0

Hence, the given vectors are perpendicular to each other.

E lectromagnetic F ield Theory

14

Example 12: Deduce a vector in X – Y plane perpendicular to 6 aɵ x + 4 aɵ y + 3 aɵ z and equal in length. →

Solution: Let A = 6 aɵ x + 4 aɵ y + 3 aɵz , the length of this vector is equal to its magnitude, that is, the length of →

→

A =| A| = ( A2x + A2y + Az2 ) = [(6)2 + (4)2 + (3)2 ] = (61)

→

→

Let B is the vector, whose length is (61) in the X – Y plane and perpendicular to A, then we have →

→

B = B x aɵ x + B y aɵ y , where | B | = (B2x + B2y ) = (61) B2x + B2x = 61

or →

...(1)

→

Further if A and B are perpendicular, then → →

A . B = 0 or (6 aɵ x + 4 aɵ y + 3 aɵz ).(B x aɵ x + B y aɵ y ) = 0 or 6 B x + 4 B y = 0 or B y = –

3 Bx 2

...(2)

Substituting this value of B y in eqn. (1), we get 9 13 2 B2x + B2x = 61 or B x = 61 or B x = ± 2 [(61 / 13)] 4 4 Substituting this value of B x in eqn. (2), we get

B y = ± 3 [(61 / 13)] Thus, the required vector is → 61 61 B = ± 2 aɵ x ∓ 3 aɵ y or 13 13

→

61 B = ± [2 aɵ x – 3 aɵ y ] 13 →

→

Example 13: Find the angle between A = 5. 8 aɵ y + 1. 55 aɵ z and B = − 6. 93 aɵ y + 4 aɵ z using both the dot product and the cross product. [C.C.S. University, Meerut, B.Tech. IV Sem. 2006] →

→

→

→

Solution: Let θ be the angle between A and B . The scalar product of A and B is → →

→ →

A . B = A . B cos θ or cos θ =

A. B AB

→

→

and A × B = AB sin θ nɵ or sin θ =

→ →

→

→

| A × B| AB

→ →

A . B = A y B y + Az Bz = 0 + (5.8) (–6.93) + (1. 55) (4) or A . B = – 40194 . + 6. 20 = – 33.994 A = ( A2y + A2z ) = (5.8)2 + (1. 55)2 = 36.0425 = 6.003 B = (B2y + B2z ) = (–6.93)2 + 42 = 480249 . + 16 = 8001 . ∴

cos θ = – →

→

→

→

→

→

33.994 = 0.7078 or θ = cos –1(0.7078) or θ ≈ 44 ° 8001 . × 6.003

A × B = aɵ x ( A y Bz – Az B y ) + aɵ y ( Az B x – Bz A x ) + aɵz ( A x B y – A y B x ) ∴ or

A × B = aɵ x [5. 8 × 4 – 1. 55 × (–6.93)] + aɵ y (.0) + aɵz (0) = (23. 2 + 10.7415)aɵ x →

→

Hence,

→

A × B = 33.9415 aɵ x ∴ | A × B | = 33.9415 sin θ =

→

| A × B| A. B

=

33.9415 = 0.7067 ∴ θ = sin –1(0.7067) = 44° or θ ≈ 44° . × 6.003 8001

Fundamentals of Coordinate Systems and Their Transformation with Vector Calculus →

→

15 →

→

Example 14: Given, A = 2 aɵ x + 3 aɵ y + aɵ z , B = 2 aɵ x + aɵ y + 3 aɵ z and C = 4 aɵ x – 2 aɵ y – 2 aɵ z . Find that C is →

→

perpendicular to both A and B. →

→

→

→

→

[UPTU, B.Tech. IV Sem. (old) 2007]

→

Solution: C is perpendicular to both A and B if A × B = C. Therefore,

aɵ x aɵ y aɵz →

→

A×B = 2

3

1

2

1

3

= aɵ x (3 × 3 – 1 × 1) + aɵ y (2 × 1 – 2 × 3) + aɵz (2 × 1 – 3 × 2) = 8aɵ x – 4 aɵ y – 4 aɵz →

→

→

A × B = 2 (4 aɵ x – 2 aɵ y – 2 aɵz ) = 2 C

∴

Hence

→

→

→

A×B =C

T riple Product Since, the cross product or vector product of two vectors is itself a vector, it can be multiplied by a third vector both scalarly and vectorially to form a triple product. There are three types of triple product.

1. Ordinary Triple Product →

→

→

The ordinary product of a vector C with a scalar product of other two vectors ( A and B ) is known as →

→ → →

ordinary triple product and expressed as T = ( A . B) C → →

→

As A . B is a scalar, this ordinary triple product is vector parallel to C. It can be easily verified that → → →

→ → →

C ( A . B) ≠ (C. A) B

2. Scalar Triple Product →

The scalar triple product is scalar and is equal to the dot product of one vector A with the cross product of →

→

→

→

→

other two vectors B and C. Mathematically scalar triple product can be expressed as A .( B × C). The scalar triple product can also be written as [ A B C]. It is equal in magnitude to the volume of the → →

→

parallelopiped formed with sides A , B and C.

Scalar Triple Product in Terms of the Components: → →

→

Let A, B and C be the three vectors whose components along principle axes X , Y , Z are

( A x , A y , Az ), (B x , B y , Bz ) and (C x , C y , Cz ). If aɵ x , aɵ y , aɵz are the unit vectors along X , Y , Z -axes, then →

→

→

A = A x aɵ x + A y aɵ y + Az aɵz , B = B x aɵ x + B y aɵ y + Bz and C = C x aɵ x + C y aɵ y + Cz aɵz

Thus, the scalar triple product

E lectromagnetic F ield Theory

16

A x A y Az →

→

→

A .( B × C) = ( A x aɵ x + A y aɵ y + Az aɵz ) . B x B y Bz C x C y Cz = ( A x aɵ x + A y aɵ y + Az aɵz ) .[aɵ x (B y Cz – Bz C y ) – aɵ y (Cz B x – Bz C x ) + aɵz (B x C y – B y C x )] Using the relations, aɵ x . aɵ x = aɵ y . aɵ y = aɵz . aɵz = 1 and aɵ x . aɵ y = aɵ y . aɵz = aɵz . aɵ x = 0, we get →

→

→

→

→

A .( B × C) = A x (B y Cz – Bz C y ) – A y (B x Cz – Bz C x ) + Az (B x C y – B y C x ) A x A y Az

→

A . ( B × C ) = Bx B y Bz

or

Cx C y

Cz

Geometrical Interpretation: →

→

→

→ →

The scalar triple product A .( B × C) represents the volume of a parallelopiped formed with sides A, B and →

C as shown in fig. 8. →

→

→

→

A .( B × C) = A ×| B × C|cos φ B×C

= A(BC sin θ) cos φ = BC sin θ( A cos φ) BC sin θ is the area of the (parallelogram OPQR) base of the →

→

→

parallelopiped and A cos φ is the projection of A along B × C, that is,

A

φ

C R

the height of the parallelopiped. Hence →

→

→

O

A .( B × C) = Base area × height = Volume of the parallelopiped.

θ B

Q

P

3. Vector Triple Product

Fig. 8

The vector triple product is a vector and is equal to the cross product of →

→

→

one vector A with the cross product of other two vectors B and C. Mathematically vector triple product can be expressed as →

→

→

A × ( B × C) →

→

→

→

→

→

→

→

A vector triple product A × ( B × C) is a vector which is perpendicular to both A and B × C . But B × C is →

→

→

→

→

→

perpendicular to the plane of B and C . Hence vector triple product A × ( B × C) must lie in the plane of B →

→

and C and perpendicular to A. →

→

→

→

→

→

It can easily be seen that A × ( B × C) ≠ ( A × B) × C →

→

→

→

→

→

As ( A × B) × C is a vector lying in the plane of A and B , and perpendicular to C.

Fundamentals of Coordinate Systems and Their Transformation with Vector Calculus

17

Vector Triple Product in Terms of Components Vector triple product is expressed as, →

→

→

→ → →

→ → →

A × ( B × C ) = ( A . C ) B – ( A . B) C

.....(1) [GBTU, B.Tech. IV Sem. 2011]

In terms of components we can write

aɵ x aɵ y aɵz A × (B × C) = ( A x aɵ x + A y aɵ y + Az aɵz ) × B x B y Bz C x C y Cz = ( A x aɵ x + A y aɵ y + Az aɵz ) × [ aɵ x (B y Cz – Bz C y ) + aɵ y (Bz C x – B x Cz ) + aɵz (B x C y – B y C x )] and using relations

aɵ x × aɵ x = aɵ y × aɵ y = aɵz × aɵz = 0 and aɵ x × aɵ y = aɵz , aɵ y × aɵz = aɵ x and aɵz × aɵ x = aɵ y

Similarly

aɵ y × aɵ x = – aɵz , aɵz × aɵ y = – aɵ x and aɵ x × aɵz = – aɵ y

On multiplying as an ordinary way, we get

A × (B × C) = aɵz (Bz C x – B x Cz ) A x – aɵ y (B x C y – B y C x ) A x – aɵz (B y Cz – Bz C y ) A y + aɵ x (B x C y – B y C x ) A y + aɵ y (B y Cz – Bz C y ) Az – aɵ x (Bz C x – B x Cz ) Az = aɵ x [B x C y A y + B x Cz Az – B y C x A y – Bz C x Az ] + aɵ y [B y C x A x + B y Cz Az – B x C y A x – Bz C y Az ] + aɵz [Bz C x A x + Bz C y A y – B x Cz A x – Cz B y A y ] ...(2) In terms of components R.H.S. of eqn. (1) can be written as → → →

→ → →

( A . C) B – C ( A. B) = (B x aɵ x + B y aɵ y + Bz aɵz ) [ A x C x + A y C y + Az Cz ] –(C x aɵ x + C y aɵ y + Cz aɵz ) ( A x B x + A y B y + Az Bz ) = [aɵ x B x ( A x C x + A y C y + Az Cz ) – aɵ x C x ( A x B x + A y B y + Az Bz )] + [aɵ y B y ( A x C x + A y C y + Az Cz ) – aɵ y C y ( A x B x + A y B y + Az Bz )] + [aɵz Bz ( A x C x + A y C y + Az Cz ) – aɵz Cz ( A x B x + A y B y + Az Bz )] = aɵ x [B x A y C y + B x Az Cz – C x A y B y – C x Az Bz ] + aɵ y [B y A x C x + B y Az Cz – C y A x B x – C y Az Bz ] + aɵz [Bz A x C x + Bz A y B y – Cz A x B x – Cz A y B y ] ...(3) Since, R.H.S. of eqns. (2) and (3) are same, so we have →

→

→

→ → →

→ → →

A × ( B × C ) = B ( A . C ) – C ( A . B)

Example 15: A parallelopiped has edges described by the vectors aɵ x + 2 aɵ y , 4 aɵ y and aɵ y + 3 aɵ z . Compute its volume. →

→

→

Solution: Here A = aɵ x + 2 aɵ y , B = 4 aɵ y and C = aɵ y + 3 aɵz

E lectromagnetic F ield Theory

18 → →

→

The volume of a parallelopiped formed by three vectors A, B and C as sides is given by their scalar triple product, that is, →

→

→

A .( B × C) A x A y Az →

→

→

A .( B × C) = B x B y

Therefore,

1 2

0

Bz = 0

4

0 = 12

0

1

3

C x C y Cz

Thus, the volume of a parallelopiped = 12 unit

Example 16: Prove the following cyclic relations in the scalar triple product →

→

→

→

→

→

→

→

→

A . ( B × C ) = B . (C × A ) = C . ( A × B) Solution: In terms of components, let →

→

→

A = A x aɵ x + A y aɵ y + Az aɵz , B = B x aɵ x + B y aɵ y + Bz aɵz and C = C x aɵ x + C y aɵ y + Cz aɵz , then A x A y Az →

→

→

A .( B × C) = B x B y

Bz

C x C y Cz Interchanging of two rows of a determinant changes its sign. Therefore,

A x A y Az

∴

By

Cx

C y Cz →

→

B x B y Bz C x C y Cz

∴

Cx →

Bx

B y Bz

A y Az = C x

C y Cz

Ax

A y Az

C y Cz

→

→

A .( B × C) = B .( C × A) A x A y Az

Similarly

B y Bz

Bz = – A x

Bx

→

Bx

→

→

→

Cx

C y Cz

= Ax

A y Az

Bx

B y Bz

→ →

→

→

→

→

→ →

→

→

→

→

A .( B × C) = C.( A × B) Hence, A . ( B × C ) = B. (C × A ) = C . ( A × B)

The above relations also concluded that in a scalar triple product the dot and cross position can be interchanged.

Fundamentals of Coordinate Systems and Their Transformation with Vector Calculus

19

C o-ordinate Systems and Transformations Electromagnetics deals directly with the field vectors. Field theory in more difficult because of the large number of variables involved with it. In the most general electromagnetic problems there are three space variables involved, and the solutions tend to become correspondingly complex. Vector and scalar fields can, in general be expressed in terms of three Cartesian or rectangular co-ordinates ( x, y, z) but often in electromagnetics it is more convenient and advantageous to express them in terms of other co-ordinates systems. The three most common coordinate systems are rectangular or cartesian (co-ordinates x, y, z), circular cylindrical (co-ordinates ρ, φ, z) and spherical (co-ordinates r, θ, φ). In addition to rectangular, cylindrical and spherical coordinate systems, there are many other systems such as the elliptical, spheroidal and paraboloidal systems. These systems are used in special circumstances. However, we will not need to deal with these systems. The rectangular (or cartesian), circular cylindrical and spherical systems are sufficient for our requirement. We shall deal with these one by one as follows:

C artesian or Rectangular Co-ordinate System Z

In this system three mutually perpendicular

Plane z = constant

co-ordinate axes are setup which are given the name

X , Y , and Z axes. A point P in space, in cartesian

Plane y = constant

z

coordinate, is specified by assigning its co-ordinate

y x

x, y, z as shown in Fig. 9. where x, y and z are the

P(x,y,z)

Y

distances from the origin to the intersection of a perpendicular dropped from the point P to the axes

X , Y , and Z axes respectively. Alternatively, the point in the space may also be considered as the intersection of three mutually perpendicular surfaces, that is,

Plane x = constant X Fig. 9

Plane x = constant; Plane y = constant; Plane z = constant. The constants being the co-ordinate values of the point. The range of the co-ordinate variables x, y, and z are:

–∞< x < ∞

–∞< y< ∞

→

–∞ < z < ∞

A vector A in cartesian co-ordinates at the point P in terms of three mutually perpendicular components

A x , A y and Az with unit vectors aɵ x , aɵ y and aɵz may be expressed as, →

A = A x aɵ x + A y aɵ y + Az aɵz

E lectromagnetic F ield Theory

20

C ircular Cylindrical Co-ordinate Systems Circular cylindrical or simply cylindrical co-ordinate are often used to those problems in which cylindrical symmetry is involved such as coaxial cables machine rotor etc. The circular cylindrical system is the three dimensional version of the polar coordinates of analytic geometry. In the two dimensional polar co-ordinates a point in the plane was located by giving its distance from origin (0) and angle θ between the line from the point to the origin and an arbitrary radial line taken

θ = 0 that is, between OP and OX in Fig. (10). A three dimensional cylindrical co-ordinate system is obtained by also specifying the distance z of the point from z = 0 reference plane (arbitrary) which is perpendicular to the line ρ = 0 Fig. (11). A point P is the intersection of three mutually perpendicular surfaces. There surfaces are a circular cylinder (ρ = constant), a plane (φ = constant), and another plane (z = constant). A point P in cylindrical co-ordinate is represented as (ρ, φ, z). Each variable is defined as

(i)

ρ is the radius of the cylinder passing through P or radial distance from Z-axis.

(ii) φ is called the azimuthal angle and is measured from the X- axis in the XY plane. (iii) z is the distance along Z-axis from XZ plane. The range of variables are:

0 ≤ρ < ∞

0 ≤ φ < 2π

–∞< z > a, a2 cos2 θ is negligibly small, so it may be neglected in comparison to r 2 . Therefore eqn. (5) takes the form

→ → 1 p.r = V = 4 π ε 0 r 2 4 π ε0 r 3 p cos θ

...(6)

If θ = 90 ° , V = 0, it means that the potential vanishes everywhere in the equatorial plane. In fact equatorial plane is equipotential line.

(i) Electric Field Intensity at Any Point Due to an Electric Dipole (In Polar Coordinates) Intensity of electric field can be expressed as negative gradient of potential, that is, →

→

E = − ∇V

...(7)

E lectrostatics

In spherical coordinates,

175

→

→ ∂V 1 ∂V ɵ 1 1 ∂V ɵ 1 ∂V ɵ ∂V ∂V ɵ θ− E = − rɵ + θ+ φ or E = − rɵ − φ θ sin θ ∂ r r ∂ θ r sin θ ∂ φ ∂ r r ∂ r ∂φ →

→

→

→

...(8)

E = Er + Eθ + Eφ

∴

The radial component of electric field E r is,

Er =

− ∂V

=−

∂r

p cos θ ∂ p cos θ =− ∂ r 4 πε0 r 2 4π ε0

2 p cos θ −2 3 or E r = r 4 π ε0 r 3

...(9)

The transverse component of electric field intensity Eθ is,

Eθ = – and

p sin θ 1 ∂V 1 ∂ p cos θ =– = r ∂θ r ∂ θ 4 π ε0 r 2 4 π ε0 r 3

or E θ =

p sin θ 4 π ε0 r 3

...(10) ...(11)

Eφ = 0

Substituting the values Er , Eθ and Eφ from eqns. (9), (10) and (11) in eqn. (8), we get →

E=

→

or

E=

2 p cos θ 4 π ε0 r

3

rɵ +

p sin θ ɵ θ+0 4 π ε0 r 3

p [2 cos θ rɵ + sin θ θɵ ] 4 πε0 r3

...(12)

The magnitude of total electric field is given by

E=

E=

or

E=

( E2r

Eθ2 )

+

p 4 π ε0 r 3 p 4 π ε0 r3

2 p cos θ = 3 4 π ε0 r

2

p sin θ + 3 4 π ε0 r

(4 cos2 θ + sin2 θ)

(3 cos2 θ + 1)

→

...(13)

→

The angle α which E makes with the radius vector OP is,

tan α =

∴

Eθ Er

=

2

p sin θ / 4 π ε0 r 3

2 p cos θ / 4 π ε0 r 3

1 α = tan−1 tan θ 2

=

1 tan θ 2

E lectromagnetic F ield Theory

176

(ii) Electric Field Intensity Due to a Dipole (In Cartesian Coordinates) →

The potential V at a point distant r from a dipole of dipole moment p is expressed as, → →

1 p. r V = 4 πε0 r 3

...(1)

→

Intensity of electric field E is related with potential as, →

E = – ∇V

...(2)

∴

→ → → 1 → ( p . r ) ∇ E = − ∇V = − 4 π ε0 r3

...(3)

or

E=−

→ → → 1 1 → → → ∇ p r + p. r) ∇ ( . ) ( 4 π ε0 r 3

→

1 3 r

...(4)

→

p = p x aɵ x + p y aɵ y + pz aɵz and r = x aɵ x + y aɵ y + z aɵz (position vector)

→ →

p . r = p x x + p y y + pz z and r = ( x2 + y2 + z2 )

∴

→ → →

∂ ∂ ∂ ∇ ( p . r ) = aɵ x + aɵ y + aɵz ( p x x + p y y + pz z) ∂y ∂ z ∂x

∴

→ → →

or

...(5)

→

∂ ∂ ∂ 1 ( x2 + y2 + z2 ) −3 / 2 ∇ 3 = aɵ x ( x2 + y2 + z2 ) −3 / 2 + aɵ y ( x2 + y2 + z2 ) −3 / 2 + aɵz r ∂z ∂x ∂y

and

=−

or

→

∇ ( p . r ) = aɵ x p x + aɵ y p y + aɵz pz = p

3 y aɵ y 3 x aɵ x 3 z aɵz − − 2 [∵ r 3 = ( x2 + y2 + z2 )3 / 2 ] 2 2 2 5 /2 2 2 2 5 /2 (x + y + z ) (x + y + z ) ( x + y2 + z2 ) 5 / 2 →

3 ( x aɵ x + y aɵ y + z aɵz ) 3r 1 ∇ 3 = − =− 5 2 2 2 5 /2 r r (x + y + z )

→

...(6)

→ 1 → → → Substituting the values of ∇ ( p . r ) from eqn. (5) and ∇ 3 from eqn. (6) in eqn. (4), we get r → → → → 1 p 3( p . r) r or − E=– 4 π ε0 r 3 r5

F

→ → → → p 1 3 ( p . r ) r E( r ) = − 3 5 4 π ε0 r r →

lux Lines or Field Lines

In some charge configurations like dipole, it is difficult to find the magnitude and direction of the field easily. Their field equations are slightly tedius and are unable to visualize the field. In such charge configurations field lines are able to visualize the field. Field lines are express in the form of various equations. These equations are quite helpful in the determination of field at a point. In fact they indicate the vector direction of field in one sense. The concept of field lines can be understood with the help of following illustration.

E lectrostatics

177

Suppose a test charge placed at a point A in a field is under the action of a

Z

→

→ E

force directed along E as shown in Fig. 71. Suppose the test charge displaces from point A ( x, y, z) to B ( x + ∆ x, y + ∆ y, z + ∆ z) and travel a

B

distance ∆ L, then →

A(x, y, z)

...(1)

∆L = ∆ x aɵ x + ∆ y aɵ y + ∆ z aɵz

Y

This vector displacement is proportional to the force or to the field

X

represented as,

Fig. 71

→

...(2)

E = E x aɵ x + E y aɵ y + Ez aɵz

Two vectors represented by eqns. (1) and (2) are proportional if and only if their corresponding components are proportional by equal amount, therefore, ∆y ∆z dx dy dz ∆x or = = (∵ as ∆ x, ∆ y and ∆ z → 0) = = Ex E y Ez Ex E y Ez Field lines and the field at a point in two dimensional system is depicted in Fig. 72. Similar equations can be expressed for cylindrical and spherical coordinate system as follows: d ρ r d φ dz = = Ερ Eφ Ez

dr Er

=

r dθ Eθ

=

(In cylindrical)

r sin θ dφ Eφ

(In spherical) Fig. 72

Example 38: Obtain the general equation of flux lines which represent the field, →

E=

p

ɵ (2 cos θ rɵ + sin θ θ)

4 π ε0 r3

In spherical coordinates, where p is a dipole moment. Solution: The flux lines equation in spherical coordinates are

dr r dθ r sin θ = = Er Eθ Eφ →

In the given problem, E =

∴

Er =

2 p cos θ 4 π ε0 r

2 p cos θ 4 π ε0 r

3

3

rɵ +

, Eθ =

...(1)

p sin θ ɵ θ 4 π ε0 r 3 p sin θ 4 π ε0 r 3

and Eφ = 0

Substituting these values of E r , Eθ and Eφ in eqn. (1), we get

E lectromagnetic F ield Theory

178

dr (4 π ε0 r 3 ) 2 p cos θ

=

r dθ (4 π ε 0 r 3 ) p sin θ

=

r sin θ 0

r dθ dr dr or = 2 cot θ dθ and d φ = 0 = r 2 cos θ sin θ

or

Integrating above equations, dr ∫ r = 2 ∫ cot θ dθ

and

∫ dφ = ∫ 0

log e r = 2 log e sin θ + C and φ = constant where C is the constant of integration

log e r = log e sin2 θ + C or

r log e 2 = C sin θ

or

Fig. 73

log e (r cosec2 θ) = C

r cosec2 θ = costant

or

Hence, the flux lines are the intersection of the plane φ = constant and flux lines of intersection of surfaces

r cosec2 θ = constant. The field (or flux lines) of dipole is shown in Fig 73.

Example 39: Calculate the potential and field due to a short dipole of dipole moment 1. 667 × 10 −27 coulomb-metre at a point distant one meter from it (i) on its axis, and (ii) on its perpendicular bisector. Solution: The potential due to short dipole in polar coordinates (r , θ) is,

V =

p cos θ

P

4 π ε0 r 2

and field at (r , θ) is given by

E=

p 1 4 π ε0 r 3

(1 + 3 cos2 θ)

Here p = 1. 667 × 10 −27 C–m and r = 1 cm =10 –2 m (i) On the axis of the dipole, θ = 0°

∴

Electric potential, V =

p 9 × 109 × 1. 667 × 10 −27 1 = 4 π ε0 r 2 (10 −2 )2

or

V = 1. 5 × 10 −13 volt

Electric field,

E=

Electric potential, V =

Fig. 74

9 −27 1 2 p 9 × 10 × 2 × 1. 667 × 10 = 3.0 × 10 −11 V /m = 4 π ε0 r 3 (10 −2 )3

(ii) At the perpendicular bisector, θ = 90 °

∴

–Q

+Q

1 p cos 90 ° =0 4 π ε0 r2

P

E lectrostatics

Electric field,

E=

179

p 9 × 109 × 1. 667 × 10 −27 1 = 1.5 × 10 −11 V /m = 4 π ε0 r 3 (10 −2 )3

E nergy Density in Electrostatic Field In an electric field if a test charge is carried by an external agent from a point of lower potential to a point of higher potential, the work is done against the electric force. As a consequence, the potential energy of the system is increased. On the other hand, if a test charge is moved from a point of higher to lower potential, the work would have been done by the electric force itself. In this case, the potential energy of the system is reduced, that is there is a loss in potential energy which is supplied to the test charge. The potential energy supplied by external work on a charge when moved from a lower to a higher potential (against the electric force) is expressed as, → →

W = − Q ∫ E . dL →

where dL is the elemental distance covered by the test charge Q under the influence of electric field E. In a system of point charges, the energy is obtained by assuming that initially all the charges are at infinity and having no energy, the system is built up by taking charges one by one to the system. The total energy of the system of n charges is,

W =

n

1 Q i Vi 2 i∑ =1

where Q i is the charge of each point charge and V i is the potential due to all other charges except that of charge Q i at the point where Q i is situated.

E xpression for Energy Density or Electrostatic Energy In order to determine the expression for electrostatic energy of a system of charges, initially the amount of work required to assemble them must be computed. Let us suppose that the three point charges

(Q1, Q2 and Q3 ) are to be arranged in an initially charge free region as shown in fig. 75. The work done in bringing a charge Q1 from infinity to a point P1 in charge free region is zero, because field does not exist in the system

(W1 = 0). The work is done in bringing charge Q 2 from infinity to point P2 because it will experience a repulsive force due to charge Q1 at P1. Thus, the work done in transferring a charge Q2 from infinity to P2 is equal to the product of Q2 and potential V21 at P1 due to Q1, that is, W2 = Q2 V21. Similarly, the work is again done in bringing the charge Q3 from infinity to P3 , because, now it will experience an electric repulsive forces due to charge Q1 at

Q1 Q2 Q3

Fig. 75

P1 and charge Q2 at P2 . Therefore, the work done in bringing Q3 at P3 is equal to W3 = Q3 (V32 + V31), where V32 and V31 are the potentials at P3 due to charges Q2 and

E lectromagnetic F ield Theory

180

Q1 respectively. Hence, the total work done, WE in arranging three charges from infinity in an initially charge free region is,

WE = W1 + W2 + W3 or

WE = 0 + Q2 V21 + Q3 (V31 + V32 )

...(1)

If the three charges are brought to initially charge free region in reverse order, that is, Q3 is brought to P3

first, then Q2 at P2 and finally charge Q1 at P3 , then the total work done in this arrangement is also the same, that is,

WE = 0 + Q2 V23 + Q1 (V12 + V13 )

...(2)

Adding eqns. (1) and (2), we get

2 WE = Q1 (V12 + V13 ) + Q2 (V21 + V23 ) + Q3 (V31 + V32 )

...(3)

V12 + V13 is the sum of potentials at P1 due to charges Q2 and Q3 respectively, therefore the total potential at P1 is V1. Similarly (V21 + V23 ) is the total potential at P2 that is V2 and V31 + V32 is the total potential at P3 , that is V3 . Hence eqn. (3) becomes

2 WE = Q1 V1 + Q2 V2 + Q3 V3 or

1 (Q V + Q2 V2 + Q3 V3 ) 2 1 1

WE =

If there are n charges, then

or

WE =

1 (Q V + Q2 V2 + Q 3 V3 + ... Q n V n) 2 1 1

WE =

1 Qi Vi 2 i∑ =1

n

...(4)

If instead of point charges, the region has a continuously distributed charge distribution having volume charge density ρ v , then summation is converted into integration and Q i is replaced by ρ v dV. Therefore,

WE =

1 1 (ρ v dV ) . V or WE = ∫ ρ v V dv ∫ V 2 2 V

...(5)

Hence, eqns (4) and (5) are applicable in the determination of total potential energy present in a system of charges or in a region having a continuous charge distribution. Similar equations can be obtained for line and surface charge distributions. → →

According to the differential form of Gauss's law, ∇ . D = ρ v →

...(6)

→

where D is an electric flux density and ∇ is a del operator. Substituting this value of volume charge density, ρ v is eqn. (5), we get

Q1

E lectrostatics WE =

181 → →

1 2

∫ ∫ ∫ v ( ∇ . D)V dv

...(7)

According to vector identity →

→

→ →

→ →

→ →

→

→

→ →

∇ .(φ A) = A . ∇ φ + φ ( ∇ . A) or φ ( ∇ . A) = ∇ .(φ A) = A . ∇ φ → →

Hence,

→

→

→ →

...(8)

V ( ∇ . D) = ∇ .(V D) − D . ∇ V → →

Substituting the value of V ( ∇ . D) from vector identity (8) in eqn. (7), we get

∴

→

→

→ →

WE =

1 2

∫ ∫ ∫ v [ ∇ .(V D) − D . ∇ V ] dv

WE =

1 2

∫ ∫ ∫ v (∇ . VD) dv − 2 ∫ ∫ ∫ v (D . ∇ V ) dv

→

→

1

→ →

...(9)

According to Gauss-divergence theorem →

→

→

→

→ → ∵ ∫ ∫ s A . dS = ∫

∫ ( ∇ . VD) d v = ∫ s V D . dS v

WE =

→

1 2

→

1

→ →

∫ (V D . dS ) − 2 ∫ v (D. ∇ v) dv

→

∫ ∫ v div A d v ...(10)

→ → 1 1 Since potential varies inversely to radius r V ∝ and D varies inversely to r 2 that is, D ∝ 2 while r r

area dS varies directly with r 2 (dS ∝ r 2 ), hence the first integral of eqn. (10) must tend to zero as the surface S is allowed to approach to infinity (that is r → ∞).

∴

∫s

→

→

V D . dS ∝

r→ ∞

1 =0 r

Hence eqn. (10), becomes

WE = − But

→

→ → 1 (D . ∇ V ) dv ∫ 2 v

→

→

...(11)

→

∇ V = E and D = ε0 E

Therefore, eqn. (11) becomes

WE = +

1 2

∫ (ε0

→ →

E . E) dV

or WE =

1 ε 2 0

2

∫v E

dv Joules

...(12)

→

Electrostatic energy density for any field E is

W =

dWE 1 D2 = ε0 E2 = (J / m3 ) dv 2 2 ε0

...(13)

E lectromagnetic F ield Theory

182

E nergy Stored in Concentric Spherical Shell Let us consider a concentric charged spherical shell of inner and outer radii R1 and R 2 (R 2 > R1) and →

carrying a charge of + Q as shown in Fig. 76. The electric field intensity E at a point distant r from the centre of the shell where R1 ≤ r ≤ R2 is

E=

Q 4 π ε0 r 2

R2

→

The energy stored in an electric field E is given by

0

1 W = ∫ ε0 E2 dV 2 v

R1

The volume of the shell of radius r and width dr is

dV = 4 π r 2 dr ∴ W =

or

∴

W =

W =

1 ε 2 0

Q2 8π ε0

Q2 8π ε0

∫R

Q2 8 π ε0

1 1 − R 1 R2

R2 1

dr r

2

=

R2

Q2

1

(4 π ε0 )2 r 4

∫R

.(4 π r 2 dr )

Fig. 76

R

1 2 − r R 1

E lectrostatic Energy Stored in a Charged Capacitor Charging of a capacitor, that is, the removal of electrons from positive plate and carrying them to the negative plate against an electric force, always involves some expenditure of energy by the charging agency. This energy is stored up in the electrostatic field set up in the medium by the assembled charges. This stored energy is recovered when the capacitor is discharged. To calculate the electrostatic energy stored in a capacitor, let us consider a capacitor of capacitance C. Suppose during the process of charging, the charge at any instant on either plate of capacitor is Q. At this instant the potential difference between the plates of the capacitor is

V =

Q C

...(1)

In the process of charging, electrons are transferred from the positive plate to the negative plate until each plate acquires an amount of charge Q. Suppose during the process of charging, a charge dq is taken from positive to negative plate, that is, we are increasing the charge on the positive plate by an amount

dq by transferring negative dq charge from positive to negative plate. The potential difference between the plates at this instant will be V ′ = q / C. The work dW that has to be done against the potential difference V′ is

dW = V ′ dq

or

dW =

q C

dq

....(2)

E lectrostatics

183

Therefore, the total amount of work done in charging the capacitor from charge zero to Q is obtained by integrating eqn. (2) within the limits q = 0 to q = Q. That is,

W =∫

Q

q

0

C

dq =

1 C

Q

q2 or 2 0

W =

1 Q2 2 C

...(3)

When Q is in Coulombs and C is in Farad, the energy stored is in Joules. But

Q = CV ∴ W = U =

1 CV 2 2

...(4)

The energy is in joules provided C is in farad and V in volts. This is the electrostatic energy stored in the capacitor.

E nergy Density in a Capacitor To find the energy density and to determine the relationship between stored energy U and the electric field E, let us consider a parallel plate capacitor of plate area A with a separation d. The electric field is uniform in the region between the plates. Let us calculate the stored energy for a parallel plate capacitor for which

A C = ε0 d

and E =

V d

or

V = Ed

...(1)

where ε0 is the permittivity of free space. When a capacitor of capacitance C is charged to a potential difference of V volt, then the energy stored by the capacitor is,

V =

1 CV 2 2

....(2)

Substituting the values of C and V from eqn. (1) in eqn. (2), we get

U=

1 A 1 2 2 ε ( E d) or U = ε0 E Ad joules 2 2 0 d

where Ad is the volume of the space between the two plates. Thus, we define the energy density U E to be energy per unit volume and is given by,

UE =

total energy volume

=

U 1 = ε0 E2 joules / meter3 Ad 2

Here we have assumed that the space between the two plates is vacuum or air. If the space between the plates is filled with a dielectric material of dielectric constant K (= ε / ε0 ), then

UE =

1 1 ε KE2 = ε E2 2 0 2

Example 40: Determine the capacitance of a capacitor consisting of two parallel metal plates 0 .30 m × 0 .30 m, surface area, separated by 5mm in air. What is the total energy stored by the capacitor if the capacitor is charged to a potential difference of 500 volts? What is the energy density?

E lectromagnetic F ield Theory

184

ε A Solution: The capacitance of a parallel plate capacitor is given by, C = 0 d ε 0 = 8. 854 × 10 −12 F /m, A = 0. 3 × 0. 3 = 0. 09 m 2 , d = 5 mm =5 × 10 –3 m

Here

C=

∴

8. 854 × 10 –12 × 0 .09 5 × 10 −3

= 159. 37 × 10 −12 F

Energy stored in the capacitor,

U =

−12

159. 37 × 10 1 CV 2 = 2 2

1 1 The energy density, U E = ε 0 E2 = ε 0 2 2

V d

2

× (500)2

= 1. 992 × 10 −5 joule

500 1 = × 8. 854 × 10 −12 × −3 2 5 × 10

2

U E =0.044 joule / m3 →

Example 41: If V = x − y + xy + 2 z , find E at (1, 2, 4) and the electrostatic energy stored in a cube of side 2m centred at the origin. [UPTU, B.Tech. IV Sem. 2003, 2008] → → ∂V ∂V ∂V Solution: We know that, E = − ∇ V = − aɵ x + aɵ y + aɵz ∂y ∂z ∂x

V = x − y + xy + 2 z ∂V = 1 + y, ∂x

∴

∂V ∂V = x − 1 and =2 ∂y ∂z →

→

E = − (1 + y) aɵ x − ( x − 1) aɵ y + 2 aɵz and Eat (1, 2, 4) = − (1 + 2) aɵ x − (1 – 1) aɵ y + 2 aɵz

∴ →

E (1, 2, 4 ) = − 3 aɵ x + 2 aɵ z

Electrostatic energy,

W =

→ → 1 ε 0 E2 ∴ E2 = E . E = (1 + y)2 + ( x − 1)2 + 4 2

Energy stored in a cube of side 2 m, W =

∴

W =

1 ε ( E2 )at (2, 2, 2) 2 0

1 14 ε0 [(1 + 2)2 + (2 − 1)2 + 4] or W = ε = 7 ε0 2 2 0

Example 42: The electric field between two concentric cylindrical conductors at ρ = 0 . 01 m and ρ = 0 . 05 → 10 6 m is given by E = ρɵ V /m. Find the energy stored in a 1. 0 m length. Assuming free space and ρ

neglect fringing. [U'khand, B.Tech. IV Sem. 2011]

E lectrostatics

185

Solution: The energy stored in an electric field E is given by,

W =

∴

W =

1 ε E2 dv, In cylindrical coordinates, dv = ρ dρ d φ d z 2 ∫v 0

0 .05 1 ε 2 0 ∫ ρ = 0 . 01

=

2

106 0 . 05 1 dρ 2 π 1012 ∫ φ = 0 ∫ z = 0 ρ ρ dρ d φ d z= 2 ε0 ∫ρ = 0 . 01 ρ ∫ φ = 0 dφ ∫ z = 0 dz 2π

1

0 . 05 0 . 05 1 1012 2π [φ]0 [z] or W = 4.427 × 2 π × 1 × [log e r ] × 8. 854 × 10 –12 [log e r ] 0 . 01 0 0 . 01 2

= 4. 427 × 2 × 3.14 × 2. 303 (log10 0. 05 − log 10 0. 01) = 4. 427 × 6. 28 × 2. 3036 (−1. 301 + 2. 0) or

W = 4. 427 × 6. 28 × 2. 3036 × 0 . 699 = 44. 77 J

Example 43: A dielectric sphere of radius ' b ' and permittivity ε is situated in vacuum. It is charged 5b throughout its volume by a charge density, ρ v = , r being the distance from the centre of the sphere. r Determine the electrostatic energy of the system. Solution: The situation of the problem is shown in Fig.77. (i) For flux density outside the sphere: According to Gauss's law →

r

→

∫ s D . dS = ∫v ρ v dv,

where ρ v =

5b r

r

r

In spherical coordinates, dv = r 2 sin θ dr dθ dφ

Pv

2π

π

b D × (4 π r ) = ∫ 5 r 2 sin θ dr dθ dφ r = 0 ∫θ = 0 ∫θ = 0 r b

2

∴

=5b ∫

b r =0

rdr

π

∫θ =0

sin θ dθ

2π

∫0

b

Fig. 77

dφ

π b2 2π = 5 b . − cos θ (φ) 0 2 0

D .(4 πr 2 ) =

or

→ 10 π b3 5 b3 5 b3 5 b3 or D = rɵ b < r < ∞ [2] [2 π] = 10 π b3 or D = = 2 2 2 r2 2 2 r 4π r

→

(ii) For D inside the sphere; r varies from 0 to r . therefore,

D .(4 π r 2 ) = ∫

r 0

π

2π

∫θ =0 ∫φ =0

r b 5 r 2 sin θ dr dθ dφ = 5 b ∫ rdr 0 r

π

∫ θ =0

sin θ dθ

2π

∫0

dφ

E lectromagnetic F ield Theory

186

D .(4 π r 2 ) =

5 br 2 2

. 2 . 2 π = 10 π br 2 ∴ D =

Total energy of the system, WE = ∫

or

or

∞ 1 WE = 2 ε0 ∫ r = b

WE =

10 π br 2 4π r

2

→

or D =

5 brɵ 0 < r < b 2

1 D2 dv ε0

v2

5 ∫ θ = 0 ∫ φ = 0 2 2π

π

b3 r2

2

r 2 sin θ dr dθ dφ +

b

π

=

25 b2 4 b 8 ε0

=

b r3 25 b2 4 1 ∞ π 2π 2π b − [− cos θ] [φ] + (− cos θ)π (φ) 0 0 0 0 8 ε0 r b 3 0

=

25 b2 25 b2 3 4 b3 b3 .2 .2 π = .4π . b .2 . 2 π + 8 ε0 3 3 8ε0

50 π b5 3 ε0

=

∞

∫r = b

1 r

2

dr

50 × 3.14 b5 3 × 8. 854 × 10 −12

π

∫0

sin θdθ

2π

∫0

dφ +

b

∫0

r 2 dr

2π

∫0 ∫0 ∫0 π

∫0

sin θ dθ

5 b 2 2π

∫0

2

r 2 sin θ drdθ dφ

dφ

= 5.911 × 1012 b5 or Energy stored, WE = 5.911 × 1012 b5 joule

Example 44: A conducting sphere of radius ' a ' has a surface charge density ρ S (C / m 2 ). Calculate the electric energy stored in the system. Solution: Energy stored in the conducting sphere: Since the charge resides only on the surface of a conducting sphere, the energy density within the sphere is zero. To find the energy stored in the system, we must know about E at a distance r from the centre of sphere of radius ' a' . Now we draw a Gaussian sphere outside the conducting sphere as shown in Fig. 78. According to Gauss's Law, the electric flux coming out from the Gaussian sphere is

ψ = Q(enclosed) = ρ s × 4 π a2

r

According to Gauss's law →

→

∫sD . dS = Q(enclosed) →

D = (4 π r 2 ) = 4 π a2 ρ s or D =

∴

The energy density,

ρ s a2 r2

a

rɵ

ρs(C/m2)

→

2 D ρs a E= rɵ = ε 0 ε0 r 2

→

0

W =∫

→ → 1 1 ε0 E2 or W = ∫ E . D dv 2 v 2

Fig. 78

E lectrostatics

W =

∴

W=

187

ρ s a2 1 2 ∫ v r 2

2 4 1 ρs a 2 ε0

=

ρ rɵ . 2s r

∞

a2

rɵ dv but dv = r 2 sin θ d r dθ d φ ε0 2 4 π 2π ∞ dr 1 ρs a 2 sin θ dθ ∫ dφ = φ r dr d d sin θ θ ∫ ∫ 2 4 r =a r 0 0 2 ε0 r

1

2π

π

∫ r = a ∫0 ∫0

∞ 2 4 2π 1 ρ s a 1 π − [− cos θ] 0 [φ]0 2 ε0 r a

2 π ρs2 a3 ρ2 a4 1 joule = s (2) (2 π) or W = 2 ε0 a ε0

Example 45: A charge distribution with spherical symmetry has volume charge density, ρ0 ρv = 0

0 ≤r ≤a r >a

Calculate (i) The electric field intensity at all points. (ii) Potential at all points. (iii) Total energy stored in the electrostatic field. Solution: The situation of the problem is shown in Fig. 79. →

→

∫ s D . dS = ∫ v ρ v dv

(i) Inside the charge distribution (r < a): According Gauss's law

Inside (r < a)

In the region r < a, ρ v = ρ 0 →

→

2π

π

= Dr r 2

π

∫ θ =0

sin θ dθ

= Dr r 2 (2) .(2 π) or

2π

∫φ =0 →

a

r 2 sin θ dθ d φ

∫ D . dS = ∫ Dr dS = Dr ∫ φ = 0 ∫ θ = 0

∴

r

dφ

→

∫ D . dS = Dr (4 π r

2

Outside (r > a) r r

) Fig. 79

∫ v ρ0

r 2 sin θ d r dθ d φ = ρ0

r

π

2π

∫ r =0 ∫ θ =0 ∫ φ =0 r

2

sin θ dr dθ dφ = ρ0

r

∫ r =0 r

2

dr

π

r3 4 = ρ 0 (2). 2 π = π r 3 ρ0 3 3 ∴

→

→

∫ D . dS = ∫ v ρ v dv

or D (4 πr 2 ) =

2π

∫ θ = 0 sin θ dθ ∫ φ = 0 dφ

→ r ρ0 4 π r 3 ρ0 or D = rɵ 0 ≤ r ≤ a 3 3

E lectromagnetic F ield Theory

188 →

→

D rρ0 E= rɵ = ε0 3 ε0

∴

....(1)

0≤r≤a

(ii) Outside the charge distribution (r ≥ a) : a

Q enclosed = ∫ ρ0 dv = ρ0 ∫ v r =0 = ρ0 or

∴

a

∫ r =0 r

2

dr

π

2π

π

∫ θ = 0 ∫ φ =0

∫ θ =0

sin θ dθ

r 2 sin θ dθ dφ dr 2π

∫ φ = 0 dφ = ρ 0

1 3 a (2) . 2 π 3

4 ρ π a3 3 0

Q enclosed = ∫ ρ0 dv =

→

3 → ρ a3 → 4 D ρ0 a 3 0 ɵ or or π = ρ π ∴ . = ρ = r rɵ D dS dv D ( 4 r ) a D E = = v ∫s ∫v 3 0 ε0 3 r2 ε0 3 r2 →

→

2

...(2)

(iii) At the surface (r = a): →

E=

ρ0 a3

→ ρ a rɵ or E = 0 rɵ 3 ε0 3 ε0 a 2

(a) The potential at any point can be obtained from the relation. →

→

V = − ∫ E.d l for r ≥ a

V = −∫

ρ a3 dr = 0 2 3 ε0 3 ε0 r ρ0 a3

∫

ρ 0 a3 1 or V = − − + C1 3 ε0 r r2 dr

where C1 is the constant of integration. Its value is determined by initial conditions, that is, at r = ∞, V = 0

0 =0+ C1 or C1 = 0 ρ 0 a3

∴

V =

At r = a

ρ a3 V = 0 3 ε0 a

For r ≤ a

→ → ρ r ρ r2 V = − ∫ E . d l = − ∫ 0 dr or V = − 0 + C2 6 ε0 3 ε0

3 ε0 r

at r ≥ 0

or

ρ a2 V = 0 3 ε0

Where C2 is the constant of integration. Its value is determined by applying initial conditions that at r = a,

ρ a2 V = 0 3 ε0 or

∴

ρ a2 =− 0 + C2 3 ε0 6 ε0

ρ0 a2

ρ0 ρ a2 ρ0 a2 ρ0 a2 ρ r 2 ρ0 a2 ∴ V =− 0 + = C2 = 0 + = (3 a2 − r 2 ) 3 ε0 6 ε0 2 ε0 6 ε0 2 ε0 6 ε0

E lectrostatics

or

189

V =

ρ0 6 ε0

(3 a2 − r2 )

(b) The energy stored in electric field E is

1 1 ε E2 dv = 2 ∫v 0 2

WE =

ε0 E2 dv +

∫v

1 2 ∫v

ε0 E2 dv

r ≥a

r≤a

Substituting the value of E for both regions from eqns. (1) and (2), we get

r ρ0 ∫ θ = 0 ∫ φ = 0 3 ε0

a 1 ε 2 0 ∫r = 0

WE =

π

2π

2

r 2 sin θ d r dθ d φ +

∞ 1 ε 2 0 ∫r = a

or

WE =

ρ2 1 ε0 02 2 9 ε0

a

∫ r =0 r =

=

or

WE =

4

dr

ρ 20

r5 18 ε0 5

a 0

a5 (2) (2 π ) + 18 ε0 5

15 ε0

π

(− cos θ )0

ρ 20

4 π ρ 20 a5

2π

π

1

∫ θ = 0 sin θ dθ ∫ φ = 0 dφ + 2 ε0

Joule

2π 0

(φ)

ρ0 a3 ∫ θ = 0 ∫ φ = 0 3 ε r2 0 2π

π

ρ20 a6 9 ε20

∞

∫r = a

1 r

2

dr

2

r 2 sin θ dθ d r d φ

π

2π

∫ θ = 0 sin θ dθ ∫ φ = 0 dφ

ρ2 a6 1 ∞ 2π π + 0 − (− cos θ)0 (φ)0 18 ε0 r a

2 5 2 π ρ 20 a5 1 ρ20 a6 1 6 × 2π ρ 0 a +1 = (2) (2 π) = 5 18 ε0 a 9 ε0 5 × 9 ε0

E lectromagnetic F ield Theory

190

Q uestions Bank 1.

State and explain Coulomb's law of force. [UPTU, B.Tech. IV Sem. (old) 2007; GBTU, B.Tech., III Sem. 2010, III Sem 2012]

2.

State the word statement of Coulomb's law. [UPTU, B.Tech. IV Sem. 2007; GBTU, B.Tech., IV Sem. 2011]

3.

Explain and state Coulomb's law and its importance. Relate force with electric field intensity. [GBTU, B.Tech. III Sem. 2011]

4.

Derive an expression for intensity of electric field at a point distant r from a point charge.

5.

If Q1, Q2 , Q3 ... charges located at distances r1, r2 , r3 , ... from a point P then electric field intensity E

→

→

at point P will be E = 6.

1 4 π ε0

n

∑

i =1

Qi r 2i

rɵi [UPTU, B.Tech. IV Sem. 2006]

Discuss how will you find the electric field due to point charges at a particular point. [UPTU, B.Tech. IV Sem. 2005]

7.

Derive an expression for the intensity of electric field due to an infinite sheet of charge.

8.

Derive an expression for the intensity of electric field due to an infinitely long wire. [GBTU, B.Tech. III Sem. 2012]

9.

A circular disc of radius ' a' is uniformly charged with surface charge density ρ s . If the disc lies on the z = 0 plane with its axis along the z-axis, show that at point (0, 0, h): →

E=

ρs h 1 − zɵ 2 2 2 ε0 ( ) + a h

[GBTU, B.Tech. III Sem. 2011]

10. A thin circular disc of inner radius ' b ' and outer radius ' a' carries a uniform surface charge density ρ s . Show that the electric field intensity at a point P (0, 0, h) on z-axis when a → ∞ is, →

E=

ρs h 1 zɵ 2 ε0 (b2 + h2 )

Using above expression show that →

E=

ρs h 2 ε0

→ ρ 1 1 − zɵ, when b → 0 and E = s zɵ, when b → 0 and a → ∞ 2 2 2 ε0 h (a + h )

11. Find the field of a sheet of charge in which symmetry is about a plane? [U'Khand B.Tech. IV Sem. 2011] 12. Relate electric flux density and electric field.

[GBTU, B.Tech. III Sem. 2011]

13. Describe electric flux density and dielectric constant.

[GBTU, B.Tech. IV Sem. 2010]

14. What is Gauss's law? State and prove it in a particular coordinate system. [UPTU, B.Tech. IV Sem. 2003, IV Sem. 2005]

E lectrostatics

191

15. State and prove Gauss's law and discuss its application.

[UPTU, B.Tech. IV Sem. 2002, IV Sem. 2004]

16. What is Gauss's law ? Apply Gauss law to find electric field everywhere due to uniformly charged spherical shell of radius r.

[UPTU, B.Tech. IV Sem. (C.O.) 2006]

17. An infinite long line charge of uniform density ρ L C/cm is situated along the z-axis. Obtain electric field intensity due to this charge using Gauss's law.

[UPTU, B.Tech. IV Sem. 2008]

18. Is Gauss's law useful finding electric field vector of a finite line charge? Explain. [UPTU, B.Tech. IV Sem. (old) 2007]

19. Use Gauss's law to calculate the electric field intensity due to a uniformly charged sphere at (i) an external point, (ii) at an internal point and (iii) at the surface. 20. What do you mean electric potential? Explain it. →

[GBTU, B.Tech. IV Sem. 2010] →

21. Relate electric flux ψ and electric flux density D with electric field E .

[UPTU, B.Tech. IV Sem. 2006]

22. Calculate the potential at any point between two grounded semi-infinite parallel electrodes separated by a distance ' b ' and a plane electrode at potential V.

[UPTU, B.Tech. IV Sem. 2006]

23. Derive an expression for potential difference at any point between spherical shells in terms of applied potential using laplace equation.

[GBTU, B.Tech. IV Sem. 2011]

24. Derive the electric field intensity at a distance r from an electric dipole. [GBTU, B.Tech. III Sem. 2011] 25. Find the electric field strength at any point due to an infinitesimal electric dipole. [UPTU, B.Tech. IV Sem. (C.O.) 2006]

26. Derive energy density in electrostatic field.

[GBTU. B.Tech. III Sem 2012]

27. Discuss the energy stored in electric field.

[UPTU, B.Tech. IV Sem. 2004]

28. Find mathematical expression for electrostatic energy in terms of field quantities. [UPTU, B.Tech. IV Sem. (old) 2007]

29. Charge distributed throughout a volume V with density ρ v give rise to an electric field with energy 1 1 content, WE = ∫ ρ v VdV . Show that its equivalent W = ∫ ε E2 dV , where ε is the permittivity of 2 2 the medium.

[UPTU, B.Tech. IV Sem. 2007]

30. What is electrostatic energy? Discuss the energy stored in a capacitor which is charged to a voltage. [UPTU, B.Tech. IV Sem. 2005]

31. Determine the energy stored in the electric field in a concentric spherical shell. [UPTU, B.Tech. IV Sem. 2006]

E lectromagnetic F ield Theory

192

nsolved Numerical Problems 1.

A charge Q2 = 121 × 10 9 coulomb is located in vacuum at P2 (− 0.03, 0.01, − 0.04). Find force on Q2 due to Q1 = 100 µ C at P1 (0.03, 0.08, 0.02). All distances are in meters.

2.

Four 3 pC charges are at the corners of a 1-m square. The two charges at the left side of the square are positive. The two charges on the right side are negative. Find the field E at the centre of the square, ε r = 1.

3.

A thin non-conducting rod of length l has a charge + Q uniformly distributed over it. Calculate the electric field intensity at a point on its right bisector at a distance y from it. Express the result in terms of Q, y and l. How the result is modified when l → ∞ ?

4.

(i)

A circular line charge of density ρ 0 coulomb/m of radius a is lying in the x - y plane with its centre at the origin, calculate the potential at a point (0, 0, h).

(ii)

Calculate the potential at a point (0, 0, h) due to a circular disc of radius a having a surface charge density of ρ s C / m 2 with its centre at the origin. Calculate the field at point P (0, 0, z).

5.

A hollow sphere is charged to 12 µC of electricity. Find the potential: (a) at its surface, (b) inside the sphere, (c) at a distance of 0.3 m from the surface. The radius of the sphere is 0.1 m.

6.

Determine the electric potential at the point at which E is zero when point charges of +3 × 10 −6 coulomb and +7 × 10 −6 coulomb are located at points (0, 0) and (0.5, 0).

7.

What is the potential at the centre of a square having charges of +2, + 1, − 2 and +3 in terms of 10 −8 C each, if side of the square is 1.41 m?

8.

A charge + 10 −9 coulomb is located at the origin in free space and another charge Q at (2, 0, 0). If the X-component of electric field at (3, 1, 1) is zero, calculate the value of Q. Is the Y -component zero at (3, 1, 1) ?

9.

A non-conducting ring of radius 0.5 m carries a total charge of 111 . × 10 −10 coulomb distributed →

non-uniformly at its circumference producing an electric field E everywhere in space. Calculate the value of line integral −∫

0 → ∞

→

E . d l (l = 0 being centre of ring)

10. A charge of −0.3 µC is located at A(25, − 30, 15), in cm, and a second charge of 0 .5 µC is at →

B (−10, 8, 12) cm. Find E at (a) the origin, (b) P (15, 20, 50) cm. →

11. Find | E| at P (3, 1, 2) for the field of: (a) two coaxial conducting cylinders, V = 50 volts at r = 2 m and V = 20 V at r = 3 m; (b) two radial conducting planes, V = 50 volts at φ = 10 ° and V = 20 Vat φ = 30 °. 12. A proton is at a distance of 10 Å from an infinite plane conductor. Calculate the force experienced by the proton and the work done in moving it to infinite distance away from the conductor.

E lectrostatics

193

13. Given the volume charge density ρ v = − 2 × 10 7 ε0 √ x C /m 3 in free space, let V = 0 at x = 0 and V = 2V at x = 2.5 mm. At x = 1 mm, find (a) potential V, and (b) E x . →

14. Find | E| at P (3, 1, 2) for the field of two radial conducting planes for which V = 50 volts at φ = 10 ° , and V = 20 volts at φ = 30 ° . 15. Given a 60 µC point charge located at the origin, find the total electric flux passing through: (a) that portion of the sphere r = 26 cm bounded by 0 < θ < π / 2 and 0 < φ < π / 2; (b) the closed surface defined by r = 26 cm and z = ± 26 cm; (c) the plane z = 26 cm. →

16. Calculate D in rectangular coordinates at point P (2, − 3, 6) produced by: (a) point charge Q A = 55 mC at Q(−2, 3, − 6); (b) a uniform line charge λ LB = 20 mC / meter on the X-axis; (c) a uniform surface charge density σ sc = 120 µC /m 2 on the plane z = −5 m.

17. A one core lead sheathed cable has a conductory core of 0.5 cm diameter and the lead sheath has inside diameter 1.5 cm The insulating material is rubber. At what voltage will the insulation break down? Given that rubber has a dielectric strength of 400 kV / cm. 18. A metallic sphere of radius 10 cm has a surface charge density of 10 nC/m 2 . Calculate electric energy stored. → 10 −4 19. Charge lies on the circular disc r ≤ 4 m, z = 0 with density ρ s = C /m 2 . Determine E at r

r = 0, z = 3 m. 20. The electric field between two concentric cylindrical conductor at r = 0.01 m and r = 0.05 on is given →

by E = (105 / r ) rɵ V / m, fringing neglected. Find the energy stored in 0.5 m length. Assume free space. 21. Two dipoles with dipole moments −10 aɵz n C / m and 18aɵ z n C/m are located at points (0, 0, − 2) and (0, 0, 3) respectively. Find the potential at origin. 22. A uniform line charges of ρ e = 3µ C / m lies along the z-axis, and a concentric circular cylinder of

radius 2 m has ρ s = (−1.5 / 4 π) µ C /m 2 . Both distributions are infinite in extent with z. Use Gauss's →

law to find D in all regions.

E lectromagnetic F ield Theory

194

nswers to Unsolved Numerical Problems 1. −54 aɵ x − 63 aɵ y − 54 aɵz 3. E =

E=

q 2

2

2 π ε0 Y (4 y + 2 )

2. 153 mV /metre 4. V = p

ρL 2 π ε0 Y

→ Eρ

ρ0 a 2 ε0

(a2 + h2 )

ρ z = s 2 ε0

1 1 − 2 2 z + ( ) a z

5. 108 × 104 Volt,108 × 104 Volt 27 × 104 Volt

6. 343.8 × 103 Volt

7. 360 Volt

8. − (3 5 /2 × 10 −9 /113 /2 ) C, No

9. 2 volt

10. 92.3 aɵ x − 77.6 aɵ y − 105.3 aɵz kV/ m, 32.9 aɵ x + 5.94 aɵ y + 19.69 aɵz kV / m

11. 23.4 V/ m, 27.2 V/ m

12. 5.76 × 10 −11 N (attractive) 5.76 × 10 −20 J

13. 0.302 volt, – 555 V/ m

14. 27.2 V / m

15. 7.5 µC, 60 µC, 30 µC

16. 6.38 aɵ x − 9.57 aɵ y + 19.14 aɵz µC / m 2 , − 212 aɵ y + 424 aɵz µC / m 2 , 60 aɵz µ C / m 2

17. 329.4 kV

18. −94.76 × 10 −6 J

19. 1.51 aɵz MV / m

20. 0.224 J

21. − 40.5 Volt

22.

→

0.477 µC rɵ 2 , 0 < r < 2, m r

→

0.239 µC rɵ 2 r > 2 m m r

D= D=

qqq

Theory

Numerical

Solved Examples

Questionnaire

195

Unit-2 S ection

B E lectric F ield in M aterial S pace

I

S

ntroduction o far we have dealt with interactions between charged particles or bodies without referring to their properties or with the cases in which electric field is produced exclusively either by charges in

different types of distributions in which charges were held stationary at fixed positions or by free charges on the surface of the conductor. From these discussions we have learnt some basic properties of electrostatic field. We have ignored problems involving material space. From the point of view of electrostatics, we can classify all substances according to the case with which electric charge move in them. Thus, charge can be transferred from point to point in metals but not in rubber or dry cotton. The metal is a conductor of electricity. These are substances that contain an unlimited supply of charges that are free to move about through the material. In fact, ordinarily, the electrons are not associated with the particular nucleus, but roam around at will. Thus the metal is conductor of electricity whilst the other type is described as an insulator or dielectric. In dielectrics all charges are attached to specific atoms or molecules. An ideal dielectric material is one which has no free charges and does not permit the passage of electric charge. However, in insulators local microscopic displacements may take place. Actually there are no perfect insulators. Usually insulators like ebonite, quartz, mica, porcelain etc have so vanishingly small conductivity that they behave as if they were perfect insulators. In fact all material media are composed of molecules and atoms. The molecules of the dielectric are certainly affected by the presence of electric field. The electric field causes the displacement of positive and negative parts of each molecule from their equilibrium position in opposite directions. From the macroscopic point of view, over all effect

E lectromagnetic F ield Theory

196

of the field is visualized as a displacement of entire positive charge in the dielectric relative to the negative charges. This displacement of positive and negative charges of dielectric under the influence of electric field is called polarization. The polarization of dielectric depends on the total electric field in the medium, but a part of the electric field is produced by the dielectric itself. This electric field of dielectric modify the free charge distribution on conducting bodies this in turn will change the electric field in the dielectric. In this chapter we develop few methods for handling this situation. Before coming to the main task, we will apply electrostatic laws and methods to some of the materials. After the introduction of current, current density and continuity equation, we shall discuss conducting material and develop ohm's law in both its microscopic and macroscopic forms. Having both conductors and dielectrics, we may put them to form capacitors. The importance and utility of Laplace's and Poisson's equations will also discuss in this chapter. Boundary conditions and electrical image method for solving electrostatic problems will also be introduced here.

P roperties of Materials Various experimental investigations lead to the conclusion that materials can be broadly classified into the following two classes:

(i) Conductors or metals

(ii) Insulators or dielectrics

Conductors and insulators can easily be distinguish on the basic of free and bound electrons. Electric charges are free to move through the body of the conductor. These mobile negative charge electrons in the metal are known as free electrons. In fact the atoms of the metallic conductors are very loosely attached to the rest of the atoms. The atoms in metal are very closely packed together in a regular pattern. Because of this close packing of atoms in metals the outer electrons of the atom do not remain attached to the atoms but are capable of free movement throughout the volume of the metal. Metals possess very low resistivity (ρ ~10 – 2 – 10 – 8 Ω m ) or high conductivity (σ ~102 − 108 sm –1). On the other hand, in insulator, the electrical charges are bound and they can hardly move within the insulator. Like metals, free electron movement does not occur in insulators or dielectrics. The charge in them are called bound charge. An insulator does not have such free movable charges as each electron stays near its parent nucleus. Insulators have high resistivity (ρ ~1011 − 1019 Ω m) or low conductivity

(σ = 10 −11 − 10 −19 sm –1). The conductivity of a material depends the temperature and frequency. Following are the main properties of conductors and insulators. In charging a conductor if electrons are removed, conductor becomes positively charged and thus its potential increases and if added, conductor becomes negatively charged and its potential decreases. The dielectric constant of conductors in electrostatics is infinite, whereas insulators have finite dielectric constant. Conductor is an equipotential surface, that is, potential at its surface everywhere is same. Electric intensity inside a conductor is zero, whereas outside, near the surface of the conductor, it is ρ / ε0 .

197

Charge resides on the outer surface of a conductor. However, the distribution of charge on the surface of a conductor is usually nonuniform, whereas the charge given to an insulator is localised. Conductors conduct electricity while insulator does not. Insulators prevent flow of current through them. When an insulator is placed in an electric field, the electrons may be able to move backwards and forwards about their equilibrium position but they are unable to leave the vicinity of their atoms. There are certain materials whose electronic properties are intermediate between those of metals and insulators. These materials are called semi-conductors. Semi-conductors like germanium, silicon, etc have resistivity (10 −4 to 0 .5 ohm. meter) between good conductors like aluminium, copper (resistivity

~1.7 × 108 Ω − m) and insulators like mica, glass (resistivity ~9 × 1011 Ω m). On the basic of band theory of solids, conductors, semi-conductors and insulators are described as follows: In case of conductors, there is no forbidden band and the valence band and conduction band overlap each other. Hence, a large amount of free electrons are available for electric conduction. In case of insulators, the forbidden energy gap is very wide. Due to this reason electrons cannot jump from valence band to conduction band. Hence, ordinarily there is no flow of current in insulators. In semi-conductors, the forbidden band is small. When a small amount of energy is supplied, the electron can easily jump from valence band to conduction band. When the temperature is increased, the forbidden band is decreased so that some electrons are liberated into the conduction band.

C onvection and Conduction Currents – →

Conduction Current All conductors contain the array of tightly bound (+ve) ions which are surrounded by electrons which can move about, quite freely due to thermal agitation. But their average velocity will be zero, in the

–

– →

– →

+ Fig. 1

absence of any external field. When the ends of the conductors are connected to a battery, that is, an electric field is applied within the conductor as shown in Fig.1, electrons →

→

will experience a force ( F = − e E) in the direction opposite to the direction of the field which accelerate the electrons. Therefore, when the electrons were in free space, it would be continue to accelerate under an external field. The drift of the electrons in a conductor due to an external field cause an electron to flow. The flow of electrons constitute an electric current. The current flowing in a metal conductor is called conduction current. The flow of charge in a conductor, under the influence of an external field, from one place to the other or the state of motion of the charge is called conduction current. The rate of flow of charges through a macroscopic surface or charge flowing per second in a conductor is the measure of conduction current. For the steady time independent current the symbol is I and i(t) for time varying currents. If in a conductor dQ coulomb of charge flows in dt second, then the conduction current I is,

I=

dQ Q or I = dt t

E lectromagnetic F ield Theory

198

The unit of current is coulomb/sec or ampere. By definition, the current I is a scalar quantity, and when we refer to the direction of current, it is supposed to be the direction of the current density vector. By convention, the direction of a current is opposite to the direction of flow of free electrons which produce current. When the direction of motion of the charge does not change with time, the current is said to be direct current (d.c.), but when the direction of charge motion reverses periodically and regularly the current is said to be alternating current (a.c.). The current in a conductor is the characteristic property of the particular conductor. All metals are good conductors of electricity. The substance having very few electrons are bad conductor (or insulator) of electricity. The flow of current takes place not only in metals but also in certain liquids and gases. In liquid conductors electron conduction or the charge is carried by ions, both positive and negative ions. In gaseous conductors the charge is carried by electrons and positive ions. In semi-conductors the charge is carried by electrons and holes, the holes behave like a positive charges.

Convection Current The current we have discussed so far is conduction current which arises in response to an electric field imposed on a conductor. This current represents the drift motion of charge carriers in a medium, the medium as a whole, may be and usually at rest. But it is also possible for charge masses, like electrons or ions to transport the charge directly from one point to another without any medium. This type of current is known as convection current. Thus, the flow of charges from one point to another without any medium is called convection current. A beam of electrons in a vacuum tube or in a cathode ray tube is a convection current. Convection current does not involve conductors and consequently does not satisfy ohm's law like a conduction current. Convection currents are important to the subject of atmospheric electricity. In fact upward convection currents in thunderstorms are sufficient to maintain the normal potential gradient of the atmosphere above the earth.

Current Density The electric current I which is the flow of electric charges is a macroscopic quantity of a particular conductor. Current density on the other hand is a microscopic quantity, characteristic of a point inside the conductor rather than a conductor as a whole. The current density is more fundamental than current itself, it is because of the fact that current density is a point parameter. The current density at any point inside the conductor is defined as a vector quantity whose magnitude is equal to the current through an infinitesimal area surrounding that point; and perpendicular to the direction of flow of charge and whose direction is along the direction of flow of positive charge at that point or →

direction in which current passes through that point. Current density is denoted by the symbol J . Consider a conductor of arbitrary shape of cross-sectional area A and suppose the current be distributed uniformly across the conductor. Then we define current density J , as,

J=

I A

...(1)

199

This is the magnitude of the current density for all point on the cross-section of the conductor. Usually, the distribution of current across a cross-section in non-uniform. In such cases consider an infinitesimal →

surface area dS surrounding the point across which the current dI is supposed

→

dS

to be uniform (Fig. 2) and is given by, →

→

dS

→

d I = J . dS

→

J

...(2)

P

The total current I through the surface S is obtained by taking surface integral of eqn. (2), as

I =∫

→ S

→

Fig. 2

...(3)

J . dS

→

From eqn. (3) current I is defined as the flux of the current density vector J through a given area.

C onduction and Convection Current Densities Conduction Current Density Because of the random motion of electrons in the absence fo any field, the resultant motion is zero, hence no current flows. When an electric field is applied to a conductor an electric current begins to flow. As electron is being a negative charged particle, the force acting on it under the influence of electric intensity →

E is,

→

→

F = − eE

Due to this force, electrons move in a specific direction. The directional motion of the free electron is called a drift. The average velocity gained by electrons during this drift motion is termed as drift →

velocity ( v d ). The electron drift is in a direction opposite to that of applied field. During the accelerated motion, the electron collides with the lattice of one atom to other. Suppose m be the mass of electron and →

t the mean time per collision. The loss of momentum in time t is m vd . The average rate at which free electron loses its momentum in collision is →

m vd t According to Newton's law, Average rate in momentum = Applied force mvd et that is, = − eE or vd = – E t m or vd = − µ e E where, mobility of electron, µ e = et /m

...(1) ...(2)

If n be the number of electrons per unit volume, then the electron charge density ρ v is, ρ v = − ne

...(3) →

In a conductor or in a conducting medium, the conduction current density J is given by →

→

J = ρ v vd

...(4)

E lectromagnetic F ield Theory

200 →

Substituting the values of vd from eqn. (2) and ρ v from eqn. (3) in eqn. (4), we get →

→

J = + ne µ e E or

→

→

J =σE

where σ = ne µ e is known as the conductivity of the conductor.

σ = ne . e t / m or σ =

ne2t m

Convection Current Density →

Here we shall relate current density J with the velocity of the volume charge density ρ v at a point. To simplify the situation, let us suppose that the elemental volume ∆V of surface area ∆S is located with its edges along the coordinate axes as shown in Fig. 3. ∆Q = ρv ∆V

Z

∆Q = ρv ∆V

∆x

Y

Y

X

∆L

∆S

∆L

∆S

X

(a)

(b) Fig. 3

and that the velocity of volume charge is only along x-axis as shown in fig. 3. If ∆Q is the charge on the volume element ∆V, whose surface area is ∆S and length ∆ L, then The charge ∆Q on the elemental volume ∆V is given by

∆ Q = ρ v ∆V = ρ v ∆S . ∆L

...(1)

where ρ v is the volume charge density

Suppose, in the time interval ∆t, the element of charge has moved a distance ∆x as shown in fig 3 (b). The resultant current is given by ∆I =

∆Q ∆t

= ρ v ∆S

∆x or ∆ I = ρ v ∆S v x ∆t

where v x is the x-component of velocity In terms of current density J x , we have J x = ρ v v x or

→

→

J = ρv v

This clearly shows that the charge in motion constitute a current. This type of current is called convection →

→

current and J or ρ v v is the convection current density.

201

Drift Velocity of Charge Carriers in a Conductor According to modern theory, a metal is a lattice of fixed positively charged ions in which large number of free electrons are moving randomly. The randomly moving free electrons inside the metal collide with the fixed ions or lattice and follow a zig-zag path. In the absence of any electric field due to this random motion, the number of electrons cross from left to right and from right to left are equal so the net current through a cross-section is zero. When an electric field is applied, inside the conductor, each free electron experience an electric force which accelerate the motion of electron. However, the velocity of the electron does not increase continuously because free electron continuously loses energy gained by external field in repeated collisions with the fixed ions of the conductor in the form of heat. Thus, applied field does not provide an accelerated motion to the electrons, but simply gives them a small constant velocity. This is called drift velocity of the electrons. Therefore, the drift velocity is the average uniform velocity acquire by free electrons inside a metal by the application of an applied electric field which is responsible for current through it.

Relation between Drift Velocity of Electrons and Current Density →

As soon as an electric field is established in the metalic wire, free electrons begin to move with velocity vd in the direction opposite to the field as shown in fig. 4.

E – –

Suppose A is the area of cross-section of wire and n be the number of free electrons per unit volume of the wire and each having charge e. If in time dt, the electron advances a distance l with drift velocity vd , then

– –

vd

– –

Fig. 4

l = vd dt The number of electrons contained in a volume Al is equal A vd dt. The number of electrons crossing the length l of the wire in time dt is same as the number of electrons contained in volume l A.

∴ Number of electrons = Volume × no. of electrons per unit volume = A vd dt . n The charge passing through the cross-section of wire in time dt

dQ = number of electrons × charge on each electron dQ = An vd dt e Thus, the current I =

or

dQ ne Avd dt = = ne A vd dt dt

I = ne Avd

This is the relation between current and drift velocity of electron. ne vd A I The current density, J= = A A ∴ J = ne vd →

This is the relation between current density J and drift velocity vd .

E lectromagnetic F ield Theory

202

Continuity Equation The law of conservation of charge states that total charge of any system remains conserved or constant if no external charges are introduced in it. This law holds for all charge transformations occurring in nature. The mathematical representation of this law of conservation of charge in differential form is called continuity equation. When the flow of charge is distributed throughout a three dimensional region, we describe it by the →

volume current density J defined by the equation →

→

J = ρv v

...(1) →

→

where v is the velocity of charge at a point where the (mobile) volume charge density is ρ v and J the current per unit area perpendicular to flow or the charge passing through unit area in unit time. Therefore, the total current I emerging from the closed surface S which encloses the volume V is given by

I=∫

→

J . nɵ dS = ∫

S

→ S

→

...(2)

J . dS

From equation (1), we have →

I = ∫ ρ v ( v . nɵ) dS

...(3)

S

Because the charge is conserved the current emerging from the closed surface S in unit time is always equal to the current entering the volume V in the same unit time. Thus, the reduction of charge per unit time inside the volume V must be equal to the current represented by eqn. (3), that is, ∂Q I=– ∂t

Q = ∫ ρ v dV

But

V

I=−

∴

∂ ∂t

∫ V ρv

or I = − ∫

V

∂ ρv dV ∂t

...(4)

The minus sign indicates that outward flow decreases the charge left in volume V. Equating equations (2) and (4), we get

∫S

→

→

J . dS = − ∫

V

∂ ρv dV ∂t

...(5)

According to Gauss-divergence theorem, the surface integral may be transformed into volume integral, as

∫S

→

→

→

J . dS = ∫ (div J ) dV

...(6)

V

Therefore, equation (5), may be rewritten as,

∫V (div

→

J ) dV = − ∫

V

∂ρv dV ∂t

or

∫V div

→

J +

∂ρv dV = 0 ∂t

...(7)

Since equation (7) is true for all volumes, therefore the integrand in this equation must vanishes, that is, →

div J +

∂ρv =0 ∂t

or

→ →

∇. J +

∂ρ v =0 ∂t

...(8)

203

This is the mathematical statement of law of conservation of charge or continuity equation between →

current density J and the volume charge density ρ v at a point. Continuity equation is one of the basic relation applicable in electromagnetic theory. ∂ρ For steady current v = 0, therefore continuity equation (8) reduces to ∂t → →

∇. J = 0

...(9)

→

→

The current density J is related with an electric field E applied to a conductor as →

→

J = σE

...(10)

where σ is the electrical conductivity of the conductor. →

In term of displacement vector D, equation (10) may be written as → σ → D J = σ = D ε ε

→

...(11)

where ε is the permittivity of the medium. →

Substituting the value of J from equation (11) in equation (8), we get →

∂ρ σ → ∇. D = − v ε ∂t

or or

or

σ → → ∂ρ v ∇. D = − ε ∂t

∂ ρv σ ρv = − ε ∂t

→ →

(∵ ∇. D = ρ v )

∂ρ v σ + ρv = 0 ∂t ε

...(12)

Equation (12) is the another form of continuity equation. The solution of equation (12) is of the form, ρ v = ρ0 e − (σ / ε) t or ρ v = ρ 0 e − t / τ

...(13)

where ρ0 is the net volume charge density at t = 0 It is clear from equation (13) that the volume charge density ρ v decays exponentially with a time constant

τ = ε / σ, also known as relaxation time. Relaxation time is the time taken for the drift velocity to decay to 1/e of its initial value. At t = τ,ρ has decayed to 36.8% of its initial value. The time constant τ for a conductor is extremely small, of the order of 10 −19 sec. This reveals that the free charge is distributed evenly over the surface of a conductor and cannot remain within the conductor.

Example 1: Given the total current in circular conductor of radius 4mm if the current density varies according to J =

10 4 A / m2 . r [GBTU, B.Tech. IV Sem. 2011]

Solution: The current I and current density J is related as

E lectromagnetic F ield Theory

204

I=∫

∫S

→

→

J . dS = ∫

∫ S J dS cos 0 ° = ∫ ∫ S JdS

104 A / m2 , and dS = r d r d φ r

Here

J=

∴

I=∫

2π

0 .004

∫ r =0

φ =0

2π

104 r d r d φ = 104 r 0 .004

I = 104 (φ)0 .(r )0

2π

0 .004

∫ φ = 0 dφ ∫ r = 0

dr

= 104 × 2 π × 0.004 = 80 π

I = 251. 2 Amp

or

Example 2: The vector current density is given by →

J =

4 r2

cos θ rɵ + 20 e −2 r sin θ θɵ − r sin θ cos φ φɵ A / m2

→

(a) Find J at r = 3, θ = 0 and θ = π (b) Find the total current passing through the spherical cap r = 3, 0 ≤ θ ≤ 20 ° , 0 ≤ φ ≤ 2 π in the rɵ direction. Solution: (a) At r = 3, θ = 0, φ = π

Current density, or

→

J =

→

J =

4

cos 0 ° rɵ + 20 e −2 ×3 sin 0 ° θɵ − 3 sin 0 ° cos π φɵ

32

4 rɵ = 0 . 444 rɵ A / m 9

(b) Total current I in rɵ direction is →

I=∫ or

I=∫

→ 4 J . dS = ∫ 2 cos θ rɵ + 20 e −2 r sin θ θɵ − r sin θ cos φ φɵ . ( r 2 sin θ dθ dφ rɵ) S r

S 2π

φ =0

θ = 20 °

∫θ = 0 °

20 ° 4 2 2 cos θ r sin θ dθ dφ = ∫ θ =0 r

2π

∫φ = 0

4 cos θ sin θ dθ dφ (∵ θɵ . rɵ = φɵ . rɵ = 0)

I =4 =4

20 °

∫0

20 °

∫0

cos θ sin θ dθ sin 2θ 2

2π

dθ [φ]0

2π

∫0

dφ 20 °

cos 2θ = 4 π − 2 0

= 2 π[− cos 40 ° + cos 0] = 2 π[−0. 7660 + 10 . ] or

I = 1. 47 Amp

205

Example 3: Given a current density 10 3 J = 2 cos θ θɵ A / m2 r π , 0 . 001 ≤ r ≤ 0 . 080 m. 4

In spherical coordinates, find the current crossing the conical strip, θ =

Solution: Current, I = ∫

or

→

→ J . dS , Here dS = r sin θ d r d φ θɵ

103 ɵ I = ∫ 2 cos θ θɵ .(r sin θ dr dφ θ) S r = 103

or

→ S

cos θ sin θ dr dφ = 103 r

∫S

I = 103 sin

π π cos 4 4

0 .080

∫r = 0 .001

r = 0 . 080

∫r = 0 . 001

1 dr r

2π

∫0

1 dr r

2π

∫φ = 0

sin θ cos θ dφ ∵ θ = π 4

dφ

0 . 080

103 × 0.7071 × 0.7071 [log e r ] 0 . 001 . 2 π = 314 . × 103 × 2.3036 [log10 0.080 − log10 0.001] = 3.14 × 2.3036 × 103 [log10 (80 × 10 −3 ) − log10 (10 × 10 – 4 )] = 3.14 × 2.3036 × 103 [log10 80 − 3 − log10 10 + 4] = 3.14 × 2.3036 × 103 [1.9031 − 3 − 1 + 4] = 3.14 × 2.3036 × 103 × 1.9031 or

I = 1. 376 × 10 4 A

Example 4: In a cylindrical conductor of radius 3mm, the current density varies with distance from the axis according to J = 10 3 e −400ρ (A /m2 ) Find the total current. →

Solution: Current I = ∫ J . dS , Here dS = dρ d φ S

∴

I=∫

0 .003 ρ =0

2π

∫0

I = 2 π × 103 ∴

I=−

or

I=

103 e − 400ρ dρ d φ = 103 0 . 003

∫ρ =0

2 π × 103 400

e −400ρ dρ ∫

2π 0

dφ

e −400ρ dρ put − 400 ρ = t ∴ − 400 dρ = dt

–1 . 2 t

∫0

0 .003

∫ρ = 0

e dt = −

2 × 3.14 × 103 −1. 2 2 × 3.14 × 103 (e − e0 ) = – 400 400

2 × 3.14 × 103 × 2.32 = 10 . 97 Amp 400 × 3.32

1 − 1 . 3 32

E lectromagnetic F ield Theory

206

→

Example 5: Show that equation of continuity div J +

∂ = 0 is contained in Maxwell's equation. ∂t

or Starting from Maxwell's equations, establish the equation of continuity. Solution: According to Maxwell's fourth equation →

→

→

curl H = J +

∂D ∂t

Taking divergence of both sides, we get → → → ∂ D div curl H = div J + ∂ t →

But div curl H = 0 → → → → ∂D ∂ D or div J + div J + div = 0 ∂ t = 0 ∂ t

∴

→

div J +

or

→ ∂ (div D) = 0 ∂t

(∵ space and time operations are interchangeable)

→

From Maxwell's first equation, div D = ρ, where ρ is the charge density. →

div J +

∴

∂ρ =0 ∂t

Example 6: Find the relaxation time for silver. Its conductivity σ is equal to 6.17 × 10 7 Siemen / m. Show that the volume charge density ρ v decays to 36. 78% of its initial value at time equals to its relaxation time. Solution: Time constant or relaxation time may be expressed as ε ε τ= = 0 σ σ

Here

ε0 = 8.854 × 10 −12 C2 / N- m 2 and σ = 6.17 × 107 Siemen /m

∴

τ=

8.854 × 10 −12 6.17 × 107

= 1. 435 × 10 −19 sec

The volume charge density ρ v is expressed as

ρ v = ρ0 e − t / τ , where τ is the time constant.

According to the given problem, ∴

t=τ ρ v = ρ0 e − (τ / τ)

ρ0 = 36.78 × 10 −2 ρ 0 e That is, volume charge density decays to 36. 78% of its initial value. =

207

C onductors Conductors are the substances which contain plenty of free electrons. These charge carriers are free to wander throughout the conducting material. These carriers respond to almost very feeble electric fields, and they continue to move as long as the field exists. These free charge carriers carry the electric current when a steady electric field is maintained in the conductor by an external agency. In actual practice there are no perfect conductors, but many substances come surprisingly close. According to the band theory of solids, to form a conductor atoms Partially 2p are brought close together. As the atom approaches each other, filled there is profound modification in energy levels. The closeness of conduction atoms, within the atomic dimensions, results in the intermixing of band electrons in the neighbouring atom. Due to this intermixing the number of permissible energy levels increases. Hence instead of 2s Valence single energy level, there will be bands of energy levels, each band band consisting of very numerous closely spaced discreat energy levels. The electrons with the highest energy levels, the valence electrons, 1∆ are located in the valence band. The next higher permitted band Beryllium or next higher empty band into which electrons can pass is termed (1s22s2) as conduction band. The conduction band and valence band are Fig. 5 separated by a region or gap known as forbidden band or energy gap. In conductors, the valence band itself is the conduction band. In a conductor there is a partially filled band above completely filled lower bands. A partially filled band may be the result of overlapping of a completely filled band and an empty band. Suppose an electric field is applied across a piece of such conductor (solid sodium). The electrons in the partially filled valence band easily acquire additional energy to move to the higher empty energy levels within the same band, without crossing any energy gap. The additional energy is in the form of kinetic energy, and moving electrons constitute an electric current. In Fig.5, the energy bands of beryllium in which there is an overlap of lower energy levels of the empty 2p band with the upper energy levels of the completed 2s band is shown. Thus, a partially filled valence energy-band is a feature of conductors. At 0° k, the conductors exhibit no energy gap between the valence and conduction bands. The insulator show a large energy gap and the semiconductor has only small energy gap as shown in Fig. 6.

Empty Conduction Band Filled Valence Band Conductor

Empty Conduction Band

Empty C.B.

Energy Gap

Energy Gap

Filled V.B.

Filled V.B. Semi-conductor

Insulator Fig. 6

E lectromagnetic F ield Theory

208

C onductor in an Electric Field When a conductor is placed in an external electric field established between the plates of a charged capacitor, the free negative charge carriers that is, free electrons in the conductor move toward the surface facing the positive plate, leaving immovable positive ions or positive charge on the other surface facing the negative plate. This motion of free electrons continues until the field set up by the induced charges is equal and opposite to the original external field. Thus inside the conductor, the resultant field is everywhere reduced to zero. In the gap between the conductor and the plates of the capacitor, the field is the same as it was before the conductor was inserted. Fig. 7(a) shows the original field between the plates of a capacitor in the absence of conductor. In Fig. 7(b) all the lines of force leaving the positive plate terminate on the induced negative charges on the left face of the conductor. An equal number of lines originated at the induced positive charges on the right face of the conductor and terminate on the negative charges on the other plate of the capacitor. The induced charges on the faces of the conductor are equal and opposite to the original charges on the plates of the capacitor. Thus, the electric field inside the conductor is zero.

Fig. 7

Under steady state condition the charge density ρ in the interior of the conductor is zero. This idea can be explained with the help of Gauss's law. According →

→ →

to Gauss's law if E = 0 within the conductor, then ∇ . E = 0 = charge density (ρ/ ε), that is, charge density is also zero within the conductors. →

→

We know that electric field can be represented as negative gradient of potential, that is, E = − ∇ V , →

therefore, when E = 0, the potential (V ) will be constant within the conductor. That is, in the steady flow of current, the surface of the conductor will be an equipotential surface.

O hm's Law in Point Form and Resistance of a Conductor When a conductor is placed in an external electric field, due to the drift motion of free electrons in the conductor, conduction current flows. As electrons move, they encounter some damping forces called →

resistance. It is experimentally observed that a metal at constant temperature, the current density J is linearly proportional to the applied electric field, that is →

→

→

→

J ∝ E or J = σ E

...(1)

The constant of proportionality is called conductivity (σ ) of the materials in mhos per m or siemens per m. The equation (1) is called Ohm's law in point form.

209

In 1826 G. Simon ohm experimentally established a relation between A

the voltage V over a length l of a conductor and the current I through the conductor in terms of a parameter, called resistance R is,

V = IR

I

J

...(2)

Thus, according to Ohm's law, the potential difference V across the ends of a conductor is equal to the product of current I and resistance R. R is independent of I and is the characteristic of a conductor. →

V=l E Fig. 8

For the cross-sectional area A of a block of conductor current density J is expressed as, →

I A and the resistivity ρ r is given by RA , where l is the length of the block ρr = l It is well known that the conductivity σ of a conductor is reciprocal of resistivity ρ r , that is, J =

σ=

l 1 = ρ r RA

R=

or

l σA

...(4)

Potential across the ends of block of length l is V = lE Substituting the values of V from eqn. (5) and R from eqn. (4) in eqn. (2), we get lE =

l I σA

or

I = σAE

→

J =σE

This is the required Ohm's law in point form.

R esistance and Power Resistance The resistance R of a conducting material can be derive from the Ohm's law as follows: →

The total current passing through a surfaces in terms of current density J is

I=∫

→ S

→

J . dS

The potential difference V is related with the electric field E, as → →

V = ∫ E . dl

The resistance of a conductor of non-uniform cross-section is

...(5)

...(6)

Substituting the value of I from eqn. (6) in eqn. (3), we get →

...(3)

E lectromagnetic F ield Theory

210

V R= = I

or

R=

∫

→

→

∫

E. d l

→

∫ S J . dS

→

→

E. d l →

→

∫ σ E. dS

The resistance of a conductor of uniform cross-section is also obtained from Ohm's law as follows: Consider a block of conductor of length l and cross-sectional area 'A' across the end of which a potential difference ' V ' is applied. The direction of flow of current I is the same as the direction of electric field E →

produced in the block. The magnitude of E is, V | E| = l

...(1)

For a conductor of uniform cross-section, I J= A According to the point form of Ohm's law

...(2)

J = σE Comparison of eqns. (2) and (3), gives, I σE = A From eqn. (1), we have σV I = l A V l Hence = R= I σA or where ρ r =

R=

...(3)

lρ r A

1 , is resistivity of the material. σ

Power and Joule's Law When a metallic conductor is placed in an electric field, free electrons begin to move with drift velocity vd . If ρ v be the volume charge density, then the electric force experience by the charge in volume dV is given by →

→

d F = ρ v dV E

...(1) →

→

If in time dt, the charges move through a distance d l with drift velocity vd , then →

→

d l = vd d t

...(2) →

The work done by the force when charges move through a distance d l is →

→

dW = d F . d l

...(3)

211 →

→

Substituting the value of d F from eqn. (1) and d l from eqn. (2) in eqn. (3), we get → →

→ →

dW = ρ v dV E . vd d t = ρ v vd . E dV dt → →

→

→

dW = J . E dV dt (∵ ρ v vd = J )

or

...(4)

The work done per unit time is the corresponding power, therefore dP =

dW → → = J . E dV dt

...(5)

If P is the power density or power per unit volume, then → →

d P = P dV = J . E dV →

→ →

P = J.E

or

...(6)

Thus, the power associated with volume V is, → →

P = ∫ ( J . E ) dV

...(7)

V

This is known as integral form of Joule's law. For a linear conductor of a uniform cross-sectional area, we have from eqn. (5) dP → → = J.E dV → →

dP = J . E dl dS

or

where dl is the elemental length and dS, the elemental area of the conductor, then → →

P = ∫ dP = ∫ J . E dl ds = ∫ J Ed ldS P = ∫ Edl

or

l

∫e Edl = V

But

∫S

and

JdS

∫ S J dS = I

P = VI = I 2 R

∴

This is again Joule's law and states that the rate of power dissipated in the form of heat varies as the square of current (I 2 ) in a linear conductor.

Example 7: An aluminium wire of diameter 0 . 24 cm is connected in series with a copper wire of diameter 0 .16 cm. The wires carry an electric current of 10 amp. Find out: (a) Current density in aluminium wire, (b) drift velocity of electrons in the copper wire (Given: Number of electrons per cubic meter volume of copper = 8. 4 × 10 28 ) Solution: (a) The radius of the aluminium wire, r =

Hence, area,

D 0. 24 = = 0.12 cm =0.12 × 10 –2 m 2 2

A = π r 2 = 3 .14 × (0 .12 × 10 −2 )2 = 4.52 × 10 −6 m 2

E lectromagnetic F ield Theory

212

∴ (b)

Current density, J =

I 10 = = 2. 22 × 10 6 Amp . / m2 A 4.5 × 10 −6

Area of cross-section of the copper wire, A′ = π r 2 = π × (0.08 × 10 −2 )2 = 2.0 × 10 −6 m 2 I 10 = = 5 × 106 Amp./m 2 A′ 2.0 × 10 −6

∴ Current density in copper wire, J ′ =

The drift velocity of electrons in copper wire, vd =

J′ ne

Here n = 8.4 × 1028 electrons/m 3 and charge on electron, e = 16 . × 10 −19 C vd =

∴

5 × 106 × 1.6 × 10 −19

28

8.4 × 10

= 3. 72 × 10 −4 m/s

Example 8: An aluminium conductor of conductivity σ = 3. 82 × 10 7 mho / m is 40m long and has a circular cross-section with a diameter of 20 mm. If there is a dc voltage of 1. 5V between ends, find. →

(i) magnitude of current density, | J | , (ii) current, (iii) power dissipated →

→

Solution: (i) The magnitude of current density, | J | = σ| E|

The length of the conductor, d = 40 m and V = 1.5 volt

(ii)

→

V 1.5 = = 0.0375 V /m d 40

∴

| E| =

Thus,

| J | = σ | E| = 3.82 × 107 × 0.0375 = 1. 432 × 10 6 A /m 2

Current,

→

I=

So, ∴

→

V ρl l and R = = , where A = π r 2 = π (10 × 10 −3 )2 R A σA R= I=

40 3.82 × 10 × π × (10 × 10 −3 )2 7

1.5 3.335 × 10 −3

or

R = 3. 335 × 10 −3 ohm

= 449. 77 Amp.

(iii) Power dissipated, = I 2 R = (449.77)2 × 3.335 × 10 −3 = 674. 65 Wat

B oundary

Conditions in Static Electric Field (or at Conductor Surface in Free Space) →

→

For many electrostatic problems, it is important to know how the field vectors E and D change in crossing a boundary under static electric field conditions. The simple relations that exist between the tangential and →

→

normal components of two vectors E and D are known as boundary conditions.

213 →

Let AB be a boundary between a conductor and free space as shown in Fig. 9. The external electric field E is divided into tangential component Et and normal component En as shown in Fig. 9. Similarly, electric →

flux density D is divided into two components, one tangential (Dt ) and other normal (Dn) to the conductor surface. Tangential components of E, that is, Et →

and of D, that is Dt should be zero, otherwise static

Free space a a′

conditions could not be maintained. Tangential force will disturb the static condition. Therefore for static

∆W

En b′

Et

c

d

conditions, the tangential components Et and Dt should be zero. To prove this condition mathematically, draw a

E

b

Conductor

A

B

Fig. 9

Gaussian surface abc d in a plane perpendicular to boundary as shown in Fig. 9. The work done in moving a charge along the rectangular path must be zero, that is,

∫ or

∫

→

→

→

→

E.d l = 0 E.d l = ∫

b → a

→

c→

→

d→

∫b E.d l + ∫c

E. d l +

→

E.d l +

a→

∫d

→

...(1)

E.d l

Since electric field inside the conductor is zero, therefore the contributions of the sides cd, b′c and a′ d is zero because they lie inside the conductor. Thus eqn. (1) reduces to

∫

→

→

E.d l = ∫

b →

a

→

E.d l +

b→

→

b′ →

∫b

b′ →

= ∫ E.d l +

∫b

a

→

E.d l + →

E. d l +

c →

∫b′

a →

∫ a′

→

E.d l +

d→

∫c

→

E.d l +

a′ →

∫d

→

E.d l +

a →

∫a ′

→

→

E.d l

...(2)

E.d l

∆h ∆h = d a′ = a′a = 2 2 → → ∆h ∆h + En = Et ∆W E . d l = Et ∆W + (− En) 2 2

Suppose ad = ∆h = bc, Therefore bb′ = b′ c = ∴

∫

But E t ∆W = 0 ∴

E t = 0 because ∆W ≠ 0

...(3)

The tangential component of electric flux density,

D t = ε0 Et = 0

...(4)

Thus, the tangential components of E and D at the boundary of conductor and free space is zero.

∆S ∆h Free space

→

→

To obtain the conditions for normal components of E and D consider a infinitesimal Gaussian cylinder as show in Fig. 10. Apply Gauss's law to this cylinder, we have

∫S ∫

→

Fig. 10

→

D . dS = Q(enclosed) →

→

→

→

D . dS = ∫ D . dS + upper plane surface

Conductor

...(5) →

→

→

→

→

→

∫ D . dS + ∫ D . dS + ∫ D . dS = Q lower plane surface

curved surface in f ree space

curved surface in conductor

...(6)

E lectromagnetic F ield Theory

214

The lower plane surface of the Gaussian cylinder and curve surface in conductor are situated in the →

conductor, therefore their contributions to the integral is zero, because the electric field E inside the conductor is zero. Thus, integral represented by eqn. (6) is reduces to, →

→

→

→

→

→

∫S D. dS = ∫ D . dS + ∫ D . dS = Q upper plane surface

...(7)

curved surface in free space

Since the height of cylindrical surface is negligible compared to its diameter, the contribution of curved surface is vanishingly small. Therefore eqn. (7) further reduces to →

→

→

→

∫ D . dS = ∫ D . dS = Q = Dn ∆S

...(8)

upper plane surface

Thus,

Dn ∆S = Q = ρ s ∆ S, where ρ s is the surface charge density Dn = ρ s

or

...(9)

The normal component of electric field intensity is, Dn = ε0 En = ρ s En =

or

ρs ε0

...(10)

Thus, the magnitude of normal component of electric field intensity is

1 times the surface charge ε0

density of the conductor. Since no work is done in moving a charge on the surface of the conductor, the conductor surface is equipotential. Hence, the direction of electric field and electric flux density is normal to the surface. π π Example 9: A potential field is given as V = 40 e −4 x sin 3 y cos 4 z volt. A point P 0 . 2, , is 12 24 →

situated at the conductor-free space boundary. Find (a) V , (b) E, (c) En, (d) Et and (e) ρ s , at this point. Solution: (a) V = 40 e −4 x sin 3 y cos 4 z

π π 3π 4π , = 40 e −4 × 0 . 2 sin cos ∴ Vat 0. 2, 12 24 12 24

= 40 e − 0 . 8 sin (b)

→

→

We know that, E = −∇V or E = − Here, ∴

40 π π cos = × 0.7071 × 0.8660 = 11. 01 Volt 4 6 2 .225

∂V ∂V ∂V aɵ x − aɵ y − aɵz ∂x ∂y ∂z

V = + 40 e −4 x sin 3 y cos 4 z ∂V ∂V = 160 e −4 x sin 3 y cos 4 z and = 120 e −4 x cos 3 y cos 4 z ∂x ∂y

215

∂V = − 160 e − 4 x sin 3 y sin 4 z. ∂z →

E = − 160 e − 4 x sin 3 y cos 4 z aɵ x − 120 e − 4 x cos 3 y cos 4 z aɵ y + 160 e −4 x sin 3 y sin 4 z aɵz

So →

Eat 0. 2,

π π , 12 24

= − 160 e −0 .8 sin

π π π π π π cos aɵ x − 120 e − 0 . 8 cos cos aɵ y + 160 e − 0 . 8 sin sin aɵz 4 6 4 6 4 6

= − 40 e − 0 . 8 [4 × 0.7071 × 0.8660 aɵ x + 3 × 0.7071 × 0.8660 aɵ y – 4 × 0.7071 × 0.5 aɵz ] =− Eat

P

40 [2.45 aɵ x + 1.837 aɵ y − 1.4142 aɵz ] = − 17.978 [2.45 aɵ x + 1.837 aɵ y − 1.4142 aɵz ] 2.225

= − 44. 046 aɵ x − 33. 025 aɵ y + 25. 424 aɵ z →

→

| Eat P| = (−44.046)2 + (−33.025)2 + (25.424)2 = 3677.08 ∴ | Eat P | = 60 . 64 V /m (c)

At the surface of the conductor only normal component exists ∴

(d)

→

At the surface of the conductor tangential component is zero ∴

(e)

→

| E n| = | E P| = 60 . 64 V /m

Et = 0 →

Dn = ρ s = ε0 | E| So

ρ s = 8.854 × 10 −12 × 60.64 C /m 2 or ρ s = 536. 9 × 10 −12 C /m2

Example 10: Given the electric potential, V =

10 sin θ sin φ r2

volt

(a) Find the equation of the conductor surface on which V = 10 volt. → π π (b) Find E at point P r, , on the conductor surface. 2 2

(c) Find ρ s on the surface at point P. Assume ε = ε0 adjacent to the surface. Solution: (a) V = 10 Volt

10 =

10 sin θ sin φ r2

or

r2 = sin θ sin φ

Eqn. (1) is the equation of the conductor surface at which V = 10 volt (b)

We know that in spherical coordinates → ∂V ∂V ɵ 1 ∂V ɵ 1 rɵ + φ θ+ E = − ∇V = − ∂ ∂ r r r sin ∂φ θ θ

→

Here

V=

10 sin θ sin φ r2

...(1)

E lectromagnetic F ield Theory

216

20 sin θ sin φ ∂V 10 cos θ sin φ ∂V ∂V 10 sin θ cos φ , and =− = = ∂r ∂θ ∂φ r3 r2 r2

∴

→

20 sin θ sin φ ɵ 10 cos θ sin φ ɵ 10 cos φ ɵ θ+ E = − − r+ φ r3 r3 r3

∴

r 2 = sin θ sin φ = sin

But,

∴

→

E at P r =1, π , π 2 2

π π sin = 1 2 2

[From eqn. (1)] or r = 1 metre.

π π π 20 sin π sin π 10 cos 10 cos sin 2 2 2 2 2 ɵ θ+ φɵ = − − rɵ + 3 3 3 1 1 1 →

= – [−20 rɵ + 0 + 0] So E at P = 20 rɵ V /m (c)

→

We know that, Dn = ρ s = ε0 | E| But

→

(ρ s ) at P = ε0 | E|at P = 8.854 × 10 −12 × 20 ∴ ρ s = 177.08 × 10 −12 C /m2

Dielectrics Materials such as glass, ceramics, ebonite, porcelain, polymers, rubber, quartz, mica, dry cotton etc. are non-conducting materials. They prevent flow of current through them. When the main function of non-conducting materials is to provide electrical insulation, they are called insulators. When non-conducting materials in an electric field act as a source of storage of electric charges, that is, when charge storage is the main function of the materials, then, the materials are called dielectrics. In fact dielectric is a material in which all the electrons are tightly bound to the nuclei of their atom and no free electron is able to migrate so that the conductivity is idealy zero. Although a field applied to an insulator may produce no migration of charges, but the bound charges can be treated as a source of electrostatic field. In an electric field insulators can be viewed as a free space arrangement of microscopic electric dipoles which are composed of positive and negative charges whose centres do not coincides. They are bound by atomic and molecular forces and are able to displace in response to an external field. They are called bound charges. The importance of dielectrics in electrical phenomena was first observed by Faraday. He kept a dielectric slab between the plates of a parallel plate capacitor and observed that the charge on the capacitor is increases abruptly by the introduction of dielectric between the plates and consequently the capacitance of the capacitor increases significantly. The main characteristic of dielectric is its ability to store electric energy. The storage of energy takes place by means of shift in the relative positions of the bound positive and negative charges against the molecular and atomic forces.

217

D ielectric An Atomic View Polar and Non-polar Dielectrics The actual mechanism of charge displacement in the presence of electric field differs in the various dielectric materials. On the basis of charge displacement dielectrics are broadly divided into two groups, namely, non-polar and polar dielectrics. In fact molecule is a neutral system in which the sum of all positive and negative charges is equal to zero. All positive charges (protons) of a molecule may be replaced

±

without adding any error by an equivalent single positive charge located at the actual centre of gravity of all positive charges. Similarly all negative charges (electrons) may be replaced by a single equivalent negative charge located at the actual centre of gravity of all negative charges. The magnitude positive and negative charges at their respective centre of gravity are equal. In a molecule the centres of gravity of positive and negative

(a) –

+ (b)

charges may or may not coincides [Fig. 11(a) & (b)].

Fig. 11

Molecules, in which centre of gravity of positive charges and centre of gravity of

negative charges coincides [Fig. 11(a)], do not possess a permanent dipole moment. On the other hand, if their centres of gravity do not coincide in space [Fig. 11(b)], the molecule possess a permanent dipole moment. A non-polar molecule is one in which the centre of gravity of positive charges coincides with the centre of gravity of the negative charges. A nonpolar molecule has obviously a zero electric dipole moment. Molecules having symmetrical structure, like H2 , N2 , CO2 , CH4 , CCl4 , C6 H6 , C6 H12 , CS2 etc. are nonpolar. Atoms, due to their spherical symmetry, are also nonpolar. Materials made up of nonpolar molecules are called nonpolar dielectrics. The permittivities of nonpolar dielectrics are low and in the range from 1 to 2.2. A polar molecule in one in which the centre of gravity of the negative charges is separated from the centre of gravity of positive charges by a finite distance. Thus, polar molecules have permanent displacement existing between the centres of gravity of the positive and negative charges. Hence in polar molecules each pair of charges acts as a dipole. Normally the dipoles are randomly oriented throughout the interior of the metal and the action of external field is to align these molecules, to some extent in the direction of the field. Molecules like HCl, CHCl3 , C2 H5 OH, C6 H5 NO2 , C6 H5 Cl,CO, H2 O, NH3 etc. are polar molecules. In HCl molecule, the electron of H atom lies more toward Cl atom than toward H atom. Hence, the H end of the molecule is positive and Cl end is negative the HCl molecule is therefore, an →

electric dipole having a dipole moment p directed from Cl atom to H atom Fig.12(a). Similarly carbon mono oxide (CO) molecule is a dipole having a

+ H

→

p (a)

–

+

Cl

C

Fig. 12

– →

p (b)

O

E lectromagnetic F ield Theory

218

→

dipole moment p from O atom to C atom [Fig. 12(b)]. Materials made up of polar molecules are called polar dielectrics.

Non-Polar Dielectric in An Electric Field Let us first observed the effect of electric field on a nonpolar dielectric. Suppose a slab of

+

nonpolar dielectric is placed in the uniform

–+ –+ –+

–+ –+ –+

parallel plate charged capacitor as shown in Fig.13. Electric field causes a force to be

–+ –+ –+

exerted on each charged particle within the

–+ –+ –+

molecule. The positive particle pushed in the

–+ –+ –+

oppositely, so that the positive and negative

+

–

–+ –+ –+

electric field E set up between the plates of a

direction of field and the negative particles

–

+

–+ –+ –+ → E0

parts of each molecule are displaced from their

E

equilibrium positions in opposite directions. The displacements are necessarily minute as

×

E′

–

–

E′

(a)

the elastic restoring forces within the atom

+ Slab

(b) Fig. 13

tend to keep it in its normal neutral configuration. Thus each atom or molecule of the dielectric acquires an induced dipole moment which will produce an electric field of its own. The development of induced dipole moments in the dielectric in the presence of electric field is called polarization of dielectric. The entire dielectric can be replaced by a configuration of a large number of such infinitesimal electric dipoles. Which produces an electric field of its own even in the absence of any free charge. The induced dipole moments and the polarization disappear when the electric field is removed. In this way the slab of a polar dielectric becomes electrically polarized. Although the dipoles exists throughout the volume of the dielectric but in interior region positive and negative charges of the adjacent dipoles cancel each other's effect, so only the dipoles near the surface are important. The net effect is the appearance of the negative charge on the inner side of one face of the slab and an equal positive charge on the inner side of opposite face of the slab produce their own →

→

→

electric field E′ [Fig.13(b)] which opposes the external field E0 . Hence the resultant field E (= E0 − E ′ ) →

within the dielectric is smaller than external field E0 , but in the same direction [Fig.13(a)]. Hence we concluded that when a dielectric is placed in an external electric field, the field within the dielectric is weakened, but not reduces to zero like a conductor.

Polar Dielectric in An Electric Field Now we consider the behaviour of a polar dielectric in the presence of external electric field. We have seen that polar dielectric has permanent dipoles or dipole moments. In the absence of an electric field, the permanent dipole moments of the molecules are distributed randomly in all directions in such a way that no net dipole moment is observed in the dielectric [Fig. 14(a)]. When an electric field is imposed on the dielectric, there is a tendency for the permanent dipoles to align themselves in the direction of field [Fig. 14(b)] although, because of thermal motion, the departure from random orientation is very small but the alignment becomes more and

219

more perfect as the electric field is increased or the temperature is decreased. Thus a net dipole moment is produced in the polar dielectric. –

–

+

–

–

+

–

+ –

+

+

– +

–

–

– +

+

(a)

+

+

–

–

– +

– +

+

–

– +

+

+

–

–

–

+

+

–

+

+

–

+

–

+

+

–

–

+

+

–

(b)

Random orientation with Σ → p=0 Fig. 14

Polarization in Dielectrics Electrons cloud of charge –Q

It is well known that insulators or dielectrics (or non-conductors) are materials whose electrons are tightly bound to the nucleus by strong electrostatic forces and there is no free electron even under the influence

±

of an external electric field. In nonpolar dielectrics the molecules, which

Nucleus with charge +Q

are usually diatomic and composed of two atoms of the same type, may be represented as a positive nuclei of charge + Q surrounded by a symmetrically distributed electrons cloud of charge − Q, as shown in

Fig. 15

(Fig. 15) In the absence of any external field, the centres of gravity of positive and negative charges coincide (Fig. 15). When such molecules are

+Q

subjected to an external field, they experience an electric force which

+

pushes the positively charged nucleus slightly in the direction of field and the centre of negatively charged electron cloud in the opposite direction.

d →

p

–

As soon as the centre of the electron cloud shift away from the nucleus, a

–Q – +

Fig. 16

strong restoring electrostatic for of attraction is generated between the nucleus and the electron cloud which tends to restore the displaced charges to their original position. At equilibrium, the internal restoring force balances the external field. In this situation the centres of positive and negative charges remain separated and molecule behaves like a dipole (Fig. 16). The process of displacement of centres or centres of gravity of the positive and negative charges by a distance, under the action of an external field to which they are subjected or the process of forming tiny dipoles in a dielectric, is called dielectric polarization and the molecules are said to be polarized. Suppose + Q and − Q charge is displaced by a distance ' d ' (Fig. 16) under the action of external force, then each molecule form a dipole whose dipole moment is,

E lectromagnetic F ield Theory

220 →

→

p = Qd

→

where d is vector pointing from the negative to positive charges. For a molecular dipole, the numerical value of Q is equal to the charge on electron, that is, 1. 6 × 10 −19 coulomb and the distance (d) between the centres of positive and negative charges is of the order of →

molecular dimension (10 −10 m). Hence, the numerical value of dipole moment ( p ) of a molecular dipole is, →

p = 1.6 × 10 −19 × 10 −10 ≈ 1. 6 × 10 −29 cm.

If there are N molecules or N dipoles in a volume ∆V of the dielectric, then the total dipole moments in an elemental volume ∆V is given by →

→

→

N

→

Q1 d1 + Q2 d2 + Q3 d 3 + ... + Q n d N =

→

∑ Qi d i

i =1

→

Thus the dipole moment per unit volume or the dielectric polarization p is, N

→

∑ Qid i

→

P = lim

i =1

∆V → 0

∆V

→

or P =

Np ∆V

C – m −2

In polar dielectrics, the molecules have permanent dipole moments. Normally in the absence of any external electric field these molecular dipoles are randomly oriented in all directions throughout the dielectrics, and change direction constantly because of the thermal motion of the molecules. In this condition, the average dipole moment over any volume element is zero. In the presence of field, the molecular dipoles has a tendency to rotate about their axis of symmetry to align themselves in the direction of field. Although, because of the thermal motion, the departure from a random arrangement is very small. The tendency of a polar dipole to rotate about their axis of symmetry to align themselves in the direction of applied field is called orientational polarization. If there are N dipoles per unit volume and we are considering a volume ∆V, then there are N∆V dipoles and the total dipole moment is obtained by the vector sum, as →

P total =

N∆V →

∑

Pi

i =1

→

The dipole moment per unit volume or dielectric polarization vector P is →

N∆V → 1 ∑ Pi ∆V → 0 ∆ V i =1

P = lim

221

Field due to a Polarized Dielectric Let us consider a piece of polarized dielectric having a large number of lined →

up microscopic dipoles with dipole moment P per unit volume. Suppose A be

A p

r

the point outside the polarized dielectric at which the potential and field

polarized dielectric

produced by the piece of polarized dielectric is desired. To find the potential at P entire dielectric is divided into a large number of elemental volume dv. →

→

Fig. 17

The dipole moment of the dipoles in this elemental volume dv is P = P dv. →

We know that the potential at P distant r from single dipole p is given by, →

1 rɵ. p dV = 4 πε0 r 2

→

1 rɵ . P or dV = dv 4 πε0 r 2

...(1)

The total potential V at point P is determined by integrating eqn. (1) throughout the volume v, that is, →

rɵ . P 1 dv V = ∫ V 4 πε0 r2

or

V =

1 4 πε0

→ rɵ

∫ v P . r2

...(2)

dv

From the vector calculus, we know that, → 1 rɵ rɵ 1 grad = 2 or ∇ = 2 r r r r Thus eqn. (2), becomes,

V =

→ → 1 1 P . ∇ dv r 4 πε0 ∫ v →

→

→

...(3) →

→

→

From the vector identity, div (φ A) = φ div A + A . grad φ or A . grad φ = div (φ A) − φ div A or

→ → 1 → 1 → → → → → 1 → → A . ∇ φ = ∇ .(φ A) − φ ( ∇ . A) or P . ∇ = ∇ . P − ( ∇ . P ) r r r

→ →

...(4)

→ → 1 Substituting the value of P . ∇ from eqn. (4) in eqn. (3), we get r

1 V= 4 πε0

→ → P 1 → → ∫ v ∇ . r − r ( ∇ . P ) dv

→ → 1 1 → → ∇ . P dv − 1 V = ( ∇ . P ) dv ∫ ∫ v v r r 4 πε0 4 πε0

or

...(5)

From divergence-Gauss theorem, we know that → → → P P → ∫ v ( ∇ . A) dv = ∫ S A . dS or ∫ V ∇ . r dv = ∫S r . dS → →

→

→

...(6)

E lectromagnetic F ield Theory

222

→ → P Substituting the value of ∫ ∇. dv from eqn. (6) in eqn. (5), we get v r V=

1 4 πε0

1 V= 4 πε0

or

→

∫S

→

P . dS 1 − r 4 πε0

→

∫S

→ →

1

∫v r ( ∇ . P ) dv

...(7)

→ →

P .^ n 1 dS − r 4 πε0

∫v

∇. P dv r

...(8)

Here S is the surface bounding the volume v of the dielectric and nɵ is the unit outward normal vector at the surface of the polarized piece of dielectric. →

→ →

→

The quantities P . nɵ and − div P (or – ∇. P ) appearing in the above integrals are two scalar functions →

obtained from the polarization P and have the dimensions of charge per unit area and charge per unit volume respectively. They are given special symbols ρ →

ρ

SP

ρ

SP

and ρ

vP

SP

and ρ

vP

. Therefore,

= P . nɵ and ρ

vP

→

= − div P

are known as polarization (or bound) surface charge density and bound volume charge

density respectively. The surface density of bound charge is given by the (normal) component of polarization vector normal to →

→

the surface (that is, P . nɵ), while the volume density of bound charge (that is, − div P) is a measure of the non-uniformity of the polarization inside the material. Thus the effect of polarization is to produce → →

accumulations of bound charge, ρ

= − ∇ . P within the dielectric and ρ

vP

SP

→

= P . nɵ on the surface.

The field due to polarization of medium is just the field of this bound charge. Therefore, eqn. (8) in terms of surface density of bound charge (ρ

(ρ

vP

SP

) may be expressed as,

V =

1 4 πε0

∫S

ρ

SP

r

dS +

1 4 π ε0

∫v

) and volume density of bound charge

ρ

vP

r

dv

...(9)

→

Electric field E is obtained by using relation →

E = − grad V

Substituting the value of V from eqn. (9), we get ρ ρ → 1 1 vP SP dS + E = − grad dv ∫ ∫ 4 πε0 v r 4πε0 S r → 1 1 1 1 ρ ρ grad dS − or grad dv E=− r r 4 πε0 ∫ S SP 4 πε0 ∫ v v P We know that,

rɵ 1 grad = – 2 r r

...(10)

223

1 Substituting this value of grad in eqn. (10), we get r →

E=

1 4 πε0

∫S

ρ

SP 2

r

rɵ

dS +

1 4 πε0

∫v

ρvP rɵ r2

dv

Eqns. (9) and (11) revealed that the potential (and field) at an external point due to polarized piece of →

dielectric can be determind by assuming it as equivalent to bound surface charges of density ρ S (= P . nɵ) and bound volume charges density ρv (= − div P). In terms of ρ S and ρv the total polarization charge of a polarized dielectric is, →

→

QP = ∫ ρ S dS + ∫ ρ v dv = ∫ P . nɵ dS + ∫ (− div P ) dv S v S v

...(11)

From the Gauss-divergence theorem,

∫S Thus,

→

→

A . nɵ dS = ∫ (div A) dv v

→

∫ S P . nɵ dS = ∫ v (div

→

...(12)

P ) dv

Replacing the first surface integral of eqn. (10) by volume integral using eqn. (11), we get →

→

QP = ∫ (div P ) dv − ∫ (div P ) dv = 0 v

v

Thus, the total polarization charge is zero. It is true because in the biginning we have assumed that the dielectric as a whole is electrically neutral.

L aws of Electrostatic Field in Presence of Dielectrics The electric field derived from Coulomb's law of electrostatic forces in air is not of universal validity. Coulomb's law has been proved experimentally for air or vacuum. It has been realized that the law is not true for dielectric media. It is because of the fact that, there is no uniform polarization in dielectrics. It is due to either dielectric itself is not uniform or due to the variation of field in uniform dielectrics. Hence laws of electrostatics need reconsideration in case of dielectrics. →

Let us consider a case in which polarization P is not uniform. When an unpolarized piece of dielectric is placed in an electric field, a certain amount of net charge will pass through a given element of area dS due →

to the polarization process. The amount of this charge will be equal to the product of the component of P →

normal to dS and the magnitude of dS, that is, P . nɵ dS. The displacement of the charge can result in a →

volume charge density, if P in non-uniform. The total charge displaced out of any volume V by the polarization is

→

∫S P . nɵ dS, where S is the surface which bounds the volume. An equal excess charge of the

opposite sign is left behind. If it is Q, then

E lectromagnetic F ield Theory

224

Q = −∫

→

P . nɵ dS

S

...(1)

If ρ vP is the polarization charge density within the volume V, then Q=∫ ρ

dv

...(2)

dv

...(3)

vP

v

Comparison of eqn. (1) and (2) gives −∫

→ S

P . nɵ dS = ∫ ρ

v vP

According to Gauss-divergence theorem surface integral can be replace by volume integral, as → → → → → ∵ ∫ S A . nɵ dS = ∫ v div A dv ∫S P . nɵ dS = ∫ v ( ∇ . P ) dv With this substitution eqn. (3) takes the form → →

− ∫ ( ∇ . P ) dv = ∫ ρ

v vP

v

or

dv

→ →

− (∇ . P) = ρ

...(4)

vP

Therefore, polarization charge density ρ

vP

→

is expressed as the negative divergence of the

polarization density P. According to Gauss's law in free space, we have → → ρ ∇ . D = ρ v or ∇ . E = v ε0

→ →

...(5)

where ε0 is the permittivity of free space. Similar law (Gauss's law) for dielectric medium can be obtained by assuming that a dielectric region contains free charge density in addition to bound charge density. The free charge might consist of electrons on a conductor or ions embedded in the dielectric materials. Within the dielectric, the total charge density can be expressed as,

ρ total = ρ

vP

+ ρv

...(6)

where ρ v is the volume charge density in free space. In this condition Gauss's law is expressed as, → → ρ v + ρ vp ∇. E = ε0

...(7)

Substituting the value of ρv P from eqn. (4) in eqn. (7), we get → →

ρ − ∇. P ∇. E = v ε0

→ →

or

→

→

→

→

→ →

or ∇ . ε0 E = ρ v – ∇ . P

→

∇ .(ε0 E + P ) = ρ v

...(8) →

For Parentheses term is designated by new vector field D, as →

→

→

D = ε0 E + P

Thus eqn. (8) is modify as,

...(9)

225 → →

∇ . D = ρv

...(10)

→ →

∇. D = ρ

or Simply,

...(11)

Eqn. (11) represents the general form of Gauss law, it is applicable to a dielectric medium as well as in free space. Eqn. (11) represents the first of the Maxwell's four fundamental equations of electromagnetics.

Example 11: The electric field intensity in polystyrene (ε r = 2.55 ) filling the space between the plates of a parallel plate capacitor is 20 kV / m. The distance between the plates is 2. 0 mm, χ e = 1. 55, calculate, (i) D, (ii) P, (iii) the surface charge density of free charge on the plates, (iv) the surface density of polarization charge and (v ), the potential difference between the plates. Solution: (i) We know that, D = ε0 ε r E = 8. 854 × 10 −12 × 2. 55 × 20 × 103

D = 451.55 × 10 −9 C /m 2 = 451. 55 nC / m2

or (ii)

P = χ e ε0 E = 1.55 × 8.854 × 10 −12 × (20 × 103 ) = 274.47 × 10 −9 C / m 2 = 274. 47 nC /m2

(iii) ρ s = Dn = 451. 55 nC /m2 →

(iv) ρ sp = P . nɵ = P = 274.47n C /m 2 (v)

V = E . d = 20 × 103 × 2.0 × 10 −3 = 40volt

D ielectric Constant (Relative Permittivity) of a Medium Dielectric constant ε r which is also called relative permittivity is the chief characteristic of a dielectric. When a slab of dielectric material is placed in a uniform electric field E0 set up between the plates of a charged parallel plate capacitor carrying real charges + Q and − Q, then opposite bound charges (smaller than real charges) appear on the surfaces of the slab of a dielectric material facing the charged plates. The charges induced on the two faces produce their own field in a direction opposite to the external field. Thus, the effect of the dielectric is to weaken the field. The ratio of the electric field in free space to →

→

that in the dielectric is called dielectric constant ' K '. Therefore, if E0 is the field in free space and E →

→

that of in a dielectric ( E ∝ E0 ), then →

|E | K = →0 | E|

...(1)

→

→

The electric field E0 at a point distant r in free space due to charge Q is given by, →

E0 =

→

1 Qr 4 πε0 r 3

→ 1 and | E0| = 4π ε

Q

2 0 r

...(2)

E lectromagnetic F ield Theory

226 →

→

Similarly, the electric field E at a point distant r in a medium of absolute permittivity ε due to charge Q is, →

→

E=

1 Qr 4 πε r 3

→

and | E| =

1 Q 4 πε r 2

...(3)

→

→

Substituting the values of | E0 | and | E| from eqns. (2) and (3) in eqn. (1), we get K =

ε ε0

Thus, the dielectric constant K of a dielectric is defined as the ratio of the absolute permittivity of the dielectric (ε ) to the permittivity of the free space (ε0 ). Absolute permittivity ε of a dielectric is the product of relative permittivity ε r and permittivity of free space (ε0 ), that is, Total permittivity or Absolute permittivity, ε = ε0 ε r

Homogeneous, Linear and Isotropic Media A medium is said to be homogeneous if its characteristics such as mass, density, molecular structure etc do not vary from point to point and having the same nature throughout. On the other hand, if the characteristics of the medium varies from point to point, then medium is said to be inhomogeneous or hetrogeneous. →

A medium is said to be linear with respect to an electric field if the electric flux density D is →

→

→

proportional to the electric field E. This is true for free space, where D = ε0 E, here permittivity ε0 is constant. In most of the media permittivity is constant. The materials for which permittivity is not constant, the media is called non-linear. An isotropic material is one which has similar properties in different directions or whose properties are independent of direction. Usually material whose molecular structure is randomly oriented will be isotropic. Crystaline media and certain plasmas which has directional characteristics are said to be nonisotropic or anisotropic.

Electrical Susceptibility We have seen that in an electric field a dielectric gets electrically polarized, that is, its molecules become electric dipoles. The induced electric dipole moment per unit volume is called the electric →

→

polarization vector P. The polarization vector P at a point depends on the total electric field at that point, that is, the field due to the primary sources of the external field and the field due to dipoles to which the polarized dielectric is equivalent. Experimental evidence has shown that in general there is no →

→

simple straight forward dependence of P on E. However, for dielectric which is almost ideal (having no free electrons), homogeneous (having the same nature throughout) and isotropic (having similar properties →

→

in all directions), the polarization P is proportional to the total field and in the same direction as E, that is,

227 →

→

P∝ E

→

→

P = χe E

or

The dimension less proportionality constant χ e is called electric susceptibility and is the characteristic of every dielectric.

Relation between Electric Susceptibility and Dielectric Constant →

Electric displacement D in a dielectric is expressed as, →

→

→

D = ε0 E + P

...(1)

→

where E is the electric field within the dielectric and ε0 the permittivity of free space. If ε is the absolute permittivity of the dielectric medium, then →

→

D=εE

We know that,

→

...(2)

→

P = χe E

→

...(3)

→

Substituting the values of D and P from eqns. (2) and (3) in eqn. (1), we get →

→

→

ε E = ε0 E + χ e E ε = ε0 + χ e ε But = K, where K is the dielectric constant of the medium ε0 ε χ χ ∴ = 1 + e or K = 1 + e ε0 ε0 ε0 or

Dielectric Strength When a dielectric is subjected to an external electric field, dielectric undergo stretching and a strain is produced on the elastic property of the molecule, which polarizes molecules by forming induced dipoles upto certain limits of the field intensity, which is called dielectric strength. Upto that limit of field all the electrons are still bound to the nucleus and no electrons are free from the molecule. But if the external field is increased above this limit the electrons can not remain bound and become free from the molecular bindings. As a result of this breakdown the dielectric becomes conducting and free electrons starting drifting towards the positive polarity while the positive ions remain near the negative polarity of the power source. With the further increase in the field sparking will start. This is called dielectric breakdown. Therefore, the dielectric strength of a dielectric may be defined as the maximum value of external electric field (or potential) that can be applied to the dielectric without its electrical breakdown or rapturing or the dielectric strength is the limiting electric field intensity above which a breakdown occurs and the charge storage property of the dielectric disappears. The dielecric strength depends on the duration of time for which dielectric is placed in an electric field and on the thickness of the insulating material.

Gauss's Law in Dielectrics The Gauss's law in dielectric states that "the flux of electric displacement vector through a closed surface is equal to the total free charge enclosed by the surface". Mathematically, it is expressed as,

E lectromagnetic F ield Theory

228 →

→

∫ D . dS = q To prove the law, let us consider a parallel plate capacitor having free

A +q

charges + q and − q on its plates of area A. We imagine the space between the

1+

–q –

B

plates initially filled by air or vacuum. Now draw a Gaussian surface ABCD with wall AD inside the plate and wall BC in space Fig. 18. The flux through

AD is zero because it is well within the conductor plate where electric field is zero (field inside the conductor is zero).

→

E0

Electric flux through the surface AB and CD is also zero, because the outward normal to the surfaces are perpendicular to the field E0 (that is,

→

→

E0 . dS = E0 dS cos 90 ° = 0). Therefore, the flux through the entire Gaussian

surface is equivalent to the flux through the surface BC only which is, →

→

The net charge enclosed by the Gaussian surface is + q. Thus, by Gauss's law → → q q ∫ E0 . dS = ε0 or E0 ∫ dS = ε0 E0 A =

or

q ε0

or

E0 =

– C Gaussian Surface

1+ D

∫ E0 . dS = E0 ∫ dS = E0 A

Fig. 18

q

...(1)

ε0 A

dielectric material of dielectric constant K. The electric field polarizes the dielectric and a layer of − q ′ charges is induces on one inner side of the dielectric and a layer of + q ′ charges is induces on the other inner

+q ′

–q′

Now, suppose the space between the plates is filled completely by a

A+q

1+ –

–q + –

B

side by the induced dipoles as shown in Fig. 19. These induced

F

→

charges produce their own field which opposes the external field E0 . →

→

Let E be the resultant field within the dielectric. The new flux through

E

the surface ABCD is,

∫

→

→

E . dS = ∫ EdS = E ∫ dS = EA

...(2)

and the net charge enclosed by the surface ABCD is q − q ′. Thus, by Gauss's law, → → q − q′ ...(3) ∫ E . dS = ε0 q − q′ or ...(4) E= ε0 A E ...(5) By definition, K = 0 or Ε0 = KE E

1+ – D

C

+ – Dielectric

Gaussian Surface Fig. 19

Comparing eqns. (1) and (5), we get E=

q Kε0 A

...(6)

229

Now comparing eqn. (4) and (6), we get q q − q′ = Kε0 A ε0 A q − q′ =

or

q K

...(7)

Substituting the value of q − q ′ from eqn. (7) in eqn. (3), we get

∫ or

∫ Kε0

→

→

→

→

E . dS =

q Kε0

E . dS = q →

...(8)

→

Kε0 E = D

But

Therefore, eqn. (7), takes the form →

→

∫ D. dS

=q

...(9)

Thus is the required expression for Gauss's law in dielectric.

Three Electric Vectors →

→

→

Electric field strength, E, Dielectric polarization vector P and electric displacement vector D are three electric field vectors. We shall discuss each, of them as follows:

→

1. Electric Field Strength E The electric field strength at a point in an electric field may be defined as the force acting on a unit →

positive charge placed at that point. The magnitude of E is measured as force per unit charge at that →

point and the direction of E is the direction of force on a unit positive charge along the outward normal from the fixed charge responsible for field. →

→

If a probe charge q0 experiences a force F at a point in an electric field, then the electric field strength E at that point is defined as, →

E=

→

F q0

→

2. Dielectric Polarization P When a dielectric is subjected to an electric field, its molecules become electric dipoles and dielectric is said to be electrically polarized. The polarized dielectric being equivalent to a large number of tiny electric dipoles situated in vacuum. The electric dipole moment per unit volume of the dielectric material is termed as dielectric polarization. Since the electric dipole moment is a vector, the dielectric →

polarization is also a vector represented by P .

E lectromagnetic F ield Theory

230

→

If p is the dipole moment of each molecular dipole and n is the number of molecules per unit volume, then →

the electric polarization P is given by, →

→

P = np

Let us consider a slab of homogeneous, isotropic dielectric of thickness ' t ' be

– q′ +q′ – + E′

+q +

placed perpendicular to a uniform electric field developed between the

–q –

plates of a charged parallel plates capacitor having charge + q and − q as shown in Fig. 20. The electric field between the plates polarizes the dielectric so that a large number of tiny dipoles are created throughout the dielectric slab. Let − q ′ and + q ′ be the bound charges induced on the end

→

faces of the dielectric as shown in Fig. 20. The dipole moment of the slab as

E0

E

whole is q ′ t. Therefore, the magnitude of the dielectric polarization is given by →

P =

Dipole moment of the slab Volume of the slab

q′ t

=

At

=

q′ A

But q′/ A is the surface density ρ′SP of the induced charge or bound polarization charges. Thus,

+

–

+

–

t Dielectric Slab Fig. 20

→

P = ρ′ SP →

The direction of polarization vector P is from negative (− q ′ ) to positive (+ q ′ ) induced charges.

→

Electric Displacement D We have seen that when a dielectric is placed in an electric field, the field within the dielectric is →

→

weakened. If E0 is the applied field and E is the induced electric field within the dielectric, then the →

resultant field E within the dielectric is →

→

→

E = E0 − E′

...(1)

that of the (bound) Let ρ s be the surface density of free charges on the capacitor plate and ρ SP polarization charges on the dielectric. The magnitude of field due to ρ S and ρ are, SP ρSP ρs and E′ = E0 = ε0 ε0 Substituting these values of E0 and E′ in eqn. (1), we get ρ ρ E = s − SP ε0 ε0 But, ρ

SP

∴

→

ε0 E = ρ s − ρ

SP

or

ε s = ρ0 E + ρ

SP

...(2)

= P (Polarization Vector) ε0 E + P = ρ s

...(3)

231 →

The quantity ε0 E + P = D, called displacement vector D. Its magnitude is equal to the surface density of free charges. Thus, D = ρs →

∴

→

→

D = ε0 E + P

→

→

...(4) →

Direction of D is the same as the direction of E and P

→

Relation between Dielectric Polarization P and Dielectric constant K: From eqn. (3) we know that

P = ρ s − ε0 E But ρ s = ε0 E0 , where E0 is magnitude of applied external field P = ε0 E0 − ε0 E E = ε0 E 0 − 1 E

∴

But

E0 = K, where K is the dielectric constant E

∴

P = ε0 E( K − 1)

or

→

→

P = ε0 ( K − 1) E

This expression revealed that in free space ( K = 1), the polarization is zero. The constant ε0 ( K − 1) = χ e and is called electric susceptibility. ∴

→

→

P = χe E

Example 12: A dielectric slab of thickness b = 0 .50 cm and dielectric constant K = 7 is placed between the plates of a parallel plate capacitor of plate area A = 100 cm2 and separation d = 1. 0 cm. A potential difference V0 = 100 volt is applied with no dielectric present. The battery then disconnected and the dielectric slab inserted. Calculate the three electric vectors E, D and P in the dielectric (ε0 = 8. 9 × 10 −12 C2 /N – m2 ). Solution: The intensity of electric field between the plates of a capacitor when no dielectric is present, is given by

E0 =

V0 100 = = 1 .0 × 104 V /m d 1.0 × 10 −2

When the dielectric slab is inserted between the plates of the capacitor, the electric field intensity is reduced to 1.0 × 10 E E= 0 = = 1. 43 × 10 3 V/m K 7.0 The electric displacement or electric flux density D in the dielectric is

D = Kε0 E = 7.0 × 8.854 × 10 −12 × 1.43 × 103 or

D = 8. 86 × 10 −8 C /m2 →

The magnitude of electric polarization vector P in the dielectric is,

E lectromagnetic F ield Theory

232

P = ( K − 1) ε0 E = (7 − 1) × 8.854 × 10 −12 × 1.43 × 103 = 6 × 8.854 × 10 −12 × 1.43 × 103 or

P = 7. 59 × 10 −8 C /m2

Example 13: The dielectric constant of helium at 0 °C and 1 atmospheric pressure is 1. 000074. Calculate the dipole moment induced in each helium atom when the gas is in an electric field of 1 volt/m. Given: ε0 = 8. 85 × 10 −12 C2 /N – m 2 and molecular density of helium = 2. 69 × 10 25 molecules / m3 at N.T.P. Solution: The electric polarization P of a dielectric placed in an electric field E is given by

P = ( K − 1) ε0 E If p is the dipole moment induced in each helium atom and n be the number of helium atoms per m 3 , then induced dipole moment per unit volume, that is, P is, P = np = ( K − 1) ε0 E or

p=

∴

p=

( K − 1) ε0 E n (1.000074 − 1) × 8.85 × 10 −12 × 1 2.69 × 1025

= 2. 43 × 10 −41 coulomb-meter

Example 14: Find the polarization in a dielectric material with (ε r = 2.8 ), when D = 3 × 10 −8 C /m2 . Solution: The polarization P in a dielectric medium is →

→

P = (ε r − 1) ε0 E

and

→

→

→

→

→

|D| D = ε E = ε0 ε r E or | E| = ε0 ε r 3 × 10 −8

→

∴

| E| =

So

| P| =

→

8.85 × 10 −12 × 2.8

= 1.2106 × 103

(2.8 − 1) × 8.85 × 10 −12 × 1. 2106 × 10 3 8.85 × 10 −12 × 2.8

= 1 . 928 × 10 −8 C / m2

Example 15: The electric susceptibility of a material is 35. 4 × 10 −12 C2 / N – m 2 . What are the values of dielectric constant and the permittivity of the material? Solution: The dielectric constant K of a material is related to χ 0 , as

K =1 + or

35.4 × 10 −12 χe =1 + ε0 8.85 × 10 −12

K =1+ 4 = 5

The permittivity,

ε = K ε0 = 5 × 8.85 × 10 −12

or

ε = 44. 25 × 10 −12 C2 / N – m2

233

E lectrostatic Boundary Conditions at Dielectric-Dielectric Surface →

(a) Boundary Condition for Electric Vector E

→

→

For the solution of many electrostatic problems, it is necessary to know how the field vectors E and D change in crossing a boundary between two media. The two media may be two dielectric with different properties or a dielectric and a conductor. The simple relations that exist between the tangential and →

→

normal components of two electrostatic vectors E and D are known as boundary conditions. Let AB be a plane boundary between two media ‘1’ and ‘2’ with permittivities ε1 and ε2 and dielectric constants K1 and K2 respectively (Fig. 21). To find the boundary →

con di tion for elec tric vec tor E, let us con sider a rectangular path abcd as shown in Fig. 21. Let the lengths ab and cd be each ∆l, and the lengths da and bc be →

infinitesimally small. Since the electrostatic field E can be expressed as the negative gradient of potential, the → →

line integral ∫ E . d l for any closed path is zero. That is

∫

→ →

E. d l = 0

Fig. 21

...(1)

Considering the closed path abcd, we have

∫

→ →

E . dl = ∫

b → → a

E. d l +

c → →

∫b

E. d l +

d → →

∫c

E. d l +

a → →

∫d

...(2)

E. d l

Since the parts bc and da of the rectangular path are negligible, we may write

∫ →

→ →

E . dl = ∫

b a

→ →

E2 . d l +

d → → E1 . d l

∫c

c → → ∵ ∫ b E . d l +

a → →

∫d

E . d l = 0

→

Since E1 t and E2 t are tangential components of the electric intensity in the two media or tangential to the paths ab and cd, we have

∫

→ →

E . d l = E1t ∆l − E2 t ∆l = 0 E1t = E2t

or →

...(3)

Thus, the component of E tangential to the boundary is same in each side of the boundary, that is, the →

tangential component of E is continuous across the boundary between two dielectrics. Special Case: If medium ‘1’ is a conductor, that is, boundary lies between a dielectric and a conductor, then →

→

E1t = 0 inside the conductor. Since E1t is zero, eqn. (3) becomes E2t = 0

E lectromagnetic F ield Theory

234

It means that, if a boundary lies between a dielectric and a conductor, then only normal →

component of electric field intensity E will exist and its direction will be perpendicular to the surface of the conductor.

→

(b) Boundary Condition for Displacement Vector D →

To find the boundary condition for D, we draw a small pill box shaped Gaussian surface, with its curved surface normal to the boundary, one plane end just above and the other plane end just below the boundary as shown in Fig. 22. Suppose ∆S be the area of each plane end which is also the area of the boundary enclosed by the pill box. The free charge enclosed by the cylinder (or pill box) is equals ρ S ∆S, where ρ

S

be the surface density of free

charge at the boundary surface. Let us apply Gauss’s law in dielectrics to the Gaussian surface (pill box). According

Fig. 22

to Gauss’s law the total flux through the pill box is equal to the total free charge enclosed by the pill box. That is,

∫ ∫

→

→

D . d S = ρS ∆ S

→ →

...(4)

→ →

∫ D. d S +

D. d S =

upper plane end

→ →

→ →

∫ D. d S + ∫ D. d S

lower plane end

curved surface

→

→

→

→

The flux of electric displacement through the plane ends of the pill box are D1. ∆ S and D2 . ∆ S , while that through the curved surface is vanishingly small (because the height of the box is negligible compared with its diameter). Therefore, eqn. (4), becomes →

→

→

→

D1. ∆ S + D2 . ∆ S = ρ ∆ S

...(5)

S

If D2 n and D1n are the normal components of the electric displacements in the two media, then eqn. (5) reduces to − D1n∆S + D2 n∆S = ρ ∆ S S

or

D 2n − D1n = ρ

...(6)

S

It is obvious from eqn. (6) that at the boundary between two dielectric media, the difference of normal components of electric displacement is equal to the surface charge density ρ

S

at the

boundary that arises due to the presence of free charges on the boundary surface. Thus, the →

discontinuity in the normal component of electric displacement D, while passing from one medium to another, arises due to the free charges on the boundary. If there is no free charges on the boundary, that is, ρS = 0 , then eqn. (6) becomes

235

D 2n = D1n

...(7) →

Thus, in the absence of any free charges on the boundary, the normal component D is the same →

on each sides of the boundary, that is, the normal component of D is continuous across a charge free boundary between two dielectrics. Special Case : If medium ‘1’ is a conductor, that is, boundary lies between a dielectric and a conductor, →

→

→

→

then D1 = 0 (as D = ε E and E in a conductor is zero.). Hence eqn. (7) becomes D1n = ρS →

That is, if the boundary lies between a dielectric and a conductor, then the normal component of D in a dielectric just outside a conductor is equal to the free surface charge density at the boundary. These boundary conditions, help us in finding the field on one side of the boundary quickly if we know the field on the other side.

R efraction of Electrostatic Lines of Force at the Boundary between two Dielectric Media

Suppose AB be a plane bound ary be tween two ideal, homogeneous isotropic dielectric media ‘1’ and ‘2’ with permittivities ε1 and ε2 and dielectric constants K1 and K2 respectively as shown in Fig. 23. A flux line (or field line) moving in one medium and reaching the boundary will be refracted into the other medium in a similar way as light wave. The angle of refraction that the field line makes with the normal to the boundary can be obtained with the help of the boundary conditions. Fig. 23

For an isotropic dielectric medium in which polarization vector →

→ →

→

→

→

P is always parallel to E, D is also parallel to E. Let us consider that the electrostatic vectors E1 and E2 →

→

(or D1 and D2 ) make angles θ1 and θ2 respectively with the normal to the boundary. →

Applying the first boundary condition that the normal component of D is continuous across the charge free boundary between two dielectric media, that is, D1n = D2 n or But ∴

...(1)

D1 cos θ1 = D2 cos θ2 D1 = ε1 E1 and D2 = ε2 E2 ε1 E1 cos θ1 = ε2 E2 cos θ2

...(2)

E lectromagnetic F ield Theory

236

→

Applying secondary boundary condition that the tangential component of E is continuous across the boundary, that is,

or

E1 t = E2 t

...(3)

E1 sin θ1 = E2 sin θ2

...(4)

Dividing eqn. (2) by eqn. (4), we get ε1 cot θ1 = ε2 cot θ2 or Since ε = ε0 K, we may write

∴

...(5)

tan θ1 ε = 1 tan θ2 ε2 tan θ1 K = 1 tan θ2 K2

...(6)

tan θ1 ε K = 1 = 1 tan θ 2 ε 2 K2

...(7)

Here, it is clear that the flux line bends closer to the normal when passes from higher to lower dielectric medium. Relation (7) is similar to Snell’s law for the refraction of light at the interface between two media. Hence, they are known as laws of refraction of electrostatic lines of force. Special Case : If medium ‘1’ is a conductor, then E1 = 0. Hence eqn. (4) becomes E2 sin θ2 = 0 or

θ2 = 0

...(8)

This states that, the electrostatic lines of force leave a conducting surface normally. Example 16: An electric field in medium 1 whose relative permittivity is 7 pasess into a medium of relative permittivity 2. If E makes an angle of 60° with the normal in the first medium, what angle does the field makes with normal in the second dielectric medium ? [GBTU, B. Tech. IV Sem. 2011]

Solution: Given ε1 = 7, ε2 = 2, θ1 = 60 ° , θ2 = ? According to snell's law in electrostatics, tan θ1 ε = 1 tan θ2 ε 2 ∴

7 tan 60 ° = tan θ2 2

or

tan θ2 =

2 2 . × tan 60 ° = × 17321 7 7

θ2 = tan –1 (0.4948) = 26 °18 ′

Example 17: The electric field strength in a mass of porcelain (relative permittivity = 6) in air is 1000 V/m. At the inner surface of the porcelain the field makes an angle of 45° to the normal and emerges into the air. Find the angle of emergence of the external field and its magnitude.

237

Solution: Let θ1 be the angle made by the field with the normal ε1, the relative permittivity and E the electric field in porcelain. Here θ1 = 45 ° , ε1 = 6, E1 = 1000 V/m, and ε2 = 1 ∴

tan θ1 ε tan θ1 6 or tan θ2 = = 1 = tan θ2 ε2 1 6

or

tan θ2 =

tan 45 ° 1 = , ∴ θ2 = tan –1(01666 . ) 6 6

θ2 = 9 °27 ′

or For magnitude of E2 , we have

sin θ1 E2 = sin θ2 E1

or E2 = E1

E2 = 1000 ×

or

=

sin θ1 sin θ2

1 1 sin 45 ° = 1000 × × sin 9 °27′ . 2 016452

1000 = 4308.48 V/m 1414 . . × 016452

Example 18: Two isotropic dielectric media are separated by plane boundary with medium 1 having relative permittivity ε r1 = 4 and medium 2 having relative permittivity ε r2 =6. Find the angle θ1 which E →

in medium 1 makes with the normal to the boundary 2 if E in medium 2 is (a) Incident normal to the boundary, and (b) Incident at an angle 60° with the normal to the plane boundary. Solution: Let θ1 be the angle which E in medium 1 makes with the normal to the boundary 2 and θ2 be the angle which the normal to the boundary makes with E in medium 2, then ε r ε0 εr tan θ1 ε = 1 = 1 = 1 , Given ε r1 = 4 and ε r2 = 6 tan θ2 ε2 ε r2 ε0 ε r2 tan θ1 4 = tan θ2 6

∴ (a)

or tan θ1 =

2 tan θ2 3

→

Since E in medium 2 is incident normal on the boundary that is, θ2 = 0, it follows that θ1 is also →

zero and therefore, E in medium 1 is also normal to the boundary. (b)

→

E in medium 2 is incident at an angle 60° with the normal to the plane boundary, that is, when θ2 = 60 ° ∴ or

tan θ1 =

2 2 = 11547 tan 60 ° = × 17321 . . 3 3

θ1 = tan–1(11547 . ) = 49 °6' '

E lectromagnetic F ield Theory

238

P oisson's and Laplace's Equations Poisson’s and Laplace’s equations are the fundamental equations which are important in more than one field of science and engineering, wherever field exists. In fact electrostatics is the study of Laplace’s and Poisson’s equations. These equations are very useful mathematical relations for the calculations of electric fields and potentials that cannot be computed by using Coulombs’s law and Gauss’s law in many electrostatic problems. Laplace's equation applies to the particular case where all charges are distributed on the surface of the conducting bodies so that volume charge density is zero at all points. We shall obtain these Laplace and Poisson’s equations as follows: →

According to the differential form of Gauss's law, the divergence of electric field E at any point is equal to 1/ε0 times the volume charge density at the point. That is , →

div E =

ρv ε0

...(1)

→

The electric field E may also be expressed as negative gradient of potential V. That is, →

→

E = − grad V = − ∇ V

→

Substituting this value of E in eqn. (1), we obtain div (− grad V ) =

ρv ε0

→ →

or

∇. ∇V = −

ρv ε0

∇2 V = –

ρv ε0

...(2)

Eq. (3) is known as Poisson’s equation for a homogenous region and the operator ∇2 as Laplacian and ∇2 in cartesian coordinates may be expressed as, ∇2V =

∂2 V ∂x

2

+

∂2 V 2

∂ y

+

∂2 V ∂ z2

Thus, in cartesian coordinates Poisson’s equation is ∇2 V =

∂2 V ∂x

2

+

∂2 V ∂y

2

+

∂2 V 2

∂z

=−

ρv ε0

...(3)

For a charge free region, where ρ v = 0, the Poisson’s equation reduces to ∇2 V = 0 This is called Laplace's equation. Laplace’s equation is applicable to those electrostatic problems in which the entire charge resides on the surface of the conductor or entire charge is concentrated in the form of point charges, line charges or surface charge densities at singular locations. It is also applicable if the region between the conductors is filled with one or more homogeneous dielectrics.

239

Laplacian, ∇2 in eqn. (4), is a pure differential operator and involves differentiation with respect to more than one variables, hence Poisson’s and Laplace’s equations are the second order partial differential equations. If the dependence of charge density function ρ (x, y, z) and the certain boundary conditions are known, then Poisson’s equation may be solved for the electrostatic potential V. The form of Laplace’s and Poisson’s equations in different coordinate system is different. The use of appropriate form depends upon the symmetry of the electrostatic problem. In cartesian coordinates Laplace’s equation is expressed as, ∇2V =

∂2 V ∂x

2

+

∂2 V 2

∂ y

+

∂2 V ∂ z2

=0

...(5)

The form of ∇2V in cylindrical and spherical coordinates may be obtained by using the expressions for gradient and divergence in these coordinate systems (as obtained earlier). For reference, the Laplacian in cylindrical coordinates (ρ, φ, z) is expressed as, ∇2V =

2 2 1 ∂ ∂V 1 ∂ V ∂ V + ρ + 2 2 ρ ∂ ρ ∂ ρ ρ ∂ φ ∂ z2

...(6)

Similarly, in spherical coordinates (r , θ, φ) it is expressed as, ∇2V =

∂2 V 1 ∂ 2 ∂V 1 1 ∂ ∂V + r + θ sin ∂ r r 2 sin θ ∂ θ ∂ θ r 2 sin2 θ ∂ φ2 r2 ∂ r

...(7)

Solutions of Laplace’s and Poisson’s Equations in One Dimension Several methods have been developed for solving Laplace’s and Poisson’s equations. The first and simplest method is that of direct integration. The method of direct integration is applicable only to those problems which are one dimensional or in which potential field is a function of only one of the three coordinates of a system. When potential is a function of two or three dimension of a coordinate system, advance mathematical knowledge is required. Solutions of Laplace’s equation are called Harmonic functions. Laplace’s equation does not by itself determine potential V. It requires suitable set of boundary conditions. Any solution to Laplace’s equation (or Poisson’s equation) which also satisfies the boundary conditions must be the only solution that exists. It is unique. This is called uniqueness theorem and states that Laplace’s (or

Poisson’s) equation can have only one solution which

satisfies the boundary conditions of the given region. The uniqueness theorem ensure that once we find any method of solving Laplace’s or Poisson’s equation subject to given boundary conditions, we have solved our problem once and for all. Poisson’s equation is used in problems involving volume charge densities, such as semiconductor diode, transistor model, vacuum tubes, magnetohydrodynamic energy conversion and ion propulsion etc.

(i) Solution of Laplace's Equation in One Dimension: Suppose potential V depends only on one variable, say, x . The Laplace’s equation becomes, ∇2V =

∂2 V ∂x 2

=0

...(1)

E lectromagnetic F ield Theory

240

The partial derivative in eqn. (1) may be replaced by an ordinary derivative, because V is independent of y and z coordinates. Therefore, d2 V d x2

=0

...(2)

Multiplying both sides by 2 dV /d x and integrating, we get dV =A dx

...(3)

where A is unknown constant or integration constant. Further integration of eqn. (3) gives V = Ax + B

...(4)

where B is another constant. The values of constants A and B can be determined only by applying boundary conditions. As the field varies only with x, the potential V is a constant if x is a constant or in other words the equipotential surfaces are describe by setting x constant. Equipotential surfaces are parallel planes normal to the x-axis. For instant, suppose boundary conditions of the problem are: V = V1 at x = x1 and V = V2 at x = x2 Applying these boundary conditions to eqn. (4), we get and

V1 = A x1 + B

...(5)

V2 = A x2 + B

...(6)

Subtracting eqn. (6) from eqn. (5), we get V1 − V2 = A ( x1 − x2 ) or

A=

V1 − V2 x1 − x2

...(7)

Substituting the value of A from eqn. (7) in eqn. (5), we get V1 = or or

V1 − V2 x1 − x2

x1 + B

B ( x1 − x2 ) = ( x1 − x2 ) V1 − (V1 − V2 ) x1 B=

V2 x1 − V1 x2 x1 − x2

...(8)

Putting the values of A and B from eqns. (7) and (8) in eqn. (4), we get

or

V=

V1 − V2 V x − V1 x2 x+ 2 1 x1 − x2 x1 − x2

V=

V1( x − x2 ) − V2 ( x − x1) x1 − x2

...(9)

Eqn. (9) is the required solution of Laplace’s equation that satisfies the boundary conditions in one dimension.

241

As an illustration, consider the case of a parallel plate capacitor in which the potential of one plate is zero and that of other plate which is at distance d from first plate is V0 , then the boundary conditions are: V2 = 0 at x2 = 0 and V1 = V0 at x1 = d, then from eqns. (7) and (8), we get V A = 0 and B = 0 d Substituting these values of A and B in eqn. (4), we get the solution of Laplace’s equation as V x V= 0 d The field between the plates is obtained as →

E = − grad V = −

→

∂V ∂x

V0 aɵ x and d → V D = − ε 0 aɵ x d E =−

∴ ∴

→

→

→

D by, D = ε E

(ii) Solution of Poisson’s Equation in one Dimension ρ Suppose the potential V depends only on one variable, say, x. The Poisson’s equation ∇2V = − v in ε one dimension becomes, ∇2V =

∂2 V ∂ x2

=−

ρv ε

...(1)

Integrating eqn. (1), we get

∫

∂2 V ρv ∂ x2 d x = ∫ − ε d x ∂V ρ =− v x+ A ∂x ε

...(2)

where A is unknown or integration constant. Again integrating eqn. (2), we get ∂V

ρ v x − A d x

∫ ∂ x d x = − ∫ ε V= –

ρv x 2 + Ax+B ε 2

...(3)

where B is another constant. Eqn. (3) represents the solution of Poisson’s equation in one dimension that satisfies the boundary conditions. The constants A and B is determined by applying boundary conditions.

Electric field and Potential Distribution near a Semiconductor Junction: Using Poisson’s Equation Let us assume a p– n junction between two halves of a semiconductor bar extending in the x-direction. We shall assume that the region for x < 0 is doped with p-type impurities and that the region for x > 0 is

E lectromagnetic F ield Theory

242

with n-type. It is well known that a charge distribution in the neighbourhood of the semiconductor junction is of the form, ρ v = 2 ρ v0 sech ( x / a) tanh ( x / a)

...(1)

Which has the maximum charge density, ρ vmax = ρ v0 that occurs at x = 0 .881a. Let us apply Poisson’s equation to find the potential distribution and electric field near the junction. There is no variation of potential in y- and z-directions, so the problem is one dimensional. One dimensional Poisson’s equation (∇2V = − ρ v / ε) may be expressed as d2V dx

2

=−

2 ρ v0 ρv x x sech tanh =− a a ε ε

...(2)

Integration of eqn. (2) gives 2 ρ v0 dV =− ε dx

∫

2 ρ v0 x x sech tanh dx = − a a ε

sinh ( x / a)

∫ cos2 h ( x / a) d x

1 dx = dt a Substituting these values in the above equation, we get 2 ρ v0 a dt 2 ρ v0 a −1 dV =− = t + A ε ∫ t2 ε dx 1 dV 2 ρ v0 a ∴ = + A dx ε cosh ( x / a) Putting cosh ( x / a) = t

∴

sinh ( x / a) .

x dV 2 ρ v0 a sech + A = a ε dx

or

...(3)

where A is the constant of integration. Thus the intensity of electric field E x is Ex = −

2 ρ v0 a dV x sech − A =− dx ε a

...(4)

Since no net charge density and electric field exist far from the junction, we have as x → ± ∞, E x → 0. Therefore, A = 0 Hence, eqn. (4) becomes Ex = −

2 ρ v 0a ε

sech

x a

Substituting A = 0 in eqn. (3), we get dV 2 ρ v0 a = sech ε dx

x a

Integrating above equation, we get V=

2 ρ v0 a ε

∫

x sech dx a

...(5)

243

or

V=

2 ρ v0 a

=

4 ρ v0 a

ε ε

∫ e− x / a + e x / a dx

∫ e − x / a (1 + e2 x / a ) 1 x /a e dx = dt a

ex /a = t ∴

Put

V=

∴

or

4 ρ v0 a ε

e− x + e x ∵ cosh ( x) = 2

2 dx

or

dx e

−x /a

= a dt

a dt

∫ 1 + t2

=

4 ρ v0 a2

V=

4 ρ v0 a2

ε

tan −1(t) + B tan −1(e x / a ) + B

ε

...(6)

where B is another constant To evaluate B, let us arbitrarily assume zero reference of potential (V = 0) at the centre of the junction, x = 0. Therefore, eqn. (6) becomes 4 ρ v0 a2

tan −1(1) + B ε 4 ρ v0 a2 π or B=− ε 4 Substituting this value of B from eqn. (7) in eqn. (6), we get 4 ρ v0 a2 4 ρ v0 a2 π tan −1(e x / a ) − V= ε ε 4 4 ρ v0 a2 π or V= tan −1 (e x / a ) − ε 4 0=

...(7)

...(8)

The total potential difference V0 across the junction is obtained from eqn. (8) as 4 ρ v0 a2 π π V0 = Vx → ∞ − Vx → − ∞ = + 4 4 ε V0 =

or

2 π ρ v0 a2

ε If S is the area of the junction cross-section, then the total positive charge Q is given by ∞ ∞ x x Q = S ∫ ρ v dx = S ∫ 2 ρ v0 sech tanh dx = 2 ρ v0 Sa sech a a 0 0

∴

x ∞ a 0

Q = 2 ρ v0 a S

or From eqn. (9), a =

...(9)

V0 ε 2 π ρ v0 Q = 2 ρ v0

V0 ε .S 2 π ρ v0

or Q = S

2 ρ v0 εV0 π

...(10)

Since the total charge is a function of the potential difference, we use circuit terms for the determination of capacitance C as ,

E lectromagnetic F ield Theory

244

I= or

dQ d = (CV0 ) dt dt dQ C= dV0

or

dV dQ =C 0 dt dt

...(11)

Differentiating eqn. (10) with respect to, V0 , we get 2 ρ v0 ε 1 −1 /2 dQ . V0 =S dV0 π 2 ∴

C =

ρv 0 ε 2π V0

...(12)

.S

Eqn. (12) indicates that the capacitance varies inversely as the square root of the potential difference across the junction. It means that a higher potential difference causes a greater separation of the charge layers and a small capacitance. Substituting the value of V0 from eqn. (9) in eqn. (12), we get 1 /2

ρ v0 ε C= 2 2 π (2 π ρ v0 a / ε)

S

or

C=

εS 2πa

...(13)

Equation (13) revealed that junction behaves as a parallel plate capacitor with a plate separation of 2 π a.

Uniqueness Theorem for Electrostatics Uniqueness theorem is a tool which makes it possible for us to conclude that there exists only one solution to a Laplace's or Poission's equation which satisfy boundary conditions. We can not solve any of these equations until boundary conditions are imposed. Boundary conditions determine the solution and particular choice of conditions used, depends on situation. Uniqueness theorem states that ''any solution to Laplace's or Poission's equation which satisfies the same boundary conditions must be the one and only one solution irrespective of the method used. It is unique, that is, it is only the possible solution''. In another way, uniqueness theorem states that ''If we are given distribution of charges within a given region of space and if also the potentials at the boundaries of this region are known, then there is one and only one solution to the Laplace's or Poission's equation for the electric potential'' A consequence of uniqueness theorem is that if we have a given electrostatic problem and we can replace the problem by an easier one but with the same charges and boundary conditions then the solution of our new easier problem is also the solution of our initial harder problem. For a general electrostatic problem involving charges and conductors, it is clear that if we are given either the potential at the surface of each conductor or charge carried by each conductor (plus the charge density throughout the volume) then we can uniquely determine the electric field. Proof: Consider a volume V bounded by same surface S. Suppose, we are given the charge density ρ throughout V and the value of scalar potential φS on S. Let us assume that the solution is not unique and there are two potentials φ1 and φ2 which satisfy.

245

∇2 φ1 = –

ρ ρ and ∇2 φ2 = – ε0 ε0

...(1)

φ1 = φS , φ2 = φS on surface S

throughout V and

We can form the difference between these potentials φ3 = φ1 – φ2

...(2)

The potential φ3 clearly satisfies ∇2 φ3 = 0

...(3)

throughout V and φ3 = 0 on S According to vector field identify →

→

→

div( SA) = S div A + A . grads

...(4)

→

If A = grad φ3 and S = φ3 div (φ 3 grad φ3 ) = φ 3 div (grad φ 3 )+(grad φ3 ).(grad φ 3)

then

→

→

→

∇.(φ 3 ∇ φ3 ) = φ 3 ∇2 φ 3 + ( ∇ φ 3 )2

or

...(5)

Thus according to Gauss's-divergence theorem →

∫v (div A) dv = ∫S →

→ →

A. dS →

→

∫v [ ∇.(φ3 ∇φ3 )] dV = ∫S (φ 3 ∇ φ 3 ) . dS

....(6)

or From eqn. (5), we have 2

∫ v [φ3 ∇

→

→

φ3 + (∇φ 3 )2 ] dV = ∫ (φ 3 ∇ φ 3 ) . dS S

...(7)

From eqn. (3), ∇2 φ3 = 0 throughout V and φ3 = 0 on surface S, therefore eqn. (7) reduces. 2

∫v (∇φ3 )

dV = 0

...(8)

→

where (∇ φ 3 )2 is positive definite quantity. There are in general two reasons why an integral may be zero. Either the quantity (∇φ3 )2 itself is zero throughout the volume. or we could have positive and negative contributions for various regions inside the volume which cancel one another out. In this situation first reason must be applicable as (∇φ 3 )2 is positive definite, it follows that φ3 = constant throughout V. However we know that φ3 = 0 on surface S, so we get

φ3 = 0 throughout volume V

In other words

φ1 – φ2 = 0 or φ1 = φ2

throughout V and on S. Hence our initial assumption that φ1 and φ2 are two different solutions of Poisson's equation satisfying same boundary conditions turns out to be wrong. Thus the solution of Poisson's equation is unique, no matter how contrived the derivation, than this is the only possible solution.

E lectromagnetic F ield Theory

246

Example 19: Let V = 2 xy2 z 3 and ε = ε0 . Given point P(1, 3, – 1) . Find V at point P also find out if V satisfies Laplace equation. [UPTU, B.Tech. IV Sem. 2008]

Solution: Given V = 2 xy2 z3 Vat P(1, 3, – 1) = 2 × 1 × 32 × (–1)3 = –18 According to Laplace equation ∇2V = 0 or ∂2 V ∂x

2

∂2 V 2

∂y

∂2 V ∂z2 ∴

∂2 V ∂x

2

∂V ∂ = ∂x ∂x

+

∂2 V 2

∂y

+

∂2 V ∂z2

=0

∂ ∂ 2 3 2 3 ∂x (2 xy z ) = ∂x (2 y z ) = 0

=

∂ ∂x

=

∂ ∂ ∂V ∂ ∂ 2 3 (4 xyz3 ) = 4 xz3 = (2 xy z ) = ∂y ∂y ∂y ∂y ∂y

=

∂ ∂ ∂V ∂ ∂ 2 3 (6 xy2 z2 ) = 12 xy2 z (2 xy z ) = = ∂z ∂z ∂z ∂z ∂z

∇2V = 4 xz3 + 12 xy2 z ∇2 V at P = 4 × 1 × (–1)3 + 12 × 1 × (3)2 × (–1) (∇2V )at P (1,3, –1) = – 4 – 108 = –112

Example 20: If a potential V = x2 yz + Ay3 z , (i) find A so that Laplace's equation is satisfied (ii) with the value of A, determine electric field at (2, 1, –1). [GBTU, B.Tech. III Sem. 2010]

Solution: (i) The Laplace's equation is, ∇2V = 0 ∴

Here

V = x2 yz + Ay3 z

→ → ∂V ∂ ∂V ∂V ∂ ∂ + aɵz + aɵz . aɵ x + aɵ y + aɵ y ∇2V = ∇. ∇ V = aɵ x ∂z ∂ y ∂ x x ∂ y z ∂ ∂

∇2V =

∂ ∂V ∂ ∂V ∂ ∂V + + ∂x ∂x ∂y ∂y ∂z ∂z

∂V ∂V ∂V = x2 y + Ay3 = 2 xyz + 0, = ( x2 z + 3 Ay2 z), ∂z ∂x ∂y ∴

∇2V =

∂ 2 ∂ 2 ∂ (2 xyz) + ( x z + 3 Ay2 z) + ( x y + Ay3 ) = 0 ∂z ∂y ∂x

2 yz + 0 + 6 Ayz + 0 = 0 or

6 A = –2

or

A =

–2 –1 = 6 3

247

(ii)

Electric field E and potential V is related as ∂V ∂V ∂V aɵ x + aɵ y + aɵz E = – grad V = – ∂y ∂z ∂x E = – [2 xyzaɵ x + ( x2 z + 3 Ay2 z) aɵ y + ( x2 y + Ay3 ) aɵz ]

∴ Substituting A = –

1 , we get 3 1 E = – 2 xyzaɵ x + ( x2 z – y2 z) aɵ y + ( x2 y – y3 ) aɵz 3

E at point (2,1, – 1) is Eat (2,1, –1) = –[ 2 × 2 × 1(–1) aɵ x + {22 × (–1) – 12 × (–1)} aɵ y + {22 × 1 – or

Eat (2, 1,–1) = 4 aɵ x + 2 aɵ y –

1 × 1} aɵz ] 3

11 aɵ z 3

Example 21: Find the Laplacian from the following: (i) grad u = 2 rz cos 2 φ rɵ – 2 rz sin 2 φ φɵ + r2 cos 2 φzɵ (ii) grad v = 10 sin2 θ cos φrɵ + 10 sin 2θ cos φ θɵ – 10 sin θ sin φ φɵ [GBTU, B.Tech. III Sem. 2011]

→ →

Solution: (i) Laplacian operator is, ∇2 u = ∇. ∇ u In cylindrical coordinates Laplacian operator is, ∇2 u = or

1 ∂ ∂u 1 ∂2 u ∂2 u + r + r ∂r ∂r r 2 ∂φ2 ∂z2

∇2 u =

1 ∂ ∂u 1 ∂ 1 ∂u ∂ r + + r ∂r ∂r r ∂φ r ∂φ ∂z

....(1) ∂u ∂z

...(2)

In the given problem 1 ∂u ɵ ∂u ∂u grad u = φ + zɵ = (2 rz cos 2 φ) rɵ – (2 rz sin 2φ) φɵ + (r 2 cos 2 φ)zɵ rɵ + ∂z ∂r r ∂φ Hence, comparison of L.H.S. and R.H.S. gives, 1 ∂u ∂u ∂u = –2 rz sin 2 φ and = 2 rz cos 2 φ, = r 2 cos 2 φ r ∂φ ∂r ∂z substituting these values in eqn. (2), we get 1 ∂ 1 ∂ ∂ 2 (r [2 rz cos 2 φ]) + (–2 rz sin 2 φ) + (r cos 2 φ) ∇2 u = ∴ r ∂r r ∂φ ∂z 2 1 or ∇2 u = (2 rz cos 2 φ) – (2 rz2 cos 2 φ) + 0 r r = 4 z cos 2 φ – 4 z cos 2 φ = 0 (ii) In spherical coordinates Laplacian operator is, 1 ∂ 2 ∂v 1 1 ∂ ∂v ∂2 v ∇2 v = 2 sin θ + 2 + 2 r ∂θ r sin2 θ ∂φ2 r ∂r ∂r r sin θ ∂θ or In the given problem

∇2 v =

∂ 1 ∂v ∂ 1 ∂ 2 ∂v 1 1 ∂v 1 r sin θ + + ...(3) r sin θ ∂θ r ∂θ r sin θ ∂φ r sin θ ∂φ r 2 ∂r ∂r

E lectromagnetic F ield Theory

248

grad v =

∂v 1 ∂v ɵ 1 ∂v ɵ rɵ + φ = (10 sin2 θ cos φ)rɵ + (10 sin 2θ cos φ) θɵ θ+ ∂r r ∂θ r sin θ ∂φ

−(10 sin θ sin φ) φɵ Comparison of L.H.S and R.H.S. gives ∂v 1 ∂v 1 ∂v = 10 sin2 θ cos φ, = 10 sin 2θ cos φ and = – 10 sin θ sin φ ∂r r ∂θ r sin θ ∂φ Substituting these values in eqn. (3), we get 1 ∂ 2 1 ∂ (r [10 sin2 θ cos φ]) + (sin θ [10 sin 2θ cos φ]) ∇2 v = 2 r sin θ ∂θ r ∂r 1 ∂ (–10 sin θ sin φ) + r sin θ ∂φ =

2

+

10 sin θ cos φ 10 cos φ [cos θ sin 2θ + 2 cos 2θ sin θ] – r sin θ r sin θ

r 20 sin2 θ cos φ 10 cos φ 10 cos φ = + [2 cos2 θ + 2 cos 2θ] − r r r 10 cos φ 10 cos φ 2 2 2 ∇ v= (2 sin θ + 2 cos θ + 2 cos 2θ – 1)= (2 + 2 cos 2θ – 1) r r 10 cos φ ∇2 v = (1 + 2 cos 2θ ) r

or

Example 22: Given V =

2 r .(10 sin2 θ cos φ)

cos φ ρ

(i) Determine the volume charge density at ρ (0 . 5, 60 °and1). (ii) Find the electric field intensity E at ρ (0 . 5, 60 ° and1). [U' Khand B. Tech. IV Sem. 2011]

Solution: (i) According to Poisson's equation, ∇2V =

ρv , where ρ v is the volume charge density. ε0

Poisson's equation in cylindrical coordinates is, 1 ∂ ∂V 1 ∂2 V ∂2 V ρ + 2 =– v ρ + ρ ∂ρ ∂ρ ρ2 ∂φ2 ε ∂z In the given problem, V = ∴

cos φ ρ ρv 1 ∂ ∂ 1 ∂2 (cos φ / ρ) ∂2 cos φ –1 cos + 2 ρ ρ φ =− + 2 2 ε ρ ∂ρ ∂ρ ρ ∂z ∂φ ρ

(

)

ρ 1 ∂ ρ cos φ 1 1 . cos φ = – v – – ρ ∂ρ ε0 ρ2 ρ2 ρ ρ 1 cos φ 1 – 3 cos φ = – v = 0 ε0 ρ ρ2 ρ ∴

ρv = 0

249 →

→

(ii) We know that E = – ∇V →

∂V 1 ∂V ɵ ∂V zɵ φ– ρɵ – ∂ρ ρ ∂φ ∂z ∂ cos φ 1 ∂ cos φ ɵ ∂ cos φ =– ρɵ – φ– zɵ ∂ρ ρ ∂z ρ ρ ∂φ ρ cos φ sin φ = 2 ρɵ + 2 φɵ + 0 ρ ρ cos 60 ° 0.866 ɵ sin 60 ° ɵ 0.5 ρɵ + φ = φ=– ρɵ + 2 2 2 (0. 5)2 (0. 5) (0. 5) (0. 5) ρɵ 0.866 ɵ = + φ 0. 5 0. 25

E=–

∴

→ E at(0.5,60 °,1)

→ E at(0.5,60 °,1)

∴

= 2ρɵ + 3.464 φɵ

Example 23: State whether following potential function satisfy Laplace's equation: (a) V = Cxyz

(b) V = Crφz

(c) V = Cr θ φ

and

[UPTU., B. Tech. IV Sem. 2002]

Solution: (a) In cartasian coordinates the Laplacis equation is ∂2 V ∂2 V ∂2 V ∇2V = 2 + 2 + 2 = 0 ∂z ∂y ∂x Here V = Cyz, where C is constant Hence Similarly

∂V ∂2 V =0 = Cyz and ∂x ∂x 2 ∂2 V ∂2 V = 2 =0 ∂y2 ∂z

Therefore ∇2V = 0. Hence relation (a) satisfy Laplace equation. (b) In cylindrical coordinates Laplace equation is expressed as, ∇2V = Here Hence, Similarly, Substituting the values of

1 ∂ ∂V 1 ∂2 V ∂2 V + 2 r + r ∂r ∂r r 2 ∂φ2 ∂z

V = Cr φ z ∂V 1 ∂ ∂V 1 ∂ 1 = Cφ z, therefore ( r C φ z) = C φ z r = ∂r r ∂r ∂r r ∂r r ∂V ∂2 V ∂V ∂2 V and and 0 = =0 = Crz and = φ C r ∂φ ∂z ∂φ2 ∂z2 1 ∂V ∂2 V ∂2 V and in eqn. (1), we get , r 2 r ∂r ∂φ ∂z2 1 ∇2V = Cφz + 0 + 0 ≠ 0 r

Hence, Laplace equation is not satisfy. (c) In spherical coordinates, Laplace eqn. is,

...(1)

E lectromagnetic F ield Theory

250

∇2V = Here Hence,

Similarly; ∴

1 ∂ r 2 ∂r

∂V ∂ 1 1 ∂2 V 2 ∂V + 2 sin θ + 2 r 2 ∂θ r sin θ ∂φ2 ∂r r sin θ ∂θ

V = Crθφ 1 ∂ 2 ∂V 1 ∂ 2 ∂ 1 ∂ 2 (Crφθ) = 2 (r C θ φ) r = r r ∂r ∂r r 2 ∂r ∂r r 2 ∂r 1 2Cθφ = 2 Cθφ . 2 r = r r ∂V = Crφ ∂θ ∂ 1 1 Crφ cos θ Cφ ∂ ∂V = (sin θCrφ) = 2 co t θ = sin θ r ∂θ r 2 sin θ ∂θ r 2 sin θ ∂θ r sin θ

Again ∴

∂V ∂2 V = Cr θ and =0 ∂φ ∂φ2 2 Cθφ Cφ ∇2V = + cot θ + 0 ≠ 0 r r

Hence, Laplace equation is not satisfy. Example 24: Consider a parallel plate capacitor occupying planes x = 0 and x = d and is kept at a potential V = 0 and V = Vd respectively. Find the potential, electric field between the plates, surface charge densities at the two plates and capacitance of the capacitor using Laplace’s equation. Assume that plates are kept in air. Solution: The potential distribution between the plates, and capacitance of a parallel plate capacitor, Using Laplace’s equation: Consider a parallel plate capacitor occupying planes x = 0 and x = d as shown in Fig. 24. Let left hand plate is at zero potential (V = 0) and the right hand plate at potential (V =) Vd . Let us apply Laplace’s equation to find the potential distribution and electric field between the plates, where the region between the plates is filled with air, that is, ε = ε0 .There is no variation of potential in the y- and z-directions, therefore the problem is one dimensional. Laplace’s equation (∇2V = 0) for one dimension reduces to, d2V dx

2

=0

Y

...(1)

V=0

V=Vd

Second derivative of the potential (V ) would be zero when its first derivative is equal to a constant say A. Thus, we have dV ...(2) =A dx or

dV = A dx

...(3)

Integration of eqn. (3), gives V = Ax + B where B is another constant.

...(4)

X x=0

x=d Fig. 24

251

We can determine the values of the constants A and B by applying boundary conditions given in the problem. The boundary conditions of the problem are: V = 0 at x = 0 and V = Vd

at

x=d

Applying first boundary condition to eqn. (4), we get 0 = 0 + B or B = 0

...(5)

Applying second boundary condition to eqn. (4), we have Vd = Ad + B

A=

or

Vd d

Substituting the values of A and B from eqns. (5) and (6) in eqn. (4), we obtain V V= d x 0> ρ (ρ2 + z2 )3 /2 ≈ z3 Therefore eqns (6) and (7) take the form Iρ2

→

H=

2 z3

→

aɵz and B =

µ0 Iρ2 2 z3

aɵz

If there are N turns in the loop, then I is replaced by IN. Therefore eqns. (6) and (7) for N turns are reduce to I N ρ2

→

H =

2

2 3 /2

2(ρ + z )

aɵ z

µ0 INρ2

→

and B =

2(ρ2 + z 2 )3 /2

aɵ z

Example 1: Find the magnetic field intensity at point P2 (4, 2, 0 ) due to a current carrying conductor →

(I dl = 2πaɵ z A - m) placed at P1 (0 , 0 , 2 ) . →

Solution: According to Biot-Savart law, magnetic field intensity dH at a point distant R from a current →

element I dl is given by →

dH =

→

→

I dl × R 4 π R3

301

If the co-ordinate of point P2 is (( x2 , y2 , z2 ) and P1 ( x1, y1, z1), then →

R = ( x2 – x1) aɵ x + ( y2 – y1) aɵ y + (z2 – z1) aɵz

In the given problem P2 (4, 2, 0) and P1 (0, 0, 2), therefore →

R = (4 − 0)aɵ x + (2 – 0) aɵ y + (0 – 2)aɵz

or

→

R = 4 aɵ x + 2 aɵ y – 2 aɵz

→

| R | = R = (4)2 + (2)2 + (–2)2 = 24 = 4.9

∴

→

→

a z × aɵ y ) – 4 π (aɵz × aɵz )] I dl × R = (2 πaɵz ) × (4 aɵ x + 2 aɵ y – 2 aɵz ) = [8π (aɵz × aɵ x ) + 4 π (^ = (8πaɵ y – 4 πaɵ x ) (∵ aɵz × aɵ x = a y , aɵz × aɵ y = – aɵ x and aɵz × aɵz = 0) →

8πaɵ y – 4 πaɵ x

dH =

or

dH = 0.017 aɵ y – 0.0085 aɵ x A / m

3

4 π × (4.9)

=

2 aɵ y – aɵ x

∴

11765 . →

→

or dH = –0 .085 aɵ x + 0 .017 aɵ y A / m

→

Example 2: A current element I dl = 2 π (0 .6 aɵ x – 0 .8 aɵ y )µ A is situated at a point (4, – 2, 3 ) . Find →

incremental field ∆H at a point (1, 3, 2 ) . Solution: According to Biot-Savart law, →

→

∆H=

→

I dl × R 4 π R3

→

, Here I dl = 2 π (0.6 aɵ x − 0.8aɵ y )

→

R = ( x2 − x1) aɵ x + ( y2 − y1) aɵ y + (z2 − z1) aɵz = (1 – 4) aɵ x + (3 + 2) aɵ y + (2 – 3) aɵz →

→

R = –3 aɵ x + 5 aɵ y − aɵz and | R | = R = (–3)2 + (5)2 + (–1)2 = 35

or →

→

I dl × R = 2 π (0.6 aɵ x − 0.8aɵ y ) × (–3 aɵ x + 5 aɵ y − aɵz ) = 2 π [–0.6 × 3 (aɵ x × aɵ x ) + 0.6 × 5 (aɵ x × aɵ y ) – 0.6 × 1(aɵ x × aɵz ) + 0.8 × 3( aɵ y × aɵ x ) – 0.8 × 5(aɵ y × aɵ y ) + 0.8 aɵ y × aɵz ] = 2 π [3.0 aɵz + 0.6 aɵ y − 2.4 aɵz + 0.8 aɵ x ] = 2 π [ 0.8 aɵ x + 0.6 aɵ y + 0.6 aɵz ] {∵ aɵ x × aɵ y = aɵz , aɵ y × aɵz = aɵ x , aɵz × aɵ x = aɵ y , aɵ x × aɵ x = aɵ y × aɵ y = aɵz × aɵz = 0} →

2 π [0.8aɵ x + 0.6 aɵ y + 0.6 aɵz ]

0.8aɵ x + 0.6 aɵ y + 0.6 aɵz

∴

∆H =

or

∆H = (1932 . aɵ x + 1449 . aɵ y + 1449 . aɵz ) × 10 –3 µ Α/ m

or

∆H = (1932 . aɵ x + 1449 . aɵ y + 1449 . aɵ z ) nA / m

4 π (35 )3 / 2

=

414 .125

→

→

E lectromagnetic F ield Theory

302

Y

Example 3: The conducting triangular loop in fig. 9. carries a current of →

10A. Find H at (0 , 0 , 5 ) due to side 1 of the loop.

1 3

[GBTU, B.Tech III Sem 2011]

2 1

(0,0)

1

X

2

Fig. 9

Solution: The situation of the problem is shown in fig. 10. If the side 1 of the triangular loop subtends angles α 2 and α1 at P

P

→

(0,0,5)

then the magnetic field intensity H at P according to Biot-Savart law is given by, →

I (cos α 2 – cos α1) φɵ 4 πρ

ρ=5

OA 2 2 = = 2 2 2 AP (OP) + (OA) (5) + (2)2

α1

H= From fig. 10 cos α1 = 90 °

and cos α 2 =

α2

2 , ρ = 5 and I = 1. A 29

or

cos α 2 =

∴

H1 =

or

H1 = – 0.0591 aɵ y A / m

2

Fig. 10

10 10 × 2 2 (– aɵ y ) – 0 φɵ = 4 × 314 . × 5 29 20 × 314 . 29 or

A

10A 1

O

(∵ φɵ = aɵ x × aɵz = – aɵ y )

H1 = – 59.1aɵ y mA / metre

Example 4: A 12 cm long straight filamentary conductor carrying a current of 6 amp is placed along positive z - axis with one end at origin. Find out the magnetic field intensity at point (0 , 6, 6 ) . Z

Solution: The situation of the problem is shown is fig. (11). OA is the filamentary conductor carrying a current of 6 amp.

(0, 0, 12)

the length of the perpendicular Q P is equal to Y coordinate

A

of P, that is, ρ = 6 cm. If α 1 and α 2 are the angles subtended by the ends of the conductor at P, then according to Biot- Savart Law, the magnetic field intensity at P is given by →

H= Here, ∴

α2

Q(0, 0, 6)

ρ = 6 cm α1

I [sin α 2 – sin α1] φɵ 4π ρ

α 2 = 45 ° and α1 = – 45 °

(0, 0, 0)

→

6 H= [ sin 45 ° – sin (– 45 ° )] φɵ 4 × 314 . ×6 6 × 2 × 0.7071 6 ×2 = sin 45 ° = 4 × 314 . ×6 4 × 314 . ×6

Y

0 X

P(0, 6, 6)

Fig. 11

303 →

H = 0.1126 φɵ

or →

H in cartesian coordinates can be expressed as, →

H x = H. aɵ x = 01126 . φɵ . aɵ x = 01126 . (– sin φ) [∵ φɵ . aɵ x = – sin φ] H x = –01126 . sin φ

or

y 6 φ = tan –1 = tan –1 or φ = tan –1 ∞ = 90 ° x 0

But ∴

H x = –01126 . sin 90 ° = –0.1126 A / m

Similarly,

. cos φ H y = H. aɵ y = 01126 . φɵ . aɵ y = 01126

→

φ = 90 °

But

[∵ φɵ . aɵ y = cos φ]

∴ H y = 0.1126cos90 ° = 0 →

Example 5: Find magnetic field intensity H at the centre of a square current loop of side 'a' meter, if a →

current of I amp is passed through it. What would be the value of H if a = 8 m and I = 10 amp. Solution: The situation of the problem is shown in fig. 12. The contribution from each sides are equal. Therefore, the magnetic field →

→

H1 =

I (cos α 2 – cos α1) φɵ 4π ρ

∴

H1 = =

or ∴

→

H1 = →

From symmetry,

H1 = →

C

a α1

ρ a

B

Fig. 12

I I [cos 45 °– cos (180 ° – 45 ° )] φɵ = [cos 45 °+ cos 45 ° ] φɵ 2π a 2π a I 2 ɵ I ɵ φ= φ 2 πa 2 2 πa →

H = 4 H1 =

→

α2

I [cos 45 °– cos 135 ° ] φɵ 4 π( a /2 )

or

4I ɵ φ 2πa

Here a = 8 m and I = 10 amp ∴

A

I

P

a

Here, α 2 = 45 ° , α1 = 135 ° and ρ is the length of perpendicular drawn from a P on current element AB = 2 →

a

D

intensity H at P, at the centre of the square due to side AB will be

10 = 0.2815 A /m 2 × 314 . ×8

H = 4 × H1 = 4 × 0. 2815 = 1.126 A /m

→

H1 =

I φɵ 2πa

E lectromagnetic F ield Theory

304

Example 6: A circular loop located on x2 + y2 = 9 , z = 0 carries a directed current of 10 amp Determine →

H at (0 , 0 , 4 ) . [UPTU, B.Tech IV Sem 2006] →

Solution: The magnetic field intensity H at a point P(0, 0, z) on z-axis due to a circular loop of radius ρ carrying a current I is given by, Iρ2 aɵz

→

H at (0,0,z) =

2(ρ2 + z2 )3 /2

weber/m 2

The equation of a circular loop of radius ρ is x2 + y2 = ρ2 ∴

→

H at (0, 0, 4) =

∴

ρ2 = 9

10 × (3)2 aɵz 2 3 /2

2

2[(3) + (4) ]

ρ =3

or =

10 × 9 3

2 × (5)

→

aɵz or H at (0, 0, 4) = 0.36 aɵ z A / m

Example 7: A conductor in the form of equilateral triangle of sides d carries a current of I amp inscribed in →

a circle of radius R. obtain an expression for the magnetic field intensity H at the centre of the circle. Solution: The situation of the problem is shown in fig. 13. The angle made by each side of the →

equilateral triangle at the centre O is 2 π /3. The magnetic field intensity H at the centre of the circle O due to one side of a triangle, say AB will be →

H AB =

α 2 = 90 °− π / 3 and 180 °– α1 = 90 °− π / 3

From Fig. 13

→

H AB =

I [cos(90 °− π / 3) − cos(90 °+ π / 3)] 4 πρ =

From ∴

C

α1 = 90 ° + π / 3

or ∴

I [cos α 2 − cos α1], where ρ = OL 4 πρ

I d

I π π 2I π sin + sin = sin 3 4 πρ 3 3 4 πρ

ρ π ∠AOL, = cos R 3 →

H AB =

O R

π/ 3 π/ 3

α2

π or ρ = R cos 3

A

π I tan 3 2 πR

α1

ρ

L

B

Fig. 13

The total magnetic field intensity at the centre (0) of an equilateral triangle when each sides carry current I is, →

H=

3I π tan 2 πR 3

or

→

H=

0 .627 I R

305

Example 8: Find the expressions of magnetic field intensity and magnetic induction at a point (ρ, φ, z ) due to an infinitely long straight filament carrying a current I in z − direction.

z Solution: Fig. 14. shows an infinity long straight filament in free space and carrying a current of I amp. and is along z − axis. Suppose P(ρ, φ, z) be a point where magnetic field intensity and magnetic induction is to be determined. Let us assume that the filament is

I O

→

made up of large number of small current elements and let I dl be one

O

→

such element at a point A on the filament. If current element I dl is at

ρ ρ

a distance R from P, then from Biot-Savart law, the magnetic field →

→

Idl A

intensity dH at P due I dl is →

→

dH = →

P R

P

A

R

Fig. 14

...(1) →

→

→

→

From fig. 14(b)

AP = AO + OP or R = – zaɵz + ρ ρɵ or R = ρ ρɵ − zaɵz , | R | = R = ρ2 + z2

and

| R | = R = ρ2 + z2

∴

ρ

– zaz

→

1 I dl × R 4 π R3 →

←

→

→

→

I dl × R = (Idz aɵz ) × (ρ ρɵ – zaɵz ) = I ρ dz (aɵz × ρɵ ) – Iz dz (^ a z × aɵz ) = I ρdz φɵ (∵ aɵz × ρɵ = φɵ and aɵz × aɵz = 0) →

→

Substituting the values of I dl × R and R in eqn. (1), we get →

dH =

Iρ dzφɵ 1 2 4 π (ρ + z2 )3 /2

....(2)

The total magnetic field intensity at P due to entire infinitely long filament is obtained by integration of eqn. (2). within the limits + ∞ to – ∞ as, ɵ ∞ → → Iφ ρ dz H = ∫ dH = ∫ 2 – z = ∞ 4π (ρ + z2 )3 /2 z = ρ tan θ

put

→

I φɵ

∴ dz = ρ sec2 θ dθ ρ2 sec2 θ dθ

π/2

I φɵ

π/2

ρ2 sec2 θ dθ

∴

H=

4 π ∫ −π / 2

or

H=

→

I φɵ π / 2 I φɵ I π π I π π/2 cos θ dθ = [sin θ] sin + sin φɵ = sin φɵ = ∫ –π / 2 2 4 π ρ −π / 2 4 πρ 4 π ρ 2 2 2 πρ

or

H=

→

I ɵ φ 2πρ

→

→

We know that, B = µ0 H

2

2

2

3 /2

(ρ + ρ tan θ)

→

Therefore, the magnetic field induction B would be → µ I B = 0 φɵ 2π ρ

=

4π

∫ −π / 2

ρ3 sec3 θ

E lectromagnetic F ield Theory

306

Example 9: A circular loop of wire of radius a lying in X Y plane with its centre at origin carries a current →

→

of I amp in + φ direction. Find the value of H(0 , 0 , z ) and H (0 , 0 , 0 ) . Z

Solution: Fig. 15. shows a circular wire of radius 'r' lying in X – Y plane with its centre at origin. Let the point P at which the →

magnetic field intensity H is to be determined, is at a height h metre

P(0, 0, z)

along the Z −axis. The Z −axis is passing through the centre 'O ' of

R′ a

z

→

←

the current loop. The position vector of point P is r = h aɵz . Suppose

dl

→

→

r

B

I dl is a one of the differential current element along a circular loop at point A. Since the differential length varying only about the

←

R Y

(0, 0, 0) 0

circumference in cylindrical coordinates I dl = I a d φ φɵ .The position

ρ

r′ = a

→

A

ɵ where a is the radius of the the circular loop. vector of A is r ' = aρ, The magnetic field intensity at P due to a differential current →

I X

element I dl at A, according to Biot-Savart law, is given by, →

→

dHA =

Fig. 15

→

I dl × R

...(1)

4 π R3

According to vector diagram OAP, →

∴

→

→

→

AP = – OP + OA or R = – aρɵ + h aɵz

or →

and

→

| R | = R = a2 + h2

→

I dl × R = (Ia dφ φɵ ) × (– aρɵ + haɵz ) = – Ia2 dφ (φɵ × ρɵ ) + Iah dφ (φɵ × aɵz ) = Ia2 dφ aɵz + Iah dφ ρɵ →

(∵ – φɵ × ρɵ = aɵz and φɵ × aɵz = ρɵ )

...(2)

→

Substituting the values of I dl × R from eqn. (2) and R in eqn. (1), we get →

dH A =

I adφ(a aɵz + hρɵ )

...(3)

4 π ( a2 + h2 )3 / 2

Now consider another differential current element at B which is diametrically opposite to A as shown in →

fig. 15. The current element at B is represented as, I dl = Iadφ (– φɵ ) = – Iadφ φɵ . The position vector of B is →

R ' = – aρɵ . In this case. →

→

→

R ' = aρɵ + hρ z and I dl × R ' = a aɵz – hρɵ ∴

→

d HB =

Ia dφ (a aɵz − hρɵ ) 4 π (a2 + h2 )3 /2

...(4)

307

The total magnetic field intensity at P due to current elements at A and at B is →

→

→

dH = dH A + dHB

...(5)

→

→

Substituting the values of dH A from eqn. (3) and dHB from eqn. (4) in eqn. (5), we get Ia2 dφ aɵz

→

dH =

...(6)

2 π (a2 + h2 )3 /2

→

The magnetic induction H at P(0, 0 z) due to entire circular loop of wire is obtained by integration of eqn. (6) within the limits φ = 0 to φ = π, that is, →

→

H = ∫ dH = ∫

φ =0

2

2 3 /2

2 π (a + h )

Ia2 aɵz π

→

H=

or

Ia2 dφ aɵz

φ =π

2 π (a2 + h2 )3 /2

or

→

H=

=

Ia2 aɵz 2

π

2 3 /2

2 π (a + h ) Ia2

2 (a2 + h2 )3 /2

∫ φ =0

dφ

aɵz

...(7)

→

Magnetic field intensity H (0, 0, 0) at origin is obtained by putting h = 0 in eqn. (7), as →

H(0, 0, 0) =

I aɵ z 2a

Example 10: Using Biot - Savart law, calculate the magnetic field induction due to a current carrying solenoid. Solution: (i) At an Axial point: Consider a long solenoid of radius 'a' meter and carrying a current of I amp. A solenoid is a uniform cylindrical coil wound on a non-magnetic frame. Suppose N be the total number of turns on the solenoid. If l meter is the length of the solenoid, then the number of turns 'n' per unit length will N/l. Let P be the point on the axis of the solenoid (Fig.16) at which the magnetic field induction is to be determined. Let us suppose that the solenoid is made up of a number of circular loops of wire consider one such loop AB of width dx. The number of turns in this coil is ndx. If x be the distance of the point P from the centre O of the coil, then the magnetic induction dB due to this elementary coil is given by, dB =

µ0 (ndx) I a2 2(a2 + x2 )3 /2

weber/meter2

...(1)

Let δ θ be the angle subtended by the elementary coil at P and r be the distance of coil from P, then from ∆ ABC, we have

∴

sin θ =

BC BC and in ∆BPC, = δθ r AB

sin θ =

rδθ dx

or

dx =

rδθ sin θ

From ∆ APO, ( AO)2 + (OP)2 = ( AP)2

A dx B θ C a O

...(2)

or r 2 = a2 + x2 or r = (a2 + x2 )1 /2

x

δθ

r θ

P

Fig. 16

∴ r 3 = (a2 + x2 )3 /2 ...(3)

E lectromagnetic F ield Theory

308

Substituting the values of dx and r 3 from eqns. (2) and (3) in eqn. (1), we get µ0 n (r δθ / sin θ)Ia2

dB =

From

∆ APO

2 r3

a = sin θ r

∴ dB =

µ0 n I δθ a2 2 sin θ r 2

dB =

or

µ0 n I δθ sin2 θ 2 sin θ

or

dB =

...(4)

1 µ n I sin θδθ 2 0

...(5)

The magnetic field induction B at P due to whole solenoid is obtained by integrating eqn. (5) between the limits θ = θ1 to θ = θ2 , where θ1 is the semi-vertical angle subtended by first turn on P and θ2 that subtended by last turn on P as shown in fig. 17. Therefore B=∫

θ2 θ1

dB = ∫

θ2 θ1

1 µ nI sin θdθ. 2 0

θ2 1 1 θ µ0 nI ∫ sin θ d θ = µ0 nI [– cos θ] 2 θ1 θ1 2 2

=

P

or

B=

µ0 nI [cosθ1 − cosθ2 ] 2

...(6)

and

H=

B nI = [cos θ1 − cos θ2 ] µ0 2

...(7)

→

θ2

θ1

Fig. 17

→

In vector form B and H are expressed as →

B =

µ0 nI (cosθ1 − cosθ2 )aɵ x 2

→

H=

and

nI (cosθ1 − cosθ2 )aɵ x 2

When point P is well inside a very long solenoid, then θ1 ~ – 0 ° and θ2 ~ – 180 ° . Hence from eqn. (6), we have B=

µ0 nI 2

(cos 0 °− cos 180 ° ) =

µ0 n I 2

[1 – (–1)]

or

B = µ0 n I

...(8)

(ii) At the ends of the solenoid: If the point P is at the end of last turn (fig. 18.), θ1 ~ – 0° and θ2 ~ – 90 °. Thus from eqn. (6), B=

or

µ0 n I

[cos 0 °– cos 90 ° ] = 2 µ nI weber/m2 B= 0 2

µ0 n I 2

[1 – 0]

Fig. 18

(iii) At the centre of a solenoid of Finite length: l/2

(a 2 +

l 2/4) 1/2

l/2

θ1 l/2

θ1

a

θ2 P

Fig. 19

θ2

309

When the point P is at the centre of a solenoid of length l, that is, P is at a distance l /2 from either end of the solenoid as in fig. 19. In this case l /2

cos θ1 =

and

2 l2 a + 4

cos(π – θ2 ) = –

l /2 2 l a + 4 2

or cos θ1 =

=–

l 2

(4 a + l2 )1 /2

l 2

2 1 /2

(4 a + l )

or

cos θ2 = –

l 2

(4 a + l2 )1 /2

Substituting these values of cos θ1 and θ2 in eqn.(6), we get or

B=

µ0 n I l l 2 2 1 /2 + 2 2 1 / 2 2 (4 a + l ) (4 a + l )

or

B=

µ0 nIl

2

(4a + l2 )1 /2

Example 11: A single turn circle coil of 50 metres in diameter carries current 28×10 4 amp. Determine the magnetic field intensity H at a point on the axis of the coil and 100 meter from the coil. The relative permeability of free space surrounding the coil in unity. [GBTU, B.Tech. IV Sem. 2011] →

Solution: The magnetic field intensity H due to a circular coil of radius ' a ' at a point distant z from the centre is H=

NIa2 2

2(a + z2 )3 /2

weber/m2

where N is the number of turns in the coil,

If the point lies on the axis of the coil, that is, z = 0, then H= Here N = 1, a = ∴

D 50 = = 25 m 2 2 H=

NIa2

=

2 a3 and

NI 2a I = 28 × 104 amp.

28 × 104 = 1.12 × 10 4 Amp / m 25

At a distance 100m from the coil, z = 100m. Therefore H=

= or

Ia2 2(a2 + z2 )3 /2

=

28 × 104 × (25)2 2 [(25)2 + (100)2 ]1 /2

28 × 104 × 25 × 25 28 × 104 × 25 × 25 = 2 × 103.08 2 (10625)

H = 84.88 × 10 4 Amp/m

E lectromagnetic F ield Theory

310

→

Example 12: Find the magnetic field intensity H due to a sheet of width W and having a surface current density k.

k=

total current I = width W

k

I = kW

or

...(1)

Consider an elemental length dl along the length of the uniform sheet, then →

W

Solution: Consider a uniform current sheet of width W having a surface current density K as shown in fig. 20. The surface current density is defined as the total current flowing per unit width of the sheet perpendicular to flow, that is,

I Fig. 20

→

I = dl = k . ds

...(2)

where ds is the area. Now, according to Biot-Savart's law. the magnetic field intensity is expressed as, →

dH =

→

→

I dl × R 4 πR 3

→

→

k ds × R

=

...(3)

4 πR 3 →

The total magnetic field intensity H can be determined by integrating eqn. (3) by taking surface integration, that is, →

→ →

→

H ∫ dH = ∫ We know that,

k.R

S

4 πR 3

...(4)

ds

Idl = Jsdl = Jdv

(∵ dv = sdl)

where J is called volume current density, defined as current per unit area perpendicular to flow. ∴ so

→

→

→

I dl = kds = Jdv →

→

H=∫

→

J dv × R

v

4 πR 3

→ →

→

or H = ∫

J×R

v

4 πR 3

dv

Example 13: A copper wire 0.254 cm in diameter carries a current of 50 ampere. Find the magnetic field induction B at the surface of the wire. The permittivity constant, µ0 = 4 π × 10 −7 weber/amp-m. →

Solution: The magnitude of magnetic field induction B at a point distant d from a straight wire carrying a current of I amp is given by, B=

µ0 I 2πd

311

At the surface of the wire, d = r , the radius of the wire ∴

B=

Here r =

µ0 I 2πr

0 .254 = 0 .127 cm = 0 .127 × 10 −2 m, I = 50 amp and µ0 = 4 π × 10 −7 Weber/amp-m 2

∴

B=

4 π × 10 −7 × 50 2 π × 0 .127 × 10 −2

= 7. 87 × 10 −3 Weber / m2

Ampere's Circuital Law Ampere’s circuital law in magnetostatics is analogous to Gauss’s law in electrostatics. It states that ‘‘the →

line integral of magnetic field intensity H about any closed path is exactly equal to the net current enclosed by that path, that is,

∫

→

→

H . dl = I

...(1)

The positive current is taken in the direction of advancement of right handed screw turned in the direction in which the closed path is traversed (fig. 21a).

Z →

H

I

→ J

dl A

Y

→ H Path

Outward (a)

(b)

X Fig. 21

For the more general case, where the current density is non-uniform, we have

∫

→ →

→ →

H. dl = I = ∫ J . dS

...(2)

S

→ →

where ∫ J . d S is the total current passing through the closed curve. Eqn. (1) is also an integral form of Ampere’s law.The law can be applied inside or outside a conductor, but total current enclosed by the path should be known. →

Ampere’s circuital law is frequently used to calculate H in highly symmetric situation. The standard current configurations which can be easily handled by Ampere’s law are, infinite straight lines, infinite planes, infinite solenoids, toroids etc.

E lectromagnetic F ield Theory

312

→

Usually, following two conditions must be fulfilled for the accurate determination of H by using Ampere’s circuital law. 1.

→

At each point on the closed path, magnetic field intensity H must be either tangential or normal to the path.

2.

→

In the case where H is tangential to the path, it must be equal at all points of the path. →

In term of magnetic induction vector B, Ampere’s circuital law states that the line integral of magnetic →

induction vector B around any closed path is equal to µ0 (permeability of free space) times the total current crossing the surface boundary by the closed path. That is,

∫

→ →

B . d l = µ0 I

In differential form, Ampere’s law may be expressed as →

→

curl B = µ0 J

or

→

I →

curl H = J

To prove the law, let us consider the field intensity at a point P distant ρ from a long straight filamentry wire carrying a current I as shown in fig. 22. Now consider the circular path of radius r centred

→ H

δθ O ρ

→ P dl

→

on this current carrying wire. The field intensity H at any point P on the circular path is I H= 2πρ

Fig. 22

...(1)

For every point on the circular path of radius r the field intensity has the same magnitude as given by →

eqn. (1) and is parallel to the tangent to the path. Therefore, the line integral of field intensity H around the circular path centred on the current carrying wire is given by,

∫

→ →

H . d l = ∫ H dl = =

Thus, we have

∫

1 dl 2 πρ ∫

1 ×2πρ = I 2πr

→ →

H. dl = I

Applications of Ampere's Law (a) Magnetic Field Due to a Long Straight Current Carrying Filamentry Wire Let us consider a long straight filamentry wire carrying a steady current I (Fig. 23). From the symmetry of the wire it is obvious that the magnetic lines of force are concentric circles centred on the wire. In order →

to find the magnetic field B at a point P, we draw a circle passing through the point P with centre at O on

313

the wire. The radius of the circle is OP = ρ. By symmetry, all points on the circle are equivalent and hence →

the magnitude of field B should be same at all these points and is directed tangentially to the circle →

→

→

everywhere. Moreover, B and dl are always directed along the same direction, therefore line integral of B along the boundary of the circular path is → →

∫

B . dl = ∫ dl = Z

→ B

→ →

∫

or

∫ dl

B . dl = B .2πρ

I O

From Ampere's law, we have B . d l = µ0 I

C

B .2 πρ = µ0 I B=

or

→

B=

In vector form

Fig. 23

µ0 I 2π ρ I µ0 ɵ φ tesla 2 πρ →

Similarly, the magnetic field intensity H is given by, → → ∵ H = B µ0

→

I ɵ φ A/m H= 2 πρ

(b) Magnetic Field Induction of a Solenoid A solenoid is a coil of wire wound on the surface of a hollow cylinder of a card-board or china-clay and has a length very large compared to its diameter (Fig. 24). It is experi mentally observed that when a current passes through such a solenoid the magnetic field outside is very small compared with that inside. Further the line of induction inside the solenoid are straight and parallel showing a uniform magnetic field except near the edges. (Fig. 25).

I0

Fig. 24

To determined the magnetic field at a point inside a current carrying solenoid consider a closed rectangular path abcd in which the side ab is parallel to the axis and sides bc and da are very long so that the side cd is far from the solenoid (Fig. 25) and thus the field at this side is negligible. As the solenoid is long and the rectangle is not too near the either ends, the field is at right angles to the sides bc and da. Thus, the line →

integral of B along the closed rectangle abcd is

∫

→ →

B . dl = ∫

b → → a

B . dl +

→ dl

→ →

∫ ∴

r

c → →

∫b

B . dl +

d → →

∫c

B . dl + ∫

a → → d

B . dl

...(1)

Fig. 25

P

E lectromagnetic F ield Theory

314 c → →

Now,

∫b

and

∫c

a → → d

∫

B. d l = ∫

(∵ along bc and da, B. dl = Bdl cos 90 ° = 0)

→

B. d l = 0

→ →

→ →

B. d l = 0

d → →

∴ →

B . dl = ∫

(∵ B is zero far outside the solenoid)

b → → a

...(2)

B. d l

→

As B is parallel to d l along ab, we have b → →

∫a

B . d l =∫

b a

B dl = B

b

∫a

dl or

b → →

∫a

B . d l = B l, where l = ba

Therefore, eqn. (2) becomes

∫

→ →

B . d l = Bl

...(3)

abcd

Applying Ampere's circuital law to the closed rectangular path abcd, we have

∫

→ →

B . d l = µ0 I

...(4)

If n be the number of turns per unit length along the length of the solenoid, then nl turns cross the rectangle abcd. Each turns carries a current I0 . Therefore, the net current crossing the rectangle abcd equals to n l I0 ∴

∫

→ →

B . d l = µ0 n l I0

...(5)

Comparison of eqns. (3) and (5) yields, Bl = µ0 nl I0 and

B = µ0 n I 0

or H = nI0

...(6) ...(7)

→

It is clear from eqn. (6) that the field B is independent of the length and diameter of the solenoid and is uniform over the cross-section of the solenoid. If the solenoid is wrapped on a core of material of permeability µ or relative permeability µ r , then B = µ nI 0 = µ 0 µ r nI0

(c) Magnetic Field of a Toroid (or Endless Solenoid) If a solenoid is bent round in the form of a closed ring and their ends are joined, we get a toroid. Toroid may also be constructed by closely winding uniform turns of wire over a non-conducting ring. The magnetic field in such a toroid may be obtained by using Ampere's circuital law. In order to find the magnetic field at a point P inside the toroid carrying a current I0 at a distance ρ from the centre O, we draw a circle through the point P concentric with the toroid as shown in fig. 26. By →

→

symmetry the direction of the field B is everywhere tangential to the circle and the magnitude of B is same at all points on the circle.

315 →

R

Therefore, the line integral of magnetic field B along the circular path of radius ρ is

∫

→ →

B. d l = ∫ B dl = B

∫ dl = B (2 π ρ)

P

ρ

If the total number of turns in the toroid is N, then the current crossing the area bounded by the circle is NI0 .

I0

Applying Ampere’s circuital law on the circle of radius r, we have

∫

→ →

B . dl = µ0 NI0 B=

or

or

2 π ρ B = µ0 N I0

Q

→ B

µ 0 NI 0

Fig. 26

2π ρ →

Thus, the magnitude of magnetic induction B varies inversely with ρ. If l be the mean circumstance of the toroid, then l = 2 π ρ, therefore, B=

µ0 N I0 l

The magnetic field at a point outside the toroid such as R is zero, because each turn of winding passes twice the area enclosed by the circle through R, carrying equal currents in opposite directions. So the total current within the circle is zero. Similarly, at an internal point, like Q the field is also zero because the circle passing through Q encloses no current. Thus, the field of a toroid is confined only within the core.

(d) Magnetic Field Intensity Due to Long Hollow Conducting Cylinder Containing Uniform Current Density Z

Let us consider a long hollow cylinder with an internal and external radii 'a ' and 'b' respectively as shown in fig. 27. If I is the total current of the cylinder, then for uniform current density in the cylinder, the current within the radius ρ is directly proportional to the area within the radius. By

b

a

I

→

symmetry, the magnetic field intensity will be along φ axis. To find H draw

Y

a cylindrical surface of radius ρ. According to Ampere's circuital law. → →

X

∫ H. dl = I(enclosed) There may be three possibilities.

Fig. 27

(i) ρ ≤ a, In this case, we consider closed circular surface C1 as shown in fig.28. According to Ampere circuital law. → →

∫ H1. dl = I(enclosed) There is no current is enclosed in a circle of radius ρ (ρ ≤ a). Therefore, or Hφ ∫ dl = 0 or Hφ (2 πρ) = 0 Hφ = 0 1

C1

1

1

Therefore, inside the hollow cylinder, the magnetic field intensity Hφ is zero. 1

E lectromagnetic F ield Theory

316

closed circular surface C1

(ii) For a1 ≤ ρ ≤ b. In this case closed circular surface is in between the surface of the hollow cylinder as shown →

in fig 29. Therefore to find H. Apply Ampere's Circuital

a

O

b

outer surface

ρ

law as

∫C

2

Inner surface

H2 dl = I(enclosed), where I(enclosed) is the Fig. 28

current enclosed by the closed surface C2 . I(enclosed) for the closed surface C2 is obtained as follows: The current per unit area of the hollow cylinder =

closed circular surface C2

I π ( b2 – a2 )

∴ I(enclosed)= Current per unit area × area of the closed

a

O

b ρ

surface =

Hence,

∫C

2

H2 dl =

∴

Hφ (2 πρ) =

or

Hφ =

I π (b2 − a2 )

I (ρ2 − a2 )

Fig. 29

(b2 − a2 )

I (ρ2 − a2 ) (b2 − a2 ) I (ρ2 − a2 ) b2 − a2

2

2

. π (ρ2 − a2 ) =

I(ρ2 − a2 ) 2 πρ (b2 − a2 )

closed circular surface C3

(iii) For ρ ≥ b. In this case closed circular surface C3 is outside the hollow cylinder (Fig. 30), therefore the current enclosed in the surface is the total current flowing in the cylinder, that is, I. Now applying Ampere circuital

aO

b ρ

law in this case as

∫C

3

or or

H3 dl = I(enclosed) = I

Hφ (2πρ) = I Hφ =

I 2π ρ

Fig. 30

317

(e) Magnetic Field Intensity Due to Uniform Sheet of Surface Current Density: Z

Consider a uniform conducting sheet of width L and having a surface current density k (that is, charge is flowing over a surface). Let the sheet be placed in x − y plane, hence is this plane z = 0 as shown in fig. 31. when the charge flows over a surface, we describe it by the surface current density k. the surface current density is defined as the total current flowing per unit width of the sheet perpendicular to flow. That is, I ...(1) k = A/m L

conducting sheet

Y

L

I

current is passing along +y-direction

X

Fig. 31

If dl is the incremental path length along L, then eqn. (1) can be expressed in integral form as. I = ∫ k dl

...(2)

The current sheet may be imagine to be divided up into a number of current carrying filaments. suppose one such filament is as shown in fig. 32. It is clear from fig. 32. that all the z-components of

I

I ←

Direction of H

→

H along z-axis will be cancelled. Therefore, Hz = 0.

Fig. 32

→

Further, there will be no components of H along y − axis because filaments are placed along y − axis. Therefore, H y = 0. Hence, only the components of →

H along x-axis, that is, H x exists only. In order to determine H x , we consider a closed loop as shown in fig. 33. Now apply Ampere's circuital law to this loop as, → →

2 → →

∫ H. dl = ∫1

H. dl + ∫

3 → →

4→ →

1 → →

2

3

4

2

Hx L

The components of H along z −axis are zero, that is →

1

Hx

components of H along 1 → 2 and 1 → 2 and 3 → 4 are zero. Therefore, → →

∫ H. dl = 0 + H x L + 0 + H x L = I(enclosed) = k L 2 Hx = k

or

Hx =

k 2

If nɵ is the unit vector normal to the current sheet, then →

H=

Z

H. dl + ∫ H. dl + ∫ H. dl = I

→

or

3

1→ ɵ k× n 2

The field above and below he sheet shall

1→ 1→ k × nɵand – k × hɵ respectively. 2 2

Y X

4 Fig. 33

E lectromagnetic F ield Theory

318

(f) The Magnetic field of A Plane Current Sheet: Let us imagine a plane very thin current sheet ABCD having a surface current

Js

→

density J S as shown in fig. 34. By →

Js

→

P

B

L

vector B would be parallel to sheet and

Q

R

A

perpendicular to J S .

∫

B

→ →

H. dl =

ABCDA C

∫

∫

→ →

H. dl +

A A

→ →

H. dl =

B

∫

Fig. 34

C

∫

D

→ →

H. dl +

B → →

H. dl = 0

∫

A

→ →

H. dl +

C

∴

∫

→ →

H. dl =

ABCDA

D

(1) (2) Two parallel current sheets (b)

Plane current sheet (a)

Now applying Ampere's circuital law to plane sheet ABCD which is perpendicular to the plane of the current sheet and also normal to the direction of current. Therefore,

or

←

D

symmetry, the lines of magnetic induction

Here

Js

C

∫

→ →

H. dl

D B

∫

A

→ →

H. dl +

D → →

∫

H. dl

C

→ →

∫ H. dl = 2 HL = I

ABCD

But I = J S L

∴ 2 H = JS

or

or

H =

B=

JS 2

µ0 J S 2

For the infinite very thin plane current sheet. This value of magnetic induction does not depend on the distance of the point of observation from the current sheet. When there are two parallel very thin current sheets of equal current density but of opposite directions as shown in fig.34(b), then the magnetic field induction between the sheets is B=

µ0 J S µ0 J S + = µ0 J S 2 2

B=

µ0 J S µ0 J S =0 – 2 2

and that of outside points is

319

→

→

(g) Curl of Magnetic Induction B or H or Ampere’s Law in Integral and Differential Forms Let us consider a closed path C bounding a surface S through which currents are flowing (Fig. 35). A general steady current distribution is

J

J

J

J

→

described by a current density vector J which is constant with time,

S

however it varies over the surface. The total current enclosed by this closed path is given by I=∫

→ S

→

J

dl

C

...(1)

J . dS

J

Fig. 35

→

where d S is the small elemental area vector inside the closed path C →

where the current density is J . →

According to Ampere’s circuital law, the line integral of the magnetic induction B around any closed path is, → →

∫

B . dl = µ0 I

...(2)

where I is the current enclosed by the closed path. Eqn. (2) is also an integral form of Ampere’s law. Substituting the value of I from eqn. (1) in eqn. (2), we get

∫

→ →

B . d l = µ0 ∫

→ → S

...(3)

J . dS

But from Stroke’s theorem, we have

∫

→ →

B . dl = ∫

→

S

→

...(4)

(curl B) . d S

Comparing eqns. (3) and (4), we obtain

∫S or

∫S

→ →

curl B . dS = µ0

→

→

∫S

→ →

...(5)

J . dS

→

(curl B − µ0 J ) . dS = 0

As the surface is arbitrary, we have →

→

curl B − µ0 J = 0 or

→

→

curl B = µ0 J

...(6)

→

In term of magnetic field intensity H eqn. (6) may be expressed as →

→

curl H = J

→

→

(∵ B = µ0 H)

...(7)

Relations represented by eqns. (6) and (7) show that the magnetic flux density at a point, is obtained →

from the given value of J at that point by integration. This is the Maxwell's equation for magnetostatics. In fact eqn. (7) or eqn. (6) is a restricted form of Maxwell's electromagnetic field equation. Later we shall see how this equation has to be generalized further by including extra term to account for the effects of time varying magnetic fields.

E lectromagnetic F ield Theory

320

D ivergence of Magnetic Induction B →

According to Biot-Savart law, the magnetic induction B due to a current loop at a point, with position →

→

vector r , with respect to a current element dl is given by, →

µ I B= 0 4π

→

∫

→

dl × r

...(1)

r3

Taking divergence of eqn. (1), we have →

µ I div B = 0 4π

∫

→ → dl × r div 3 r

...(2)

From the vector identity →

→

→

→

→

→

div(A × B) = B . curl A − A . curl B We have

→ → → → → r r ∇ . d l × 3 = 3 . curl d l − d l . curl r r

→

→

But

...(3)

→

curl d l = 0

As d l is a constant vector, ∴

→ r r3

→ → → r ∇. d l × 3 = − d l . curl r

→

→ r r3

→ r 1 curl 3 = − curl grad r r

...(4) → ∵ grad 1 = − r r r3

...(5)

We know that curl grad S = 0 ∴

→ r curl 3 = 0 r

Hence, from eqn. (4), we have r → div d l × 3 = 0 r

...(6)

→ → r Substituting this value of div dl × 3 from eqn. (6) in eqn. (2), we obtain, r →

div B = 0 →

Thus, the divergence of magnetic flux density B is always zero. This is the Maxwell's equation for magnetostatics.

321

→ Example 14: A current sheet with surface current density k is given by k = k Iɵz Am –1, where k is a

constant coincides with the XZ plane as shown in figure 36. Find a general relation for flux density. [UPTU, B.Tech. IV Sem 2006]

Y a

b

d

c

l

I X

z Fig. 36

Solution: In this problem the current is in Iɵz direction and therefore z − component of magnetic field →

intensity H is absent. Further due to symmetrical distribution of current with respect to a point at which →

→

→

H is derived, y – component of H will get cancelled. The only x–component of H, that is H x will be present. Applying Ampere's circuital law to the loop abcd, we have b → →

∫c

H. dl + ∫

a → →

d → →

c → →

b

a

d

H. dl + ∫

H. dl + ∫

H. dl = I

→

H above the plane at a point is equal and opposite at a symmetrical point just located below the plane O + Hl + O + Hl = I or 2 Hl = I

So

→

H =

or

I k. Iɵz or = 2l 2l

→

→

B =µ H =

µ k. Iɵz 2l

→

Example 15: The magnitude of field vector H at a radius of 1 meter from a long conductor is 2 amp/meter. Find the current in the wire. Solution: According to Ampere’s circuital law, the current in the wire is given by

∫ or or

→ →

H. dl = I

∫ H dl = H ∫ dl = I

or I = H × 2πr

I = 2 × 2 π × 1 = 4π amp

Example 16: A cylindrical conductor of radius 10 –2 m has an internal magnetic field. → ρ ρ2 H = (4. 77 × 10 4 ) – φɵ A/m What is the total current? 2 – 2 3 × 10

Solution: According to Ampere circuital law

E lectromagnetic F ield Theory

322 → →

→

I = ∫ H. dl, Here dl = ρ d φ φɵ ∴

ρ ρ2 I = ∫ 4.77 × 104 – φɵ . ρdφ φɵ – 2 2 3 × 10

or

I = 4.77 × 104

∫

2 ρ2 ρ3 ρ3 4 ρ d = × φ 4 77 10 – . – –2 –2 2 3 × 10 2 3 × 10

2π

∫0

dφ

(10 –2 )2 (10 –2 )3 2π Here ρ = 10 –2 m ∴ I = 4.77 × 104 – 2 3 × 10 –2 10 – 4 10 – 4 I = 4.77 × 104 × 2 × 3.14 – 3 2

or

I=

4.77 × 104 × 2 × 314 . × 10 – 4 Amp 6

or I = 4.992 Amp

Example 17: J = 10 2 sin θ ^ r A /m2 , in spherical coordinate. Find current crossing the spherical shell of radius (r = 0 . 02 m ). Solution: We know that,

∴

S

I=∫

θ =0

I=∫ or

→ →

I=∫

J . ds

π π

θ =0

→

→

Here J = 102 sin θ rɵ and dS = r 2 sin θ dθ dφ rɵ

2π

∫ φ =0 (10 sin θ rɵ) .(r 2π

∫ φ =0 10 r

I = 102 × (0.02)2

2 π

∫0

2

sin θ dθ dφ rɵ)

sin2 θ dθ d φ =102 r 2

π

2

∫θ = 0 sin

θ dθ

2π

∫0

dφ

1 – cos 2θ 2π dθ [φ]0 2 π

= or

π 102 × (0.02)2 sin 2θ .2 π ∫ (1 – cos 2θ) dθ = 100 × 0 .02 × 0 .02 × π × θ – 0 2 0 2

I = 100 × 0.02 × 0.02 × π × π or I = 0 . 394 Amp

Example 18: A circular conductor of radius ρ0 = 1cm has internal field →

H = where a =

ρ 10 4 1 sin(aρ )– cos(aρ ) φɵ A/m a ρ a2

π . Find the total current in the conductor. 2ρ0

Solution: According to Ampere circuital law

323 → →

→

I = ∫ H. dl → →

Here dl = ρd φ φɵ

4 ρ 104 1 ɵ ɵ = 10 1 sin(aρ) – ρ cos (aρ) ρdφ ρ ρ ρ sin( ) – cos ( ) φ . φ φ a a d ρ a2 ρ a2 a a

∴

H. dl =

∴

→ → π πρ ρ 1 H. dl = 104 ρ – sin cos dφ 2 2ρ0 (π / 2ρ0 ) 2ρ0 (π / 2ρ0 )

π ∵ a = 2ρ0

→ → 4ρ2 104 × 4 ρ20 2ρ2 π π ( H. dl)ρ=ρ0 = 104 20 sin – 0 cos dφ = dφ 2 2 π π2 π

∴

I=

∫

→ →

H. dl =

104 × 4 × ρ20 π

r =ρ0

2

2π

∫0

dφ

Substituting ρ0 = 1cm = 10 –2 m I=

∴

104 × 4 × (10 –2 )2 π2

.2π =

104 × 4 × 10 –4 × 2 or I = 2 . 548 amp 314 .

Example 19: Find the current density in the following: →

→

(ii) H y = k sin x aɵ y where k is a constant.

(i) Hz = k xy aɵ z . Solution: (i) We know that

→

→

→

→

Here H = Hz = k x yaɵz

aɵ x J = ∂ / ∂x 0

aɵ y aɵz → ∂ ∂ ∂ / ∂y ∂ / ∂z or J = aɵ x (kxy) + aɵ y (kxy) ∂y ∂x 0 kxy

→

∴

→

J = ∇× H

→

J = kxaɵ x − kyaɵ y

or →

→

(ii) H y = k sin xaɵ y = H ∴

or

aɵ x J = ∂ / ∂x 0

→

aɵ y ∂ / ∂y k sin x

aɵz ∂ / ∂z 0

→

→ ∂ ∂ J = aɵ x (k sin x) – 0 + aɵ y (0 – 0) + aɵz (k sin x) – 0 or J = k cos x aɵ z ∂x ∂y

→

Example 20: Obtain an expression for magnetic induction B at a distance r from the axis of a long cylindrical wire of radius a and carrying a current I0 .

E lectromagnetic F ield Theory

324

Solution: The section of the long cylindrical wire perpendicular to its length is shown in fig. 37. Let us consider that the current is uniformly distributed over the cross-section of the wire. Draw a circle of →

radius r around the axis of the wire. By symmetry, the direction of B at any point on the circumference of the circular path is tangential to it and its magnitude is constant. Therefore, from Ampere’s law. → →

∫

B . d l = µ0 I0

→ →

∫ or

→ B

...(1)

→ B

B . dl = ∫ B dl = B ∫ d l

dl

B (2 π r ) = µ0 × (current enclosed by the path) ...(2)

r a , In this case the total current I0 passes through the whole

a

r>

a

There may be three possibilities,

section of the wire. B (2 π r ) = µ0 I0 or B =

∴

µ 0 I0 2π r

...(3) Fig. 37

(2) r < a, In this case the fraction of current that passes through the →

enclosed path contributes to B.

∫

∴

B dl = B (2 π r ) = µ0 I

Here I = current density × area =

B=

∴

I0

π a2

(π r 2 ) =

µ0 I0 r 2 2

2 πra

I0 r 2 a2

or B =

µ 0 I0 r

...(4)

2 π a2

(3) At r = a, that is at the surface of the wire, B is obtained by putting r = a in eqn. (4), B=

∴

µ0 I0 a 2

2πa

or B =

µ 0 I0 2π a

Example 21: A long, straight conductor cross-section with radius ' a ' has a magnetic field strength → → → Ir I ^ ^ for . Find within the conductor and ( r > a ) J in both the regions, = H = H φ ( r < a ) φ 2πr 2 πa2

[UPTU, B.Tech. IV Sem 2007] →

Solution: (i) For r < a, The magnetic field strength H within the conductor is given by. →

H= →

Ir 2 πa2

^

φ

The current density J inside the conductor is obtained as

...(1)

325 →

→

→

J = ∇× H →

→

In cylindrical coordinates ∇ × H is expressed as, → → 1 ∂H ∂Hφ ∂Hr ∂Hz ɵ 1 ∂ (r Hφ ) ∂Hr ɵ z – – – J = ∇× H = rɵ + z φ+ ∂z ∂r r ∂r ∂z ∂φ r ∂φ

→

→

Here H =

...(2)

→ Ir ɵ φ, that is, H is only in φ direction, therefore Hr = Hz = 0 2 2 πa

Hφ =

or

Ir 2 πa2

Therefore eqn. (2) reduces to →

→ → 1 ∂ rI r ∂ Ir J = ∇× H = 0 – rɵ + (0 – 0) φɵ + – 0 zɵ 2 2 z ∂ ∂ r r 2 π a 2 π a →

J = –0 +

or

→

(ii) For r > a,

H=

→ 1 I I 1 2 rzɵ = 2 zɵ or J = 2 zɵ 2 r 2 πa πa πa

...(3)

I ɵ φ 2πr

→ I ɵ Since only H is only in φɵ direction, that is, Hr = Hz = 0 and Hφ = φ 2πr →

J =

In this case

1 ∂ 1 ∂ rI (rHφ ) = =0 r ∂r r ∂r 2 π r →

That is, current density J is zero outside the conductor. Example 22: Determine the magnetic field from an infinitely long wire of finite radius r1 in cylindrical coordinate system. Solution: The infinitely long wire of radius r1 is shown in fig. 38. If I is the current flowing through the wire, then J=

Current density,

I

...(1)

π r12

Bφ

This current density J is uniformly distributed over the cross-section of the wire. By symmetry it is obvious that in cylindrical coordinates the magnetic →

flux density B has only Bφ component and Bφ is the component of r only.

r1

r

I

According the Biot-Savart’s law

∫

→ →

B . d l = µ0 I = µ0

∫

→

Fig. 38

→

...(2)

J . dS

In cylindrical coordinates, →

B = Bφ ^ φ, d S = r d φ d r and

dl=rdφ

E lectromagnetic F ield Theory

326

Substituting these values in eqn. (2) and applying boundary condition of the situation, we get ∴ or

2π

∫0

Bφ r

Bφ r d φ = µ0 J 2π

∫0

d φ = µ0 J

2 π r Bφ = Bφ =

∴

2π

r

∫0 ∫0

rdφdr

2π

r r d r d φ ∫0

∫0

µ0 J r 2 2

2π

∫0

...(3) 2π

or

d φ or 2 π r Bφ =

Bφ r φ 0

= µ0 J ∫

2π

0

r

r2 dφ 2 0

µ0 J r 2 .2 π 2

µ0 J r

...(4)

2

Substituting the value of J from eqn. (1) into eqn. (4), we get Bφ =

µ0 r I 2 π r12

Weber/ m2

For r ≥ r1 that is, outside the wire, the total current enclosed is I. Therefore, again from Ampere’s law, 2π

∫0

Bφ r d φ = µ 0 I or Bφ r 2 π = µ0 I Bφ =

or

µ0 I Weber/m2 2π r

Example 23: Determine the magnetic field and its curl at radius r within a copper conductor of radius r0 > r carrying current I uniformly distributed over the cross-section. Solution: Cross-sectional view of the copper conductor of radius r0 is shown in fig. 39. Since current I is uniformly distributed over the cross-section, then from Ampere’s circuital law,

∫

→ →

H. dl = I

...(1)

Since r0 is the radius of the conductor, the uniform current density J is J=

current I = Area π r02

r0 Hφ

...(2)

I I′

When r < r0 , the fraction of current that passes through enclosed

r 0

→

circular path contributes to H. If it is I ′ , then I′ = current density × area enclosed by the circular path I′ =

I π r02

r × π r 2 or I ′ = I r0

2

...(3)

Fig. 39

Therefore, for a circular path of radius r0 such that r < r0 , Ampere’s Circuital law is

∫

→ →

H . d l = I′

...(4)

327

or

∫

∴

Hφ 2 π r = I ′ =

Hφ d l = Hφ

∫

dl = Hφ (2 π r )

I r2

or Hφ =

2 r0

Ir

...(5)

2 π r02 →

In cylindrical coordinate system the curl of magnetic field H is, → → 1 ∂ Hz ∂ Hφ ^ ∂ Hr ∂ Hz ^ 1 ∂ (rHφ ) ∂ Hr ^ ∇×H= − − − r + φ+ z ∂z ∂r r ∂r ∂φ r ∂ φ ∂z Hr = 0, Hφ =

From eqn. (5),

Ir 2

2 π r0

and Hz = 0

...(6) ...(7)

Substituting the values of Hr , Hφ and Hz from eqn. (7) in (6), we get →

→

→

→

∇×H=−

∴

∇×H=

or But from eqn. (2),

1 ∂ (r Hφ ) ^ ∂ Ir ^ 1 ∂ ^ r + r + z =− r ∂z ∂r ∂ z 2 π r02 r ∂r

∂ Hφ

→ → 1 1 2rI ^ zɵ z or ∇ × H = 2 r 2 π r0 π r02

...(8)

2

I /π r0 = J →

→

∇ × H = J^ z

∴

I r2 ^ 2 π r2 z 0

or

→

→

→

∇ × H = J

M agnetic Flux Density-Maxwell's Equation →

→

Just like an electric field strength E and displacement vector D in electrostatics the magnetic field →

→

intensity H and magnetic flux density (or magnetic induction) B are the two principal field vectors in →

→

→

magnetostatics. However, H is more useful than D . While calculating H the surrounding medium is not →

taken into account, whereas for the calculation of B the medium has to be considered. In a medium of →

magnetic permeability µ, the magnetic flux density B is defined as →

→

B =µH

For free space, µ = µ 0 = 4 π × 10

−7

...(1)

H /m.

Comparing other media with free space, we define relative permeability µ r as → → µ or µ = µ 0 µ r µr = ∴ B = µ0 µ r H µ0

...(2)

→

where B is measured in Weber per square meter (Wb/m 2 ) or tesla (T) or newton / (amp-meter). →

→

→

Eqn. (2) gives a general relation between B and H . Any expression derived for H is automatically valid for

→

B when the permeability of the medium is taken into consideration. An electrostatic equation corresponding to eqn. (1) is, →

→

D =εE

...(3)

E lectromagnetic F ield Theory

328

→

→

In a dielectric medium the electrical force exerted upon a charge is proportional to E and not to D , but in →

a magnetic material the magnetic force that is exerted upon a moving charge is proportional to B not to →

→

→

H . Hence, B is the magnetic field for moving charges as E is the electrical field for free charges. →

The magnetic flux φ m across a surface may be defined in a similar manner as the electric flux φ E . If B be →

→

→

the magnetic induction at a small patch of vector area dS, then the scalar product B . d S is defined as the →

magnetic flux d φ m through the patch of area dS . That is, →

→

dφm = B . d S

→

The magnetic flux through the entire designated area S of which d S is one of the patch is, φm = ∫

→ S

→

...(4)

B . dS

→

If B is measured in teslas (T) or webers per square meter (W/m 2 ), then magnetic flux should be in webers. →

→

→

→

If B is uniform and normal to dS , that is, the angle between B and d S is 90°, then the magnetic flux φm = 0

→

→

On the other hand if B is parallel to d S, then φm = ∫ BdS = B ∫ S

S

dS or

φ m = BA

Comparison of Biot-Savart law in magnetostatics and Coulomb's law in electrostatics establishes an →

→

→

→

→

→

analogy between B and E. The relations, in free space, B = µ0 H and D = ε 0 E lead to an analogy →

→

between B and D. According to the Gauss's law in electrostatics, the total flux passing through any closed surface is equal to the charge enclosed in that surface, that is, φE = ∫

→

S

→

D . dS = Q

...(5)

The charge Q is a source of lines of electric flux and these lines start and terminate on positive and negative charges, respectively. No such source has ever been discovered for the lines of magnetic flux. The magnetic flux lines are closed and do not terminate on a "magnetic charge". For this reason Gauss's law in magnetostatics is, →

→

∫ S B . dS = 0

...(6)

→ → → → An application of Gauss divergence theorem ∫ ( ∇. B) dV = ∫ B . dS gives S v → →

∇. B = 0

...(7)

329

Eqn. (7) is the first Maxwell's equation of the magnetostatics. The integral equivalent of differential eqn. (7) is represented by eqn. (6). Eqns. (6) and (7) imply that there are no free magnetic charges or monopoles analogous to free electrical charges in electrostatics.

M axwell's

Equation for Static Fields

The complete set of differential and integral form of Maxwell's equations that apply to static electric fields (for electrostatics) and steady magnetic fields (or magnetostatics) are as follows: Differential Form

Integral Form →

→ →

→

→

→

→

→

∫ S D . dS = Q = ∫ v ρv dv

∇. D = ρ v

∫

∇ × E =0 →

→

→

→

∫ H. d l

∇×H= J

=I=∫ →

→ →

→

E. d l = 0 → S

→

J . dS

→

∫ S B . dS = 0

∇. B = 0

The explanation, physical significance and the derivation of these equations are discussed in next unit. → 2. 39 × 10 3 cos φ ρɵ A /m exists in free space. Find the magnetic flux Example 24: A radial field H = ρ

crossing the surface defined by 0 ≤ φ ≤ π / 4 and 0 ≤ z ≤ 1m →

→

→

Solution: Magnetic flux, ψ m = ∫ B . dS, Here, dS = ρ d φ d zρɵ S

∴

µ × 2.39 × 103 0 ψm = ∫ cos φ ρɵ .(ρ d φ d z ρɵ ) S ρ = µ 0 × 2.39 × 103

π /4

1

∫ φ = 0 cos φ dφ ∫ 0 d z π /4

= 4 π × 10 −7 × 2.39 × 103 [sin φ] 0

[ z]

1 0

(∵ µ0 = 4 π × 10 −7 weber / amp-meter)

= 4 × 3.14 × 10 −7 × 2.39 × 103 × 0.7071 × 1 = 21.226 × 10 −4 or

ψ m = 21. 226 × 10 −4 weber

E lectromagnetic F ield Theory

330

→

Example 25: In cylindrical coordinates B = (1 / ρ) φɵ . Determine the magnetic flux crossing the plane surface defined by 0 .5 ≤ ρ ≤ 2 .5 m and 0 ≤ z ≤ 2. 0 m. Solution: Magnetic flux crossing the surface is express as, →

→

→

ψ m = ∫ B . dS , Here, d S = dρ d z φɵ S

∴

1 ψm = ∫ S ρ

or

ψm = ∫

or

2. 5 = 2 × 1.6094 ψ m = 2 × log e 0. 5

2. 5 0 .5

2 .5 φɵ .(dρ d z φɵ ) = ∫ ρ = 0 .5

dρ ρ

2

∫ 0 dz

2

∫0

dρ dz ρ

2 .5 2 [ z] 0 .5 0

= [log e ρ]

= [log e (2.5) − log e (0.5)] × 2

or

ψ m = 3.22 weber

Example 26: A solenoid is 1. 0 meter long and 3. 0 cm in mean diameter. It has five layers of windings of →

850 turns each and caries a current of 5. 0 A. What is the value of the magnetic induction B at its centre? What is the magnetic flux ψ for a cross-section of the solenoid at its centre ? Solution: The magnetic field induction B at the centre of a long solenoid having n turns per meter and carrying a current of I amp., is given by B = µ0 nI I = 5 amp, µ0 = 4 π × 10 −7 weber /A- m and n =

Here ∴

B = 4 π × 10 − 7 ×

5 × 850 −1 m 1.0

5 × 850 × 5 or B = 2 . 669 × 10 −2 weber / meter2 1

The magnetic flux through an area is given by →

→

→

ψ B = ∫ B . dS , where B is uniform and perpendicular to the area, then S

ψ B = ∫ B dS cos 0 ° = B ∫ dS or ψ B = B . A, where A is the area Here ∴

A = π r 2 = 3.14 × (1.5 × 10 −2 ) 2 = 7.065 × 10 −4 m2 ψ B = 2.669 × 10 −2 × 7.065 × 10 − 4 or ψ B = 18. 85 × 10 − 6 weber

M agnetic Scalar and Vector Potentials →

In electrostatic, the concept of scalar electric potential was introduced because electrostatic field E is →

→

conservative (that is, for which curl E = 0). The electric field E is expressed as negative gradient of →

potential (that is, E = − ∇ V ). The solution of many tedious electrostatic problems become much

331

simplified by the introduction of electrostatic potential. The electrostatic scalar potential is a space function that depends upon the magnitude and location of charges. The similar potential function for the magnetic field can not be introduced as such because magnetic fields are non-conservative (that is, for →

which curl B ≠ 0). However, for magnetic fields it was desirable to set up magnetic potential the space →

→

derivative of which would given B or H . Corresponding to individual charges in electrostatic fields, the source of magnetic field is current element (Idl) of the circuit. The magnetic potentials would therefore depend upon these current elements. The electrostatic potential is a scalar quantity having magnitudes only. In magnetostatics the current elements have directions as well as magnitudes. Therefore, in defining magnetic potentials the direction of the source must also be taken into consideration. There are two magnetic potentials: (i) magnetic scalar potential (ii) magnetic vector potential.

M agnetic Scalar Potential (Vm) It is well known that the electric field is created due to charge distribution whereas magnetic field is created due to a current distribution. Now by analogy with electric field we define magnetic scalar → →

→

potential Vm . Just as electric potential V is related with electric field intensity E ( E = − ∇ V ), magnetic →

scalar potential Vm is related with magnetic field intensity H as, →

→

H = − ∇ Vm →

→

→

This definition must not be contrary to our previous results for the magnetic field, ∇ × H = J

→

From Ampere's law in circuital form it is obvious that the line integral of magnetic field intensity H

around any close path does not vanish unless it encloses zero current. Therefore, if we introduce a →

magnetic scalar potential in a region (analogous to electric potential) from which H would be derived this magnetic potential would not be single valued function of position unless the region were current free. For current free region, J = 0. Thus, →

→

→

curl H = ∇ × H = 0 or

→ →

∫ H. d l = 0 →

→

Now since curl of a gradient is zero, so in that case when J = 0, the magnetic field intensity H can be written as the gradient of a scalar potential (Vm ) as →

→

→

→

∇ × H = ∇ × (− ∇ Vm ) = 0

→

Thus, the vector H can be expressed as →

→

H = − ∇ Vm or

→

→

B = − µ ∇ Vm, when J = 0

E lectromagnetic F ield Theory

332

where Vm is called the magnetic scalar potential. The dimensions of Vm are obviously amperes. →

→

→

As an analogy of electric field intensity ( E = − ∇V ). However divergence of B is also zero. → →

→ →

∇ . B = −µ ∇ . H = 0

That is, →

→

µ ∇ .(− ∇ Vm ) = 0

or

∇2 Vm = 0

or

Thus magnetic vector potential Vm , satisfy an equation similar to Laplace equation, like electrostatic potential V. The difference of magnetic scalar potential between any two points can be related to the line integral of →

H as, Vm1 − Vm2 = ∫

1 → →

H. d l amp.

2

In this line integral any of the path of the magnetic fluxes can be taken between points ' 1' and ' 2 ', but that path must always be chosen in such a way that does not contain current, which mathematically means that the magnetic scalar potential Vm is a multivalued function in general and is restricted to the regions containing no current. This is the difference from electric potential V, which is a single valued function and has no restrictions like magnetic scalar potential. Thus the magnetic scalar potential is of practical utility only to derive magnetic fields in the absence of continuous current distribution. The magnetic scalar potential can not be used if fields within current carrying media are desired.

M agnetic Scalar Potential For a Current Loop Let us consider a current loop carrying a current of I amp. as shown in Fig. 40. According to Biot-Savart →

→

law, the magnetic induction B due to whole loop at an outside point P(R ) distant r from the current →

→

element Id l is given by →

B=

→

µ0 4π

∫

→

Id l × R

...(1)

R3

→

R

→

If the point of observation is shifted from P (R ) to Q (R + dR )

→

→

→

dA = dl × (–dR)

dl

→

dR

through an infinitesimal distant dR , then we may write →

B . dR =

or

µ0 4π

→

∫

→

Id l × R R3 →

→

. dR = →

µ0 I 4π

µ I B . dR = 0 4π

→

∫

∫

→

→

I

dR .(d l × R )

→

→

dR P(R)

→ →

→

→

Q(R + dR)

dω

R3 →

→

I

(dR × d l ) . R R3

→ →

→

→

→

→

Fig. 40 →

→

→

[∵ A.( B × C) = B .(C × A) = C .( A × B)]

...(2)

333

→

→

→

When the point of observation P (R ) is shifted to Q (R + dR ) the solid angle subtended by the current loop at the point under consideration (Q) will change. Let this change in solid angle is represent by dω. The similar change will also be observed if we keep the point of observation P stationary but the loop is →

displaced through (− d R ) . In this situation eqn. (2) may be modified as, →

→

µ I B . dR = 0 4π →

→

→

∫

→

→

(− d R × d l ) . R

...(3)

R3

→

→

→

But − d R × d l = d A (area traced out by the current element I d l during the displacement of loop through − d R. Therefore eqn. (3) becomes →

→

µ I B . dR = − 0 4π

→ →

∫

da .R R3

→ →

Here

da . R R3

is the change in solid angle (dω) subtended by current loop when point P is shifted to Q.

∴

→

→

B . dR =

µ0 I dω 4π

...(4)

Solid angle subtended by the current loop is a function of coordinates of field point that is, ( x, y, z). Therefore, we have dω =

∂ω ∂ω ∂ω ∂ω ∂ω ∂ω dx + dy + dz = aɵ x + aɵ y + aɵz . (aɵ x dx + aɵ y dy + aɵz dz) ∂z ∂x ∂y ∂y ∂z ∂x →

→

dω = ∇ω . dR

or

...(5)

Substituting this value of dω in eqn. (4) we get →

→

B . dR = −

or But

→ µ0 I → ∇ ω . dR or 4π

→

B=−

µ0 I → ∇ω 4π

→

→ µ Iω B = ∇ − 0 4π →

→

B = − ∇ Vm

∴ Vm =

µ0 Iω 4π

Thus, magnetic scalar potential for a current loop, Vm =

µ0 Iω 4π

E lectromagnetic F ield Theory

334

A

pplication of a Magnetic Scalar Potential: Equivalence of a Small Current Loop and a Magnetic Dipole P

Let us Consider a small rectangular current loop BCDE of

ω

area A and carrying a current of I amp. Let P be any point →

having position vector R relative to centre of loop O at which magnetic scalar potential is to be determined. Let →

us imagine that R is large compared with the dimensions

R

B

of current loop as shown in Fig. 41. The magnetic scalar potential Vm at P is given by, Vm =

µ0 Iω 4π

...(1)

where ω is the solid angle subtended by the current loop → →

BCDE at P and is equals

A. R R

C

A O

E

θ

→

A (Area vector)

I D Fig. 41

3

or

ω=

∴

Vm =

A cos θ R2 µ0 I A cos θ 4π R

2

=

µ0 (IA) cos θ 4π R2

...(2)

The eqn. (2) for magnetic scalar potential is similar to that for potential due to an electric dipole, that is, V=

1 p cos θ 4 πε0 R 2

...(3)

→

where p is the electric dipole moment. Therefore, we conclude that the magnetic field due to small current loop is equal to electric field due to an electric dipole. The comparison of eqns. (2) and (3) suggest that the term (IA) is analogous to electric dipole moment. Thus the term (IA) is defined as magnetic dipole moment, that is, m = IA Thus, a current loop of area A carrying a current I is equivalent to magnetic dipole of dipole moment, m = IA.

335

M agnetic Scalar Potential in the Region Between the Inner and Outer Conductors of a Coaxial Line Let us consider a coaxial line with inner conductor of radius ' a ' and the outer of →

radius ' b ' as shown in Fig. 42. In the region a ≤ ρ ≤ b, J = 0. The magnetic field →

b

intensity H at a point distant ρ from the centre of the cable is given by →

H=

I ɵ φ 2πρ

ρ a

...(1)

where I is the uniformly distributed current in the inner conductor.

→

Fig. 42

The magnetic scalar potential Vm in the region a ≤ ρ ≤ b is related with H as, →

→

→

→ 1 ∂ ∂ 1 ∂ ɵ ∂ or ∇ = + φɵ φ+0+0 + aɵz ρ ∂φ ∂ρ ρ ∂φ ∂z

→

1 ∂Vm → φ ρ ∂φ

H = − ∇ Vm

In cylindrical coordinates ∇ = ρɵ

∴

H=−

...(2)

Comparison of eqns (1) and (2) gives I 1 ∂Vm =− 2 πρ ρ ∂φ I

I φ + C1 2π where C1 is the constant of integration which is set equal to zero.

∫ dVm = − 2π ∫ dφ

or

∴

Vm = −

or

Vm = −

I φ 2π

M agnetic Vector Potential Since the source of magnetic field, that is, current is a vector quantity, the potential in magnetostatic →

must be a vector. If the magnetic vector potential is represented by a vector A, then it should be possible →

→

→

to obtain B or H as the space derivative of A . There are two possible space derivative operation on a →

vector, namely divergence and the curl. The divergence operation (∇.) given a scalar quantity, whereas the →

curl operation ( ∇ X ) yields a vector quantity. Hence, the curl is only space-derivative operation which can →

→

be used to design magnetic vector potential. The vector H or B may be derivable through a suitable →

magnetic vector potential A via the relation, →

→

→

H=∇×A

...(1)

E lectromagnetic F ield Theory

336

or more widely used relation is,

→

→

→

B= ∇ × A

...(2)

→

Since the divergence of magnetic field ( B) is zero, the divergence of curl of a vector is also zero, that is, →

→

→

∇ × ( ∇ × A) = 0 →

→

→

→

→

→

∇ × B = ∇ × ( ∇ × A) = µ J

But,

...(3)

→

where J is current density. →

Now curl of a curl is not zero. Therefore if we define ( A) as magnetic vector potential, then current →

→

density J will be obtained by taking the curl of this vector potential ( A) twice. By using vector identity eqn. (3) becomes →

→

→

→ → →

→

∇ × ( ∇ × ∇) = ∇ ( ∇ . A) − ∇2 A = µ J

...(4) →

To give Amper's law a simple form in terms of vector potential A, we assert that we will choose only such →

a potential A which is divergence less, that is, → →

∇. A = 0

..(5) →

This is the second condition on magnetic vector potential A . →

Thus, a vector A satisfying both conditions represented by eqns (2) and (5) is generally taken as the magnetic vector potential for any given current distribution. →

→

→

→ →

For the function A satisfying both the relations (curl A = B and ∇ . A = 0), Ampere's law or eqn. (4) takes the form

→

→

∇2 A = − µ J

...(6)

Each of the three components ( x, y, and z) of this equation is equivalent to Poisson's equation, that is, ∇2 V = −

ρ ε

...(7)

→

Thus, magnetic vector potential A satisfy an equation similar to Poisson's equation. The eqn. (6) is differential equation for vector potential which is valid for direct current or very slow varying (radio frequency) current. By analogy of the solution of the Poisson's equation (7) as V=

1 4πε

ρ

∫r

dV

The solution of vector potential equation (6) will be →

µ A = 4π

→

→

∫V

J dV R

...(8) →

where R is the position vector of a point at which A is determined. Which is the one form of solution of vector Poisson's equation (6).

337 →

Thus, we could regard eqn. (8) as the definition of A for a volume current distribution. For line and surface currents, we will have similar integrals as, →

µ 4π

→

µ 4π

A=

A =

and

→

I dl R

∫

...(9)

→

∫

K dS R

...(10)

Eqns. (8), (9) and (10) enable us to obtain the magnetic vector potential for a given volume, line and surface current distribution respectively. →

Since the curl operation implies differentiation with respect to a length, the unit of A are weber per meter. →

The magnetic vector potential A permits us to maintain at least a sort of artificial similarity and →

symmetry between the magnetic and the electric fields. Since the integrals involved in calculation of A →

are generally much simpler than those involved in a direct calculation of B using Biot-Savart law. The use of vector potential thus does simplify the calculation of the magnetic field. The magnetic vector potential is extremely useful in studing radiation from antennas, from apertures, and radiation leakage from transmission lines, waveguides and microwave ovens. The magnetic vector potential is frequently used in regions where the current density is zero or non zero.

D

erivation of Magnetic Vector Potential →

→

According to Biot-Savart's law the magnetic induction dB due to a current element Id l at a point distant R from the element is given by →

dB = →

→

→

→

µ0 Id l × R 4π R3

...(1)

If R is the position vector, R = x aɵ x + y aɵ y + z aɵz , then from vector calculous, →

→

R 1 ∇ =− 3 R R

...(2)

Therefore eqn. (1), becomes →

µ0 I → → d l × − ∇ 4π

dB =

→

µ0 I 4π

→

→

dB = or From vector identity, we know that →

1 R

→ → 1 ∇ R × d l →

→

→

→

→

...(3)

∇ × (φ A) = ( ∇ φ) × A + φ ( ∇ × A) →

∴

→

→

→

( ∇ φ) × A = ∇ × (φ A) − φ ( ∇ × A)

...(4)

→ → → → dl 1 → 1 − ∇ × ( d l) ∇ ×dl = ∇ × R R R

...(5)

or →

E lectromagnetic F ield Theory

338

→ 1 → Substituting the value of ∇ × d l from eqn. (5) in eqn. (3), we get R →

dB =

→ → µ0 I → d l 1 → ∇× d l − ∇ × ( ) 4π R R

...(6)

→

As d l is not a function of coordinates ( x, y, z) of the point at which the magnetic induction is required →

and the operator ∇ represents the differentiation with respect to these coordinates, we must have →

→

∇ × dl =0

With this substitution eqn. (6) becomes, → µ0 I → d l dB = ∇× 4π R →

...(7)

→

The total magnetic induction B at the given point by the entire closed loop carrying current I is given by →

B=∫

→

µ I dB = 0 4π

∫

→ dl ∇× R

→

...(8)

→

→

Since the ∇ X operation is independent of the integration of

dl around the closed loop, we may represent R

eqn. (8) as µ I B=∇× 0 4π

→

→

∫

→ µ I dl = curl 0 4π R

∫

→ dl R

...(9)

The quantity within the square bracket is obviously a vector quantity. It is clear from eqn. (9) that a →

vector exists such that its curl gives the magnetic induction B produced by a current carrying closed loop. →

This vector is known as magnetic vector potential A. Therefore, →

→

B = curl A

where,

→

µ0 I 4π

→

µ0 4π

A =

→

∫

dl R

For volume current distribution, A=

→

J

∫ Rz

→

and for surface current, vector potential A is expressed as, → →

µ A = 0 4π

where K is the surface current density.

→

∫

K dS R

339

M agnetic

Potential and Field Due to a Long Straight Current Carrying Wire

Let us consider a long straight filamentary wire of length L and carrying a current of I amp, placed in free space (Fig. 43). Suppose P be a point distant ρ from the wire at which magnetic potential and field are to be determined and 0 be the foot of the perpendicular from P to wire (OP = ρ). Consider a small →

current element Idz of length dz at a distance z from O. Let R be the distance →

of the element from point P. The magnetic potential, dA at P due to this current element is given by →

dA = or

→

dA =

µ0 I Idz directed along the current 4π R

µ0 I 4π

dz 2

(ρ + z2 )

aɵz [∵ R = (ρ2 + z2 )1 /2 ]

The magnetic potential due to entire length of the wire is given by →

A = aɵz

or

→

A=

µ0 I 4π

+ L /2

∫ − L /2

dz 2

(ρ + z2 )

µ0 I + L /2 aɵz [log {z + (ρ2 + z2 )}] − L /2 4π [∵ ∫

or

→

A=

dx 2

2

(a + x )

1 /2 µ0 I L L2 − log aɵz log + + ρ2 4π 2 4

= log [ x + (a2 + x2 )]

L L2 + ρ2 − + 2 4

1 /2 1 /2 4ρ2 L L2 L 2 1 1 + + + + ρ 2 2 L 2 4 µ0 I µ0 I ɵ ɵ log az log a = = z 1 /2 4π 4π 2 1 /2 L L2 L −1 + 1 + 4ρ + ρ2 − + 2 2 4 2 L

or

2 1 /2 1 + 1 + 4ρ → L2 µ0 I A= aɵz log 1 /2 4π 2 4ρ −1 + 1 + 2 L

E lectromagnetic F ield Theory

340

For a very long wire L → ∞,

ρ2 L2

> 1, we can neglect 1. Thus, →

A=

or

→ 1 + ρ2 / L2 L2 µ I aɵz log 2 2 or A = 0 aɵz log 2 + 1 4π 4π ρ ρ / L

µ0 I

→

A =

µ0 I L log ρ 4π

2

aɵz =

µ0 I . 2 L log aɵz ρ 4π

µ0 I L log aɵ z ρ 2π

This is the required expression for magnetic vector potential due to long straight current carrying wire. →

→

We know that the magnetic vector potential A is related with magnetic field induction B as →

→

B = curl A

∴

→

µ I L → µ I L B = curl 0 log aɵz = ∇ × 0 log aɵz ρ 2 π ρ 2 π ρɵ B = ∂ / ∂ρ

→

0 or or or

φɵ ∂ /∂ φ µ0 I L log ρ 2π

aɵz ∂ / ∂z 0

→

µ. I ∂ [log ( L /ρ)] B = ρɵ (0 − 0) + φɵ (0 − 0) + aɵz 2 π ∂ρ

→

B=

→

µ0 I 1 L − aɵz 2 π ( L /ρ) ρ2

B=− →

µ0 I aɵ z 2 πρ

That is, the magnetic field induction B is directed perpendicular to the plane of paper downward and has µ I magnitude 0 . 2 πρ

341

B

iot-Savart Law From Magnetic Vector Potential

We know that, →

→

→

B =∇×A →

...(1) →

where A is the magnetic vector potential and B the magnetic field induction. →

The magnetic vector potential for a volume current distribution at a point whose radius vector is R , is given by →

µ A= 0 4π

∫V

→ J R dV

...(2)

→

where J is the current density. →

Substituting the value of A from eqn. (2) in eqn. (1), we have →

→

B =∇×

µ0 4π

∫V

→ J R dV or

→

B =

µ0 4π

∫V

→ J V × dV R

→

...(3)

From vector identity, we know that →

→

→

→

→

→

∇ × (φ A) = ( ∇ φ) × A + φ ( ∇ × A)

Using this identity eqn. (3) becomes (Here φ = →

B =

→ → 1 and A = J ) R

→ 1

µ0 4π

∫ V ∇ R

→

× J +

1 → → ( ∇ × J ) dV R

...(4)

The second term of R.H.S of eqn. (4) is zero because curl operation is taken relative to the field point as →

a function of ( x, y, z) whereas current density J is taken as a function of source point. The first term becomes →

∂ 1 → ∂ 1 ɵ → 1 → ɵ ∇ × J = ( J × R) (R × J ) = − R ∂R R ∂R R

...(5)

→

where Rɵ is the unit vector along radius vector R . Now for a elementary filament of length (dl), carrying a current of I amp. and having a constant area ' a' , the volume will be dV = adl. Now substituting the values of first term from eqn. (5) and value of dV = adl in eqn. (4), we get →

B =

µ0 4π

→ 1

∫ V ∇ R

→

Now current density J × area (a) = Total current I ∴

→

µ I B = 0 4π

→

∫V

d l × Rɵ R2

→ µ × J dV = 0 4π

∂ 1 → → ( J × R ) adl R

∫ V – ∂R

E lectromagnetic F ield Theory

342

→

→

µ0 I 4π

B =

or

(d l × R )

∫V

3

R

^ R ∵ R = R

dV

This is the required Biot-Savart law.

A mpere Circuital Law from Magnetic Vector Potential →

→

The relation between magnetic vector potential A and magnetic field induction B is given by, →

→

→

→

→

→

→

→

→ → →

B =∇×A →

Using vector identity,

→

→

B = µ0 H

→

∇×H=∇×

The curl of H will be

→

and

B 1 → → → = ∇ × ( ∇ × A) µ0 µ0

∇ × ( ∇ × A) = ∇ ( ∇ . A) − ∇2 A →

→ →

The magnetic vector potential A is such that its ∇ . A = 0. Therefore by using this condition above equation becomes

→

→

∇×H=

→

→

µ 1 (−∇2 A) = − 0 J µ0 µ0

→

(∵ ∇2 A = − µ0 J )

→

∇ × H = J

or

This is the differential form of Ampere's law. →

Example 27: Given the magnetic vector potential A = −

ρ2 aɵ z weber/m, calculate the total magnetic flux 4

crossing the surface. φ=

π , 1 ≤ ρ ≤ 2 m, 0 ≤ z ≤ 5 m. 2 [GBTU, B.Tech III Sem 2010] →

→

Solution: The magnetic vector potential A is related with magnetic induction B as, →

→

→

B=∇×A

→

ρ2 aɵz is along z-axis, therefore Aρ = A φ = 0 4 In cylindrical coordinates → → ∂Aρ ∂Az ɵ ∂ ∂ ∂ 1 ∂A ∇ × A = z − Aρ zɵ − (ρAφ ) ρɵ + ρφ + (ρAφ ) − ∂z ∂φ ∂ρ ρ ∂φ ∂ρ ∂z Since A = −

→

Here A = ∴

or

− ρ2 4

aɵz , hence Aρ = A φ = 0 and Az =

− ρ2 4

→

→

1 ρ

→

→

1 ∂ ρ2 ρ ∂ ρ2 ɵ − ρɵ − − φ ρ ∂φ 4 ρ ∂ρ 4

∇×A= ∇×A=

∂Az 1 ∂Az ∂Az ɵ 1 ∂Az ɵ − 0 ρɵ + 0 − ρɵ − ρφ ρ φ + 0 = ∂φ ∂φ ∂ ρ ρ ∂ρ ρ

343 →

→

→ → 2ρ ɵ ρ φ or ∇ × A = φɵ 4 2 → → → ρ B = ∇ × A = φɵ 2

∇×A=

or or

→

dS = dρ dz φɵ

Here

→

→

→

or

ψ =∫

∫ 2 dρ dz

or

ψ=

→

Therefore flux density, ψ = ∫ B . dS and B . dS =

ψ=

or

ρ ɵ φ.(dρ d z φɵ ) 2

ρ

1 2

5

∫ z =0

dz ∫

2 ρ =1

ρ dρ =

1 5 ρ2 [ z] . 2 0 2

2 1

1 4 1 1 3 15 = 3. 75 weber 5. − = × 5 × = 2 2 2 2 2 4

Example 28: When vector magnetic potential is given by →

A =

1 r3

ɵ (2. 0 cos θ rɵ + sin θ θ)

find the magnetic flux density. [GBTU. B.Tech III Sem 2010] →

→

Solution: Magnetic flux density B is related with vector magnetic potential A as →

→

→

B=∇×A

→

→

In spherical coordinates, ∇ × A is given by →

→

∇×A=

Here Ar =

2. 0 r

3

cos θ, A θ =

sin θ r3 →

∴

→

r 2 sin θ

rɵ ∂ / ∂r Ar

r θɵ

r sin θ φɵ ∂ / ∂θ ∂ /∂ φ rA θ r sin θ A φ

and A φ = 0

→

∇×A=

→

∇×A=

or

1

1 r 2 sin θ

rɵ ∂ / ∂r Ar

r θɵ

r sin θ φɵ ∂ / ∂θ ∂ / ∂φ r Aθ 0

∂ rA θ ∂ Ar ∂ Ar ∂ − 0 r θɵ + rA θ − (0 − rɵ + r sin θ φɵ ∂φ ∂θ r sin θ ∂r ∂φ 1

2

Substituting the values of A r , Aθ and Aφ in the above equation, we get, ∴

or

→

→

→

B=∇×A=

∂ r sin θ ∂ 2 cos θ ∂ 2 cos θ ɵ ∂r sin θ rθ + 3 − r sin θ 3 rɵ + − 3 ∂ ∂ r θ r3 ∂φ ∂φ r r sin θ r r 1

2

→

1 ∂ sin θ 1 ∂ 2 cos θ ɵ B= r 3 − φ 3 r ∂r r r ∂θ r

φɵ

E lectromagnetic F ield Theory

344

1 −2 sin θ 1 −2 sin θ ɵ = − φ = 0 or 3 r r 3 r r

→

B=0

Example 29: Using vector potential, find the magnetic flux density at a point due to long straight filamentary conductor carrying current in aɵ z direction. [U'khand B.Tech IV Sem 2009]

So lu tion:

Z

Let us consider a long straight fil amentary

L

conductor of length 2 L and carries a current I in aɵz direction as

P

I

shown in Fig. 44. Suppose Idl is the current element at the

az

R

→

origin O. The vector magnetic potential A is given by →

A=

µ0 4π

∫

→ µ Idl or A = 0 R 4π

L

∫–L

L

∫– L

ρ

x

y

→

µ0 4π

Y φ

vector potential A shall exist, therefore Az =

z

Idl 0

Idz aɵz R

Since the current is along aɵz direction only, Az component of

∴

–ax a aφ y

–L X

Idz µ0 I = 2π R

–Z

dz

L

∫0

( x2 + y2 + z2 )

Fig. 44

[∵ R = ( x2 + y2 + z2 )] or

Az =

µ0 I 2π

L

dz

(∵ ρ2 = x2 + y2 )

∫ 0 (ρ2 + z2 )1 /2

Integrating by substituting z = ρ tan θ, ∴ dz = ρ sec2 θ dθ and ρ2 + z2 = (ρ2 + ρ2 tan2 θ) = ρ2 sec2 θ ρ sec2 θ dθ

∴

Az =

µ0 I 2π

or

Az =

µ0 I µ I log e (sec θ + tan θ) = 0 log e [ (1 + tan2 θ) + tan θ] 2π 2π

∫

ρ sec θ

=

µ0 I 2π

∫ sec θ dθ

L

or

or or or We know that,

z + (ρ2 + z2 ) µ I or Az = µ0 I A z = 0 log e 2π 2π ρ 0 µ0 I (log e [ L + (ρ2 + L2 )] − log e ρ] 2π µ I Az = 0 (log e 2 L − log e ρ) 2π

log e

L+

L2 + ρ2 − log e 1 ρ

Az =

Az = →

µ0 I 2 L log e ρ 2π

→

→

B=∇×A

(For L >> ρ, L2 + ρ2 ≈ L2 ) ...(1)

345

In cylindrical coordinates →

aɵ ρ ρ aɵφ aɵz 1 ∂ / ∂ρ ∂ / ∂φ ∂ / ∂z ρ Aρ ρAφ Az

→

∇×A=

In this case Aρ = Aφ = 0

ρ aɵφ 1 ∂ ( ∇ × A) = ∂ / ∂ρ ρ ∂φ 0 0 →

∴

→

aɵρ

→

→

∇×A=

or

→

→

∂Az aɵ ∂ρ φ

→

→

→

B=∇×A=−

→

B=−

or

→

B=

or

Az

∂ Az =0 ∂φ

∇×A=−

or

∂ / ∂z

1 ∂ Az 1 ∂Az ∂Az ∂Az − 0 aɵρ + 0 − aɵρ − aɵ ρ aɵφ + {0 − 0} aɵz = ρ ∂φ ∂ρ ∂ρ φ ρ ∂φ

Since Az is independent of φ, therefore, Hence

aɵz

∂ µ0 I ∂ 2 L log e Az aɵφ = − aɵφ ρ ∂ρ 2 π ∂ρ

[∵ from eqn. (1)]

µ0 I ρ 2 L . aɵ − 2 π 2 L ρ2 φ

µ0 I aɵ 2 πρ φ →

→

Example 30: The magnetic vector potential A = 10 sin θ θɵ is in spherical coordinate system. Find B at (2, π / 2, 0 ). →

→

Solution: Magnetic vector potential A and magnetic flux density B is related as, →

→

→

B =∇×A In spherical coordinates, →

→

∇×A=

rɵ

1 r 2 sin θ

∂ / ∂r Ar

r θɵ

r sin θ φɵ

∂ / ∂θ ∂/∂φ rA θ r sin θ A φ

→

Since A has only Aθ component, therefore A r = A φ = 0 ∴

or

→

→

∇×A=

→

1 r 2 sin θ →

→

B =∇×A=

rɵ ∂ / ∂r 0

r θɵ r sin θ φɵ ∂ / ∂θ ∂ / ∂φ rA θ 0

1 ∂ ∂ (rAθ ) rɵ + {0 − 0} rθɵ + r sin θφɵ (rAθ ) − 0 0 − r ∂φ ∂ r 2 sin θ

E lectromagnetic F ield Theory

346 →

→

→

∂ ( Aθ ) 1 1 ∂ rɵ + (rAθ ) φɵ r sin θ ∂ φ r ∂r

1 1 ∂ ∂ (10 sin θ) rɵ + (r .10 sin θ) φɵ r sin θ ∂φ r ∂r

B =∇×A=−

or →

Here A = 10 sin θ θɵ or A θ = 10 sin θ →

→

→

→

→

→

B =∇×A=−

∴

B =∇×A= →

→

10 sin θ ɵ φ r

→

B at (2, π / 2, 0) = [ ∇ × A] (2, π / 2, 0) =

10 × sin (π /2) ɵ φ 2

or

→

B at (2, π / 2, 0 ) = 5 φɵ

→

Example 31: Magnetic vector potential A in cylindrical coordinates is expressed as, →

A = 5 e −ρ cos φ ρɵ − 5 cos φ zɵ

→

Find magnetic flux density B at (2, π /2, 0 ) →

→

Solution: Magnetic flux density B and magnetic vector potential A is related as →

→

→

B=∇×A

...(1)

→ → 1 ∂Az ∂Aφ In cylindrical coordinates, ∇ × A = − ρɵ ∂z ρ ∂φ

∂Aρ ∂Az ɵ 1 ∂(ρAφ ) ∂Aρ + − − zɵ φ+ ρ ∂ρ ∂ρ ∂φ ∂z

...(2)

In the given problem, Aρ = 5 e −ρ cos φ, Aφ = 0 and Az = − 5 cos φ Therefore, eqn. (2) reduces to →

→ ∂ 1 ∂ ∂ ∇ ×A = (−5 cos φ) − 0 ρɵ + (5 e −ρ cos φ) − (−5 cos φ) ρ ∂φ ∂z ∂ρ

∂ 1 φɵ + 0 − (5 e −ρ cos φ) zɵ ∂φ ρ

→

or

→

5 sin φ 5 e −ρ ρɵ + (0 − 0) φɵ − (− sin φ) zɵ ρ ρ

→

→

5 sin φ 5 sin φ − ρ ρɵ + e zɵ ρ ρ

B=∇×A=

or or

→

∇×A=

or

→

B at (2, π / 2, 0) =

→

5 sin (π /2) 5 sin (π /2) −2 ρɵ + e zɵ 2 2

B at (2, π / 2, 0 ) = 2.5 ρɵ + 2. 5 e −2 zɵ

347

Example 32: Using magnetic vector potential, derive an expression for the magnetic field at P due to two long straight parallel wire carrying equal and oppositely directed currents as shown in Fig. 45. Z

Solution: Suppose two long straight conductors ' a ' and ' b ' are

(b)

(a)

I

I

separated by a distance ' d ' as shown in Fig. 45. Let P be the point at which the expression for magnetic field is determined.

y+d /2 y – d /2

We know that the magnetic vector potential at a point distance ρ 0

from the origin of a long straight wire of length 2 L is given by Az =

µ0 I 2π

2 L log e ρ

Y r2

r1

x

d

where I is the current flowing in the wire.

P y

If Az a and Az b be the z-component of magnetic vector potentials X at P, then

Fig. 45

Az = Az a + Az b Here

∴

Az a =

Az = =

or

Az =

2 L 2 L µ0 I µ I log e and Az b = − 0 log e r 2π π 2 r2 1 2 L 2 L µ0 I log e − log e 2 π r1 r2 µ0 I [log e 2 L − log e r1 − log e 2 L + log e r2 ] 2π µ0 I [log e r2 − log e r1] 2π

From Fig. 45, r1 = ( y − d /2)2 + x2

But

or ∴

or

From eqn. (2)

and r2 = ( y + (d /2)2 + x2

aɵ x → → B = ∇ × A = ∂ /∂ x 0 →

→

→

...(1)

B = ( ∇ × A) = aɵ x

aɵ y aɵz ∂ / ∂y ∂ / ∂z 0 Az

...(2)

(∵ A has only z-component)

∂Az ∂A − aɵ y z ∂x ∂y

....(3)

→

∂ µ0 I ∂ µ0 I (log e r2 − log e r1) − aɵ y 2 π (log e r2 − log e r1) ∂y 2 π ∂ x

→

µ0 I 1 ∂r2 µ I 1 ∂r2 1 ∂ r1 1 ∂r1 − aɵ y 0 − − 2 π r2 ∂y 2 π r2 ∂x r1 ∂ x r1 ∂y

B = aɵ x

B = aɵ x

2 d r1 = y − + x2 2

1 /2

∴

∂ r1 ∂y

=

1 2

...(4) (1 / 2 ) −1

2 d 2 y − + x 2

2 ( y − d /2) .1

E lectromagnetic F ield Theory

348

or

Similarly

( y − d /2) ( y − d /2) ∂r1 = = 2 2 ∂y r1 [( y − d /2) + x ∂r1 ∂x

and

1 [( y − d /2)2 + x2 ] (1 /2) −1 2 x 2

r2 = [( y + d /2)2 + x2 ]1 /2

Again from eqn. (2)

or

=

∂r2 ∂y ∂ r2 ∂x

Substituting the values of

=

=

∂r1 x x = = 2 2 ∂x [( y − d / 2) + x ] r1

1 [( y + d /2)2 + x2 ]1 /2 −1 2 x 2

...(6)

∂r2 1 = [( y + d /2)2 + x2 ](1 /2)−1 2( y + d /2).1 ∂y 2 ...(7)

r2 or

∂r2 x = ∂x r2

...(8)

∂ r1 ∂ r1 ∂ r2 ∂r2 and from eqns. (5), (6), (7) and (8) in eqn. (4), we get , , ∂x ∂y, ∂x ∂y B = aɵ x

or

or

( y + d /2)

→

or

∴

....(5)

→

B=

→

H =

µ0 I 1 ( y + d /2) 1 ( y − d /2) µ I 1 x 1 x − aɵ y 0 − − 2 π r2 r2 r1 r1 2 π r2 r2 r1 r1

µ0 I ( y + d /2) ( y − d /2)2 µ I − aɵ x − 0 2 2 2π 2π r1 r2

x x 2 − 2 aɵ y r1 r2

→

x B I ( y + d /2 ) ( y − d /2 ) I x ɵx − = − aɵ − a 2 2 y µ0 2 π 2π r 2 r22 r r1 1 2 ❍❍

349

Q uestion Bank 1.

Define Biot-Savart law and what are its application?

[GBTU, B.Tech IV Sem 2010]

2.

State and explain Biot-Savart law.

[GBTU, B.Tech III Sem 2010] →

3.

Derive an expression of flux density B for circular current.

4.

Determine the magnetic flux density B at a distance ' d ' meter from an infinite straight wire carrying

[U'khand, B.Tech IV Sem 2009, 2011]

→

current I. Also find out when the length of the wire is semi-infinite.

[GBTU, B.Tech IV Sem 2011]

5.

Explain Biot-Savart's law and Ampere's circuital law.

6.

Define Biot-Savart law and Ampere's law. A long straight conductor cross section with radius ' a' has → → I I ɵ magnetic field strength, H = r 2 aɵφ within the conductor (r < a) and H = a for (r > a). Find 2πr φ 2 πa

[U'khand, B.Tech IV Sem 2011]

→

J in both regions.

[UPTU, B.Tech IV Sem 2007]

7.

State and derive Ampere circuital law.

[GBTU, B.Tech IV Sem 2011]

8.

What is Ampere's circuital law? State and obtain the necessary relation of force due to current I in a conductor. [UPTU, B.Tech IV Sem 2005]

9.

State and prove Ampere's circuital law and derive Maxwell's equation from it. [GBTU, B.Tech III Sem 2011]

10. State and explain Ampere's circuital law of force and give its application. [UPTU, B.Tech IV Sem 2002] 11. Describe the Ampere's circuital law of magnetostatic field.

[GBTU, B.Tech IV Sem 2010]

12. State and explain the Ampere's circuital law.

[UPTU, B.Tech IV Sem 2000]

13. State and explain Ampere's force law.

[UPTU, B.Tech IV Sem (C.O.) 2006]

14. Apply Ampere's circuital law to determine the magnetic field strength at any point (i) within the core of a toroidal coil carrying current I and (ii) In the region between two very large closely spaced parallel planes carrying equal and oppositely directed currents. [UPTU, B.Tech IV Sem (C.O.) 2006] 15. Using vector potential find the magnetic flux density at a point due to long straight filamentary conductor carrying current in aɵz direction. [U'khand B.Tech IV Sem 2009] 16. Discuss and derive magnetic scalar and vector potentials.

[GBTU B.Tech III Sem 2011]

17. Find the vector potential due to long straight wire of length L and carrying a current I. Also determine the self inductance per unit length of the long solenoid. [U'khand, B.Tech IV Sem 2011] 18. State and explain the Maxwell's equation in differential and integral form. [UPTU, B.Tech IV Sem 2006, 2007, GBTU, B.Tech. IV Sem 2010, 2011, III Sem 2010, 2011]

E lectromagnetic F ield Theory

350

nsolved Numerical Problems 1.

An infinitely long straight filament carrying a current of 3 amp is placed along z-axis. Calculate the magnetic field intensity at P (1, 2, 1).

2.

A 8m long straight filamentary conductor carrying a current of 4 amp is placed along positive z-axis with one end at origin. Find out the magnetic field intensity at point (0, 4, 4). 2 2 µI πa

3.

A square of edge ' a' carries a current I. Show that the value of B at the centre is given by B =

4.

What is the flux density at the centre of a square loop of 10 turns carrying a current of 10 Amp. The loop is in air and is 2m on a side.

5.

Find magnetic field intensity at the centre of an equilateral triangular loop of side 4m carrying current of 5 Amp.

6.

A rectangular loop carrying 10 amp of current is placed on z = 0 plane. Find the field intensity at (4, 2, 0). →

7.

In the region with 0 < r < 0.5 m in cylindrical coordinate the current density is J = 4 .5 e −2 r aɵz A/m →

→

and J = 0 elsewhere. Use Ampere's law to find H everywhere. 8.

A z-directed current distribution is given by →

J z = (r 2 + 4 r ) aɵz for r < a

→

find B at any point using Ampere's circuital law. 9.

→

In cylindrical coordinate, J = 10 e −100 r aɵφ A /m for the region 0.01 ≤ r ≤ 0.02 m and 0 ≤ z ≤ 1m. Find total current crossing the intersection of this region with the plane φ = constant.

10. Find the vector magnetic field intensity in cartesian coordinates at a point P (1.5, 2, 3) caused by a current filament of 24 amp. in aɵz direction on z-axis extending from z = 6 to ∞. 11. The magnetic field intensity in the z = 0 plane is given by →

H = − y ( x2 + y2 ) aɵ x + x ( x2 + y2 ) aɵ y (A /m)

Find the total current passing through z = 0 plane in the aɵz direction inside the rectangular bounded by −1 < x < 1 and −2 < y2 . →

12. A radial field H =

3.2 × 106 r

cos φ aɵ r A /m exists in free space. Calculate magnetic flux crossing the

surface defined by − π /4 ≤ φ ≤ π /4, 0 ≤ z ≤ 2 m. →

13. A magnetic field known to be directed in a certesian or rectangular coordinate system so that H →

→

exists in the x-direction find whether ∇ × H exists in which (a) H x = constant, (b) H x = f ( y, z)

351

→

→

14. Find J if (a) H = 3 aɵ x + 7 y aɵ y + 2 x aɵz A /m →

(b) H = 6ρ ρɵ + 2ρ φɵ + 5 zɵ A /m →

15. Given flux density B = 3 x aɵ x + 5 aɵ y + z aɵz tesla. Find the magnetic flux φ m through the surface of a volume enclosed by six planes at x = 1, x = 6, y = 0, y = 4, z = 1 and z = 7m. →

16. In cylindrical coordinates B = (2 /ρ) φɵ . Determine the magnetic flux crossing the plane surface defined by 0.5 ≤ ρ ≤ 2.5 m and 0 ≤ z ≤ 2.0m. →

→

17. If the vector potential A is given as A = 5( x2 + y2 + z2 ) aɵ x . Find out flux density. 18. Within cylindrical system, the magnetic vector potential due to current density J0 is →

A = − (µ0 J0 /4) ( x2 + y2 ) aɵz . Obtain the magnetic field intensity, within the conductor. →

→

19. Find J if (a) H = 3 aɵ x + 7 y aɵ y + 2 x aɵz A /m →

(b) H = 6ρ ρɵ + 2ρ φɵ + 5 zɵ A /m 20. Magnetic field intensity in free space is given by →

H = 20 ( x aɵ x + y aɵ y ) / ( x2 + y2 ) A /m

→ →

→

(a) Show that ∇ . B = 0, (b) Find the current density J , and (c) find the scalar vector potential Vm ( x, y, z) if Vm = 0 at P (1, 1, 1) →

21. For a flux density, B =

0. 2 (sin2 φ) zɵ ρ

Calculate the magnetic flux φ m across the z = 0 plane in cylindrical coordinates for r ≤ 0.05 m. → − y aɵz + x aɵ y 22. Show that the vector potential A = is curl free at all points except origin. x2 + y2 →

23. If magnetic vector potential A = 10 ρ1. 5 zɵ wb / m in free space find →

→

→

→

(a) H (b) J and (c) show that for r = 1 ∫ H. d l = I for the circular path. 24. For a z-directed current of I amp flowing in an infinitely long conductor, verify that the magnetic → → µ I vector potential can be written as A = − k 0 log e ( x2 + y2 ). 4π

E lectromagnetic F ield Theory

352

nswers to Unsolved Numerical Problems 1.

→

H = − 0.19 aɵ x + 0.95 aɵ y

2.

H x = − 0.1126 A /m, H y = 0

4. 5.656 × 10 −5 wb / m 2

5.

6. 1.779 aɵz AT /m

. 0.297 7. 1125 (1 − e −2 r − 2 re −2 r ) aɵ φ A / m; aɵφ A /m r r

8.

r3 4 r2 a3 4 a2 aɵφ ; µ + µ + aɵ 3 φ 4 4 3

9.

1.79 aɵz AT /m

I = 23 mA

10. H = − 0.147 aɵ x + 0.1062 aɵ y A /m

11. 53.3 amp

12. 11.37 weber

13.

14.

2 −2 aɵ y amp /m 2 ; zɵ amp /m 2 ρ

→

→

∇ × H = 0,

∂ Hx ∂H x aɵ x − aɵz ∂z ∂y

15. 480 weber

→

16. 6.437 weber

17.

18. (µ 0 J0 r ) /2 aɵφ

19. −2 aɵ y A /m 2 ; 2 zɵ /ρ A /m

20. J = 0; Vm = 10 log e [2 /( x2 + y2 )]

21. 0.0314 weber

22.

→

→

At all points, ∇ × A = 0; at origin it is

23.

B = 10 (z aɵ y − yaɵz )

→

H =−

undetermind I=−

→ 22.5 15 ρ φɵ A / m; A = zɵ A /m 2 ; µ0 µ0 ρ

30 1 µ0

A

mmm

Theory

Numerical

Solved Examples

Questionnaire

353

Unit-3 S ection

B M agnetic F orces, M aterials and D evices

I

ntroduction

A magnetic field can not arises from stationary charges and can not exert any force on a stationary charge. It is well established fact that the electric charge in motion or electric current produce a magnetic field. This is confirmed by the fact that around a current carrying conductor or wire, each end of a magnetic compass needle experience a force dependent upon the magnitude of current. In electromagnetic wave theory, magnetic fields due to electric currents are of main concern and the effect of permanent magnets are of little importance. The magnetic field produce by moving charges interacts with other moving charges and produce magnetic forces. The study of magnetic force exerted by magnetic field on current elements or loops or on charged particles is important to solve problems in electronic devices. Like the electric field, the actual magnetic field inside matter fluctuates widely from point to point and instant to instant, therefore we consider the average over regions large enough to contain many atoms. Hence, the magnetic field in matter means the macroscopic magnetic field. The prominent role of magnetic material in modern technology gave the birth to industrial electronics, entertainment electronics and computer industry. The magnetism of materials is mainly a consequence of interaction of uncompensated magnetic moments of their constituent atoms or molecules. From the macroscopic point of view, in atoms of all materials, electrons revolving round the nucleus in their orbits with very high velocities, can be assumed as tiny, atomic scale current loops. In addition, these electrons also revolve about their own axis. This motion of electric charge about their own axis, can again be taken

E lectromagnetic F ield Theory

354

as equivalent to tiny closed circuits. To analyse the effects of magnetic materials in an external magnetic field, it is essential to analyse the behaviour of a single current loop. Since these loops are quite small it is reasonable to assume that they are in uniform magnetic field. The significant effect is the torque by which a uniform magnetic field acts on a current loop. The torque tend to turn the loop due to which the magnetic field of the loop increases the external magnetic field. In this chapter we consider mainly the forces and torques acting on a current carrying conductors, magnetization in materials, various boundary relations in magnetic field and the concept of self inductance and mutual inductance.

F

orces Due to Magnetic Field

There are mainly three ways in which force due to magnetic fields can be experienced.

(i) Force on a Moving Charge and a Lorentz Force →

In an electric field E charged particle Q, either stationary or in motion, experiences a force given by →

→

F = QE

The magnitude of the force is directly proportional to the product of Q and E, and its direction is the direction of the field. →

→

It is experimentally observed that a charge Q moving with velocity v in a magnetic field of flux density B

→

experiences a force whose magnitude is equal to the product of the magnitudes of charge Q, its velocity v →

→

→

and the flux density B, and the sine of the angle between v and B (fig. 1). →

Z→ → → F = Q (v × B)

The direction of the force is perpendicular to the plane containing both v →

and B. Mathematically the force may be expressed as, →

→

→

F = Q ( v × B)

→

n

F = QvB sin θ

or

Y

B

→

If θ = 90 ° or sin θ = 1, that is, when v and B are at right angles to each other, the force is maximum and is given as, Fmax = QvB

+Q θ →

v

→

→

On the other hand, if θ = 0 or sin θ = 0, that is, when v and B are parallel to each other, then F = 0. In means that when a charge is moving with a velocity →

X

→

v parallel to the magnetic field B, no force is exerted by the magnetic field on

Fig. 1

the charge Q. Again if v = 0, F = 0, that is, when the charge is at rest or stationary no force is exerted by the magnetic field.

355

Lorentz Force A charge whether stationary or in motion produces an electric field around it. If it is in motion, then in addition to electric field, it also produces a magnetic field. →

If a charged particle carrying a charge Q is moving with velocity v in a space where both electric and →

magnetic fields exists, then the force on a moving charged particle due to combined electric field E and →

magnetic field B is given by →

→

→

→

F = Q [ E + ( v + B)]

This is called Lorentz force. Thus, a Lorentz force is a combined electric and magnetic forces acting simultaneously on a moving charged particle.

Force on a Differential Current Element →

→

The force on a charge moving with velocity v in a steady magnetic field B may be expressed as the →

differential force d F exerted on a differential element of charge dQ as →

→

→

d F = dQ ( v × B)

...(1)

→

The force on a current element Id l of a current carrying wire or conductor due to the magnetic field of →

→

flux density B can be determined by using the equation of convection current density J in terms of the →

velocity v of volume charge density ρ v as, →

→

J = ρv v

...(2)

The differential element of charge, dQ may also be expressed in terms of differential volume element dV and volume charge density ρ v as, ...(3) dQ = ρ v dV Substituting this value of dQ in eqn. (1), we get →

→

→

d F = ρ v dV ( v × B) →

→

→

d F = (ρ v v × B) dV

or →

Substituting the value of ρV v from eqn. (2), we get →

→

→

→

→

d F = ( J × B) dV →

or differential force d F becomes →

d F = ( J dV × B)

...(4)

→

If K is the surface current density and dS is the differential area, then →

→

J dV = K dS

Therefore eqn. (4) is modified as, →

→

→

d F = ( K × B) dS

...(5)

E lectromagnetic F ield Theory

356 →

Now, current density J and current I is related as I J= dS and differential volume, dV = dS × dl → I ∴ J dV = .(dS × dl) = Idl dS →

→

→

Therefore, the differential force d F on current element d l by the magnetic field of flux density B is →

→

→

d F = Id l × B

...(6)

Therefore, the total force experienced by the current carrying conductor of length L placed in a magnetic field is obtained by the integration of eqn. (6) as, →

→

→

→

→

F = ∫ (Id l × B) = I ∫ d l × B

→

→

→

F = I ( L × B)

or

...(7) →

Similar equations can be obtained for surface current element K dS by integrating eqn. (5) and for volume current element Jdv by integrating eqn. (4) as →

F =

→

F =

and

→

∫S ( K dS × ∫V

→

→

...(8)

B) →

J dV × B

...(9)

Force on a Current Carrying Conductor The force on a current carrying conductor can also be determined in the following manner: →

When a conductor carrying a current, is placed in a steady magnetic field, magnetic forces are exerted on the free electrons within the conductor. But when they are moving in a conductor under the influence of the force, their (that is, the electrons) motion gets halted by the collisions with the lattice structure (or region of immobile positive ions) of the conducting material, as a result of which the force get transferred from the charges to the conductor which carries them and hence the conductor as a

i

× × × × ×P × × × × × ×

× ×

B × × × F×

× ×

× ×

vd

× ×

× ×

L

× ×

× ×

× ×

× × Q×

i

× × ×

Fig. 2

whole experience a force when placed in a magnetic field. →

Let us suppose a straight current carrying conductor of length L be placed in a uniform magnetic field B

acts perpendicular to the length of the conductor and directed into the plane of the paper. (fig. 2). Let i be the current flowing through the conductor from P to Q. Due to this free electrons move with drift →

velocity v d in a direction opposite to the direction of the current. The magnetic force acting on each moving electron in a magnetic field is given by →

→

→

F′ = e( v d × B)

→

→

→

→

[∵ F = Q ( v × B)]

Since drift velocity v d and B are perpendicular to each other, therefore, the magnitude of F is given by F ′ = e vd B

357

If n be the number of free electrons per unit volume of the conductor and A its area of cross-section, then total number of electrons in a conductor of length L is, N = nLA The total force or accumulated force on all free electrons of the conductor or the total force transmitted to the conductor of length L is given by, F = NF ′ = nLAe vd B = (n Ae vd ) BL But n Ae vd = i, that is the current in the conductor ∴

F = iBL →

If instead of perpendicular, the conductor makes an angle θ with the magnetic field of flux density B, then the force on the conductor is, F = i BL sin θ In vector form, the force is, →

→

→

F = i( L × B)

→

Where L is a displacement vector pointing along the conductor in the direction of current. →

→

The magnetic force on a conductor in a magnetic field is normal to both L and B. The direction of the force can be obtained by Fleming's left-hand rule. According to Fleming left-hand rule "If the thumb, fore-finger and middle-finger of the left hand are stretched mutually perpendicular to each other in such a way that the fore-finger points in the direction of the magnetic field and the middle-finger in the direction of the current, then the thumb will automatically point in the direction of the magnetic force on the conductor.

A mpere's Law of (Magnetic) Force Ampere (1820-1825) carried out elaborate experiments to establish a law of force between two steady line currents. He showed that the force between parallel wires carrying steady currents is purely magnetic, although, apparently, no magnetic are involved. According to the Ampere’s law of force the two steady line currents attract or repel each other with a (magnetic) force which is directly proportional to the product of magnitude of their currents and inversely proportional to the square of the distance between them, this force also depends upon the lengths and orientations of the two wires, and also upon the nature of the medium. If the two parallel wires carrying steady currents I1 and I2 , are separated by a distance r, then the force F acting between them is given by F=

µ I1 I2 Newton 4 π r2

...(1)

E lectromagnetic F ield Theory

358

The force is attractive, if the currents are flowing in same direction and repulsive if they flow in opposite directions (in the two wires).

I2

In more general case of two current loops ‘1’ and ‘2’ carrying →

→

→

I2dl2 × r21

currents I1 and I2 respectively (Fig. 3). The magnetic force dF12 on →

→

→

I2 dl2

produced by the entire current loop ‘2’, when both the current loops are in free space, is given by, →

→

d F12 = I1d l1 × B2

1

dF12

→

the current element I1 d l1 due to magnetic induction B2 at d l1

→

I1

→

→

r21

→

I1dl1

2

...(2) →

According to Biot-Savart law, the magnetic induction B2 at d l1 due to loop ‘2’ is

→

µ I B2 = 0 2 4π →

→

∫

2

Fig. 3

→

d l2 × r21

...(3)

3 r21

Substituting the value of B2 from eqn. (3) in eqn. (2), we get →

∴

d F12 →

or

d F12

µ I = I1 d l1 × 0 2 4π →

→ → d l2 × r21 3 r21

∫2

→

→

d l2 × r21 µ I I → = 0 1 2 d l1 × ∫ 3 2 4π r21

...(4)

→

The force F12 on the entire current loop ‘1’ due to current loop ‘2’, is obtained by integrating eqn. (4) over the entire current loop. That is, →

F12

µ I I = 0 1 2 4π

→

∫1 ∫2

→

→

d l1 × (d l2 × r21)

...(5)

3 r21

→

Similarly, the force F21 on current loop ‘2’ due to current loop ‘1’ is [obtained by mutually interchanging the subscripts 1 and 2 in eqn. (5)] given by →

F21

µ I I = 0 1 2 4π

→

∫2 ∫1

→

→

d l2 × (d l1 × r12 )

...(6)

3 r12

Magnetic Interaction of Two Current Elements → I2 I dl × → r 2 2

Let us consider two current carrying loops (1) and (2) of →

→

arbitrary shape as shown in fig. 4. If d l1 and d l2 are the current

→

elements in the two loops carrying steady currents I1 and I2 respectively, then the magnetic force on the current element →

→

→

1

dF12

I2dl2 r

I1d l1 due to the magnetic induction B2 produced by loop (2) at

I1

→

→

I1 dl1

2

Fig. 4

359 →

the location of current element I1d l1, when both are in free space, is given by →

→

→

d F12 = I1 d l1 × B2

...(1) →

According to Biot-Savart law, the magnetic induction B2 produced by the current loop (2) is →

µ I B2 = 0 2 4π

→

→

d l2 × r

∫2

...(2)

r3

→

Substituting this value of B2 from eqn. (2) in eqn. (1), we obtain, →

d F12 = →

d F12 =

or

→

µ0 I2 I1 4π

d l1 × ∫

→

...(3)

r3

→

∫2

→

d l2 × r

2

µ0 I2 I1 4π →

→

→

→

→

d l1 × (d l2 × r )

...(4)

r3

→ → →

→ → →

Using the vector identify, A × ( B × C) = ( A . C) B − ( A .B) C, we get →

→

→

→ →

→

→

→

→

d l1 × (dl2 × r ) = (d l1. r ) d l2 − (d l1 . dl2 ) r Therefore, eqn. (4) becomes →

dF12

µ I I = 0 2 1 4π

→

∫2

→

→

→

→

→

(d l1 . r ) d l2 − (d l1 . d l2 ) r

...(5)

r3 →

Similarly the force on current element I2 d l2 due to current loop (1) is given by (Obtained by →

→

interchanging the subscripts 1 and 2, and by replacing r by − r ), →

dF21

µ I I =− 0 1 2 4π

→

∫1

→

→

→

→

→

(d l2 . r ) d l1 − (d l2 . d l1) r

...(6)

r3

Comparison of eqns. (5) and (6) gives →

→

dF12 ≠ dF21

...(7)

→

→

→

That is, dF12 and dF21 are unequal. This is against Newton's third law from which we expect that dF12 →

must be equal to − dF21. Newton’s law is, however, not violated when we consider the force on entire loop due to the other entire loop. That is, →

→

F12 = ∫ dF12

...(8)

1

→

Substituting the value of dF12 from eqn. (5) in eqn. (8), we get →

F12

µ I I = 0 1 2 4π

→

∫1 ∫2

→

→

→

→

r3

Now considering the first term of right side of eqn. (9), we get → →

∫1 ∫2

→

(d l1 . r ) d l2 r3

→

= ∫ d l2 2

→ →

∫1

d l1 . r r3

→

(d l1 . r ) d l2 − (d l1 . d l2 ) r

→

= ∫ d l2 2

∫1

→

d l1.

→

r

r3

...(9)

E lectromagnetic F ield Theory

360

→ →

or

∫1 ∫2

→

→

(d l1 . r ) d l2

= ∫ d l2

r3

2

→ ∵ r = − grad 1 ...(10) r3 r

→ 1 ∫1 d l1 . − grad r

From Stoke’s theorem

∫ we have

→ →

→ →

A . dl = ∫ curl A . dS

1

S

→

1 → . dS r

∫1 grad r . d l1 = ∫S curl grad

But we know that curl (grad S) = 0 ∴

1 → − ∫ grad . d l1 = ∫ 1 1 r

→ → dl . r = 0 1 r3

...(11)

Substituting the result of eqn. (11) in eqn. (10), we obtain,

∫1 ∫2

→ → → (d l1. r ) d l2 =0 r3

Therefore, eqn. (9) reduces to →

µ I I =− 0 1 2 4π

F12

→

∫1 ∫2

→

→

(d l1 . d l2 ) r

...(12)

r3

Similarly, the force acting on current loop (2) due to current loop (1) will be →

→

F21 = ∫ dF21 2

→

Substituting the value of dF21 from eqn. (6) in the above equation, we have →

F21 = − →

As before,

∴

∫1 ∫2 →

F21

µ0 I1 I2 4π →

µ I I = 0 1 2 4π

Comparing eqns. (12) and (13) we obtain →

→

F12 = − F21 This is Newton’s third law.

∫2 ∫1

→

→

→

→

→

r3

→

(d l2 . r ) d l1 r3

→

(d l2 . r ) d l1 − (d l2 . d l1) r

=0 →

∫1 ∫2

→

→

(d l2 . d l1) r r3

...(13)

361

M agnetic Force Between Two Long Parallel Wires Let (1) and (2) be two parallel wires separated by a distance d and

2

1

carrying currents I1 and I2 amperes respectively, as shown in fig. 5. →

I1

The magnetic force on current element I2 d l2 of wire or conductor

I2 →

→

(2) due to a magnetic induction B12 produced by current I1 in wire

→

dl1

or conductor (1) at wire (2) is given by →

∴

→

→

dF2 = I2 d l2 × B12

...(1)

d F2 = I2 d l2 B12

...(2)

dl2

→

B21

→

→

B12

dF2

→

dF1 d

→

But, we know that the magnetic induction B12 at wire (2) produced

Fig. 5

by the current I1 in wire (1) is given by B12 =

µ0 I1 2 πd

...(3)

Substituting this value of B12 from eqn. (3) in eqn. (2), we get µ I d F2 = I2 d l2 0 1 or 2 π d

dF2 =

µ0 I1 I2 d l2 2πd

Hence, magnetic force per unit length on second wire due to current in first is dF2 µ0 I1 I2 = N/ m d l2 2πd Similarly, the magnetic force per unit length on first wire due to current in second is given by d F µ0 I1 I2 = N/ m dl 2πd Thus, the force on either wire or conductor is proportional to the products of currents. The directions of forces on two wires or conductors are obtained by applying Fleming’s left-hand rule or right-hand palm rule. The force is attractive if the currents in wires are in the same direction and repulsive if currents are in opposite directions. Example 1: A point charge q = 10 nC has velocity 5 × 10 6 m /s in the directionaɵ v = 0 . 04 aɵ x − 0 . 05 aɵ y + 0 . 2 aɵ z . Calculate the force exerted on charge when: →

(i) A magnetic induction B = aɵ x + 2 aɵ y + 3 aɵ y mT is applied. →

(ii) An electric induction E = − 3 aɵ x + 4 aɵ y + 6 aɵ z k V /m is applied →

→

(iii) When B and E are applied simultaneously.

E lectromagnetic F ield Theory

362

→

Solution: (i) The magnetic force F exerted on charge Q moving with velocity v in a uniform magnetic →

field B is, →

→

→

F = q( v × B)

→

Here v = 5 × 106 nɵ , where nɵ =

(0.04 aɵ x − 0.05 aɵ y + 0.2 aɵz [(0.04)2 + (0.05)2 +(0.2)2 ] →

v =

∴ or

...(1) =

0.04 aɵ x – 0.05 aɵ y + 0.2 aɵ y 0.21

5 × 106 (0.04 aɵ x − 0.05 aɵ y + 0.2 aɵz ) 0.21 →

→

v = 23.81 × 106 (0.04 aɵ x − 0.05 aɵ y + 0.2 aɵz ) and B = (aɵ x + 2 aɵ y + 3 aɵz ) × 10 −3 V /m →

→

v × B = 23.81 × 106 (0.04 aɵ x − 0.05 aɵ y + 0 .2 aɵz ) × (aɵ x + 2 aɵ y + 3 aɵz ) × 10 −3

∴

aɵ x aɵ y aɵz 0.04 −0.05 0.2 1 2 3

3

= 23.81 × 10 →

→

v × B = 23.81 × 103 [aɵ x (−0.05 × 3 − 2 × 0.2) + aɵ y (0.2 × 1 − 3 × 0.04)

or

+ aɵz (0.04 × 2 + 0.05 × 1)] 3

= 23.81 × 10 [− aɵ x (0.55) + aɵ y (0.08) + (0.13) aɵz ] →

→

v × B = 23.81 × 103 (− 0.55 aɵ x + 0.08 aɵ y + 0.13 aɵz )

or →

→

Substituting this value of v × B in eqn. (1), we get →

F = 10 × 10 –9 × 23.81 × 103 (−0.55 aɵ x + 0.08 aɵ y + 0.13 aɵz ) = 23.8 × 10 −5 (− 0.55 aɵ x + 0.08 aɵ y + 0.13 aɵz )

→

. F = 10 −5 (−13.09 aɵ x + 190 aɵ y + 3.09 aɵ z ) N

or

→

(ii) The force due to electric field E is given by →

→

→

F = q E ∴ F = 10 × 10 −9 (−3 aɵ x + 4 aɵ y + 6 aɵz ) × 103

→

F = 10 −5 (−3 aɵ x + 4 aɵ y + 6 aɵ z ) N

or →

→

(iii) When B and E is applied simultaneously, the combined force is, →

→

→

→

F = q E + q( v × B) = 10 −5 (−3 aɵ x + 4 aɵ y + 6 aɵz ) + 10 −5 (−13.09 aɵ x + 190 . aɵ y + 3.09 aɵz )

or

→

F = 10 −5 (−16. 09 aɵ x + 5. 90 aɵ y + 9. 09 aɵ z )N

363

Example 2: A conductor of length 3. 0 m located at z = 0 , x = 4 m carries a current of 10 amp in the − aɵ y

direction. Find the uniform magnetic field in the region, when the force on the conductor is 10 . × 10 −2 N in the direction (− aɵ x + aɵ z )/ 2 . Solution: The force experienced by the conductor of length L which carries a current of I amp and →

placed in a steady magnetic field of flux density B is →

→

→

→

→

→

→

F = ∫ Id l × B = I ∫ d l × B = I ( L × B ) →

...(1)

→

I = 10 amp, L = − 3 aɵ y , B = B x aɵ x + B y aɵ y + Bz aɵz

Here,

→

F = 1.0 × 10 −2

and →

(− aɵ x + aɵz ) 2

= 7.07 × 10 −3 (− aɵ x + aɵz )N

→

Substituting the values of I , L and B in eqn. (1), we get →

F = 10[−3 aɵ y × (B x aɵ x + B y aɵ y + Bz aɵz )] = 30 [B x aɵz − Bz aɵ x ] (∵ aɵ y × aɵ x = – aɵz , aɵ y × aɵz = aɵ x and aɵ y × aɵ y = 0)

→

F = 7.07 × 10 −3 (− aɵ x + aɵz )

But

7.07 × 10 −3 (− aɵ x + aɵz ) = 30 B x aɵz − 30 Bz aɵ x

∴

Comparing the coefficients of aɵ x and aɵz on both sides, we get − 7.07 × 10 −3 = − 30 Bz ∴ Bz = 2 .356 × 10 −4 Similarly, 7. 07 × 10 −3 = 30 B x ∴ B x = 2 .356 × 10 −4 →

B = B x aɵ x + Bz aɵz = 2 . 356 × 10 −4 (aɵ x + aɵ y ) wb /m 2

Hence,

→

Example 3: The field B = − 3 aɵ x + 4 aɵ y + 6 aɵ z mT is present in free space. Find the magnetic force exerted on a straight wire carrying a current of 10 amp in aɵ AB direction, given A(1, 1, 1) and B(2, 1, 1). →

Solution: Magnetic force d F exerted on a wire is →

→

→

d F = I (d l × B )

∴

or

→

→

d l = dx aɵ x + dy aɵ y + dz aɵz and B = − 3 aɵ x + 4 aɵ y + 6 aɵz

Here

aɵ x aɵ y → → d l × B = dx dy −3 4 →

aɵz dz 6

→

d l × B = aɵ x (6 dy − 4 dz) + aɵ y (−3 dz − 6 dx) + aɵz (4 dx + 3 dy) →

→

d l × B at A (1, 1, 1) and B (2, 1, 1) is

E lectromagnetic F ield Theory

364

→ → 1 1 d l × B = aɵ x 6 ∫ dy − 4 ∫ dz + aɵ y −3 1 1

B

∫A

or

1

2

∫1 dz − 6 ∫1

2 1 dx + aɵz 4 ∫ dx + 3 ∫ dy 1 1

= aɵ x [6 × 0 − 4 × 0] + aɵ y [−3 × 0 − 6 (2 − 1)] + aɵz [4 (2 − 1) + 3 × 0] B

∫A

or

→

→

d l × B = − 6 aɵ y + 4 aɵz →

F = I∫

∴

B A

→

→

(d l × B) = 10 (− 6 aɵ y + 4 aɵz ) mN

→

F = − 60 aɵ y + 4 aɵ z mN

or

Example 4: Two different current elements I1 dl1 = 10 −6 aɵ y A–m. at P1 (2, 0 , 0 ) and I2 d l2 =10 −6 (0 . 5 aɵ x + 0 .4 aɵ y + 0 . 3 aɵ z ) A–m at P2 (3, 3, 3 ) are located in free space. Calculate the vector force exerted on →

→

I2 dl2 by I1d l1 and on I1 dl1 by current element I2 d l2 . →

Solution: The vector force exerted on current element I2 dl2 due to the magnetic induction d B1 produced →

→

by current element I1 dl1 near I2 dl2 is given as, →

→

→

dF2 = I2 d l2 × dB1 →

→

→

The magnetic induction d B1 produced by the current element I1 dl1 near I2 d l2 is, →

µ I dl × aɵR12 dB1 = 0 1 1 4π (R12 )2 →

where aɵR12 is the unit vector along R12 , aɵR 12 is given by P1(2, 0, 0) and P2 (3, 3, 3) ∴

∴

aɵR12 =

(3 − 2) aɵ x + (3 − 0) aɵ y + (3 − 0) aɵz (1)2 + (3)2 + (3)2

→

I1d l1 × aɵR12 = (10 −6 aɵ y ) ×

(aɵ x + 3 aɵ y + 3 aɵz ) 4.36

=

=

aɵ x + 3 aɵ y + 3 aɵz 4 .36

and R12 = 19

10 −6 (− aɵz + 3 aɵ x ) 4 . 36

(∵ aɵ y × aɵ x = – aɵz , aɵ y × aɵz = aɵ x and aɵ y × aɵ y = 0) →

→ 4 π × 10 −7 10 −6 (3 aɵ x − aɵz ) 10 −13 or d B = (3 aɵ x − aɵz ) × 1 82.84 4π 4.36 ( 19 )2

Hence,

d B1 =

∴

dF2 = I2 dl2 × dB1 = 10 −6 (0.5 aɵ x + 0.4 aɵ y + 0.3 aɵz ) ×

→

→

=

→

10 −13 (3 aɵ x − aɵz ) 82.84

10 −19 [3 × 0 .5 × 0 + 0.5 aɵ y − 0.4 × 3 aɵz − 0.4 aɵ x + 0.3 × 3 aɵ y ] 82.84 (∵ aɵ x × aɵ x = aɵz × aɵz = 0 and aɵ x × aɵz = – aɵ y , aɵ y × aɵz = aɵ x )

365 →

10 −19 [0.5 aɵ y − 1.2 aɵz − 0.4 aɵ x + 0.9 aɵ y ] 82.84

or

dF2 =

or

dF2 = 1. 21 × 10 −21 [−0 . 4 aɵ x + 1. 4 aɵ y − 1. 2 aɵ z ]

→

→

Similarly the vector force exerted on current element I1dl1 due to the magnetic induction d B2 produced →

→

by current element I2 dl2 near I1 dl1 is given as, →

→

→

d F1 = I1d l1 × d B2 and Here

→

→

µ I d l × aɵ dB2 = 0 2 2 2R 21 4π (R21) →

R21 = (2 − 3) aɵ x + (0 − 3) aɵ y + (0 − 3) aɵz

or

aɵR 21 =

∴

d B2 = −

∴

dB2 = −

→

→

− aɵ x − 3 aɵ y − 3 aɵz (1)2 + (3)2 + (3)2 4 π × 10 −7 10 . 4π

−6

=−

(aɵ x + 3 aɵ y + 3 aɵz ) 19

and | R21| = ( 19 )

(0.5 aɵ x + 0.4 aɵ y + 0.3 aɵz ) × (aɵ x + 3 aɵ y + 3 aɵz ) ( 19 )2

10 −7 × 10 −6 (0.5 × 3 aɵz − 3 × 0.5 aɵ y − 0.4 × 1 aɵz + 0.4 × 3 aɵ x 19 × 4.36

+ 0.3 × 1 aɵ y − 0.3 × 3 aɵ x ) 10 −13 d B2 = − (1.5 aɵz − 1.5 aɵ y − 0.4 aɵz + 1.2 aɵ x + 0.3 aɵ y − 0.9 aɵ x ) 82.84 →

→

or

d B2 = – 1.21 × 10 −15 (0.3 aɵ x − 1.2 aɵ y + 11 . aɵz )

∴

d F2 = 10 −6 aɵ y × [−1.21 × 10 −5 (0.3 aɵ x − 1.2 aɵ y + 11 . aɵz )]

→

= − 1.21 × 10 −21 (− 0.3 × 1 aɵz + 11 . × 1 aɵ x ) or

→

d F2 = 10 −21 (−1. 33 aɵ x + 0 . 363 aɵ z ) N

Example 5: Two straight wires A and B of lengths 10 and 16 m and carrying currents of 4.0 and 5.0 ampere respectively in opposite directions, lie parallel to each other 4.0 cm apart. Compute the force on a 10 cm long section of the wire B near its centre. Solution: The magnitude of force (repulsive) on each wire per unit length is expressed as F µ0 I1I2 = l 2π d Here ∴

µ0 = 2 × 10 −7 newton/amp2 , I1 = 4 .0 amp, I2 = 5 .0 amp and d = 4 .0 cm = 4 × 10 − 2 m 2π −7 F 2 × 10 × 4 .0 × 5 .0 = = 1.0 × 10 −4 newton/m l 4 .0 × 10 −2

Force on 10 cm length of the wire B due to wire A is, F ′= F × 10 × 10 −2 = 1.0 × 10 −4 × 10 × 10 −2 = 1.0 ×10 −5 Newton

E lectromagnetic F ield Theory

366

Example 6: Calculate the resultant force on the current-carrying rectangular loop PQRS (Fig. 6) due to a current carrying long wire. Take I1 = 30 amp, I2 = 20 amp, L = 30 cm; a = 1.0 cm and b = 8 .0 cm. Solution: The given wire and the sides PQ and SR of the rectangular loop P

PQRS are parallel (Fig. 6.). The (attractive) force on the side PQ due to current I1 in the wire is, µ II F1 = 0 1 2 L 2π a Here

µ0 = 2 × 10 −7 N/ A 2, I1 = 30 amp, I2 = 20 amp. 2π a = 1.0 cm = 1 × 10 −2 m and L = 30 cm = 30 × 10 −2 m

S

I1 I2

I2

Q

L

R b

a

Fig. 6

∴

F1 =

2 × 10

−7

× 30 × 20 × 30 × 10

−2

1 × 10 −2

= 3 .6 × 10 −3 Newton (Attractive)

Force F1 is directed towards the wire. The (repulsive) force on the side SR of the rectangle due to current I1 in the given wire is µ I1I2 F2 = 0 L 2 π (a + b) Here ∴

a + b = (1.0 + 8 .0) cm = 9 × 10 −2 m F2 =

2 × 10 −7 × 30 × 20 × 30 × 10 −2 9 × 10 − 2

= 0 .4 × 10 −3 Newton (Repulsive)

Force F2 is directed away from the wire. The forces on the sides PS and QR of the rectangle will be equal and opposite, because they carry currents in opposite directions. Therefore, their resultant will be zero. Hence, the resultant force on the rectangular loop PQRS is F = F1 − F2 = 3 .6 × 10 −3 − 0 .4 × 10 −3 or

F = 3.2 × 10 −3 Newton, directed toward the wire.

Example 7: Two long filamentry wires, carrying currents I1 and I2 are placed perpendicular to each other in such a way that they just avoid a contact. Find the magnetic force on an elmental length dl of the second wire situated at a distance l from the first wire. Solution: The situation of the problem is shown in fig. 7. The magnetic field B at the site of elemental length dl due to the vertical wire carrying current I1 is, B=

µ0 I1 2πl

µ I ∵ B = 0 ...(1) 2 π d

The field B is normal to the plane of figure and in the downward direction. The magnetic force dF on the length dl of the second wire is

I1 I2 l Fig. 7

dl

367

dF = I2 dl B sin 90 ° = I2 Bdl Substituting the value of B from eqn. (1), we get or

or

dF = I2

dF =

µ0 I1 dl 2πl

µ0 I1 I2 dl l 2π

This force is parallel to the current I1 in the first wire.

Example 8: Two horizontal straight conductors, each 20 cm long, are arranged one vertically above the other and carry currents in opposite directions. The lower conductor is fixed while the upper is free to move in guides remaining parallel to the lower. If the upper conductor weights 1.20 gm what is the approximate current that will maintain the conductors at a distance 0 . 75 cm apart?

F

Solution: Suppose AB and CD be two straight conductors carrying a current I in opposite directions as shown in fig. 8.

I

C

D

mg

The upper conductor will be in equilibrium when the repulsive magnetic force F exerted on it by the lower

R

conductor AB, balances its weight mg. Let R be the equilibrium distance between AB and CD.

I A

The repulsive force on CD due to opposite current is given

Fig. 8

by F=

20cm

µ0 I 2 l 2π R

where l = 20 × 10 −2 m and R = 0.75 × 10 −2 m. At equilibrium, this force is balanced by the weight mg of the conductor CD, that is, F = mg or Here

µ0 I 2 l = mg 2π R

m = 1.20 × 10 −3 kg and g = 9.8 m /s2 I2 4 π × 10 –7 × × (20 × 10 –2 ) = 120 × 10 –3 × 9. 8 2π (0.75 × 10 –2 )

or ∴

I2 =

1.20 × 10 −3 × 9.8 × 0.75 × 10 −2 2 × 20 × 10 −7 × 10 −2

I = 2205 = 46. 95 amp

= 2205

B

E lectromagnetic F ield Theory

368

Magnetic Torque and Moment Before obtaining an expression for the torque on a current carrying conductor, we discuss the concept of torque first.

1. Concept of Torque Z

Just as force is required to produce linear acceleration in a particle, a torque (or moment of force) is required to produce rotatory motion about a fixed axis. The fixed axis is called, axis

T(Torque)

→

of rotation. Let a force F act on a particle at P whose position →

vector is r with respect to the origin '0' as shown in fig. 9. The

O

→

torque acting on the particle about the origin '0' is defined as

→

r

→

→

the vector product of r and F, That is, →

→

→

P

Y

T = r × F

Y

F θ

Fig. 9

The magnitude of torque is, T = r . F sin θ

Z

where θ is the angle between r and F and is directed →

→

T(Torque)

perpendicular to the plane containg r and F. Now let us consider the case of two forces F1 and F2 applied to an object simultaneously. Force F1 acts at a point P1 with

O

→

position vector r1 and force F2 acts at P2 with position vector → r2

→

r2

as shown in fig. 10. The torque T about the origin O is Y

given by →

→

→

→

F2 = – F1 P2

→

→

→

r21

→

P1

F1

Fig. 10

T = r1 × F1 + r2 × F2

→

Y

r1

→

F1 + F2 = 0 or F2 = − F1

Here

→

→

→

→

→

→

→

→

→

T = r1 × F1 − r2 × F1 or T = ( r1 − r2 ) × F1

∴

→

→

→

→

r21 = (r1 − r2 )

From fig. 10. →

→

∴

→

→

→

T = r21 × F1 →

The vector, r21 = r1 − r2 joints the point P2 of application of force F2 to that of point P1 of application of →

→

→

force F1 and is independent of choice of origin for the vectors r1 and r2 . Therefore, the torue is also independent of the choice of the origin, provided that the total force is zero.

369

T

orque on a Differential Current Loop Y

Let us consider a differential current loop of sides dx and dy at a point O in a magnetic field B (fig. 11). Suppose the sides of

d

→

c

dx

the loop dx and dy be parallel to X-and Y -axes so that the

U R3

I

normal to the area of the loop is along the Z-axis. Since the →

dy

loop is of differential size, the value of B at all points on the

V

loop may be taken as B0 . Since the total force on the loop is zero, we are free to consider the origin for the torque at the

B

a

O R4

dy

centre of the loop. Suppose I is the current flowing through

I

the loop.

R1 I

S a

→

The differential force dF1 on side ' ab ' of current element

xX

R2 T

b

dx Fig. 11

Idx is →

→

d F1 = I (dx aɵ x × B0 )

...(1)

→

aɵ x × B 0 = aɵ x × (aɵ x B0 x + aɵ y B 0 y + aɵz B0 z ) →

= (aɵ x × aɵ x ) B0 x + (aɵ x × aɵ y ) B0

(aɵ x × B0 ) = aɵz B 0 y − aɵ y B0 z

or

y

+ (aɵ x × aɵz ) B0 z

(∵ aɵ x × aɵ x = 0, aɵ x × aɵ y = aɵz and aɵ x × aɵz = − aɵ y )...(2)

→

∴

dF1 = Idx (aɵz B 0 y − aɵ y B0 z )

...(3)

→

For the side ' ab ' of the loop the lever arm R1 extends from origin 0 to the midpoint S of the side ' ab ' , that →

is, R1 = –

→ 1 dy aɵ y . Thus, the torque d T1 is, 2 →

→

→

dT1 = R1 × d F1 1 1 = − dy aɵ y × (Idx (aɵz B 0 y − aɵ y B0 z )} = − Idx dy (aɵ y × [aɵz B0 2 2 →

y

− aɵ y B0 z ])

1 Idx dy [(aɵ y × aɵz ) B 0 y − (aɵ y × aɵ y ) B0 z ] 2 1 [∵ aɵ y × aɵz = aɵ x and aɵ y × aɵ y = 0] = − Idx dy aɵ x B 0 y 2 → 1 ...(4) dT1 = − Idx dy B 0 y aɵ x ∴ 2 → → → 1 For the side ' cd ' the lever arm R3 = dy aɵ y and d F3 = − I (dx aɵ x × B0 ) 2 d T1 = −

or

→

Thus, the torque dT3 acting on side cd is →

→

→

dT3 = R3 × d F3 =

→ 1 dy aɵ y × [− I (dx aɵ x × B0 )] 2

E lectromagnetic F ield Theory

370

From eqn. (3)

→

aɵ x × B0 = aɵz B 0 y − aɵ y B0 z →

1 I dx dy (aɵ y × (aɵz B 0 y − aɵ y B0 z )) 2 → 1 dT3 = − I dx dy aɵ x B 0 y 2

∴

dT3 = −

or

(As in previous case)...(5)

→

For the side bc of the loop, the lever arm R2 extends from origin 0 to the mid point T of the side, that is, →

R2 =

1 dx aɵ x and the differential force on side bc of current element Idy is, 2 →

→

→

dF2 = (I dy aɵ y × B 0 ) = Idy (aɵ y × B 0)

→

= Idy (aɵ y × (aɵ x B0 x + aɵ y B 0 y + az B0 z )) = Idy [(aɵ y × aɵ x ) B0 x + (aɵ y × aɵ y ) B0 y + (aɵ y × aɵz ) B0 z ] = Idy [− aɵz B0 x + 0 + aɵ x B0 z ] (∵ aɵ y × aɵ x = − aɵz , aɵ y × aɵ y = 0 and aɵ y × aɵz = aɵ x )

d F2 = Idy (aɵ x B0 z − aɵz B0 x )

or

...(6)

→

Therefore the torque dT2 acting on side bc is →

→

→

dT2 = R2 × d F2 1 = dx aɵ x × I dy (aɵ x B0 z − aɵz B0 x ) 2 1 I dx dy [(aɵ x × aɵ x ) B0 z − (aɵ x × aɵz ) B0 x ] 2 → 1 or (∵ aɵ x × aɵ x = 0 and aɵ x × aɵz = − aɵ y )...(7) dT2 = I dx dy B0 x aɵ y 2 → → 1 For the side ' da' the lever arm R4 = − dx aɵ x and differential force, dF4 = − Idy (aɵ y × B 0) 2 =

→

Therefore, the torque dT4 acting on side da is →

→

→

dT4 = R4 × dF4 =− or

→

dT4 =

→ 1 dx aɵ x × [− I dy aɵ y × B 0] 2

1 Idx dy [aɵ x × (aɵ y × (aɵ x B0 x + aɵ y B0 y + az B0 z ))] 2

As in previous case →

dT4 =

1 I dx dy B0 x aɵ y 2

...(8)

Thus the torque contribution on the entire loop is →

→

→

→

→

dT = dT1 + dT2 + dT3 + dT4 Thus from eqns. (4), (5), (7) and (8), we have → 1 1 dT = − I dx dy B0 y aɵ x − I dx dy aɵ x B0 2 2 = − I dx dy B0 y aɵ x + I dx dy B0 x aɵ y

...(9)

y

+

1 1 I dx dy B0 x aɵ y + I dx dy B0 x aɵ y 2 2

371 →

dT = I dx dy (B0 x aɵ y − B0

or

y

aɵ x )

...(10)

→

→

B0 x aɵ y − B0 y aɵz = aɵz × B0

But,

→

(∵ B0 = aɵ x B0 x + aɵ y B 0 y + aɵz B0 z )

→

dT = I dx dy (aɵz × B0 )

∴

...(11)

But, dx dy is the area of the differential loop along aɵz direction, therefore →

dx dy aɵz = d S →

→

→

dT = I d S × B

∴

→

Now we define the product of the loop current, I and vector area d S as the differential magnetic dipole →

moment d m with unit A-m 2 . Therefore, →

→

→

→

d m = I dS

→

dT = d m × B

and

→

Since we assumed a differential current loop so that we might assume B was constant throughout the loop, it follows that the torque on a planar loop of any size or shape in a uniform magnetic field is given as →

→

→

→

→

T = IS × B= m× B

The torque on the current loop always tends to turn the loop so as to align the field produced by the loop with the external magnetic field that is causing the torque.

T

orque on a Current Carrying Loop

Let us consider a rectangular wire loop ABCD of length L and width b and carrying a current of I amp. →

Suppose the loop ABCD be placed in a magnetic field of flux density B which is directed into the plane of paper as shown in fig. 12 (a). Let the normal to the loop make an angle θ with the direction of field as shown in fig. 12 (b). →

× × × I× × × × × × × × × × I× × × F2

× × B × × × × × I× × × × × C × × × ×

B × × × × F × 1 × × ×

→

F

or m al

×A × × × ×I × × × × × × × D × × × ×

N

F4 × × × × F3 × × × ×

θ

B

b b sin θ →

F (a)

(b) Fig. 12

E lectromagnetic F ield Theory

372

The force on the side DC of the loop is, →

→

→

F2 = I (D C × B) or F2 = I bB sin θ

The force on the side BA of the loop is given by, →

→

→

F4 = I (BA × B) or F4 = I bB sin θ

→

→

The forces F2 and F4 are equal and opposite and have the same line of action. Hence they cancel out the effect of each other's. The sides AD and CB of the rectangular loop are always perpendicular, to the field direction. Therefore, the magnitude of force on AD is F1 = I ( AD) B sin 90 ° or F1 = IBL Similarly the force on the side CB is F2 = I (CB) B sin 90 ° or F2 = IBL Since the forces acting on the sides AD and CB are equal and opposite and they do not have the same line of action so they constitute a couple due to which the loop tends to rotate in the clockwise direction. Thus a torque is exerted on the loop whose magnitude is given by, T = Force × perpendicular distance of the line of action or

T = F × (b sin θ), but F = IBL

∴

T = I B Lb sin θ, but Lb = A, the area of the loop.

∴

T = I B A sin θ

The term I A is defined as the dipole moment ' m ' of the loop. Thus T = m B sin θ In vector form,

→

→

→

T =m× B

Magnetic Dipole Like current carrying conductors, magnetic field also exist around a magnetised object. In atoms of all materials, electrons revolving round the nucleus in their orbits with very high velocity can be assumed as tiny current loop of atomic dimension. In addition, these electrons also revolve about their own axis, that electrons spin about their own axis. This spinning motion of electron can again be taken as equivalent to tiny closed circuit. thus, electron itself has a magnetic moment because of this spinning motion. The spin of nucleus also produces a magnetic moment, but it is usually not significant. Furthermore, some atoms possess permanent magnetic moment while others have magnetic moments which are induced when atom is placed in an external magnetic field.

373

Torque → → T=m×B

→

Magnetic moment

→

B

→

m

→

m

+

A

θ

nucleus

× – I

–

I current (a)

(b) Fig. 13

A small current loop has associated with it a magnetic moment, m given by the product of the current, I and the area of the loop A, that is, →

→

m= I A

→

The magnetic moment vector m is perpendicular to the plane of the loop and directed upward as shown in fig. 13. →

→

When the atom is place in an external magnetic field B, the torque T tends to align the moment of current loop or atomic loop with the field and is given as →

→

→

T =m×B →

→

As shown in fig. 13(b), this torque tends to rotate the loop so that the moment vector m aligns with B. The magnitude of torque is, T = m B sin θ When alignment is achieved (θ = 0) the torque becomes zero. Like an electric dipole moment, a bar magnet of pole strength Qm and pole separation L (fig. 14), constitutes a magnetic dipole of magnetic dipole moment Qm L. When such a bar magnet is freely →

suspended in a uniform magnetic field B, the north pole (+) of the magnet experience a force, F = Qm B to the right and the south pole (−)

L 2

an equal force (Qm B) to the left. The forces acting on north pole and south pole of the bar magnet act in opposite directions and

F Torque → → F sin θ T = m × B

θ

constitute a torque which is given as, L sin θ or T = F L sin θ 2

But,

F = Qm B

∴

T = Qm BL sin θ

or

T = mB sin θ (∵ Qm L = m)

or

→

→

→

T =m×B

F sin θ F

θ

T = 2F

L 2

Axis of rotation

– Fig. 14

θ

B

E lectromagnetic F ield Theory

374

As tiny current loop and a bar magnet produce magnetic dipoles, they are analogous if they produce the same amount of torque in an external magnetic field B. Therefore, T = Qm LB = IAB where A is the surface area of the loop and I is the loop current. Qm L = IA (amp-meter2 )

∴

→

Hence, a bar magnet and a current loop experience equal torque in a magnetic field B provided their magnetic moments are equal. Example 9: What is the maximum torque on a square loop of 100 turns in a magnetic field of flux density ' B' tesla. The loop has 10 cm side and carries a current of 3 amp. What is the magnetic moment of the loop? Solution: Maximum torque on a single loop = BI A N-m Maximum torque on 100 turns of a square loop = 100 BIA Here I = 3 amp, A = 10 × 10 × 10 −4 m 2 and B = 1 Tesla Tmax = 100 × 1 × 3 × 10 −2 = 3 Newton

∴

Magnetic moment (m) = IA = 3 × 10 −2 aɵ n mA - m 2 Example 10: The rectangular coil as shown in fig. is in a magnetic field of magnetic flux density → aɵ x + aɵ y B = 0 . 01 T 2 (a)

Find the torque about z-axis when the coil is in the position as shown fig. 15. and carries a current of 5. 0 amp.

(b)

At what position of the coil torque will be zero.

z

Solution: (a) The torque on the current loop is →

→

→

I

T =m×B

0.08m

→

where m is the magnetic moment of the current loop and is given by →

–0.04m

aɵ x + aɵ y T = 0.016 aɵ x × 0.01 2

x

m = IAaɵ x = 5 × (0.08 × 0.04) aɵ x = 0.016 aɵ x amp-m

→

∴

(∵ aɵ x × aɵ x = 0 and aɵ x × aɵ y = aɵz )

→

Fig. 15

T = 1.13 × 10 −4 aɵ z N - m (b)

→

→

→

T =m×B

→

→

or T =|m| |B|sin θ nɵ →

→

Thus, torque will be zero when angle between m and B is 90° or when the coil will turn through 45°.

y

375

Example 11: A wire of length l carrying a current I, has been given the shape of a circular coil and →

suspended in a magnetic field B. Show that the torque acting on a coil is maximum when the coil has one turn only, and the maximum torque is given by T =

I bl2 4π

Solution: Let us suppose that the circular coil is made up of the wire of length l has N turns of radius R, then l = (2πR ) N or

R=

or

A=

l 2πN

l ∴ the area of the coil, A = πR 2 = π 2 πN

2

l2

4πN 2 Now, if the current in the coil is I, the coil and perpendicular to the coil makes an angle θ with the →

magnetic field B, then the torque acting on the coil is, l2 sin θ T = NIBA sin θ = NIB 4πN 2 =

IBl2 sin θ 4πN

The torque is maximum when N = 1 and θ = 90 ° , that is, the plane of the coil is parallel to field. Therefore, maximum torque,

Tmax =

I B l2 4π

Magnetisation in Materials When a magnetic material is placed in an external magnetic field, the elementary current loops or atomic magnets in the material becomes aligned parallel to the field and acquire magnetic moment. Some atom possess permanent magnetic moments while others have moments which are induced by external magnetic field. The material is then said to be magnetised. A material which has a large number of atomic magnetic dipoles aligned in one direction and distributed uniformly, is called uniformly →

magnetised material. To define a macroscopic vector quantity, the magnetisation M, we sum up →

vectorially all the dipole moments in a small volume element ∆v, and then divide the result by ∆v. If m i is the magnetic dipole moment of ith atom, then the resulting quantity, →

M = lim

∆V → 0

1 mi ∆v ∑ i

E lectromagnetic F ield Theory

376

is called the magnetic dipole moment per unit volume or simply the magnetisation. In the →

∑ mi

unmagnetised state of material, the summation

will be zero due to random orientation of the →

→

magnetic moment m i but in the presence of an external magnetic field M usually depends on the field. Thus, the magnetic moments per unit volume is defined as magnetisation M or intensity of magnetisation. →

In fact the vector function M provides a macroscopic description of atomic currents inside the material. →

Specifically M measures the number of atomic current circuits per unit volume times the average magnetic moment of each circuit. Thus, →

The magnetisation, M may also defined in term of its elementary magnetic dipole moment per unit volume as, →

→

M =Nm where N is the atomic density.

Relation Between Magnetising Current Density (J m ) and Intensity of Magnetisation (M): Since the atom in a magnetic material is equivalent to a current loop, the material is made up of large number of small current loops. For the uniform magnetisation of the material, there is no net effective current in the interior of the material. It is because of the fact that the currents in various loops tend to cancel each other's effect. On the other hand, for non-uniform magnetisation of the material, the concellation of currents in various loops will not be complete due to abrupt change in magnetisation. As a result of which more charge move downward than moving up. This non-uniform magnetisation gives rise the net current which produce a magnetic field. This, net current which comes about from non uniform magnetisation of material is called the magnetisation current. If I m is the magnetisation current, then according to Ampere's circuital law of atom, we have →

→

→

→

I m = ∫ H. d l = ∫ J m . dS S

The intensity of magnetisation in a material is, →

M = lim

∆v→ 0

Using stoke's theorem, we have

1 ∆V

→

→

→

→

→

→

→

∑ Mi i

→

→

∫ S ( ∇ × M) . dS = ∫ S J m . dS Thus,

∇ × M = Jm

That is, the magnetisation current density, J m is the curl of magnetisation. If the convectional current enclosed by a closed path be I and I m the magnetisation current, then the total current be I t (current due to both free and bound charges). Thus →

→

→

→

→

∇×H= J

→

∇× M = Jm

377 →

→

∇×

and

B → = Jt µ0

Magnetic Induction or Magnetic Flux Density (B) When a piece of magnetic material is subjected in an external magnetic field, the material becomes magnetised. The magnetism so developed in the material is called "induced magnetism" this phenomenon is called magnetic induction. When a magnetic material is megnetised by placing it in a magnetic field, the resultant field inside the material is the sum of magnetic field produced due to magnetisation of the material and the original magnetic field which is responsible for the magnetisation of the material. If B0 is the primary field in which the unmagnetised material is placed and Bm is the magnetic flux density due to magnetisation of material. Then →

→

→

B = B0 + B m

→

→

But B0 = µ0 H, the line of force crossing per unit area in a perpendicular direction due to primary →

→

magnetic field and B m = µ0 H, the line of force crossing per unit area in a perpendicular direction due to magnetisation or intensity of magnetisation. Thus, →

→

→

B = µ0 H + µ0 M →

Thus, the magnetic induction or magnetic flux density B is measured by the number of lines or induction passing per cm 2 through a plane perpendicular to their direction. The SI unit of magnetic induction is tesla (T) or weber/m 2 (Wb-m –2 ) or newton /(amp-metre) (NA –1 m –1 ). Its CGS unit is 'gauss'.

→

Magnetic Field Intensity ( H) When a magnetic material is placed in a magnetic field it gets magnetised. The capability of the magnetic field to magnetised a material is denoted by magnetic field intensity or magnetising force of the applied →

field. The magnetic field intensity H is measured by the magnetic force exerted on a unit north pole placed in a magnetic field. The intensity of magnetic field is also measured by the number of lines of →

force passing per unit area through a plane normal to their direction. Magnetic field strength H is also defined through the vector relation, →

H=

→

→

B → −I µ0 →

where B is magnetic flux density inside the material and I is the intensity of magnetisation and µ0 is the permeability of free space. The CGS unit of intensity of magnetic field is "oersted". The S.I. unit of →

H is amp/m (Am –1 ).

E lectromagnetic F ield Theory

378

Magnetic Susceptibility (χ m ) The magnetic susceptibility, x m is the dimension less quantity and is a characteristic of the media. The magnetic susceptibility describs the contribution made by a material when subjected to a magnetic field to the total magnetic flux density present. It is defined as the ratio of the intensity of magnetisation, I produced to the magnetic field intensity H. Thus

χm =

I M or H H

Since I is the magnetic moment per unit volume. The susceptibility so defined is also called volume susceptibility. The diamagnetic materials have a low negative susceptibility, paramagnetic materials have a low positive susceptibility and the ferromagnetic materials have a high positive value of susceptibility.

Magnetic Permeability (µ) The magnetic permeability (µ) of a piece of magnetic material is a measure of its conduction of magnetic lines of force through it. It is defined as the ratio of the magnetic flux density, B inside the magnetised →

material to the external magnetic field strength H , that is, B µ= H The magnetic permeability of the material is related with the permeability µ0 in free space as, µ = µ0 µr where µ r is the relative permeability which is dimension less quantity and is defined as, µ µr = µ0 Thus the relative permeability of a magnetic material is the ratio of the magnetic permeability µ of the material to the permeability of free space. In terms magnetic flux density B, the relative permeability µ r is defined as the ratio of the magnetic flux density B in a material when subjected in an external field and the flux density B0 in vacuum in the same field, that is, µr =

B B0

For a paramagnetic material µ r is greater than one for a diamagnetic material µ r is less than one. The ferromagnetic materials has higher relative permeability.

Relation Between Relative Permeability (µ r ) and Magnetic Susceptibility (χ m ) When a magnetic material is subjected in a magnetic field. It becomes magnetised. The total magnetic flux density B inside the material is the sum of flux density that would have been produced by the external magnetising field in vacuum and the flux density due to the magnetisation of material. If the →

→

intensity of magnetisation of the material is I , then the magnetic intensity H of the magnetising field is expressed as,

379

B −I µ0

...(1)

B = µ0 H + µ0 I

...(2)

H= or

But intensity of magnetisation, I is equals to the product of magnetic susceptibility χ m and magnetic →

intensity H of the magnetising field, that is, →

→

I = χm H

Again

→

→

B =µH

Substituting these values of I and B in eqn. (2), we get µH = µ0 H + µ0 χ m H or or But

µ = µ 0 (1 + χ m )

µ =1 + χm µ0 µ = µ r , relative permeability µ0

∴

µr = 1 + χm

Example 12: Find the magnetisation in a magnetic material where (a) µ = 18 . × 10 −5 h /m, H =120 A / m and (b) B = 300 µT and χ m = 15

Solution: (a) In the given problem, µ = 1.8 × 10

−5

h / m, H = 120 A / m and

[U'khand, B.Tech IV Sem 2011] µ0 = 4 π × 10 –7 w /m

We know that susceptibility χ m is related with magnetisation M as, χm =

M H

or M = χ m H

Further, χ m = (µ r − 1), where µ r is the relativity permeability, that is µ = µ r µ0 1.8 × 10 −5 µ = = 14.33 µ 0 4 × 3.14 × 10 −7

∴

µr =

Thus,

M = (µ r − 1) H = (14.33 − 1) × 120

or

M = 1599. 74 A /m

(b) We know that, B = µH or B = µ0 µ r H, but µ r = 1 + χ m = 1 + 15 = 16 300 × 10 −6 B = = 14.93 A /m µ0 µ r 4 × 3.14 × 10 −7 × 16

∴

H=

Hence

M = χ m H = 15 × 14.93 = 233. 95 A /m

E lectromagnetic F ield Theory

380

Example 13: An isotropic material has a magnetic susceptibility of 3 and the magnetic flux density, →

B = 10 y aɵ x m wb / m2 . Determine µ, J , M and H. [GBTU, B.Tech III Sem 2010]

Solution: Here χ m = 3 ∴

µr = χm + 1 = 3 + 1 = 4 µ = µ0 µ r = 4 π × 10 −7 × 4 = 16 π × 10 −7

Hence, We know that,

→

→

→

→

Here B = 10 y aɵ x mwb /m 2 = 10 −2 y aɵ x wb /m 2

J = ∇ × H,

→

→

H=

and

10 −2 y aɵ x B = = 1.989 × 10 3 y aɵ x µ 16 π × 10 −7

→

→

M = χ m H = 3 × 1.989 × 103 y aɵ x = 5. 967 × 10 3 y aɵ x →

→

→

J =∇×H=

aɵ x ∂ / ∂x

aɵ y aɵz ∂ / ∂y ∂ / ∂z

1.989 × 103 y

0

0

∂ (1.989 × 103 y) J = aɵz − ∂y

→

or

→

J = − 1. 989 × 10 3 aɵ z

or

Example 14: Magnetisation in magnetic material for which χ m = 15 is given in a certain region as →

H = 500 z 2 aɵ x A /m at z = 4 cm. Find the magnitude of →

→

→

(b) J and

(a) JT

(c) J m

Solution: (a)

The total current density JT (bound plus free charge) is →

JT But

µ = µ0 µr

∴

→

→

→

→

B → (µ H ) =∇× =∇× µ0 µ0 →

→

→

→

JT = ∇ × µ r H = µ r ( ∇ × H )

→

µ r = 1 + χ m = 1 + 15 = 16 and H =500 z2 aɵ x →

→

→

or

JT = 16 ∇ × 500 z2 aɵz A /m = 500 × 16 ∇ × z2 aɵ x

or

→ ∂ ∂ ∂ + aɵz × z2 aɵ x + aɵ y JT = 8000 aɵ x ∂z ∂y ∂x

381

∂ 2 ∂ 2 (z ) + aɵ y (z ) = 8000 − aɵz ∂z ∂y (∵ aɵ x × aɵ x = 0, aɵ y × aɵ x = − aɵz and aɵ y × aɵ x = aɵz ) →

→

JT = 8000 aɵ y (2 z) or JT = 16000 z aɵ y

or →

→

The magnitude of JT , that is | JT | at z = 4 cm or z = 0.04 m is →

| JT | = 16000 × 0.04 = 640 A /m2 →

→

→

→

(b) The total free current density, J = ∇ × H , Here H = 500 z2 aɵ x ∂ ∂ ∂ + aɵ y + aɵz × 500 z2 aɵ x J = aɵ x ∂y ∂z ∂x

→

∴

∂ 2 ∂ 2 (z ) + aɵ y (z ) = 500 − aɵz y z ∂ ∂ →

→

J = 500 × 2 z aɵ y or J = 1000 z aɵ y

→

The magnitude of | J | at z = 0.04 m is →

| J | = 1000 × 0 . 04 = 40 A /m (c)

→

The bound charge current density J m is →

→

→

J m = JT − J →

∴ That is,

→

→

| J m| = | JT | − | J| at z = 0.04 m →

| J m|at

z = 0 . 04

= 640 − 40 = 600 A / m

Example 15: Find the magnitude of magnetization in a material in which, (i) there are 8.1 × 10 28 atoms / m3 and the atoms have equal dipole moments of 17 . × 10 −33 A - m, (ii) The relative permeability is 1. 00038 and the magnetic field intensity is 0 . 31 A / m, (iii) B = 3 × 10 –5 wb / m3 and the magnetic susceptibility is − 4 × 10 − 6 . Solution: (i)

If m is the average dipole moment per unit volume and N the number of atoms per unit volume, then the magnetisation M is, M = mN amp / metre N = Atomic density = 8.1 × 10 28 atoms / m 3 and m = 1.7 × 10 −33 amp / m 2 ∴

M = 1.7 × 10 −33 × 8.1 × 1028 = 13. 77 × 10 −5 Amp / m

E lectromagnetic F ield Theory

382

(ii)

We know that M = χ m H, where χ m is the magnetic susceptibility χ m is related with relative permeability as, µ r = 1 + χ m or χ m = µ r − 1 ∴ ∴

M = (µ r − 1) H, where H = 0.31 A / m (100038 . – 1) × 0.31 or M = 0.00038 × 0.31 = 0 .1178 m /A

(iii) We know that B = µ0 H ∴ Here

M = χm

B µ0

χ m = − 4 × 10 −6 , B = 3 × 10 −5 wb/m 2 and µ0 = 4 π × 10 −7 4 × 10 −6 × 3 × 10 −5

∴

M=−

or

M = 0 . 9554 mA/m

4 × 314 . × 10 −7

= 6.9554 × 10 −3 A / m

Example 16: Region 0 ≤ z ≤ 2 m is occupied by an infinite slab of permeable material (µ r = 2 .5 ). If

→

→

→

→

B = 10 y aɵ x − 5 x aɵ y m wb / m2 within the slab, determine; (i) J (ii) J b and (iii) M.

Solution: (i)

We know that →

→

→

J = ∇ × H,

Here ∴ Hence,

→

→

B = µ H or

→

Given B = 10 y aɵ x – 5 xaɵ y and µ r = 2 . 5 →

H=

B µ

=

10 y aɵ x − 5 x aɵ y B = µ0 µ r 4 π × 10 −7 × 2.5

→

H = (0.318 y aɵ x − 0.159 x aɵ y ) × 107 → → ∂ ∂ ∂ + aɵ y + aɵz × (0.318 y aɵ x − 0.159 x aɵ y ) × 107 J = ∇ × H = aɵ x ∂y ∂z ∂x

→

∂ ∂ ∂ ∂ ( x) − aɵz (0 .318 y) + 0 .318aɵ y ( y) + 0.159 aɵ x = 107 − 0.159 aɵz ( x) x y ∂ z ∂ ∂ ∂ z ɵ ɵ ɵ ɵ ɵ ɵ ɵ ɵ ɵ ɵ (∵ ax × ax = 0, ax × a y = az , a y × ax = − az, a y × a y = 0, aɵz × aɵ x = aɵ y and aɵz × aɵ y = − aɵ x ) ∴ or or

→

J = [(− 0.159 − 0.318) aɵz + 0 + 0] 107

→

→

J = 0.477 × 107 aɵz or J = 4. 77 × 10 6 aɵ z mili-amp/m

→

J = − 4. 77 aɵ z kA / m 2

383 →

→

→

(ii) The bound current density J b = χ m J = (µ r − 1) J or (iii) We know that,

C

→

J b = (2.5 − 1) × (−4.77aɵz ) = − 1.5 × 4.77 aɵz or

→

J b = −7.155 aɵ z kA / m2

M = χ m H = (µ r − 1) H

or

M = 15 . × (0.318 y aɵ x − 0159 . x aɵ y ) × 107 × 10 −3

or

M = 4. 77 y aɵ x − 2. 385 x aɵ y kA / m

lassification of Magnetic Materials

Whenever an electric current flows a magnetic field is produced. Electron rotating around a nucleus may be regarded as tiny current loop. A small current loop has associated with it a magnetic moment. In an atom, the orbital motion and the spin of atomic electrons are equivalent to tiny current loops, individual atoms creates magnetic fields around them, when their orbital electrons have a net magnetic moment as a result of their orbital angular momentum and spin angular momentum. The magnetic moment of an atom is the vector sum of the magnetic moments of the orbital motions and spin motion of all electrons in the atom. The macroscopic magnetic properties of a material arise from the magnetic moments of its component atoms and molecules. Different materials have different characteristics in an applied magnetic field. On the basis of these characteristics, materials are classified as; diamagnetic, paramagnetic, ferromagnetic. Other classes of materials which in structure close to ferromagnetic materials but possess different magnetic effects are antiferromagnetic and ferrimagnetic.

1. Diamagnetic Materials In diamagnetic materials the orbital motion of electrons and their spin motions produce magnetic fields opposite to each other, so net magnetic dipole moment of each atom is zero. Thus, atoms do not have permanent dipole moment. In external magnetic field, the orbiting electrons either speed up or slow down in its orbit depending upon whether the induced orbital magnetic moment is antiparallel or parallel to the external applied magnetic field, but in either case the induced dipole moments form in such a way that the induced magnetic field opposes the external applied field. Therefore, the resultant magnetic induction is weakened by the presence of diamagnetic material. Thus, the materials whose atoms respond so as to reduce the magnetic field are said to be diamagnetic material. The phenomenon of reduction of resultant magnetic field inside the material medium by magnetisation is called diamagnetism and the materials which exhibit this property are called diamagnetic materials. The relative permeabilities of diamagnetic materials are slightly less than unity (µ r < 1) and their magnetic susceptibilities are negative, of the order of ~10 −5 . The materials which possess diamagnetism are: bismuth's, zinc, copper, silver, gold, lead, water, mercury, sodium chloride (NaCl), nitrogen, hydrogen etc. In fact, the magnetic moments are induced in all materials whenever they are placed in a magnetic field. Thus all materials possess the property of diamagnetism. However, in materials which have permanent dipole moments, magnetic effect due to induced dipole moments is much smaller than the effect of paramagnetism or ferromagnetism, that is why the material does not shown diamagnetism.

384

E lectromagnetic F ield Theory

2. Paramagnetic Materials In some materials the paramagnetism arise due to permanent magnetic moments created due to incomplete cancellation of orbital magnetic moments and spin magnetic moments of the electrons of the atom. In the absence of any external field, the random orientation of atoms make the overall average magnetic moment zero. But in the presence of external magnetic field a torque will act to each atomic magnet which tends to rotate the atom to align them along with applied magnetic field. Therefore, the resultant magnetic field is strengthened by the presence of these type of materials, called paramagnetic materials. Since the diamagnetic effect is still there which decreases magnetic field. Therefore, the net increase in magnetic field or magnetic induction is small. Thus, the tendency of a material to increase the magnetic field by magnetisation in materials is called paramagnetism. The relative permeabilities of paramagnetic materials are slightly greater than unity (µ r > 1) and their susceptibilities are positive, of the order of 10 −3 to 10 −5 . At low magnetic fields magnetisation I varies with H in a linear way (I ∝ H), but at very high fields the deviation from proportionality occurs. Example of paramagnetic materials are; aluminium, sodium, platinum, manganese, antimony, copperchloride, liquid oxygen, solution of salt of iron and nickel etc.

3. Ferromagnetic Materials: There are certain materials or substance the permanent atomic dipoles of which have a strong tendency to align themselves in the direction of the field in spite of randomizing nature of thermal agitation of its atoms. This property of substance is called ferromagnetism and the materials which exhibit this property are called ferromagnetic materials. A ferromagnetic materials has a magnetic moment even in zero applied field. The atoms or molecules of a ferromagnetic substances have a net intrinsic magnetic moment which is mainly due to the spin motion of the electrons. The interaction between neighbouring atomic dipoles is very strong. It is called spin exchange interaction and is present even in the absence of an external magnetic field. In fact quantum mechanical exchange interaction produce large internal fields and the neighbouring dipoles tend to align in the same direction (Fig. 16). The phenomenon of ferromagnetism arises due to both the interaction between the neighbouring atomic dipoles and the alignment of permanent dipoles in atoms or molecules that result from unpaired electrons in the outer shells. Even in the absence of an external field the ferromagnetic materials show ferromagnetism that is why permanent magnets are made from them. The ferromagnetic materials such as iron, cobalt, nickel, dysrosium, gadolinium have a perculiar electronic structure such that their outermost shell contains electrons even through the inner shell next to the outermost is still unfilled. Due to this perculiarity in structure the spin magnetic moments of ferromagnetic materials become drastically large,

Fig. 16

resulting large atomic dipole moments. The arrangement of atomic dipoles in ferromagnetic materials is shown in fig.16.

385

The relative permeabilities of ferromagnetic materials are of the order of hundreds and thousands (or µ r 0, and µ2 = 5 µ0 H /m for region (2), where z < 0. There exists a →

→

surface current of density K = 60 ^ ax − 3 ^ ay + 2^ az mT a x A /m at the boundary z = 0.For a field B1 = 2 ^ →

in region (1). Find the value of magnetic flux density B2 in region (2).

→ K

Solution: The situation of the problem is shown in fig. 21. The field in the region (1) is given as →

B1 = 2 ^ ax − 3 ^ a y + 2^ az mT

→

µ1 µ2

H2t

→

z=0

2

Fig. 21

B1n = (B1 . ^ n) ^ n = [ (2 ^ ax − 3 ^ a y + 2^ az ) .^ az ] ^ az → B1n

^ n 1

∆l

The normal component of B1 may be expressed as →

H1t

= 2aɵ z mT →

According to the boundary condition, the normal component of B at the boundary between two magnetic media is continuous that is, →

→

→

B2 n = B1n = 2 ^ az mT ∴ B 2n = 2aɵ z mT

→

Tangential component of B1 is →

→

→

→

B1t = B1 − B1n = 2 ^ ax − 3 ^ a y + 2^ az − 2 ^ az or B1t = 2 ^ ax − 3 ^ a y mT

∴

→

→

H1t =

ax − 3 ^ a y (2 ^ ax − 3 ^ a y ) × 10 −3 Tesla B1t 2 ^ = 398 .09 (2 ^ ax − 3 ^ a y) = = µ1 2 µ0 2 × 4 π × 10 −7

→

H1t = 796.18aɵ x − 1194 . 27aɵ y A / m

or →

→

→

H2 t = H1t − K × ^ n = 796 .18 ^ ax − 1194 .27 ^ a y − (60 ^ ax ) × ^ az (∵ ^ ax × ^ az = − ^ a y) = 796 .18 ^ ax − 1194 .27 ^ a y + 60 ^ ay →

H2t = 796.18aɵ x − 1134 .27 aɵ y A / m →

→

B2 t = µ2 H2 t = 5 × 4 π × 10 −7 (796 .18 ^ ax − 1134 .27 ^ ay ) = 6 . 28 × 10 −6 (796 .18^ ax − 1134 . 27 ^ ay ) = (5.0 ^ ax −712 . ^ a y ) × 10 −3 or ∴ or

→

B2 t = (5 .0 ^ ax − 7 .12 ^ a y ) mT →

→

→

B2 = B2 n + B2 t = 2 ^ az + (5 .00 ^ ax − 712 . ^ a y ) mT →

B2 = 5 aɵ x − 7.12 aɵ y + 2aɵ z mT

E lectromagnetic F ield Theory

390

Example 18: Magnetic flux line is received at an iron-air boundary at an angle of incidence 60°. Determine the angle of refraction at the boundary in air. The relative permeability of iron is 350. Solution: If θ1 is the angle of incidence and θ2 that of refraction at the boundary in air, then at the boundary tan θ2 µ r2 = tan θ1 µ r1 where µ r2 and µ r1 are relative permeabilities of air and iron respectively. µ r2 = 1, µ r1 = 350 and θ1 = 60 ° µr 1.732 1 tan θ2 = 2 tan θ1 = tan 60 ° = 350 350 µ r1

Here ∴

tan θ2 = 4 .95 × 10 −3 or θ2 = tan −1 (4 .95 × 10 −3 )

or

θ2 = 0 . 28 °

or

Example 19: When µ = µ1 = 4µ H / m, in the region-1 (z > 0), while µ2 = 7µ H / m for (z < 0 ). Moreover, →

let K = 80 ^ a x A / m on the surface, z = 0, Now if B1 = 2 ^ ax – 3 ^ ay + ^ az mT, in region1, calculate the →

value of B2 . [UPTU, B.Tech IV Sem 2003]

Solution: The situation of the problem is shown is fig. 22. The field in region (1) is given as,

→

Region I

→

B2

B1 = 2 ^ ax – 3 ^ ay + ^ az mT →

→

= [(2 ^ ax – 3 ^ a y +^ az ).(– ^ az )] (– ^ az ) = ^ az mT ∴

n

→

The normal component of B1 is, B1n = (B1. ^ n) ^ n

Region II

→

B1n = ^ az mT

According to magnetic boundary condition, the normal component of →

B at the boundary between two regions is continuous, that is, →

→

B2 n = B1n = ^ az mT The tangential component of B1 is →

→

→

B1t = B1 – B1n = 2 ^ ax – 3 ^ a y +^ az – ^ az = 2 ^ ax – 3 ^ a y mT →

∴

(2 aɵ x – 3 aɵ y )10 –3 B H1t = 1t = = 500 aɵ x – 750 aɵ y A / m µ1 4 × 10 –6

Thus,

H2 t = H1t – K × ^ n = 500 aɵ x – 750 aɵ y – (– aɵz ) × 80 aɵ x

→

→

→

→

= 500 aɵ x – 750 aɵ y + 80 aɵ y

→

B2 Fig. 22

391 →

or

H2 t = 500 aɵ x – 690 aɵ y A / m

and

B2 t = µ2 H2 t = 7 × 10 –6 (500 aɵ x – 690 aɵ y )

or

B2 t = 3.5 aɵ x – 4.69 aɵ y mT

→

→

→

∴

→

→

→

B2 = Bn + B2 t = aɵ z + 3.5 aɵ x − 4.69 aɵ y mT

I

nductors and Inductance

An inductor, sometimes called an inductance, is a conductor and is useful for storing magnetic energy just as a capacitor is used for storing electrical energy. Inductors are often connected in series and in parallel, and sometimes with capacitors and resistors. It is a practical fact that an inductor always has a certain internal resistance. Typical examples of inductors are loops, coils, solenoids, toroids, co-axial transmission lines and parallel wire transmission lines. An inductor is such a parameter in electrical circuits, the value of which depends on the physical dimensions and the type of magnetic material used.

1. Self- Inductance In order to understand the function and the affect of using an inductor in an electrical circuit, consider a solenoid coil carrying a current of I amp. We know that, when a current flows through a coil or circuit, it produces magnetic flux which in linked with the coil. If the current through the coil is changed, the flux linked with the coil also changes due to its own magnetic field. According to the Faraday's laws of induction, this change is flux produces induced e.m.f. in the coil which opposes the change in flux or change of current in the circuit. The phenomenon of producing an induced e.m.f., in a coil or in an electric circuit itself is known as self induction. The induced e.m.f., so produced in the coil or circuit is called self induced e.m.f. or back e.m.f. I

I

K

K K

Current Increasing

Current Decreasing

Fig. 23

If a coil carrying a current of I amp has N turn and φB be the magnetic flux linked with each turn of the coil. then, the number of flux linkages is NφB . If no magnetic materials are present near the coil, then the total flux linkages with the coil is proportional to the current I, That is, NφB ∝ I NφB = LI

E lectromagnetic F ield Theory

392

where L is the constant of proportionality and is called the self inductance or inductance of the coil. The inductance or self inductance is also defined as the ratio of the total flux linkage to the current which they link. L=

NφB I

This definition is applicable only for linear magnetic media. The unit of inductance is the henry (H) equivalent to one weber-turn per ampere.

2. Mutual Inductance Instead of a single coil (or circuit), if there are two coils (or circuits) very near to each other. Then the flux changes takes place is one coil due to the current change in other coil (or circuit), then mutual inductance arises.

K Key is Pressed

K Key is opended Fig. 24

Whenever the current flowing through a coil (or circuit) changes, the magnetic flux linked with a neighbouring coil also changes. Hence an emf is induced in the second coil. This phenomenon of electromagnetic induction is called mutual induction. The coil in which current changes is called the primary coil and the second in which induced emf is set up is called the secondary coil. Let a current I in a primary coil produces a magnetic flux φB is each turn of the secondary coil. If N S be the total number of turns in secondary coil, then the total flux linked with secondary coil be N S φB . This flux linkage through the secondary coil is proportional to the current I flowing in the primary coil. That is, N S φB ∝ I

or N S φB = MI

where M is the proportionality constant and is called the mutual inductance of the two coils. that is, N φ M= S B I Induced emf produced in the secondary is d(N S φB ) d e=– = – ( MI ) dt dt dI or e = –M dt e or M= dI / dt

S P I Fig. 25

393

The mutual inductance of two circuits or coils is related to their self inductances. The ratio of mutual inductance (M) of two circuits and the square root of the product of their self inductances L1 and L2 is knows as coefficient of couple k. that is, k =

M Henry L1 L2

Coefficient of compling k, lies between '0' and '1' and depends upon the geometry of the two circuits and their relative position. Usually the value of k is less than '1' (k < 1). If k = 1, then the entire flux produced by the primary coil is linked with secondary, without any leakage of flux. In this case M =

L1 L2 , this is the

maximum value of mutual inductance between two coils or circuits. If k = 0, it means M = 0, that is, there is no compling between two coils. In actual practice the value of k is less than one.

I

nductance or Self Inductance of a Long (or Infinitely Long) Solenoid

A solenoid is a coil of wire wound on the surface of a hollow cylinder of a card-board or china-clay and has a length very large compared to its diameter. When a current is passed through such a coil a magnetic field is produced inside the coil parallel to its axis. If such a long air-cored soleonid of length l meter has N, closely wound turns, then the magnetic field produced by it is. N B = µ0 I weber / meter2 l where I is the current flowing through the coil and µ0 is the permeability of free space. The magnetic flux linked with a single turn of the solenoid is, → →

φB = ∫ B. dS = B. A where

→

∫ dS = A is the area of cross- section. Assuming the field to be uniform the flux linked with each

turn of the solenoid is φB . The total magnetic flux linked with N turn of the solenoid is given by φ = NφB = NBA or

µ N2 I A N φ = N µ0 I . A = 0 l l

But, according to the definition of inductance,

or

L=

Total Magnetic flux φ µ0 N 2 IA = = current I lI

L=

µ0 N 2 A Henry l

If, The number of turn per unit length of the solenoid be n, then

E lectromagnetic F ield Theory

394

N = nl µ (nl)2 A L= 0 l L = µ0 n2 l A Henry

or or

If the solenoid is wound on a core of a constant permeability µ, then the total flux through the entire turns of the solenoid is µN 2 IA / l and the inductance becomes, L=

I

µN 2 A l

or L = µ n2 l A

nductance (or Self Inductance) of a Toroid

If a solenoid is bent round in the form of a closed ring and their ends are joined, we get a toroid. Toroid may also be called endless solenoid. The magnetic field of a toroid is confined only within the core. Fig. 26 shows a toroidal coil of mean radius R and having circular cross-section. If B be the magnetic field produced inside the coil due of flow of current I on the circular path, then according to Ampere's circuital law, → →

∫ B. dl = µ × total current enclosed by the path. = µ0 NI where N be the total number of turns and µ0 the permeability constant. → →

∫ B. dl = B × 2πR ∴

B × 2 πR = µ0 NI

or

B=

µ0 NI 2 πR

The magnetic flux linked with a single turn of the toroid is φB = B × A where A is the area of cross-section. The total flux linked with all N turns of the toroid is given by φ = NφB = NBA or

µ N 2 IA µ NI φ=N 0 A= 0 2 πR 2 πR

According to the definition of φ inductance L=

total magnetic flux φ µ N2 A = = 0 current I 2 πR

395

µ0 N 2 A Henery 2 πR

L=

or

If the toroid be wound on a care of permeability µ, then the inductance of toroid becomes. µ N2 A Henery 2π R

L=

If the toroid has a circular cross- section of radius r, then area of cross-section, A = πr 2 ∴

L=

µ N 2 (πr 2 ) . 2π Ρ

or

L=

µ N 2 r2 2R

I

nductance (or Self Inductance) of Two Parallel Wires I amp

Let us consider two parallel wires P and Q carrying a current of I amp in

P

opposite directions as shown in fig. 27, and are

x

separated by a distance d. Let O be the point in a space between

d

the wires distant x from wire P and distant d – x from wire Q. The magnetic field at O due to current carrying wires will be in

dx

r d–x

the same direction. The magnitude of resultant magnetic field at O due to both wires is given by B=

l

.O

I amp

µ0 I µ0 I µ I 1 1 + = 0 + 2 πx 2 π (d – x) 2 π x (d – x)

Q

Fig. 27

Let us consider a small region between two wires at O of length l and width dx. The magnetic flux through this small region is dφB = BA =

µ0 I 1 1 + (l dx) 2 π x (d – x)

The total flux through the entire region between the wires is φB = ∫

x =d – r x =r

dφB =

µ0 I l 2π

x =d – r

∫ x =r

1 1 x + (d – x) dx

or

φB =

µ0 I l [log e x – log e (d – x)]rd – r 2π

or

φB =

µ0 I l {log e (d – r ) – log e r – log e [d – (d – r )] + log e (d – r )} 2π

or

φB =

2µ0 I l [log e (d – r ) – log e r ] 2π

or

φB =

µ0 I l d – r log e r π

By the definition of inductance,

E lectromagnetic F ield Theory

396

φB µ0 l d – r log e = r π I

L=

I

nductance (or Self Inductance) Per Unit Length of a Co-axial Cable

Let us consider an infinitely long co-axial cable of inner radius 'a ' and outer radius ' b ' as shown in fig. 28. Suppose uniformly distributed current I is flowing in the inner conductor along +z axis. Consider a point →

distant ρ between the two conductors. The magnetic induction B between the conductors at a distance ρ is, →

B=

µ0 I aɵ (a < ρ < b) 2 πρ φ

Consider now l meter length of the co-axial cable, the magnetic flux linked with an area ds (= dρ dz aɵφ ) is → →

φB = ∫ ∴ or

S

B. ds

b → →

φB = ∫

l

0

∫a

φB = ∫

l

b

0

∫a

z

B. ds = ∫

l

0

b

∫a

µ0 I aɵ .(dρ dz aɵφ ) 2 πρ φ

µ0 I µ I d ρ dz = 0 2 πρ 2π

l

b

∫0 dz ∫a

dρ ρ

ρ I

I

b a

µ0 I b × l × [log e ρ]a 2π µ I b φB = 0 log e a 2π =

or

Fig. 28

According to the definition of inductance, Total magnetic flux φB = current I µ0 l b L= log e Henery a 2π L= ∴

Inductance per unit length of the cable is L=

µ0 b log e H / m a 2π

M utual Inductance Between two Coaxial Solenoids (or Coils) Consider a long air-cored solenoid of area of cross-section A and

Primary

Secondary (S) P

having np turns per unit length of the solenoid. Let a shorter secondary coil S having N S turns wound closely over the central portion of the primary coil P as shown in fig. 29. When a current of I amp flows through the primary of the solenoid, then the magnetic field produced inside the primary coil is. B = µ0 np I w b / m2

NS Fig. 29

397

Therefore, the magnetic flux linked with the primary of the solenoid is also linked with the secondary coil because the secondary coil S is wound closely over the central portion of the primary. Thus, the magnetic flux linked with each turn of the secondary is given by φB = BA = µ0 np IA weber Therefore, the total magnetic flux linked with N S turns of the secondary is given as. N S φB = µ0 np N S IA weber- turns By definition the mutual inductance of the two solenoids is M=

N S φB µ0 np N S IA = I I

M = µ0 n p N S A

or

Inductances in Series When two coils of inductances L1 and L2 are connected in series with large separation and a current of I amp is passed through their combination, then the total flux linked with the combination is the sum of the flux linked with each coil, that is, φtotal = L1 I + L2 I If L be the equivalent inductance of the combination, then φtotal = L I ∴

LI = L1I + L2 I

or

L = L1 + L2

This is true for the two coils which are separated by such a large distance that there is no flux linkage in any coil due to the current in other. When the separation between the coil is small, there will be a mutual inductance between them. In this situation, the resultant emf induced in the combination of coils is the sum of emf induced in the respective coils. If e is the resultant induced emf in the combination and e1 and e2 are the induced emf in the respective coils, then e = e1 + e2 dI dI dI dI – M + – L2 –M = – L1 dt dt dt dt or

e = (– L1 – L2 – 2 M)

dI dt

If the coils are so placed that the flux linking each coil due to the current in itself is in the same direction as flux linking due to the current in the other coil or if the flux is in the same sense, then e = –L

dI dt

Where L is the total inductance of the combination.

E lectromagnetic F ield Theory

398

∴

–L

dI dI = (– L1 – L2 – 2 M) dt dt L = L1 + L2 + 2 M

or

If the coils are so arranged that the flux linking each coil due to its own current is in the opposite direction to the flux linking due to the current in the other coil or if the flux linkage is in the opposite sense, then. L = L1 + L2 − 2 M

Inductances in Parallel Let two coils of inductances L1 and L2 are connected in parallel and a current I is divided between them as shown in fig. 30., then the total current, I = I1 + I2 or

dI dI1 dI2 = + dt dt dt

When the currents through the inductors were increasing, emf 's were induces in them. As the potential difference across each coil is the same, the induced emf must also be the same. e = – L1

that is,

dI1 dI = – L2 2 dt dt

I1

L1

If L is the resultant inductance, then e = –L

I2

dI dt

I

or

e dI =– L dt

or

dI dI e dI d =– = – (I1 + I2 ) = – 1 – 2 L dt dt dt dt

∴

e e e = + L L1 L2

or

1 1 1 = + L L1 L2

L2

I

Fig. 30

or L =

L1 L2 L1 + L2

If the separation is small and there is a mutual inductance M between the coils, then the resultant inductance becomes. L=

L1 L2 – M2 L1 + L2 ± 2 M

If the flux linkage is in the same sense, the sign in the denominator is positive and it is negative if the flux linkage is in opposite sense.

399

Co-efficient of Coupling To derive the expression for co-efficient of coupling in terms of mutual inductance and self inductance of two coils, it is imagined that two coils are associated with different magnetic circuits. Let us suppose that coil 1 has N1 turns with self inductance L1 and coil 2 has N2 turns with self inductance L2 If I1 is the current flowing through coil 1 and I2 through coil '2' , then L1 =

N1φ1 I1

and L2 =

N2 φ 2 I2

...(1)

where φ1 is the flux produced in coil '1' by the current I1 and φ2 that produced in coil '2' by the current I2 . If a fraction k1 of flux φ1 in coil '1' is linking with the coil '2' and similarly fraction k2 of φ2 is linking with coil '1', then Mutual inductance,

M12 =

N1φ12 I2

where φ12 is the flux linking with coil '1' produced by the coil '2'. thus, φ12 = k2 φ2 ∴

M12 =

N1k2 φ2 I2

...(2)

similarly if φ21 is the flux linking with coil '2' produced by the coil '1', then φ21 = k1 φ1 Hence

M21 =

N2 φ21 M2 k1 φ1 = I1 I1

...(3)

Multiplying eqn. (2) and (3), we get M12 M21 = Substituting

k1 k2 N1N2 φ1 φ2 I1I2

...(4)

N1 φ1 N φ = L1 and 2 2 = L2 from eqn. (1) in eqn. (4), we get I1 I2 M12 M21 = k1 k2 L1 L2

...(5)

Assuming k1 = k2 = k and M12 = M21 = M, then eqn. (5) becomes M2 = k2 L1 L2 or

M = k L1 L2 or k =

M L1 L2

where k is the coupling coefficient. Example 20: A very long solenoid with 2 × 2 cm cross-section has an iron core (µ r = 1000 ) and 4000 turns /m. If it carries a current of 500 mA. find its self inductance per meter. [UPTU, B.Tech.IV. Sem 2008]

Solution: The self inductance of a solenoid of length l, cross-sectional area A and having N turns is given by

E lectromagnetic F ield Theory

400

L=

µ N2 A l

Here µ = µ0 µ r = 4 π × 10 –7 (web/amp–m) ×1000 , N = 4000, A = 2 × 2 × 10 –4 m2 and l = 1m 4 π × 10 –7 × 1000 × (4000)2 × 2 × 2 × 10 –4 1

∴

L=

or

L = 8.038 henery

Example 21: A solenoid having a core of cross-section 4 × 10 –4 m2 , half air and half iron (relative permeability 1000), is 20 cm long. If the number of turns on it is 1000, what will be its inductance. Solution: If A1 and A2 are areas of cross-section of the air and iron cored part of the solenoid then the inductance of length l of the N turned solenoid is L= Here

µ0 N 2 ( A /2) µ N 2 ( A /2) + l l

N = 1000, l = 20cm =0.2m, A = 4 × 10–4 m2 ,µ r = 1000 and µ = µ0 µ r = 4 π × 10 –7 × 1000

∴

L=

µ0 N 2 A (1 + µ r ) 2l

or

L=

4 × 314 . × 10 –7 × (1000)2 × 1001 × 4 × 10 –4 2 × 0 .2

henery L = 1257 .

Example 22: Calculate the inductance of 10 cm long co-axial cable filled with a material for which µ r = 80 having inner and outer diameters as 1mm and 4mm respectively. Solution: The inductance per unit length of a co-axial cable of inner radius 'a ' and outer radius 'b' is. µ µ l L µ b b log e or L = 0 r log e = a a 2π l 2π Here ∴

µ0 = 4 π × 10 –7, µ r = 80, l = 10 cm =10 × 10 –2 m, b = 4 mm and a = 1mm L=

4 π × 10 –7 × 80 × 10 × 10 –2 4 log e 1 2π

= 2 × 10 –7 × 80 × 10 × 10 –2 [log e (4) – log e (1)] = 16 × 10 –7[1.3863 – 0] or

L = 22.18 × 10 –7 henery

401

Example 23: A toroidal core of fig.31. has r0 = 10 cm and a circular cross-section with a = 1cm. If the core is made of steel (µ = 1000 µ0 ) and has a coil with 200 turns, calculate the amount of current that will produce a flux of 0.5m wb in the core. [UPTU, B.Tech IV Sem 2008]

Solution: Let φB be the magnetic flux linked due to a current I and

φB

N be the number of turns in the coil, the total flux NφB linked with the coil is, NφB NφB = LI or I = L We know that, the inductance L of the toroid is L= ∴

I=

0

I

2a

µN 2 A 2 πR NφB (2 πR ) 2

µN A

r0

200 turns 0

=

φB × 2 πR µ NA

Fig. 31

Area of cross-section, A = π a2 = 314 . × (1 × 10 –2 )2 = 314 . × 10 –4 m 2 Here φB = 0.5 mwb, N = 200, µ = 1000µ0 = 10004 π × 10 –7and R = 10cm =10 × 10 –2 m 0.5 × 10 –3 × 2 π × 10 × 10 –2

∴

I=

or

I = 3.98 amp

. × 10 –4 1000 × 4 π × 10 –7 × 200 × 314

Example 24: A toroidal solenoid of 0 .4 cm 2 cross-sectional area has 1200 turns and an average radius of 10cm . A coil of 50 turns and 5cm 2 cross-sectional area is wound on a part of the toroid. Find the →

mutual inductance. Assume B to be uniform. Solution: The magnetic field produce within the air core of the toroid is B=

µ0 N p I 2 πR

where R is the average radius of the toroid and N p is the total number of turns. The secondary coil whose cross-sectional area AS is larger than the cross-sectional area Ap of the toroid, is placed in this field which is uniform and perpendicular to its area. The magnetic flux linkage through the coil is µ0 N p I Ap N S φB = N S B Ap = N S 2 πR The mutual inductance is therefore,

E lectromagnetic F ield Theory

402

M=

N p N S Ap N S φB = µ0 I 2 πR

Here N p = 1200, R = 10 cm =0.10m, N S = 50, Ap = 4 cm2 = 4 × 10 –4 m 2 and µ0 = 4 π × 10 –7wb/A–m ∴

1200 –4 M = (4 π × 10 –7) (50) × 4 × 10 2 π × 010 .

or

M = 4.8 × 10 –5 henery

Example 25: Determine the mutual inductance between an infinitely long straight conductor along y− axis and a rectangular single turn coil situated in x − y plane with its corners located at point (a,0 ), (a + d,0 ), (a, h) and (a + d, h).

Y

Solution: The relative position of the infinitely long straight conductor along y −axis and rectangular coil is depicted in fig.

(a, h)

But

dy

I

current I in the straight conductor is given by φB = MI or M =

(a + d, h) P

32. The magnetic flux linked with the rectangular coil due to the

dx

φB I

(a, 0) (a + d, 0)

→ →

φB = ∫ B. dS

The flux density produced by the conductor at a point P in the

wire

Z

Fig. 32

x − y plane is →

B=

µI aɵz 2 πx

→

For the coil placed in X – Y plane, dS = dy aɵz y= h

x =a +d

h

a +d

∫0 dy∫a

dx x

φB = ∫

∴

φB =

µI µI a+ d h.[log e (a + d) – log a] [ y]0h [log e x]a = 2π 2π

or

φB =

µ Ih a+ d log e 2π a

or

M=

φB µ h a + d log e = a 2π I

∴

M=

µh a + d log e a 2π

y =0

∫ x =a

µI µI dx dy = 2 πx 2π

∴

X

403

Example 26: An iron ring of relative permeability 100 is wound uniformly with two coils of 100 and 400 turns of wire. The cross-section of ring is 4 cm 2 and the mean circumference is 50 cm. Calculate (a) self inductance (b) mutual inductance, (c) total inductance when the coils are connected is series in the same sense and also in opposition. Solution: (a) The self inductance of an iron ring or toroid is L=

µ0 µ r 2π

N 2 A r

If L1 is the self inductance of a coil of N1 turns of wire and L2 that of a coil of N2 turns of wire, then L1 =

µ0µ r N12 A µ µ N2 A and L2 = 0 r 2 2π r 2 πr

Here, µ r = 100, A = 4 cm 2 = 4 × 10 –4 m, 2 πr = 50 cm =50 × 10 –2 m, N1 = 100, N2 = 400 and µ0 = 4 π × 10 –7 ∴

L1 =

and

L2 =

4 π × 10 –7 × 100 × (100)2 × 4 × 10 –4 50 × 10 –2 4 π × 10 –7 × 100 × (400)2 × 4 × 10 –4 50 × 10 –2

= 1.005 × 10 –3 henry = 16.08 × 10 –3 henry

Mutual inductance of two coil M= =

µ N1N2 A l 4 π × 10 –7 × 100 × 100 × 400 × 4 × 10 –4 50 × 10 –2

= 4.02 × 10 –3 henry

(c) Total inductance if the two coils are connected in series and flux of same sense, that is L = L1 + L2 + 2 M = 1005 . + 16.08 + 2 × 4.02 or

L = 25.125 mh

Total inductance if the two coils are connected in series but flux of opposite sense, that is, L = L1 + L2 – 2 M = 1005 . + 16.08 – 2 × 4.02 or

L = 9.045 mh

E nergy Stored in a Magnetic Field An inductor stores magnetic energy when a current through an inductor coil is gradually increases from zero to its maximum value I. when a current builds up in a circuit containing an inductor of inductance L. The magnetic flux linked with the circuit changes and therefore a self induced e.m.f. or back e.m.f. is induced in the circuit which opposes the growth of current. The energy is needed to over come this opposition or the work is done in building up the current in the circuit from zero to a maximum value I0 , that is

E lectromagnetic F ield Theory

404

W =∫

I0 0

Pdt joule, where P = – eI

...(1)

where e is the induced e.m.f. produced in the circuit. We know that, dI e = –L dt dI dI P = – – L I = LI ∴ dt dt I0 I0 dI dt = ∫ LI dI W =∫ LI ∴ 0 0 dt W=L

or

I02 joule 2

...(2)

This work (which is supplied to the inductor) is used to establish the magnetic field around the inductor where it is stored as potential energy U. Therefore, U =

1 L I02 2

...(3)

when the current is cut off by switching off the battery, the magnetic field collages and the energy is recovered by the circuit. We know that the magnetic flux φB linked with the inductor of inductance L is, φB = L I0 W=

∴

1 φ I 2 B 0

...(4)

From eqn. (3), we have L=

2U I02

If I0 = 1 amp., then L = 2U = 2W Therefore, the inductance of an inductor is twice the work required to establish one ampere of current in the inductive circuit.

E

nergy Density in a Magnetic Field

In order to derive an expression for energy stored per unit volume (that is, energy density) in the magnetic field established around an inductor, consider a very long air cored solenoid of length l and cross-sectional area A. When a current of I amp flows in the solenoid, a magnetic field is established which is uniform every where inside, and is negligible outside the solenoid coils. The total energy stored in the magnetic field corresponding to the current I in the solenoid of inductance L is given by,

405

U =

1 L I2 2

The inductance L of a solenoid of length l and area of cross section A is given by, L=

µ0 N 2 A l

where µ0 is the permeability constant and N is the total number of turns in it. 1 µ0 N 2 A 2 I l 2

U =

∴

If n is the number of turns per unit length of the solenoid, then the magnetic field inside the solenoid is given by N B = µ0 n I or B = µ0 I l Hence

U =

2 2 2 1 µ0 N I . Al 2 µ0 l2

or

U =

1 B2 1 µ0 NI Al Al = 2 µ0 l 2µ0

2

where Al is the volume of the solenoid in which the energy is uniformly distributed. Therefore, the energy per unit volume or energy density u in the magnetic field is. U 1 B2 (Joules/meter3 ) = Al 2 µ0

u=

The energy density in term of magnetic flux density H is given by u=

1(µ0 H)2 2µ0

u=

or

1 µ H2 2 0

If the magnetic field is not uniform every where inside the solenoid or throughout the entire volume, the total energy stored in the volume V of the magnetic field is given by U=

1 2 µ0

∫V

B2 dV

Although the above expressions are derived for the special case of solenoid, but these hold good for all configuration of magnetic field.

E

nergy Density in Terms of Magnetic Vector Potential →

The energy density or energy per unit volume of the stored energy in a magnetic field B is expressed as, u=

U 1 B2 = V 2 µ0

...(1)

E lectromagnetic F ield Theory

406

where µ0 is the permeability constant. Eqn. (1) can also be written as, U =

→ →

1 2 µ0

∫v ( B. B) dv

→

The magnetic vector potential A is expressed as →

→ →

B = ∇× A ∴

U =

1 2µ0

→ → →

∫v

B.( ∇× A) dv

....(2)

According to vector identity, → →

→

→

→

→

→

→

→ →

→ → →

→ → →

div ( A × B) = B. curl A – A. curl B →

→

B. curl A = A. curl B + div ( A × B)

or

→ → →

B.( ∇× A) = A.( ∇× B) + ∇.( A × B)

or

...(3)

→ → →

Substituting the value of B.( ∇× A) from eqn. (3) in eqn (2), we get U =

1 2µ0

∫v

→ → →

→ → →

A.( ∇× B) + ∇.( A × B)] dv

→ → → → → → 1 ∫ A.( ∇× B) dv + ∫ ∇.( A × B) dv U = v 2µ0 v

or

...(4)

According to Gauss-divergence theorem,

∫v

→

div A dv = ∫

→ → S

→ → →

A. dS → →

→

∫v ∇.( A× B) dv = ∫S ( A× B). dS

→

→

→

(∵ Here A = A × B)

....(5)

→ → →

Substituting the value of ∫ ∇.( A × B) dv from eqn. (5) in eqn. (4), we get v

U =

→ → → 1 → → → A.( ∇× B) dv + ∫ ( A × B). dS ∫ S 2 µ0 v

...(6)

→

If J is the current density, then we know that, →

→

→

→ →

→

∇× H = J or ∇× B = µ0 J

...(7)

→ →

Substituting this value of ∇× B from eqn. (7) in eqn. (6), we get U=

→ → → → 1 → A.(µ0 J ) dv + ∫ ( A × B). dS S 2 µ0 ∫ v

...(8)

407

The surface integral,

→ →

→

∫S ( A× B). dS

will become zero, as volume (v) would include all the space, then

surface tends to infinity, that is,

∫S

→ →

→

( A × B). dS = 0

S→ ∞

Thus, the eqn. (8) becomes → → 1 ( A. J ) dv ∫ 2 v

U =

→

Hence, the expression of stored energy in a magnetic field in terms of vector potential A is U =

I

→ → 1 ( A. J ) dv 2 ∫v

nductance of a Coil (or Circuit) in Terms of Magnetic Vector →

Potential A

→

The inductance L of a coil (or a circuit) in terms of energy stored in a magnetic field B is expressed as, U =

1 L I2 2

....(1)

where I is the total current flowing in the coil. L=

or

2U

....(2)

I2 →

The energy density or energy stored in a magnetic field B is given by, u= U =

or

U 1 B2 = V 2 µ0 → →

1 2µ0

∫ v ( B. B) dv

...(3)

Substituting the value of U from eqn (3). in eqn. (2), we get L=

→ →

1 µ0 I

2

∫v ( B. B) dv

....(4) →

→

The magnetic vector potential A is related with magnetic induction B as, →

→

→

B = ∇× A →

Substituting this value B from eqn (5) in eqn. (4), we get

....(5)

E lectromagnetic F ield Theory

408

1

L=

µ0 I 1

L=

or

I

2

∫v

2

∫v

→ → →

B.( ∇× A) dv

→ → →

H.( ∇× A) dv

...(6)

According to vector identity →

→

→

→ →

→

div ( A × B) = B. curl A – A. curl B →

→

→ →

→

→ → →

B. curl A = A.( ∇× B) + ∇.( A × B)

or

→ → →

→ →

→

→ →

→

H.( ∇× A) = A.( ∇× H) + ∇.( A × H)

or

...(7)

→ → →

Substituting the value of H.( ∇× A) from eqn. (7) in eqn. (6), we get L=

I

→

→ →

→

∫v [ A.( ∇× H) + ∇.( A× H)] dv

2

→ → → 1 → → → A.( ∇× H) dv + ∫ ∇.( A × H) dv 2 ∫v v I

L=

or

→ →

1

...(8)

According to Gauss-Divergence theorem →

→ →

∫ v div A dv =∫S →

→

→

→

→

A. dS →

→

→

∫ v ∇.( A× H) dv =∫S ( A × H). dS

∴

→

→

→

( Here, A = A × H)

...(9)

→

Substituting this value of ∫ ∇.( A × H) dv from eqn. (9) in eqn. (8), we get, v

→ → → 1 → → → A.( ∇× H) dv + ∫ ( A × h ). dS 2 ∫v S I

L=

The surface integral,

→

→

→

∫S ( A× H). dS = 0,

....(10)

Since the surface, S encloses the volume, v, containing all the →

→

energy stored in a magnetic field, and this needs that A and H be zero on the bounding surface. Therefore, eqn. (10) reduces to, L=

1 I

2

→ →

→

∫ v A.( ∇× H)dv

...(11)

→

If J is the current density, then →

→

→

∇× H = J

Therefore, eqn. (11) modifies as, L=

1 I

2

→ →

∫v ( A. J ) dv

...(12)

409

Since the current density exists only within the coil or conductor, the integrand is zero at all points →

outside the conductor. The magnetic vector potential is one that arises only due to the current J . Thus →

the contribution of orginal current density to vector potential A is to be ignored →

→

The magnetic vector potential A due to current density J is given as, →

→

A=∫

v

µ0 J dv 4π R

...(13)

The inductance L may therefore be expressed as, L=

1 I2

∫v

→ µ J → ∫ v 4 πR . J dv

...(14)

A slightly similar expression is obtained by considering current filaments of small cross-section for which →

→

J dv may be replaced by I dL and the volume integral by a closed line integral along the axis of the filament.

∴

L=

or

L=

1 I2 µ0 4π

∫

→ µ0 I dL → ∫ 4 πR . I dL

∫

→ → dL ∫ R . dL

Example 27: A coil of 500 turns carrying a current of 10 amp. produces a magnetic flux of 10 weber. calculate the energy stored in the magnetic field. Solution: Suppose φB be the magnetic flux linked with each turn of the coil due to a current I and N be the total number of turns in the coil. If L be the inductance of the coil, then total flux linked with the coil is, NφB = L I or L = Here ∴

NφB I

φB = 10 weber, I = 10 amp,and N = 500 L=

500 × 10 = 500 henery 10

We know that the energy stored in the magnetic field. U = or

1 1 L I 2 = × 500 × (10)2 2 2

U = 25000 joule

E lectromagnetic F ield Theory

410

Example 28: What must be the strength of a uniform electric field, if it is to have the same energy density as that possessed by a 5000 gauss magnetic field? Solution: We know that the magentic energy density uB =

1 B2 2 µ0

Here B = 5000 gauss =5000 × 10 –4 wb /m 2 uE =

The electric energy density,

1 ε E2 2 0

According to the given problem uB = uE or E2 =

or But

1 B2 1 = ε E2 2 µ0 2 0

B2 B or E = µ0 ε0 µ0 ε

1 =c ε0µ0

∴

E = cB = 3 × 108 m /s × 5000 × 10 –4 wb /m 2

or

E = 15 . × 10 8 volt/m

Example 29: A long co-axial cable consists of two concentric cylinders with radii 'a' and ' b '. A steady current I is sent through the central conductor which returns through the outer conductor. (a) Calculate the energy stored in the magnetic field for a length l of the cable. (b) Calculate the inductance of a length l of this co-axial cable. Solution: (a) Consider a long co-axial cable consisting of two concentric cylinder with radii 'a' and ' b' such that b > a as shown in fig. 33. In order to find the energy stored in the space between the two →

concentric cylinder, we first determine the magnetic field B in the space between the two cylinders at a point P. Let us consider a co-axial cylindrical shell through the point P of radius 'r' and width dr as shown in fig. 33. Applying Ampere's law for this shell as,

r

→ →

∫ B. dl = µ0 I Where I is the current enclosed by the shell. Since the magnetic field is → →

uniform and parallel to the elemental length dl, B can be taken outside the integral as, →

l b

a

→

B ∫ dl = µ0 I or B (2 πr ) = µ0 I

or

B=

µ0 I 2 πr Fig. 33

dr P

411

Therefore the magnetic energy stored in the space between the conductors is given by 1 B2 2 µ0

u=

or

u=

∴ u=

2

1 (µ0 I /2 πr ) µ0 2

µ0 I 2

8π 2 r 2

The energy dU B stored in the shell of radius r, width dr and length l is dU B = energy density × volume of elemental shell µ0 I 2

=

or

8π 2 r 2

.(2 πrl) dr

µ0 I 2 l dr 4π r

dU B =

The total magnetic energy stored in the entire space between the two cylinders is, U =

or

U =

µ0 I 2 l 4π

b

∫a

dr µ0 I 2 l b = [log e r ]a 4π r

µ0 I 2 l b log e a 4π

(b) The magnetic energy stored in a cable of length l and self inductance L, when I current flows through it is. U =

∴

L=

2U 1 L I 2 or L = 2 2 I 2µ0 I 2 l 4 πI

2

log e

µ l b b or L = 0 log e a 2π a

E lectromagnetic F ield Theory

412

Q uestion Bank 1.

Determine the magnetic force acting on a moving charge.

2.

State and explain Lorentz force and Lorentz force equation.

3.

Derive an expression for magnetic force on a differential current element as well as on a current

[UPTU, B.Tech IV Sem (old) 2007]

carrying conductor. 4.

State and explain Ampere's circuital law in electromagnetism.

5.

Find the element of force dF12 caused by the current element I2 dl2 on the current element I1dl1, and

→

→

also the force dF12 on the current element I2 dI2 caused by I1dl1. Show that they are unequal. 6.

Show that two parallel wires carrying currents in same direction attract each other while those carrying currents in opposite directions repel each other. Compute the magnitude of this force.

7.

Explain the magnetic torque and moment in details.

8.

Derive an expression for a torque on a differential current loop as well as on a current carrying loop.

9.

Discuss magnetic dipoles and magnetisation in materials.

[U'khand. B.Tech. IV Sem. 2011]

10. How the magnetic materials are classified into dia, para and ferro-magnetic materials on the basis of their behaviour in magnetic field. 11. Explain the classification of magnetic materials.

[UPTU, B.Tech. IV Sem 2008]

12. Discuss the boundary conditions for magnetic field.

[UPTU, B.Tech. IV Sem 2004, 2005]

13. Explain the boundary conditions for static magnetic field.

[U'khand, B.Tech. III Sem 2009]

→

14. Prove that the normal component of B is continuous across a boundary between two isotropic and homogeneous materials with permeabilities µ1 and µ2 . What can be said about H1n and H2 n in the above case ?

[UPTU, B.Tech IV Sem. 2003]

15. State and explain magnetic boundary conditions.

[GBTU, B.Tech III Sem. 2011]

16. Define inductance, mutual inductance and coefficient of coupling.

[GBTU, B.Tech. III Sem 2010]

17. What do you mean by self inductance and mutual inductance of a circuit. Derive an expression for the inductance of a solenoid and toroid. 18. Determine the self inductance per unit length of the long solenoid.

[GBTU, B.Tech. IV Sem 2011]

19. Calculate the self inductance per unit length of an infinitely long solenoid.

[GBTU. B.Tech. III Sem 2011]

20. How will you determine the self inductance per unit length of a co-axial cable. 21. Find an expression for self inductance of two parallel wires. 22. Derive an expression for magnetic energy in terms of field quantity. 23. What you mean by energy stored in a magnetic field.

[UPTU, B.Tech. IV Sem (old) 2007]

[UPTU, B.Tech. IV Sem 2002, 2004, 2005] →

24. Show that the energy stored in a magnetic field of strength H is

1 µH2 . 2

[UPTU, B.Tech. IV Sem 2003]

413

nsolved Numerical Problems 1.

Compute the magnetic force on a wire one meter long and carrying a current of 10 amp when placed in a uniform magnetic field of magnetic induction 1.5 wb/m2 making an angle 30° with the direction of field.

2.

A straight conductor placed along x-axis carries a current of 200 amp in positive x-direction. If a uniform magnetic field B = 1 tesla exist every where in direction parallel to the x − y plane at an angle 45° with x-axis determine the magnitude and direction of force per unit length of the conductor ?

3.

A conductor of length 2.5 m located at z = 0, x = 4 m carries a current of 12 amp in aɵ y direction. Find the magnetic flux in the region if the force on the conductor is 1.2× 10 –2 N in the direction (– aɵ x + aɵz ) / 2 .

4.

An electron is projected into a magnetic field of flux density 20 wb/m2 with a velocity of 3.0×107 m/sec perpendicular to the field. compute the magnetic force on the electron.

5.

An electron is moving with a velocity 2 aɵ x + 3 aɵ y m/sec in an electric field of intensity 3 aɵ x + 6 aɵ y + 2 aɵz and a magnetic field of 2 aɵ y + 3 aɵz tesla. Find the magnitude and direction of Lorentz force acting on the electron.

6.

Two long parallel wires separated by 2cm in air carry current of 100 amp. Find the force on 1 meter length of the conductor.

7.

A current of 20 amp flows through each of the parallel long wires which are 4 cm apart. Compute the force exerted per unit length of each wire.

8.

Two long wires, each carrying an electric current of 5.0 amp are kept parallel to each other at a separation of 2.5 cm. Find the magnitude of the magnetic force experienced by 10 cm of a wire.

9.

A circular coil of 200 turns has a mean area of 10 cm2 , and the plane of the coil makes an angle of 30° with uniform magnetic flux density of 1.2 tesla. Determine the torque experienced by the coil if it carries a current of 50 amp.

10. Find the maximum torque on an 85 turn, rectangular coil, 0.2 m by 0.3m carrying a current of 2 amp in a magnetic field of 6.5 tesla. 11. A current of 1 amp is flowing through a coil of radius 0.5 m and 100 turns. Calculate the magnetic moment of the coil. 12. A specimen of iron is uniformly magnetised by a magnetising field of 500 amp/meter. If the magnetic induction in the specimen is 0.2 wb / m 2 , find the relative permeability and the susceptibility. 13. A 200 turn, rectangular coil, 0.3 m by 0.15 m with a current of 5 amp is in a uniform magnetic field →

B = 0.2 tesla. Find the magnetic moment and the maximum torque.

E lectromagnetic F ield Theory

414

14. A magnetising field of 1600 A/m produces a magnetic flux of 2.4 × 10 −5 wb in an iron bar of cross-sectional area 0.2 cm 2 . Calculate permeability and susceptibility of the bar. 15. An iron rod of 0.2 cm 2 cross-sectional area is subjected to a magnetising field of 1200 A / m. The susceptibility of iron is 599. Find the permeability and magnetic flux produced. 16. The horizontal component of flux density of the earth's magnet, that is, field is 1.7 × 10 −5 wb / m 2 . What is the horizontal component of magnetic intensity ? 17. Calculate the self inductance of a coil of 1000 turns, if a current of 6 amp produce a magnetic flux of 6000 Maxwell through the coil. (Given 1 Weber = 108 Maxwell). 18. Long solenoid of length 1 meter, radius 2.5 cm having number of turns 500. Calculate the inductance of the solenoid. 19. A solenoid of length 16 cm has 1280 turns and its area of cross-section is 10 cm 2 . A coil of 1000 turns is wound closely on the middle part of the solenoid. Calculate the mutual inductance between the solenoid and the coil.

nswers to Unsolved Numerical Problems 1.

2.

1414 . N, Along Z - axis

4.

9.1 × 10 −11 newton

−19 (2 aɵ x + aɵz ), 215 . × 10 –18 N 5. 9.6 × 10

6.

0.1 newton

−3 7. 2 × 10 N

8.

2.0 × 10 −5 N

3.

9.

7.5 N 1.2 × 10 −2 (aɵ x + aɵz ) 30 2

10.39 aɵz N - m

10. 66.3 N - m

2 11. 78.5 amp- m

12. 318.5, 318.5

2 13. 45 amp- m , 9 N - m

−4 2 14. 7.5 × 10 N / A , 596

−4 –5 15. 7.536 × 10 T – m /A,1.81 × 10 weber

16. 13.5 amp / m

−3 17. 10 henery

18. 0.6 mh

−3 19. 10.05 × 10 henery

mmm

4 Waves and Applications & Electromagnetic Wave Propagation

Section-A : Waves and Applications Section-B : Electromagnetic Wave Propagation

Theory

Numerical

Solved Examples

Questionnaire

417

Unit-4 S ection

A W aves and A pplications

I ntroduction n the foregoing units we have seen how charges at rest produces electric fields (or electrostatic fields) and moving charges produces static magnetic fields (or magnetostatic fields). In these units, the electric and magnetic phenomena were treated as independent. The only interconnection between the two lies in the fact that the current which produces magnetic fields are basically electrical, being charges

I

→

→

→

→

in motion. In the static cases, electric field vectors E and D, and magnetic field vectors B and H are independent pairs. In a conducting medium static electric and magnetic fields may both exist; static electric field may cause current to flow which in turn may give rise to a static magnetic field. But electric field is completely determined from the static electric charges or potential distribution, however, the magnetic field in no way determines the electric field. Nearly, independent nature of electric and magnetic phenomena disappears when we consider time dependent problems. According to Faraday, the time varying magnetic field produces an electric field and according to Maxwell, a changing electric field produces a changing magnetic field. Thus, the two →

→

→

→

→

equations, ∇ × E = 0 and ∇ × H = J , valid for static fields need modification for time dependent problems because there is a mutual dependence of electric and magnetic field vectors. Due to mutual dependency of electric and magnetic fields, we must speak electro-magnetic fields, rather then electric or magnetic fields. A complete set of relations giving the connection between the charges at rest (Electro

418

E lectromagnetic F ield Theory

statics), charges in motion (Current electricity), electric and magnetic fields (Electro-magnetism) were derived theoretically and summarized in four equations by Maxwell and are called Maxwell’s equations. In fact Maxwell’s equations are the back bone of electrodynamics. The success of these equations are remarkable, it includes the fundamental operating principles of all large scale electromagnetic devices such as motors, electronic computers, microprocessors, cyclotron, television, microwave radar, transmitting systems, receiving systems etc. Classical electromagnetic theory constituted by Maxwell predicts that accelerated charges produce electromagnetic waves which travel at the speed of light. A changing electric field produces a changing magnetic field and that in turn produces an electric field which produces a magnetic field and so on, in this way some kind of energy transfer is started. Energy will be transferred from the electric to the magnetic field and vice-versa, indefinitely. Thus, electromagnetic waves transport electric and magnetic energies from one point to another. The behaviour of an electromagnetic wave depends on its own and characteristics of the medium in which it is propagated. The amount of reflection and refraction of an electromagnetic wave depends on the nature of the interface of the two media in which it is propagated. The experimental discovery of Oersted in 1820, that a moving charge produces static magnetic field in the surrounding space, inspired Faraday to think that if a current could produce a magnetic field then a magnetic field should be able to produce a current. Faraday worked in this direction intermittently over a period of about ten years and finally got success in 1831, when he discovered that if a magnet is moved in the vicinity of a coil or a coil is moved relative to the fixed magnet a current was induced in the coil, as indicated by the deflection in the galvenometer attached with the circuit. He also observed that the deflection in the galvenometer is increased with the increase of velocity of the magnet relative to the coil or with the increase of velocity of the coil relative to the magnet. The current exists as long as there is a relative motion between the magnet and the circuit. The emf, and current so produced are called induced emf, and induced current. The phenomenon is known as electromagnetic induction. Using the concept of magnetic field lines and magnetic flux, Faraday arrived at a relation between the changing magnetic flux and induced emf which is known as Faraday’s law of electromagnetic induction.

F araday's Law Faraday’s experimental conclusions on electromagnetic induction were framed in the form of two laws which are known as Faraday’s laws of electromagnetic induction and are stated as follows:

First Law: Whenever the magnetic flux linked with a closed circuit changes, an induced emf is set up in the circuit whose magnitude, at any instant, is proportional to the rate of change of magnetic flux φB linked with the circuit.

419

Thus, if φ B is the magnetic flux linked with a circuit at any instant t and e the induced emf produced, then according to first law. e∝

d φB dt

Second Law: This law defines the direction of induced emf, or current in the circuit and stated as ‘‘the direction of induced emf, or current in the circuit is such that it opposes the change in flux that produced it. This law is know as Lentz’s law. Though the direction of induced emf was determined by Faraday, but it was expressed in the form of law by Lentz. The current in the closed circuit is induced by changing magnetic flux. The induced current itself produces a magnetic field and hence a magnetic flux. The flux produced by induced current may or may not have the same sign as original flux. If a current is induced by increasing flux, it will reduce the original flux. On the other hand, if a current is induced due to a decreasing flux, it will increase the original flux. Mathematically, the two laws in a combined form may be expressed as, e=−

d φB dt

If the closed circuit has N turns or is a tightly-wounded coil of N turns, then the induced emf, is given by e = −N

d φB d = − (Nφ B ) dt dt

Deduction of Faraday's Laws A

Let us consider a closed circuit consisting of two parallel conducting rails separated by a distance l and connected at one end by a battery E. At the other end a sliding bar AB of length E →

→ B

→

l

F

l is free to slide with a velocity v as shown in fig. 1. Let the →

magnetic flux density B act normal to the plane of the figure

B

→

δx

and directed into it. The rails, the bar and the field B are in three mutually perpendicular directions.

Fig. 1

When a steady current I is passed round the circuit by means of battery E in a direction shown, then the current carrying bar AB experiences a mechanical force, the magnitude of which is, F = IBl

→

→

→

[∵ F = I ( l × B)] ...(1)

E lectromagnetic F ield Theory

420

The direction of the force is given by the Fleming’s left-hand rule. As soon as the bar moves under this force, the magnetic flux through the circuit changes and induced emf, or induced current is established in the circuit. The current through the circuit therefore changes. Suppose the current is I while under the force the bar moves through a small distance δ x in a small time interval δ t. The work done by the force F during the displacement δ x of the bar AB is given by δ W = F δ x = (I Bl) δ x

...(2)

During the displacement of bar through a distance δ x, the energy drawn from the battery is EIδt. This energy is used up partly in overcoming the resistance (say R) of the circuit and partly in doing the work δ W in displacing the bar. Therefore, EI δ t = I 2 Rδt + I Blδ x or

E = IR + Bl

or

I=

...(3)

δx δt

E − Bl (δ x /δt) R

...(4)

Eqn. (4) indicates that the battery emf, E is opposed by a generated emf, Bl (δ x /δ t). This generated emf, represents the induced emf, in the circuit. Therefore, e = −Bl

∂x ∂t

...(5)

l δ x represents the area ( δ A) swept by the bar AB in time δ t. Therefore, the change in magnetic flux δφ B linked with the circuit in time δ t is, δφ B = BδA = Blδ x or

δ φB δt

=

B lδ x ∂t

From eqn. (5), we have e=−

∂ φB ∂t

e=−

d φB dt

In the limit δ t → 0

This is Faraday’s law of electromagnetic induction.

421

I ntegral and Differential Forms (or Vector Forms) of Faraday's Law Let us suppose that a magnetic field is produced by a current

→

S

E

carrying coil (or stationary magnet) and there is a closed loop C of an arbitrary shape which encloses a surface S in the field (Fig. 2). If →

→

B is the magnetic flux density near the circuit, then the magnetic →

dS

flux dφ B through a small area dS of the surface S is given by → →

d φ B = B . dS and

→ ...(1)

B

C

the total flux through the entire area S is φB = ∫

→ →

Fig. 2

...(2)

B. dS

S

→

dl

→

As the magnetic flux is changed, an electric field E is induced around the circuit. By definition the emf →

induced in the closed circuit is equal to the line integral of the electric field E induced around the circuit. That is, → →

e = ∫ E. d l

...(3)

According to Faraday’s law, the induced emf in the circuit is equal to the negative time-rate of change of magnetic flux linked with the circuit. That is, e=−

d φB dt

...(4)

Thus, from eqns. (3) and (4) we have → →

e = ∫ E . dl = −

d φB dt

...(5)

Substituting the value of φ B from eqn. (2) in eqn. (5), we get → →

e = ∫ E. d l = −

→ → d B. dS ∫ dt S

...(6)

Eqn. (6) is the integral form of the Faraday’s law. It states that ''the line integral of the electric field around any closed circuit is equal to the negative time-rate of change of the magnetic flux through the circuit''. When the circuit is stationary the time derivative can be taken inside the integral, where it becomes a partial time derivative. Thus, eqn. (6) for stationary circuit reduces to →

e = −∫

S

∂B → . dS ∂t

Thus for a stationary circuit the Faraday’s law in integral form may be written as,

...(7)

E lectromagnetic F ield Theory

422 →

→ →

∫

E . dl = − ∫

∂B → . dS ∂t

S

...(8)

Applying Stoke’s theorem to the closed line integral for transforming line integral into surface integral. That is,

∫

→ →

→

→

→

E . d l = ∫ ( ∇ × E) . dS S

Therefore, eqn. (8) becomes

∫S

→

→

→

→

( ∇ × E) . dS = − ∫

S

∂B → . dS ∂t

...(9)

where the surface integral may be taken over identical surfaces. As the surface is arbitrary, that is, the above equation holds for any surface area. Therefore, from eqn. (9), we have →

→

→

→

→

∂B → ( ∇ × E) . dS = − . dS ∂t and

→

→

→

→

∂B ∇× E =− ∂t

dB or curl E = − dt

...(10)

Eqn. (10) is the differential form of Faraday’s law. The negative sign in Faraday’s law [eqn. (10)] shows that the direction of induced emf is such as to oppose the change that produced it (Lentz's law). Therefore, if we increase the flux through the circuit, the induced emf produces current in such a direction as to reduce the flux.

T ransformer and Motional Electromotive Forces According to Faraday's law, the induced emf. e, developed in an electric circuit is equal to the time rate of change of magnetic flux linked with the circuit, that is, e=−

dφ B

...(1)

dt →

→

The magnetic flux through an elementary area dS will be B . dS and through the entire circuit is →

→

φB = ∫ B . dS S

...(2)

The change in magnetic flux induces electric field around the circuit. The line integral of the electric field gives the induce emf in the closed circuit as, → →

e = ∫ E . dl From eqns. (1), (2) and (3), we get

...(3)

423 →

→

e = ∫ E .d l = −

→ → d (∫ B . d S ) dt S

...(4)

Eqn. (4) reveals that both electric and magnetic fields are incorporated in time varying field and they are interrelated. The flux through the circuit can be changed in a number of ways. The circuit may remain stationary →

while magnetic field B changes with time. The circuit itself may change position, the magnetic field remains static. The circuit and field both are changing. We shall discuss each case one by one. →

(i) Stationary Conducting Loop in Time-Varying Magnetic Field ( B ) (TRANSFORMER EMF): →

When a stationary conducting loop is subjected in a time varying magnetic field B , the time derivative of R.H.S. of eqn. (4) can be taken inside the integral, where it becomes partial derivative as, →

→

→

e = ∫ E .d l = − ∫

S

∂B → .dS ∂t

...(5)

The emf produced by a time changing magnetic field (or by the time varying current) in a stationary circuit or loop is called transformer emf, since it is due to transformer action. Eqn (5) is called transformer induction equation. →

(ii) Moving Conducting Loop in a Static Magnetic Field ( B ) (MOTIONAL EMF): →

Let us consider a conducting loop placed in a stationary uniform magnetic field B , perpendicular to the page and directed into it as shown in fig. 3. When the loop is set in motion towards right with a uniform →

velocity v perpendicular both to its own length and to the magnetic field, every charged particle in it experience a magnetic force given by, → Fm

→

→

= q ( v × B)

...(1)

Where q is the charge on the charged particle (like free electrons and positive (ions). This force gives rise to an induced emf, at right angle to its velocity. This induced emf creates an voltage difference between the two ends of the conductor, the magnitude of which depends on how induced emf is oriented with respect to the conductor. So induced electric field intensity is given by. ∴

→

Ee =

Hence, the emf induced in the loop is

→

→ → Fm = ( v × B) q

...(2)

E lectromagnetic F ield Theory

424

→

→

→

→

→

e = ∫ Ee . dl = ∫ ( v × B ) . dl

...(3)

Thus, when a conductor or a loop is moved in a magnetic field emf is generated in it. This type of emf is called motional emf because it is due to motional action of the circuit. The magnitude of motional emf e = vBl This emf is maximum. According to Stoke's theorem,

∫

→

→

→

→

→

E . d l = ∫ ( ∇ × E) . d S S

→

Substituting the value of E from eqn (2), we get, →

∫S (∇ ×

→

→

→

→

→

→

E) . d S = ∫ ∇ × ( v × B ) . d S S

→

→

→

→

→

( ∇ × E ) = ∇ × ( v × B)

or

(iii) Moving Loop in Time Varying Field: →

Let us consider a case in which the conducting loop is in motion as well as the magnetic field B is function of time and position. In this case both transformer emf and motional emf are induced. When a stationary conducting loop is placed in a time changing magnetic field B, the transformer emf induced in the circuit and is given by, →

e = −∫

∂B → .dS S ∂t

...(1) →

On the other hand when a conducting loop is moving in a static magnetic field B , the motional emf is induced in the loop and is given by, →

→

→

e = ∫ ( v × B ) . dl

...(2)

→

Where v is the velocity with which the loop is moving. →

→

When the conducting loop is moving with uniform velocity v in a time varying magnetic field B , then both type of emf is induced in the circuit due to variation of magnetic field as well as due to its own motion. Therefore, the total emf induces in the loop is obtained by combining eqns. (1) and (2) as, →

→

e = ∫ ( v × B) . dl − ∫ By applying stoke's theorem, we get

→

→

S

∂B → . dS ∂t

425

→

→

→

→

→

→

→

→

→

e = ∫ E . dl = ∫ ( ∇ × E) . dS = ∫ ( v × B) . dl − ∫ S

→

→

→

→

→

∫ S ( ∇ × E) . dS = ∫ S ( ∇ × ( v

or

→

→

→

→

→

→

→

→

× B)) . dS − ∫

∇ × E = ∇ × ( v × B) −

or

S

∂B → . dS ∂t

S

∂B → . dS ∂t

→

∂B ∂t

Motional E.M.F. from Lorentz Force Suppose a conducting rod PQ is resting on a U-shaped stationary conductor →

S

and is immersed in a uniform magnetic field B, perpendicular to the plane of the paper directed downward as shown in Fig. 4. When the rod is set in →

motion towards the right with uniform velocity v , perpendicular to its own →

length and to the magnetic field B, due to the motion of the rod, the free →

R

→

electrons in the rod experienced a magnetic force Fm = (q ( v × B)), the direction of which is such that it takes the free electrons towards the end Q of the rod leaving positive charge near the end P of the rod. Thus, there is an accumulation of negative charge at Q and positive charge at P. The potential difference so developed between the ends P and Q of →

the rod produces an electric field E in a direction opposite to that of magnetic force Fm . Due to the continuous accumulation of charges, the strength of electric field gradually increases. When the electric →

→

force becomes equal to the magnetic force on the free electrons in the rods that is, when F e = F m or →

→

q Ee = − q( v × B), then the motion of free electrons in the rod ceases. Now as the rod moves with uniform velocity on a U-shaped stationary conductor, a closed circuit PQRS is available to the electrons where they drift along the path QRSPQ. Thus an electric current flows along QPSRQ (Fig.4). As a result of this current the excess negative charge at Q neutralise and thus the charges →

at P and Q are reduced so that electric field E is weakened. Now magnetic force acting on electrons dominates which causes further movement of free electrons towards Q end of the rod. So long the rod continues its motion, there is a continual drifting of electrons and hence a continual current. Hence, a moving rod in a magnetic field is a source of emf induced within it. This emf is induced in the rod due to its motion, that is why it is called motional induced emf. The magnitude of this motional induced emf in the circuit is equal to the work done in bringing a unit →

positive charge near the circuit. Therefore, if F is the resultant force acting on a charge q, then e=

work 1 → → = ∫ F . dl charge q

Now, the resultant force acting on charge q is the sum of electric and magnetic forces, that is,

E lectromagnetic F ield Theory

426 →

→

→

F = Fm + Fe

→

∴ But

e= 1 q

→

→

→

→

F m = q( v × B) and F e = q E

where

∫

→

→

F e . dl =

∫

→ 1 1 → F m . dl + ∫ q q

∫

→

→

F e . dl

→ →

E . dl, which is the line integral of an electrostatic field (which is conservative) around a → →

closed path and thus ∫ E . dl = 0 ∴

e= →

1 q

→

∫

→

F m . dl

→

In rod PQ, the angle between F m and dl is zero, therefore →

→

F m . dl = Fm dl cos 0 ° = Fm dl e=

So

1 q

→

0

∫l

Fm dl

→

→

→

F m = q( v × B) or | F m| = Fm = qvB

But

0

∴

e = vB ∫ dl

or

e = − v Bl

l

Example 1: An area of 1 . 0 m 2 in the z = 0 plane is enclosed by a filamentary conductor. Find the induced emf, when

→

B = 0 .10 cos 10 3 t [(aɵ y + aɵ z )/ 2 ] Tesla

Solution: The voltage induced in the filamentary conductor is, e=−

d φB dt

=−

d dt

→

→

∫s B .d S

Since the loop is in z = 0 plane, that is, in the x − y plane, then →

dS = aɵz dS ∴

→

→

B . dS = 0.10 cos 103 t [(aɵ y + aɵz ) / 2 ] . aɵz dS = 0.10 cos 103 t dS / 2 d dt

∫s

0.10

cos 103 t dS = −

So

e=–

or

e=−

or

e = 70 . 71 sin 10 3 t volt

0.10 2

2

0.10 d (cos 103 t 2 dt

× 103 (− sin 103 t) × 1.0 =

∫sdS)

0.10 × 103 × 1 2

sin 103 t

427

Example 2: A rectangular loop with sliding conductor is shown in fig 5. →

|∈x→|

The magnetic flux density B is normal to the plane of the loop and is uniform everywhere. The sliding conductor moves with a uniform velocity →

→

v . The magnitude of flux density B vary harmonically with time as

→

B = B0 cos ω t. Calculate the total induced emf in the rectangular loop.

Solution: In this case, motion of the conductor, that is, motional emf and time variation of flux density B, that is, transformer emf both are involved. Therefore, total emf induced in the loop is,

Here

→

→

→

∂B → e = ∫ ( v × B) . d l − ∫ . dS = em0 + etran s ∂t →

→

→

∂B = − B0 , ω sin ω t ∂t

B = B0 cos ω t, therefore,

→

∂B → = −∫ . dS = + B0 ω sin ω t s ∂t

Thus,

etran

But

∫ dS = xl

∴

etran = B0 ω sin ω t ( xl) →

→

→

∫ dS

→

em = ∫ ( v × B) . dl = vBl + vB0 cos ω t . l

and ∴

e = em + etran = vB0 l cos ω t + B0 ω xl sin ω t e = B0 l [v cos ω t + x ω sin ω t ]

or

→

Example 3: Consider the loop of fig. 6. If B = 0 . 5 aɵ z wb / m2 , R = 20 Ω, l = 10 cm and the rod is moving with a constant velocity of 8 aɵ x m / s, find: (i) The induced emf in the rod

v

(ii) The current through the resistance

l

(iii) The motional force on the rod (iv) The power dissipated by the resistance. [UPTU, B.Tech IV Sem 2007] →

Solution: (i) Since the rod is moving with a constant velocity v in →

a stationary magnetic field B, the induced emf, e, in the rod of length l is given by

→

→

→

e = ∫ ( v × B) . d l Here,

→

→

v = 8 aɵ x , B = 0.5 aɵz and l = 10 cm =0.1m

6

E lectromagnetic F ield Theory

428

→

(ii)

→

→

v × B = 8 aɵ x × 0.5 aɵz = − 4 aɵ y and d l = aɵ y dy

∴ Hence

e = − ∫ 4 aɵ y . aɵ y d y = −4 ∫ dy = − 4 × 0.1

or

e = – 0 . 4 volt

l

The current through the resistance of 20 Ω ie =

0. 4 e =− = 0 . 02 amp R 20

(iii) Motional force, Fm = IBl = 0.02 × 0.5 × 0.1 = 10 −3 N Fm = 1m N

or

(iv) The power dissipated by the resistance, P = I 2 R = (0.02)2 × 20 P = 0 . 008 Watt

or

Example 4: A straight conductor of 0 . 25m in length lies on the X-axis with one end at origin as shown in Fig. 7. The conductor is subjected to →

a magnetic field B = 0 . 04 aɵ y tesla and velocity v = 5 sin 10 3 t aɵ z m / s. Determine the motional electric field intensity and emf induced in the conductor. Solution: We know that

→

→

→

x= 0.25m

7

X

Ve = q Ee = q ( v × B) ∴

O

→

Ee = v × B →

→

Here v = 5 sin 103 t aɵz and B = 0.04 aɵ y Hence, the motional electric field intensity, Ee = 5 sin 103 t aɵz × 0.04 aɵ y or

Ee = 0.2 sin 103 t (− aɵ x )

or

Ee = −0 . 2 sin 10 3 t aɵ x

Therefore, emf induced in the conductor is given by Ve = ∫

0 .25

0 0 . 25

→

→

→

( v × B) . dl,

(– 0.2 sin 103 t aɵ x ) . dx aɵ x = − 0.2 sin 103 t

∴

Ve = ∫

or

Ve = − 0.2 sin 103 t ( x)0

or

Ve = − 0 . 05 sin 10 3 t volt

0

→

Here dl = dx aɵ x

0 . 25

= − 0.2 × 0.25 sin 103 t

0 . 25

∫0

dx

429

Example 5: Find the voltage V0 of the circuit shown in fig. 8.

Fig. 8

Solution: The situation of the problem is shown in fig. 9. Magnetic flux in the first loop, that is, φB = B1 A1 = 1.0 sin (2 π × 60 t) × 10 −3 (0.3 × 0.4) (T ) 1

or

φB = 0.12 × 10 −3 sin (2 π × 60 t) 1

Fig. 9

Magnetic flux in the second loop is, φB = B2 A2 = 10 . sin (2 π × 60 t) × 10 −3 × (0.3 × 0.3) T 2

or

φB = 0.09 × 10 −3 sin (2 π × 60 t) 2

The magnitude of induced emf in the first loop is dφB 1 = 012 . × 10 −3 + 2 π × 60 cos (2 π × 60 t) Ve1 = dt or

Ve1 = 0.12 × 10 −3 × 2 × 3.14 × 60 cos (2 π × 60 t)

or

Ve1 = 45.216 × 10 −3 cos (2 π × 60 t) V = 45.216 cos (2 π × 60 t) mV

The magnitude of induced emf in the second loop is, dφB d 2 Ve = = 0.09 × 10 −3 sin (2 π × 60 t) 2 dt dt = 0.09 × 10 −3 × 2 π × 60 cos (2 π × 60 t) or

Ve2 = 33.91 × 10 −3 cos (2 π × 60 t), V = 33.91 cos (2 π × 60 t) mV

Current flowing in the first loop is given by

or

Ve1 = I (R1 + R2 ) = I (100 + 150) Ve I= 1 250

Voltage developed across 100 Ω resistance Ve3 = I × 100 =

Ve1 250

× 100

E lectromagnetic F ield Theory

430

45.216 cos (2 π × 60 t) = 18.08 cos (2 π × 60 t) 2. 5

or

Ve3 =

∴

V0 = Ve2 + Ve3 = 33.91 cos (2 π × 60 t) + 18.08 cos (2 π × 60 t)

V0 = 51.99 cos (2 π × 60 t ) mV

Example 6: Determine the total electromotive force (emf) induced in the rectangular loop of area A has uniform magnetic flux density B normal to the plane of the loop. The magnetic flux density is varying harmonically with respect to time and is given by B = B0 cos ωt Solution: The situation of the problem is shown in fig. 10. In this problem rectangular loop is stationary while magnetic flux density B is changing with time. Thus, according to transformer induction equation the induced emf, in the rectangular loop is, →

∂B → d φB . ∂S e=− = −∫ S ∂t dt = −∫

S

= − B0 ∫

∂B ∂ (B0 cos ω t) dS = − ∫ dS S ∂t ∂t S

(− sin ωt) ω dS

= B0 ω sin ωt or or

∫S

Fig. 10

dS

e = B0ω sin ωt . A, where A is the area of the loop e = B0 ω A sin ω t

Example 7: A ring of radius 10 cm and resistance 2 ohm is placed in a magnetic field of 0.5 weber/m2 which is perpendicular to the plane of the ring. If the magnetic field is decreasing at the rate of 0.1 weber/m2 per second, find the values of the induced emf and the induced current in the ring. →

Solution: For the changing field B, the time rate of change of magnetic flux is d φB d dB = (BA) = A dt dt dt Here A = πr 2 = 3 .14 × (10 × 10 −2 )2 = 3 .14 × 10 −2 m2 and ∴

dB 0 .1 weber/m2 -sec = dt 1

d φB = 3.14 × 10 −2 × 0.1 = 3.14 × 10 −3 weber/sec dt

According to Faraday’s law, the magnitude of induced emf produced in the ring is d φB = 3 .14 × 10 −3 volt = 3.14 mV e= dt If R is the resistance of the loop, then induced current is e 3 .14 i= = = 1. 57 milli-amp R 2

431

Example 8: A wire of length 1 meter moves at right angles to its length at a speed of 100 m/sec in uniform magnetic field 1 weber/m2 which is also acting at right angles to the length of the wire. Calculate the induced emf, in the wire when the direction of motion is (i)

at right angles to the field.

(ii) inclined at 30° to the direction of the field. →

Solution: Since the rod is conducting, it contains electrons. When it is moving with a velocity v in a →

uniform magnetic field B, a force acts on each charge Q of the rod, which is given by →

→

→

F = Q ( v × B)

This force pushes negatively charged electrons towards one end of the rod and thus the other end of the rod becomes positively charged. This process is continue till the accumulation of excess charges at the ends establishes an electric field E in the rod such that the net force on each charge carrier become zero. That is, →

→

→

→

→

→

QE + Q ( v × B) = 0 or E = − ( v × B) Thus, the emf, induced between the two ends of the rod is l → →

l

0

0

e=∫ ∴ emf, induced, (i)

→

E. d l = − ∫ →

→

→

→

( v × B) . d l = vBl sin θ

→

e = ( v × B) . l

emf, induced in the wire when the direction of motion of wire is at right angle to the field is e = vB sin 90 ° l = vBl e = 100 × 1 × 1 = 100 volts

(ii)

When the direction of motion of the wire inclined at 30° to the direction of the field, the induced emf, is v Bl e = vB sin 30 ° l = 2 100 × 1 × 1 e= = 50 volts ∴ 2

D isplacement Current The concept of displacement current was first introduced by Maxwell purely on theoretical ground. To resolve the paradox of the charging capacitor and inadequacy of Ampere's magnetic law for time varying field. Maxwell postulated that it is not only the current in a conductor that produces a magnetic field, but a changing electric field in a vacuum, or in a dielectric, also produces a magnetic field. It means that a changing electric field is equivalent to a current which flows as long as the electric field is changing. This equivalent current in vacuum or dielectric produces the same magnetic effect as an ordinary or conduction current in a conductor. This equivalent current is known as displacement current. Maxwell

E lectromagnetic F ield Theory

432

→ → ∂D redefined that the current density in the Ampere's magnetic circuital law is not J , but J + . He ∂t →

→

called ∂ D / ∂t the displacement current density and J as conduction current density.

Modified Ampere's Law and Displacement Current The idea of displacement current originated from the study of discharge of a condenser leads to a modification in Ampere’s circuital law. The modification and generalization of Ampere’s law dφ by the introduction of an additional term µ0 ε0 E to the right dt of standard Ampere’s law is one of the major contributions of Maxwell’s to the electromagnetic theory. To establish the concept of displacement current, let us consider

Fig. 11

the process of charging of a parallel plate capacitor through a series circuit containing a cell E, a resistor R, a capacitor C and a switch S as shown in fig. 11. When the switch S is pressed circuit is closed and a current flows in the circuit as a result of which charge starts accumulating gradually on the plates of the capacitor. The current in the circuit decreases as the charge grows on the plates. When the capacitor is fully charged to the e.m.f of the cell, the current stops. There is no actual flow of electrons in the space between the plates during charging. In fact free electrons are flowing into one plate and forcing electrons of the second plate of the capacitor to come out and take part in continuing current in the given hit. In other words, we can say that there is a discontinuity of conduction current in the space between the plates. If we place a compass needle in the space between the plates, the needle deflects. This indicates that there is a magnetic field between the capacitor plates even though there is no motion of charge. This indicates that there must be some other source of magnetic field in the gap. The other source is nothing but the changing electric field between the plates. Let us consider a plane surface S1 and a hemispherical surface S2 around a capacitor plate as shown in fig. 11. Let both the surfaces are bounded by the same closed path l. Now apply Ampere’s circuital law to both surfaces. If Ampere’s law is applied to the surface S1, then

∫S

1

→

→

B . d l = µ0 i

...(1)

Because during the process of charging, current i has been flowing through plane surface S1. But, if it is applied to the hemispherical surface S2 , then

∫S

→

→

B .d l = 0

...(2)

2

Because current density j or current i is zero at all points on S2 . Equations (1) and (2) contradict each other and thus both cannot be correct. Maxwell resolved this paradox of the charging capacitor by adding one more term on the right hand side of equation (1). The

433

new term is based on the ideology that a changing electric field is a source of magnetic filed in the gap between the capacitor plates (during charging) and is equivalent to a displacement current that can be denoted by id . The condition of continuity of the current requires the displacement current id to be equal to ε0 dφ E /dt, where φ E is the flux of the electric field through an area bounded by the closed curve and ε0 is the permittivity of the free space. Therefore, if there exists an electric current as well as changing electric field, the resultant magnetic field or Ampere's circuital law is modified as,

∫

→

→ d φE B . d l = µ0 i + ε0 = µ0 (i + id ) dt

...(3)

Let us now show that the displacement current id in the gap between the plates is equal to the conduction current i flowing in the connecting wires. Suppose Q be the charge collected on the capacitor plates at any instant t, then the electric field developed between the plates will be E=

Q ε0 A

...(4)

where A is the surface area of each plate. Differentiating equation (4), with respect to ‘ t’, we get dE 1 dQ 1 dE = = i or i = ε0 A ε0 A dt ε0 A d t dt

...(5)

By definition, the displacement current id is d φE id = ε0 dt

...(6)

We know that the electric flux, φ E = E A dE ∴ id = ε0 A dt

...(7)

From equations (5) and (7) it is clear that id = i Thus, the displacement current in the gap is identical with the conduction current in the connecting wires. In fact the displacement current does not mean an actual current that flows through the capacitor. It is simply an apparent current which represents the rate at which charge flows from one electrode to another in the external circuit. Equation (7) may also be written as, →

→

dD id = A dt

(∵ D = ε0 E) ..(8)

In term of displacement current density eqn. (8) may be written as →

id d D = A dt

or

→

Jd =

→

dD dt

...(9)

E lectromagnetic F ield Theory

434

In a capacitor of a lossy dielectric material (Fig. 12) of conductivity σ and permittivity ε, the total current is the sum of conduction current and displacement current. A capacitor of a lossy dielectric material is equivalent to a circuit which is a combination of a resistance R and a capacitor C in parallel (Fig. 13). The conduction current flows through the resistance part of the circuit and displacement current through pure capacitor part. Thus, the total current in a capacitor of lossy dielectric is,

Fig. 12

Fig. 13

I = iR + id If we divide the respective currents by the area of the capacitor plate A, we get an equation in current density terms. →

→

→

J = j

That is,

→

R

+ jd

→

where j R and j d are conduction and displacement current densities respectively.

Modified Ampere's Law and Displacement Current Density For static electromagnetic fields, Ampere's magnetic circuital law (in vector form) states that →

→

→

∇ × B = µ0 J

...(1)

→

where J is the current density. Taking divergence of above equation, we get →

→

→

→

→

∇ .( ∇ × B) = ∇ .(µ0 J )

From the vector identity, the divergence of curl of any vector field is always zero. or

→

→

div curl B = µ0 div J = 0 →

div J = 0

or

...(2)

But the equation of continuity states that → →

∇. J = −

∂ρ v ∂t

where ρ v is the volume charge density

...(3)

435

Eqns. (2) and (3) contradict each other and thus both can not be correct. Maxwell resolved this disputatious situation by adding one more term on the right hand side of eqn. (1). Let us add unknown →

term J d to eqn (1), thus →

→

→

→

∇ × B = µ0 ( J + J d )

...(4)

Taking divergence of both sides of eqn. (4), we get →

→

→

→

→

→

→ →

→ →

∇ .( ∇ × B) = µ0 ( ∇. J + ∇ . J d ) ∇ .( ∇ × B) = 0

But

→ →

∴

→ →

∇. J d = − ∇. J

...(5)

Using eqn. (3), we get → →

∇. J d =

∂ρ v ∂t

...(6) →

The volume charge density ρ v is related with electric displacement D, through Gauss's law as → →

∇ . D = ρv

...(7)

Substituting the value of ρ v from eqn. (7) in eqn. (6), we get →

→ →

∇. J d

→

∂ → → → ∂D = ( ∇ . D) = ∇ . ∂t ∂t

Because ∇ is independent of t →

∴

Jd

→

∂D = ∂t

Thus, the additional term that must be added to eqn (1) is: →

→

Jd

∂D = ∂t

and is called displacement current density. Thus, the modified Ampere's magnetic circuital law is → → → → ∂ D ∇ × B = µ0 J + ∂ t →

→

→

→

∂D ∇ × H = J+ ∂t

or

Displacement Current from Displacement Current Density The displacement current id is obtained from the following relation: →

→

Substituting J d =

→

→

id = ∫ J d . dS S

∂D in eqn (1) we get ∂t

...(1)

E lectromagnetic F ield Theory

436 →

id = ∫

→

→

S

∂ = ∂t

∂D → . dS ∂t →

→

∫S D . dS

But, displacement vector, D = ε0 E id = ε0

∂ ∂t

id = ε0

dφE dt

∴ →

→

∫S

→

→

E . dS

But, electric flux φE = ∫ E . dS S

∴

→

→

Ratio of Conduction Current Density, J c and Displacement Current Density J d : →

If σ is the conductivity of the medium, then conduction current density J c is given as, →

→

Jc = σE

...(1)

→

where E is the electric field intensity. →

E = E0 e jωt

In phasor form,

→

J c = σ E0 e j ω t

∴

...(2)

→

The displacement current density, J d is given as →

→

Jd =

→

dD d → d = (ε E) = (ε E0 e j ω t ) dt dt dt

J d = ε E( jω) e jωt

or →

...(3)

→

The ratio of J c and J d is →

Jc

→

=

Jd

→

or

Jc

→

Jd

=

σ E0 e jωt

jε E ω e j ω t

=

σ j εω

σ Jc = εω Jd

Maxwell's Equation A theory describing the interactions between charges, currents, and electric and magnetic fields was developed by James Clerk Maxwell in 1864 in the form of four fundamental equations called Maxwell’s equations or Maxwell’s field equations or electromagnetic field equations. Maxwell’s equations representing the experimentally achieved fundamental laws of electromagnetism may be stated in the integral as well as in differential forms as follows:

437

Integral form

1.

∫

→

→

E. dS =

Q or ε0

∫

→

→

E . dS =

1 ε0

∫

Differential form →

div E =

ρ dv

ρ ε0

(for symmetric distribution of charge) 2.

→

→

→

→

→

∫ B . dS = 0

3.

∫

4.

∫ B .d l

E.dl =−

→

→

div B = 0 ∂ φB ∂t

or

∫

→

→

→

E.d l = − ∫

→ → ∂B → . dS (∵ φ B = ∫ B . dS ) ∂t

→ → ∂ φE or ∫ B . d l = µ0 = µ0 i + ε0 ∂t →

→

(∵ i = ∫ j . d S

→

∫ j

and

→

→

∂B curl E = − ∂t

→ ∂ E → → → . dS ∂ E curl B = µ0 j + ε0 ∂t ∂t → → = ∫ E. dS )

+ ε0 φE

Maxwell’s equations summarized the whole theoretical contents of classical electrodynamics. These Maxwell’s equations are written for the fields in vacuum where no material medium is present, in the →

presence of electric charge of density ρ and electric current (that is, charge in motion) of density j . They →

→

show how the electric field E and the magnetic induction field B in vacuum are related to the electric →

charge density ρ or charge Q and electric current density j or current i. The quantities ε0 and µ0 are called the permittivity and permeability of vacuum respectively. They appear in equations for the equality of system of units. Each Maxwell’s equation is one of the experimentally observed fundamental phenomenon in electricity and magnetism.

P hysical Significance →

Maxwell’s first equation (div E = ρ / ε0 ) represents the Gauss’s law in electrostatics for the static charges, which states that the electric flux through any closed hypothetical surface is equal to 1/ε times the total charge enclosed by the surface. It can be derived from Coulomb’s law. →

Maxwell’s second equation (div B = 0) expresses Gauss’s law for magnetism (or magnetostatics). It can →

→

be written in analogy to first Maxwell equation simply by replacing E by B and φ E by φ B . It states that the net magnetic flux through any closed surface is zero. Since a magnetic monopole does not exist, any closed volume always contain equal and opposite magnetic poles (North pole and South pole), leading to a net magnetic pole strength zero. It also signifies that the number of magnetic lines of flux entering into any region is equal to the line of flux leaving it or in other words, magnetic lines of flux are continuous.

E lectromagnetic F ield Theory

438 →

→

Third Maxwell equation (curl E = – ∂ B / ∂t) is the Faraday’s law in electromagnetic induction. It states that induced electromotive force around any closed surface is equal to the negative time rate of change of magnetic flux through the path enclosing the surface.Therefore, it signifies that an electric field is produced by a changing magnetic flux. →

Fourth Maxwell’s equation [curl B = µ0 (j + ε0 ∂E / ∂t )] represents the generalized form of Ampere’s law as extended by Maxwell for the inclusion of time varying electric fields. It is valid for both steady (when the charges are at rest or in electrostatic conditions) and non-steady (when charges are in motion or current flows) states. It states that the magnetomatic force around a closed path is equal to the sum of conduction current (µ0 i) and displacement current (ε0 ∂ φ E / ∂t) through the surface bounded by that path. This signifies that a conduction current as well as changing electric flux, produces a magnetic field. A close examination of Maxwell’s third and fourth equations reveal that the Maxwell’s modified form of Ampere’s law contains a term µ0 i while Faraday’s law does not. The absence of similar term in Faraday’s law is justified because free magnetic pole do not exist. The term ε0

d φE , known as displacement current in fourth Maxwell’s equation, signifies that magnetic dt

field can also be produced by a changing electric field. Since µ0 ε0 ~1.1 × 10 −17sec2 /m2 is a very small quantity, the term ε0µ0

d φE will not contribute significantly unless (dφ E /dt) is extremely large. Therefore, dt

displacement current term could be detectable only when electric flux changes very rapidly. Hence, fourth Maxwell’s equation gives the new concept of generation of magnetic field by displacement current associated with the field. This only conception is responsible for overwhelming success of the electromagnetic theory. In the absence of any free charges or currents or in the empty space, that is, when Q or ρ and i or j are zero, the Maxwell’s equations take the form:-

∫

→

→

→

→

→

→

→

→

E . dS = 0

∫ B . dS = 0 ∫

E. d l = − ∫

∫ B.dl

div E = 0

;

div B = 0

→

∂B → . dS ∂t

;

→

= µ0 ε0 ∫

∂E → . dS ∂t

→

;

;

→

→

→

∂B curl E = − ∂t →

→

∂E curl B = µ0 ε0 ∂t

For magnetic and electric fields in a material medium, Maxwell’s equations have to be modified corresponding to the phenomena of electric polarization and magnetization. In material medium both

439

free charges and polarization charges act as sources of electric field, and magnetization is an additional →

source of magnetic field B. Here we are considering only those material media in which electric →

→

polarization is directly proportional to the electric field ( P ∝ E) and the magnetization to the magnetic →

→

field ( I ∝ B) or in the homogeneous linear media. Maxwell’s equations for such material are obtained by simply replacing ε0 and µ0 by ε and µ respectively in Maxwell’s equations in the absence of any material medium. The quantities ε and µ which are the characteristic of the material are called permittivity and permeability of the material medium respectively. The quantities ε and µ are related with their corresponding quantities in vacuum as ε = kE ε0 and

µ = kM µ0

where kE is the dielectric constant and kM the relative permeability of the material medium. Therefore to express Maxwell’s equations in material medium we have to modify first and fourth equations of Maxwell accordingly. With these modifications Maxwell’s equations in material medium may be expressed as, →

→

→

→

→

∫ D . dS = ∫ ρ dv

; div D = ρ

∫ B . dS = 0

; div B = 0 or

∫

→

→

E. d l = − ∫ →

→

∫ H. d l = ∫

→

∂B → . dS ∂t

→

→

div H = 0

→

→

∂B ; curl E = − ∂t

→

→

µ∂ H or curl E = − ∂t

→ → → → → ∂D ∂ D → j + ∂ t . d S ; curl H = j + ∂ t

In static field, the Maxwell’s equations reduce to → →

∇. E =

ρ ε

→ →

∇. B = 0

→

→

→

→

∇ × E =0 →

∇×H= J

→

→

∂ E ∂B because, a static field is one which does not change with time. It means that for static fields = =0 ∂t ∂t The Maxwell’s equations discussed so far are all in MKS system of units. In CGS Gaussian system of units these are written for the field in vacuum, in the presence of electric charge of density ρ and electric current (that is, charge in motion) of density J as →

div D = 4πρ →

div B = 0

→

→

1 ∂B curl E = − c ∂t →

→

4π → 1 ∂ D curl H = − J + c c ∂t

E lectromagnetic F ield Theory

440

D erivativons of Maxwell's Equations →

→

1. Maxwell’s First Equations, div D = ρ or div E = ρ / ε 0 :When a thin dielectric sample is placed in a uniform electric field its molecules get polarized. Thus, a dielectric sample in an electric field contains free charge that we embedded as well as polarisation charge or bound charge due to the polarisation of its molecule. Let us suppose an imaginary and arbitrary closed surface S located inside a such a dielectric. If ρ and ρ p are the free and bound charge densities respectively at a point in a small volume element dV, then for such a medium Gauss’s law may be expressed as →

∫S

→

E . dS =

1 ε0

∫V

(ρ + ρ P ) dV

...(1)

where ε0 is the permittivity of fee space. But, we know that the polarisation charge per unit volume or polarisation change density →

→

ρ P = − div P ,

where P is the electric polarisation.

Substituting this value of ρ P in equation (1), we get

∫S ∫S

or

→

→

E . dS =

→

1 ε0

→

ε0 E . dS = ∫

V

According to Gauss-divergence theorem,

∫V

→

(ρ − div P ) dV →

ρ dV − ∫

V

∫S

→

div P dV

→

A . dS = ∫

V

→

div A dV , the surface integral can be changed

into volume integral, as

∫V →

→

div (ε 0 E ) dV = ∫

V

ρ dV − ∫

V

→

div P dV or

→

→

∫V

→

div (ε 0 E + P ) dV = ∫

V

→

→

→

ρ dV ...(2)

→

But the quantity ε0 E + P is termed as electric displacement vector D or D = ε0 E + P Therefore, equation (2) becomes

∫V

→

div D dV = ∫

V

ρ dV or

∫V

→

(div D − ρ) dV = 0

...(3)

→

Since the equation (3) is true for all volumes, hence the integrand, div D − ρ must vanish, that is, →

→

div D − ρ = 0 or div D = ρ or

→ →

→ →

∇ . D = ρ or ∇. E = ρ / ε0

This is the required Maxwell’s first equation. In free space volume charge density is zero, that is, ρ = 0 →

→ →

→ →

Therefore, Maxwell’s first equation in free space is reduces to div D = ∇ . D = 0 or ∇. E = 0

441

→

→ →

2. Maxwell’s Second Equation, div B = 0 or ∇ . B = 0 Magnetic field lines due to circular current in vacuum are continuous and form closed loops or go off to infinity. This fact is also true in presence of magnetic material because the effect of the magnetisation of material is same as that of a (fictitious) current circulating around the material. Hence, under all circumstances, the number of magnetic lines of force entering any arbitrary closed surface enclosing a volume is exactly the same as that leaving it. In other words, the net magnetic flux of magnetic induction →

B through any closed Gaussian surface is always zero. That is, φB = ∫

→ S

→

B . dS = 0

...(1)

The above equation implies that there are no isolated magnetic poles to serve as source or sink for the →

lines of magnetic induction B. The equation (1) is analogous to Gauss’s law in electrostatics. →

The equation (1) can also be written in differential form by using Gauss-divergence theorem to B. Hence,

∫S

→

→

B . dS = ∫

V

→

div B dV = 0

...(2)

where V is the volume enclosed by the surface S. or

∫V

→

div B dV = 0

...(3) →

Since the equation (3) is true for all volumes, therefore, the integrand, div B must vanish, that is, →

div B = 0 This is the required Maxwell’s second equation. It is true in free space as well as in material medium.

3. Maxwell’s Third Equation or Faraday’s Law of Electromagnetic Induction → → → → B ∂ curl E = ∇ × E = ∂ t According to Faraday’s law of electromagnetic induction, induced e.m.f. around a closed circuit is equal to the negative time-rate of change of magnetic flux linked with the circuit. Mathematically this result can be expressed as e=−

dφ B dt

...(1)

This conclusion is found to be independent of the way in which the flux is changed and the shape of the circuit. Let us suppose that a magnetic field is produced by a stationary magnet or current carrying coil. Let C be a coil or closed loop of arbitrary shape

Fig. 14

E lectromagnetic F ield Theory

442 →

(Fig.14) which encloses a surface S in the field. If B is the magnetic field induction at any instant at a →

point P in the field, then the magnetic flux linked with an infinitesimal area dS around P is, →

→

δφ = B . dS

The total magnetic flux linked with the coil or loop, φ B = ∫ δφ = ∫

→ S

→

...(2)

B . dS

But changing magnetic flux induces an electric field around a circuit. By the definition of induced e.m.f., we have e=∫

→

→

...(3)

E.d l

C

→

→

where E is the electric field produced due to changing magnetic field B Substituting the values of e and φ B from equations (3) and (2) in equation (1), we get

∫C

→

→

E.d l = −

d dt

∫

→

→

...(4)

B. dS

This is the integral form of Faraday’s law. The flux linked with any circuit may change in number of ways. The circuit itself may change the →

position though the magnetic field B remain unchanged or the magnetic field may change while circuit remains fixed. In case of stationary circuit, the time derivative may be taken inside, where it becomes a partial derivative. Therefore, equation (4) becomes

∫

→

→

→

E.d l = − ∫

∂B → . dS ∂t

S

...(5)

According to Stroke’s theorem the line integral may be transform into surface integral as

∫

→

→

E.d l = ∫

→

→

curl E . dS

S

→ → → → ∫ curl A . dS = ∫ A . d l S

...(6)

Comparing equations (5) and (6), we get

∫S

→

→

→

curl E . dS = − ∫

S

∂B → . dS or ∂t

∫S

→ → ∂B → curl E + ∂ t . dS = 0 →

Since the equation (7) is true for all surfaces, therefore the integrand, curl E + →

→

...(7)

→

∂B , must vanish, that is, ∂t

∂B curl E + =0 ∂t →

or

→

→

→ → ∂B ∂B or ∇ × E = – curl E = − ∂t ∂t

This is the required Maxwell’s third equation in free space and differential form of Faraday's law.

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443

4. Maxwell’s Fourth Equation or curl

→ → → ∂ D H = J + ∂t

or curl

→ B =µ0

→ J

+ ε0

→ ∂ E ∂t

Let consider a region in which there is a steady flow of current, that →

is, the current density J does not change with time, however its value may change from place to place. Now consider the closed path in the region (Fig. 15) bounded by the surface S. The total current -I enclosed by this path is the flux of current density J through the surface S bounded by closed loop, that is,

I=∫

→

→

...(1)

J . dS

S

According to Ampere’s circuital law, the line integral of the →

Fig. 15

magnetic induction B around any closed path is

∫

→

→

→

→

B . d l = µ0 I = µ0

→

∫S

→

...(2)

J . dS

From Stroke’s theorem

∫

B.dl = ∫

→

S

→

...(3)

curl B . dS

Comparing equations (2) and (3), we get

∫S or

∫S

→

→

→

curl B . dS = µ0 ∫ →

→ S

→

J . dS

→

(curl B − µ0 J ) . dS = 0

...(4)

As the surface is arbitrary, therefore integrand of equation (4) must vanish, that is, →

→

→

→

curl B − µ0 J = 0 or curl B = µ0 J

...(5)

Now, let us consider the case in which the charge distributions and electric fields are changing with time. ∂ρ Suppose we have a charge distribution with charged density ρ ( x, y, z, t) such that ≠ 0. ∂t

For time dependent charge distribution the equation of continuity is → → ∂ρ ∂ρ = 0 or div J = − div J + ∂t ∂t ∂ρ is non-zero, this implies that Since ∂t

...(6)

→

div J ≠ 0

...(7)

Taking divergence of equation (5), we get

→ → → 1 1 div J = div curl B = div (curl B) = 0 µ0 µ0

or

→

div J = 0

→

(∵ div curl B = 0) .

...(8)

The contradiction between equations (7) and (8) shows that equation (5) representing Ampere’s law is valid only for steady current and is incorrect or incomplete for time varying fields. Hence, for time varying fields Ampere's law must be modified to include the time dependent term.

E lectromagnetic F ield T heory

444

To make the Ampere’s law valid for time varying currents, Maxwell modified Ampere’s law by postulating that just as changing magnetic field produces an electric field (Faraday’s law) a changing electric field also produces a magnetic field. Therefore, a changing electric field is equivalent to a current, called displacement current id which flows as long as electric field is changing. On the analogy with Faraday’s law, Maxwell included the term of the form →

→

∂E curl B = µ0 ε0 ∂t

...(9)

→

Hence, the expression for curl B for changing electric field becomes →

→

→

curl B = µ0 J + µ0 ε0

∂E ∂t

...(10)

To verify the correctness of addition of the new term, taking the divergence of both sides of equation (10) that is, → → → → → ∂ E ∂ ...(11) div (curl B )= div (µ 0 J )+ div µ0 ε0 = µ0 div J + µ0 ε0 div E ∂t ∂t →

Since div (curl B )= 0, therefore, equation (11) becomes →

div J = − ε0

→ ∂ (div E) ∂t

→

But, according to Poisson’s equation, div E =

...(12)

ρ ε0

Therefore, equation (12) becomes →

div J = −

∂ρ ∂t

This is the well known continuity equation described in equation (6) hence, the supposition was correct. Equation (10) may also be expressed as → → ∂ E curl B = µ0 J + ε0 ∂ t →

→

→

...(13)

→

→

or →

→

→

Since, div (curl B ) = 0, therefore, div ( J + J d ) = 0 or

→

→

div J d = − div J =

∂ρ ∂t

According to the differential form of Gauss’s law, we have →

div D = ρ ∴

→

∂E curl B = µ0 ( J + J d ), where J d = ε0 ∂t

→

→ ∂ div J d = (div D) = div ∂t

→ ∂ D ∂ t

...(14)

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445

→

→

∂D Jd = ∂t

This gives

Therefore, modified form of Ampere’s law [equation (14)] may also be expressed as → → → → → → ∂D ∂ D curl B = µ0 J + or curl H = J + ∂t ∂t

→

→

(∵ B = µ0 H)

This is required fourth Maxwell’s equation. In free space, the current density J =0 Therefore, the fourth Maxwell’s equation in free space is reduces to →

→

∂D curl H = ∂t

Maxwell's Equations in Integral Form The four Maxwell’s equations of electromagnetic field theory in differential form may be expressed as, → →

∇.D = ρ

...(1)

→ →

∇.B =0

and

→

→

→

→

...(2) →

∂B ∇× E=− ∂t

...(3) →

→

∂D ∇×H= J + ∂t

...(4)

These equations can be transformed into the integral form simply by taking surface integrals and volume integrals in the following manner: Taking the volume integral of eqn. (1) over a volume v enclosed by a closed surface S, we get

∫V

→ →

( ∇ . D) dv = ∫

V

ρ dv

...(5)

The volume integral can be transformed into surface integral with the help of Gauss-divergence theorem as

∫V

→ →

( ∇ . D) dv = ∫

Therefore, eqn. (5) becomes

→

→

S

→

D . dS

...(6)

→

∫S D . d S = ∫V ρdv

...(7)

This is the integral form of first Maxwell’s equation. Similarly, taking volume integral of eqn. (2) over a volume v enclosed by the surface S, we obtain

∫V

→ →

( ∇ . B) dv = ∫

From Gauss-divergence theorem

V

(0) d v

...(8)

E lectromagnetic F ield T heory

446

∫V

→ →

( ∇ . B) dv = ∫

→

→

...(9)

B . dS

S

Therefore, eqn. (8) takes the form →

→

∫S B . d S

=0

...(10)

This is the integral form of second Maxwell’s equation. Taking surface integral of both sides of eqn. (3) over an open surface S, we obtain

∫S

→

→

→

( ∇ × E) . dS = − ∫

S

→ → ∂ B ∂ t . dS

...(11)

The surface integral can be converted into line integral by using Stoke’s theorem as

∫S

→

→

→

→

→

( ∇ × E) . dS = ∫ E . d l

...(12)

Therefore, eqn. (11) becomes

∫

→

→

→

E . d l = −∫

S

∂B → .dS ∂t

...(13)

This is the integral form of third Maxwell’s equation. Similarly, taking surface integral of both sides of equation (4), over the surface S, we get

∫S

→

→

→

( ∇ × H) . dS = ∫

Using Stroke’s theorem, we obtain

∫S

→

→

→

S

→

→ → ∂ D → J + ∂ t . dS →

( ∇ × H) . dS = ∫ H . d l

...(14)

...(15)

Therefore, eqn. (14) becomes

∫

→ → ∂ D → H.d l = ∫ J + .dS S ∂t →

→

...(16)

This is the integral form of fourth Maxwell’s equation. Thus, equations (7), (10), (13) and (16) are the Maxwell’s equations of electromagnetic field theory in integral form.

Equations for Harmonically M axwell's Sinusoidally Varying Fields)

Time Varying Fields (or

If the electric and magnetic fields vary harmonically with time, the Maxwell’s four equations of electromagnetic field theory change their forms. The harmonically time varying electric field may be expressed as

E = E0 cos (ω t + φ)

...(1)

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447

where E0 is the magnitude of peak value of electric field E at a point in a coordinate system, ω the angular frequency and φ the phase angle in radian or degree. A complex quantity e j(ω t + φ) can be expressed in terms of real and imaginary parts by Euler’s identity as

e j(ω t + φ) = cos (ωt + φ) + j sin (ωt + φ), [where j = √ (− 1)]

...(2)

Real part of eqn. (2), may be expressed as Real part of e j(ω t + φ) = Re e j(ωt + φ) = cos (ωt + φ) In term of Real part eqn. (1) may be written as E = Re [ E0 e j(ω t + φ)] = Re [ E0 e j ωt e jφ ]

...(3)

Dropping Re and suppressing e jωt from eqn. (3), we obtain E = E0 e jφ

...(4)

jφ

E (= E0 e ) is called phasor. Taking general form of phasor as E = E0 e j(ω t + φ)

...(5)

Differentiating eqn. (5) partially with respect to ' t', we obtain ∂E = j ω E0 e j(ω t + φ) ∂t or

∂E = j ωE ∂t

...(6)

Thus, the partial derivative with respect to time of a phasor is equivalent to the product of jω with corresponding phasor. →

For sinusoidal electrostatic field, the displacement vector D is,

D = D0 e j ω t

...(7)

The partial differentiation of eqn. (7), with respect to ' t' gives ∂D = jωD0 e jωt ∂t or

∂D = j ωD ∂t

...(8)

Similarly, harmonically time varying magnetic field B may be expressed as B = B0 e jω t and its partial derivative in term of phasor is expressed as ∂B = jω B ∂t

...(9)

Thus, the Maxwell’s equations for harmonically time varying electric and magnetic fields in differential and integral forms may be obtain by replacing ∂E / ∂t, ∂D / ∂t and ∂B / ∂t by their corresponding phasor

E lectromagnetic F ield T heory

448

terms from equations (6), (7) and (8) as j ω E, j ω D and jωB respectively. Thus, the differential and integral forms of Maxwell’s equations for fields varying harmonically with time may be expressed as, Differential form

Integral form

→ →

∫S

∇. D =ρ

→ →

→

→

∫

∇ × E = − jωB

→

→

→

→

→

→

→

∇ × H = J + jω D or ∇ × H = (σ + j ω ε) E

→

→

→

∫ H.d l = ∫ ( J

→

→

D . dS = ∫ ρd v v

∫S

∇. B =0

→

→

→

→

B . dS = 0

→

E. d l = − jω ∫

→

→

+ j ω D) . dS or

→ S

→

→

B . dS

→

∫ H.dl

= (σ + j ω ε) ∫

→ S

→

E . dS

Example 9: A parallel plate capacitor is being charged. Show that the displacement current across an area in the region between the plates and parallel to it is equal to the conduction current in the connecting wires. Solution: The electric field across an area A in the region between the plates is given by Q E= ε0 A where Q is the charge accumulated at the positive plate. The flux of the field through an area A is Q Q φE = E . A = .A = ε0 A ε0 The displacement current is given by

id = ε0

d φE dt

= ε0

d Q dQ = d t ε0 d t

dQ But is the rate at which the charge is transported to the positive plate through the connecting wires, dt that is, it is the conduction current. Therefore, id = i

Example 10: An a.c. voltage source is connected across a parallel plate capacitor. Verify that the displacement current in the capacitor is the same as the conduction current in the connecting wires. Solution: Let the a.c. voltage source connected across the plates of a capacitor is

V = V0 sin ω t The conduction current flowing in the wire dV i= C = CV0 ω cos ωt dt The capacity of parallel plate capacitor of an area of cross-section A and plate separation d is

...(1)

...(2)

Waves and A pplications

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449

A d

C=ε

Electric field E in the dielectric medium between the plates is V V0 sin ω t E= = d d V0 sin ω t ∴Electric displacement, D = ε E = ε d

...(3)

→

The displacement current,

id = ∫

or

id =

or or

S

∂D → ∂D . dS = ∫ dS ∂t ∂t

ε V0 ω cos ω t . A d ε A id = V ω cos ω t d 0 id = CV0 ω cos ωt

...(4)

Comparing equations (2) and (4), we obtain

∴

id = i

Example 11: For a conducting medium, σ = 58 × 10 6 Siemen/m, ε r = 1, find out the conduction and displacement current densities if the magnitude of electric field intensity E is given by

E = 150 sin (1010 t ) volt/m →

→

Also calculate the frequency at which| J d | = | J c| . Solution: The conduction current density, J c = σE Here

∴

σ = 58 × 106 S/m and E = 150 sin (1010 t) Volt/m

J c = 58 × 106 × 150 sin (1010 t) = 8. 7 × 10 9 sin (1010 t ) A /m2 →

→

→ ∂ The displacement current density, J d = = (ε E) ∂t ∂t

∂D

∂E ∂ = ε0 ε r E ∂t ∂t

∴

Jd = ε

or

J d = 8 .854 × 10 −12 × 1

∂ [150 sin (1010 t)] ∂t

= 8 .854 × 10 −12 × 150 × 1010 cos (1010 t)] or

Jd = 13.28 cos (1010 t ) Amp /m2 →

For harmonically time varying fields, D = D0 e i ω t →

∴ Thus,

→ → ∂D = j ω D = jω ε E ∂t

Jd = j ω ε E

E lectromagnetic F ield T heory

450

If ω is the angular frequency at which J c = J d , then J c = J d or σ E = ω ε E σ σ σ or f = ω= = ε 2 π ε 2 π ε0 ε r

or

∴

58 × 106

f =

2 × 3 .14 × 8 .854 × 10 −12 × 1

= 1. 043 × 1018 Hz

Example 12: In a material for which σ = 5 S / m and ε r = 1, the electric field intensity is

E = 250 sin (1010 t ) V /m. Find conduction and displacement current densities and the frequency at which both have equal amplitude. [GBTU, B.Tech III Sem 2012]

Solution: The conduction current density J c = σ E Here σ = 5 S / m and E = 250 sin (1010 t)

Jc = 5 × 250 sin (1010 t) = 1250 sin (1010 t ) A / m

∴

∂E ∂t ∂ J d = ε [250 sin (1010 t)] = ε0 ε r 250 × 1010 cos (1010 t) ∂t

The displacement current density, J d = ε

∴

J d = 8. 85 × 10 −12 × 1 × 250 × 1010 cos (1010 t) = 22.125 cos (1010 t ) A / m For harmonically time varying field, J d = jω εE →

→

If ' f ' is the frequency at which | J d| =| J c|, then

σ E = ω ε E or ω = ∴

f =

σ σ = = 2π f ε ε0 ε r

σ 5 = = 8 .996 × 1010 Hz 2 π ε 2 × 3 .14 × 8.85 × 10 –12

Example 13: A conductor of circular cross-section of radius 2mm carries a current ic = 2. 5 sin (5 × 10 8 t ) µ A. What is the amplitude of the displacement current if σ = 35 Ms / m and ε r = 1. Solution: The ratio of the conduction current J c to the displacement current J d is given as,

σ σ Jc = = J d ω ε ωε0 ε r Here, σ = 35 × 106 S / m, ω = 5 × 108 , ε0 = 885 . × 10 −12 and ε r = 1 →

∴

Jc

→

Jd

=

35 × 106 5 × 10 × 8.85 × 10 −2 × 1 8

= 79.09 × 108

The conduction current density is related with conduction current as

Waves and A pplications

`

451

→

J c = ic /(Area)

2

Here, area, A = πa = π (2 × 10

−3 2

) and i = ic sin ω t ic = 2.5 × 10 −6 amp

∴

2.5 × 10 −6

→

∴

Jc =

3.14 × (2 × 10 −3 )2

Hence,

Jc = 79.09 × 108 Jd

or

Jd =

Jc

79.09 × 108

=

= 0.199

0.199 79.09 × 108

or Displacement current, J d = 2. 516 × 10 −11 amp / m2 Example 14: (a) Show that the ratio of the conduction current density and the displacement current density is σ / ω ε, for the applied electric field ( E = Em cos ω t ), by assuming µ = µ0 . (b) What will be the amplitude ratio when the applied field ( E = Em e − t / τ ) when τ is real ? Solution: (a) We know that, the total current density J is the sum of conduction current density, J c and displacement current density J d , that is,

→

∂D

J = J c + J d = σE + ∂t In the given problem, E = Em cos ωt ≈ Em Real e jω t

∴

∂ J c = σ Em Real e jω t and J d = (ε E) ∂t

or

J d = ε jω Em Real e jω t

∴

σEm Real e jω t Jc σ = = j t ω iωε Jd j ε ω Em Real e or

(b)

σ Jc = ωε Jd

Given E = Em e − t / τ

εE ∂ (ε Em e − t / τ) = − m e − t / τ and J c = σE = σEm e − t / τ ∂t τ

Here

Jd =

∴

στ Jc or =− ε Jd

στ Jc = Jd ε

E lectromagnetic F ield T heory

452

Example 15: Using the Maxwell's relation → → ∂ D curl B = µ0 J + ∂t →

→

prove that div D = ρ, where ρ is the charge density. Solution: Taking divergence of both sides of given equation, that is, → → → ∂ D div (curl B) = div µ0 J + ∂t

But ∴

or

→

→

→

→

div curl B = ∇ .( ∇ × B) = 0 → → ∂D =0 µ0 div J + ∂t

→ ∂D div J = − div ∂t →

→

→ ∂ (div D) ∂t → ∂ρ According to continuity equation, div J = − ∂t → ∂ρ ∂ − = − (div D) ∴ ∂t ∂t

or

div J = −

or

div D = ρ

(∵ ∇ is independent of time t)

→

→ →

Example 16: Using Maxwell's equation obtain continuity equation ∇ . J = − ∂ρ / ∂t. Solution: Maxwell's fourth equation is →

→

→

→

∂D ∇×H= J + ∂t

or

→

∇ × B = µ0

→

→

∂D J + µ0 ∂t

→

→

→

→

D = ε0 E

But ∴

→

→

∇ × B = µ0

→

∂E J + µ0 ε0 ∂t

Taking divergence of both sides, we get

→ → → ∂E div ( ∇ × B) = µ0 div J + µ0 ε0 div ∂t

But

→

→

→

→

→

div (∇ × B) = ∇ .( ∇ × B) = 0

Waves and A pplications

`

453

→ ∂ (div E) = 0 ∂t → → ∂ div J = − ε0 (div E) ∂t → ρ From another Maxwell's equation, div E = ε0

∴

∴

→

→

µ0 div J + µ0 ε0

→

div J = −

(∵ ∇ is independent of t)

∂ρ ∂t →

Example 17: If the magnetic field vector H has only aɵ z component given by Hz = 3 x cos β + 6 y sin γ and →

if the field is invariant with time, what is the expression for current density J . Solution: We know that, →

→

→

→

∂D ∇×H= J + ∂t →

→

Since the field is invariant, therefore ∂ E / ∂t or Thus, Here

∴

or or or

→

∂D is zero. ∂t

→

∇×H=J

→

H = Hz = 3 x cos β + 6 y sin γ, Hy = 0, and H x = 0 aɵ x ∂ ∇×H= ∂x 0

→

→

→

→

aɵ y aɵz ∂ ∂ ∂y ∂z 0 3 x cos β + 6 y sin γ →

J = ∇ × H = aɵ x

∂ ∂ (3 x cos β + 6 y sin γ) + aɵ y 0 − (3 x cos β + 6 y sin γ) ∂y ∂x

→

J = aɵ x 6 sin γ − aɵ y 3 cos β

→

J = 6 sin γ aɵ x − 3 cos β aɵ y

Isotropic and Anisotropic Medium A medium whose physical properties are independent of direction or properties are same in all directions is called isotropic medium. On the other hand a medium in which certain physical properties are different in different directions. Such as wood. The strength of wood along the grain differs from the perpendicular to the grain. Single crystals that are not cubic are an isotropic with respect to some physical properties, such as transmission of electromagnetic waves.

E lectromagnetic F ield T heory

454

Q uestions Bank 1.

Derive Faraday law of Induction. Explain the concept of transformer and motional electromotive force. Discuss the relevance of anisotropic media.

2.

[GBTU, B.Tech III Sem, 2012] →

→

Show that the Faraday's law of electromagnetic induction can be expressed as ∇ × E = − down its integral form.

→

∂B ∂t

. Write

[UPTU, B.Tech I Sem 2001]

3.

Explain Faraday's law of electromagnetic induction.

4.

Explain Faraday's law and Maxwell's equations.

5.

What is Faraday's law? Give the statement and prove it.

[UPTU, B.Tech IV Sem 2005]

6.

Explain Faraday's law and its application.

[UPTU, B.Tech IV Sem 2002]

7.

What do you mean by displacement current and conduction current?

8.

Explain the concept of displacement current and show how it led to the modification of the Ampere's law.

9.

[C.C.S. University Meerut, B.Tech IV Sem 2003, 06] [U'Khand, B.Tech IV Sem 2011]

[UPTU, B.Tech I & II Sem 2002, I Sem 2003, II Sem 2007]

Discuss the modification of Ampere's law made by Maxwell's taking displacement current in consideration. Explain the displacement current and its implication.

10. Write the Maxwell's equation. Give physical interpretation of Maxwell's equation. [UPTU, B.Tech IV Sem 2005]

11. State and prove Maxwell's equation and give their physical interpretation. 12. Write four Maxwell's equations in differential form.

[UPTU, B.Tech IV Sem 2004]

[UPTU, B.Tech IV Sem (C.O.) 2006]

13. Write the differential form of Maxwell's equations. Are all four Maxwell's equations independent? 14. Write down the word statement of Maxwell's equations.

Waves and A pplications

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455

nsolved Numerical Problems 1.

A sliding conductor of length l is situated in x − y plane. The conductor is moving with velocity v = v0 aɵ x where v0 is constant. Find the induced emf.

2.

A conductor 30cm long rotates about one end at 1000 r.p.m. in a plane perpendicular to a magnetic field of strength 0. 5 wb / m 2 . Find the emf induced in it.

3.

A flat square coil of 10 turns has sides of length 12cm. The coil rotates in a magnetic field whose flux density is 0. 025 wb / m 2 . What is the angular velocity of the coil if the maximum emf, produced is 20 milli-volt.

4.

A coil of 100 turns and having an average diameter of 0. 4 meter is placed perpendicular to a uniform magnetic field of flux density 0. 5 wb / m 2 . The field is (i) reduced to zero, (ii) reverse in direction, (iii) rotated through 90° in 0. 2 sec. Compute the emf induced in the coil in each case.

5.

A rectangular loop of length a = 1 meter and width b = 0. 8 meter is placed in a uniform magnetic field. →

Determine the maximum value of induced emf, if the magnetic flux density B = 0.1 ω b /m 2 is constant and the loop rotates about the axis with a frequency of 50 Hz. →

→

6.

Consider B = B0 e kt aɵz for cylindrical region ρ < b. Find electric field intensity E using Faraday's law.

7.

A closed coil having 50 turn, area 300 cm 2 , is rotated from a position where its plane makes an angle of 45° with a magnetic field of flux density 2. 0 w b /m 2 to a position perpendicular to the field in a time of 0.1 sec. What is the average emf induced in the coil.

8.

What is the maximum emf induced in a circular coil of 4000 turns of an average radius 12 cm rotating at 30 rev/sec in the earth's magnetic field, where the flux density is 0. 5 × 10 − 4 ω b /m 2 ?

9.

A copper disc of radius 10 cm rotates 20 times per second with its plane perpendicular to a uniform magnetic field. If the induced emf between the centre and the edge of the disc is 3.14 m volt. Calculate the field.

10. A parallel plate capacitor with plate area A and the separation between the plates d, is charged by a constant current I. Consider a plane surface of area A / 2 parallel to the plates and drawn symmetrically between the plates. Find the displacement current through the area. 11. A parallel plate capacitor having plate area A and plate separation ' d ' is joined to a battery of emf, e′ and interval resistance R at t = 0. Consider a plane surface of area A /2 parallel to the plates and situated symmetrically between them. Find the displacement current through this surface as a function of time. 12.

→ → E Do the fields; E = E0 sin x sin t aɵ y (V /m) and H = 0 cos x cos t aɵz (A / m) satisfy Maxwell's µ

equation ?

E lectromagnetic F ield T heory

456

nswers to Unsolved Numerical Problems 1.

V0 B0 l

2.

3. 0.88 rev / sec

4.

(i) 31.4 volt (ii) 62.8 volt (iii) 31.4 volt 1 B ke k tρaɵφ 2 0

0.375 volt

5.

25.12 volt

6. −

7.

8.8 volt

8. 1.70 volt

9.

5.0 m w b /m 2

e ′ − td / ε AR e 11. 2k

10. I / 2 →

12.

→

Since µ0 ε0 = 1, Hence E and H are not satisfying the Maxwell's equations vvv

Theory Numerical Electromagnetic W ave P ropagation

Solved Examples

Questionnaire

457

Unit-4 S ection

B E lectromagnetic W ave P ropagation

I ntroduction

T

h e most remarkable achievement of Maxwell's equations was the prodiction of existence of electromagnetic waves. When Maxwell's equations are combined we get various wave equations that

predict the existence of electromagnetic waves propagating with the velocity of light in different media. In general, wave means propagation of energy or informations. Electromagnetic energy has three fundamental characteristics: They travel at high speed; assume the properties of waves and radiate outward from a source. In this present section, we shall focus our attention towards the simplest type of wave, that is, uniform plane electromagnetic waves in various media. In fact a changing electric field produces a changing magnetic field which in turn generates an electric field and so on with a resultant propagation of energy. It means that a change in either field produces the other field. Maxwell achieved from his equations that variation in electric and magnetic fields would lead to a wave consisting of fluctuating electric and magnetic fields perpendicular to each other and also perpendicular to the direction of propagation of the wave. These waves which can propagation in space even without any material medium are called electromagnetic waves. A uniform plane electromagnetic wave is one in which the electric field E is a function of one space coordinate (say x ) and time coordinate (t) only and is independent of other two space →

→

coordinates (say y and z). The electric and magnetic field vectors E and H lie in a plane perpendicular to the →

→

direction of propagation of wave. In addition, the amplitude and phase of E and H are constant or uniform

E lectromagnetic F ield Theory

458

over the plane in which they lie. The uniform plane wave is a transverse electromagnetic wave or TEM wave. The behaviour of a wave depends on its own and the characteristics of the medium in which they are propagating. First of all we shall derive wave equations and their solutions in free space (where conductivity is zero, that is, σ = 0, ε = ε0 and µ = µ0 ); lossless or perfect dielectrics (where conductivity σ = 0, ε = ε r ε 0 , µ = µ0 or σ > ω ε). For the power consideration in a uniform plane wave we shall develop poynting theorem. The orthogonal → →

relationship between E, H and direction of propagation is always true for a uniform plane wave. The →

→

directions of E and H within a plane perpendicular to the direction of propagation may change. Thus a complete description of electromagnetic wave needs a statement of field vector orientation as a function of time, at a fixed position in space, in addition to the parameters like wavelengths, phase velocity and power, that is, the wave polarisation. Any polarisation state can be described in terms of mutually perpendicular components of the electric field and their relative phasing. The wave polarisation will be discuss in detail. The study of effect of discontinuity in the medium of propagation shall also be consider in this part of study. We shall investigate the behaviour of electromagnetic waves in different combinations of different types of media such as dielectric, conductors and combination of these. We shall restrict to non-magnetic dielectrics. We consider ideally thin, infinitely plane interface between the two linear, isotropic and homogeneous media.

E lectromagnetic Wave Equations

in Different Media & Their Solutions

The general equations for electromagnetic wave propagation in a homogeneous, linear medium is one of the most important consequences of Maxwell’s equations. A medium is homogeneous if its properties do not vary from point to point and a medium is isotropic if its properties are same in all directions. To demonstrate the existence of waves in the electromagnetic field and to derive vector wave equations describing the propagation of waves, we start with Maxwell’s field equations in material medium which are as follows. → →

∇.D = ρ

...(1)

→ →

∇. B =0

→

→

∂B ∂t

→

...(2) ...(3)

→

→

curl H = J + together with Ohm’s law

∇. H = 0

→

∇× E=− →

or

→ →

∂D ∂t

→

J =σE

....(4)

...(5)

where σ is the electrical conductivity of the medium. For linear and isotropic medium,

→

→

D=εE

...(6)

Electromagnetic W ave P ropagation →

459

→

B =µH

and

...(7)

and for homogeneous medium, the dielectric constant ε and magnetic permeability µ are constants. →

The wave equation for E is derived by taking the curl of eqn. (3) as → → → → → ∂ B ∇ × ( ∇ × E) = − ∇ × ∂t → → Substituting B = µ H in eqn. (8), we get →

→

→

∇ × ( ∇ × E) + µ →

→

→

...(8)

∂ → → ( ∇ × H) = 0 ∂t

Putting the value of curl H (or ∇ × H) from eqn. (4), we obtain → ∂ → ∂ D ∇ × ( ∇ × E) + µ =0 J + ∂ t ∂ t → → Substituting the values of J and D in the above equation from eqns. (5) and (6), we get → → → ∂ → ∂ → ∇ × ( ∇ × E) + µ (ε E) = 0 σE+ ∂t ∂t

or

→

→

→

→

→

→

→

∇ × ( ∇ × E) + µ σ

→

∂E

+ µε

→

∂t →

→

→

→ → →

∂2 E ∂ t2

=0

...(9)

→ → →

Using vector identity, A × ( B × C) = ( A . C) B − ( A . B) C in eqn. (9) we get → → →

→

→

∇ ( ∇ . E) − ∇2 E + µ ε

→

∂2 E

+µσ

∂2t

∂E =0 ∂t

...(10)

→

Using eqn. (1) and substituting the value of D from eqn. (6), we get → →

→

∇ . D = ρ or

→ →

∇. E =

or

→ →

→

∇ .( ε E) = ρ

ρ ε

Substituting the above value of ∇. E in eqn. (10), we obtain →

→

→ ∂2 E ∂E ρ ∇ − ∇2 E + µε 2 + µ σ =0 ε ∂t ∂t

→

2

→

→

∇ E = µε

or

∂2 E ∂ t2

→

→ ρ ∂E + µσ + ∇ ε ∂t

Similarly by taking the curl of eqn. (4) and using eqns (5) and (6), we have →

→

→

→ →

→

→

∇ × ( ∇ × H) = ∇× J + ∂ / ∂t ( ∇ × D)

→

→

Substituting the values of J and D from eqn. (5) and (6), we get → → →

→

→

→

∇ ( ∇ . H) – ∇2 H = σ ( ∇ × E) + ε

∂ → → ( ∇ × E) ∂t

...(11)

E lectromagnetic F ield Theory

460 → →

→

→

Substituting the values of ∇ . B from eqn. (2) and ∇ × E from eqn. (3) in the above equation, we get →

∂B ∂2 B –ε 2 ∂t ∂t

→

∂2 H

∇2 H = µ ε

or

→

→

– ∇2 H = – σ

→

∂ t2

+ µσ

→

∂H ∂t

...(12) →

Equations (11) and (12) are the general wave equations in terms of electric and magnetic field vectors, E →

and H which governing the electromagnetic field in a homogeneous, linear medium, whether the medium →

is conducting or non-conducting. However, the electric vector E plays the dominant role in the propagation of electromagnetic energy from one point to another. The first right hand side term of →

→

wave equation (11) in term of E arises due to electric displacement vector D, the second due to conduction current and third due to space distribution of volume charge. (i)

For a homogeneous isotropic dielectric or non-conducting medium in which the disturbance propagates with the same velocity in all directions. That is, ρ = 0 and σ = 0 Therefore, eqns. (11) and (12) for non-conducting medium reduce to 2

→

∇ E = µε

and

→

∇2 H = µ ε

→

∂2 E

...(13)

∂ t2 →

∂2 H

...(14)

∂ t2

(ii) For a conducting medium in which there is, no net charge within a conductor because the charges reside only on the surface. That is, ρ = 0, but σ ≠ 0. Therefore, eqns. (11) and (12) for conducting medium reduce to →

→

→

∂2 E

∂E ∇ E = µ ε 2 + µσ ∂t ∂ t 2

and

2

→

∇ H =µε

→

∂2 H ∂ t2

...(15) →

∂H + µσ ∂t

...(16)

(iii) For a free space in which there are no charges, that is, no conduction current or , ρ = 0, σ = 0, ε = ε0 and µ = µ0 . Thus free space may be consider as perfect dielectric, Therefore, eqns. (11) and (12) for free space reduce to →

∇2 E = µ 0 ε 0 and

→

∇2 H = µ 0 ε 0

→

∂2 E

...(17)

∂ t2

→

∂2 H ∂ t2

...(18)

Thus, the wave equations (11) and (12) are used to study the propagation of electromagnetic waves in different media.

Electromagnetic W ave P ropagation

461

For the solution of wave equations we shall adopt the method of complex variable.The time →

dependence of the field E is taken to be e − j ω t , therefore, →

→ →

E (r , t) = ES (r ) e − j ω t

...(19)

→

ES ( r ) is in general complex so that the actual electric field which is obtained by taking real part of → →

eqn. (19) is proportional to cos (ωt + φ), where φ is the phase of ES ( r ). If eqn. (19) is the solution of eqn. (15), eqn. (19) must satisfy eqn. (15). Therefore, using eqn. (19) and dropping common factor e − jωt , we get →

→

∇2 ES + ω2 ε µ ES + j ω σ µES = 0 [∵ (∂E / ∂t) = − jωES (r ) e − jωt and (∂ 2E / dt 2 ) = − ω2 ES (r ) e − jω t ] →

→

...(20)

→ →

Here the spatial electric field ES depends on the space coordinates, that is, ES = ES ( r ) For plane electromagnetic waves it is convenient to use →

→→

→

ES = E0 e j k. r

...(21)

→

where k is the propagation wave vector defined as →

k =

→ 2π ^ ω ^^ n = n, n is a unit vector along propagation vector k . λ v

Thus, in the light of eqn. (21), eqn. (19) may be expressed as →

→ →

→

E (r, t) = E0 e j( k . r

− ω t)

...(22)

Here E0 is complex amplitude and is constant in space and time.

E lectromagnetic Wave Equations in Free Space Let us derive the wave equation for an electrically free space or vacuum, containing neither free charge →

nor conduction current that is, the charge density ρ = 0, the current density J = 0 and ε = ε0 , µ = µ0 . →

→

→

→

Hence D = ε0 E and B = µ0 H. In fact free space may be treated as perfect dielectric. With these conditions, the Maxwell’s equations reduce to → →

∇.D =0

...(1)

→ →

→

→

∇. B =0 ∇. H =0

→

→

∇× E=−

→

...(2)

→

∂B ∂t

...(3)

→

→

∂D ∇×H= ∂t →

Taking curl of eqn. (3) and substituting B = µ0 H, we get

...(4)

E lectromagnetic F ield Theory

462 →

→

→

∂ → → ∂ → → ( ∇ × B ) = − µ0 ( ∇ × H) ∂t ∂t

∇ × ( ∇ × E) = − →

→

→

→

→

Using eqn. (4) and substituting D = ε0 E, we get, ∇ × ( ∇ × E) = − µ0 ε0 →

→

→

→ → →

∂2 E ∂ t2

→ → →

Using vector identity, A × ( B × C) = ( A . C) B − ( A . B) C, we obtain → → →

2

→

→

∇ ( ∇ . E) − ∇ E = − µ0 ε0

→ →

→ →

2

∂2 E ∂ t2

→

∇. D = 0 or ∇ . E = 0 we get, ∇ E = µ0 ε0

Since

→

→

∂2 E ∂ t2

...(5) →

→

Similary for plane wave equation in term of H , taking curl of eqn. (4) and substituting D = ε0 E, we get →

→

→

→ ∂ → → ∂ → ( ∇ × D) = ( ∇ × ε0 E) ∂t ∂t Using above vector identity and eqns. (2) and (3), we get,

∇ × ( ∇ × H) = →

– ∇2 H = ε0 →

→

∂ → → ( ∇× E) ∂t

Using eqn (3) and substituting B = µ0 H, we get →

– ∇2 H = –µ0 ε0 →

→

→

∂2 H ∂ t2

Thus, the plane wave equations for H and B are 2

→

∇ H = µ0 ε0 2

→

...(6)

∂t2

→

→

∇ B = µ0 ε0

and

→

∂2 H

→

∂2 B ∂t2

...(7) →

Substituting E = D / ε0 in eqn. (5), we get plane wave equation for D as, →

→

∇2 D = µ0 ε0

∂D

...(8)

∂t2 →

→

The equations (5), (6), (7) and (8) represents wave equations governing electromagnetic fields E and H in free space. Equations (5) to (8) are vector equations of identical form and if compared with general (classical) wave equation of the form 1 ∂2 u ∇2 u = 2 v ∂ t2 gives the velocity of propagation of electromagnetic wave through free space as, v=

1 1 = −7 µ0 ε0 4 π × 10 × 8 .85 × 10 −12 (∵ ε0 = 8 .854 × 10 −12 C 2 / N- m2 and µ 0 = 4 π × 10 −7 N- sec 2 /C 2)

= 2.998 × 10 8 m/sec = c

Electromagnetic W ave P ropagation

463

Thus, the electromagnetic waves in free space propagate with the velocity of light. This established that, the light waves are also electromagnetic. Therefore, equations (5) to (8) take the form →

∇2 E = 2

→

∇ H =

and

→

1 ∂2 E →

1 ∂2 H c2 ∂ t2

or

2

→

∇ B =

→

1 ∂2 B c2 ∂ t2

...(10)

→

→

∇2 D = µ0 ε0

P

...(9)

c2 ∂ t2

∂D ∂t

...(11)

ropagation of Uniform Plane Electromagnetic Waves in Vacuum

An electromagnetic wave is said to be a uniform plane wave if it propagates only along one space coordinate and is independent of other two space coordinates. the amplitude and phase of their field →

→

vectors E and H are uniform or constant over the whole plane perpendicular to the direction of propagation. Thus, a wave which is independent of x and y, and is a function of z and t only is said to be a uniform plane wave. Therefore for a uniform plane wave travelling in z − direction, we have →

→

→

→

E = E(z, t) and H = H (z, t) →

...(1)

→

The partial derivatives of components of E and H with respect to x and y is zero. That is ∂E y ∂H y ∂E x ∂H x = 0 and = 0 and = 0, =0 ∂y ∂y ∂x ∂x Hence, Maxwell's first equation for plane wave reduce to, → ∂E div E = 0, that is, z = 0 or Ez is constant in space ∂z

...(2)

Similarly, second equation yields →

div H = 0, that is ,

∂Hz = 0 or Hz is constant in space ∂z

...(3)

Equations (2) and (3) show that Ez and Hz cannot be a function of z which is the direction of propagation. It means that Ez and Hz are constant in z − direction. Thus, they represent static components and consequently, no part of wave equation. Therefore, we shall have, →

Ez = Hz = 0

→

...(4)

Hence, the field vectors E and H have only x and y components, that is, →

E = aɵ x Ex + aɵ y E y

→

H = aɵ x H x + aɵ y H y

and →

→

...(5) ...(6)

Therefore E and H do not have any z-component, the z − direction being the direction of propagation, both these field vectors are perpendicular to the direction of propagation. So, Maxwell's electromagnetic waves are purely transverse.

E lectromagnetic F ield Theory

464

Third Maxwell equation for free Space is

→

→

→ → ∂B ∂H or ∇ × E = – µ0 ∂t ∂t In determinant form above equation is expressed as aɵ x aɵ y aɵz → →

∇× E = –

→

→

∇× E =

or

∂ ∂x Ex

∂ ∂y Ey

→

→

(∵ B = µ0 H for free space) ...(7)

∂H y ∂ ∂H x ∂Hz = – µ0 [ aɵ x ] + aɵ y + aɵz ∂z ∂t ∂t ∂t Ez

∂H y ∂E y ∂E y ∂E x ∂H x ∂E ∂Hz ∂E ∂E – + aɵz + aɵ y aɵ x z – = – µ0 aɵ x + aɵ y x – z + aɵz ∂z ∂y ∂t ∂t ∂z ∂x ∂t ∂x ∂y

For uniform plane wave Ez = 0,

∂ ∂ = =0 ∂x ∂y

Thus, eqn (8) for uniform plane wave reduces to ∂E y ∂H y ∂E x ∂H x ∂Hz – aɵ x – aɵ y µ0 – aɵz µ0 + aɵ y = – aɵ x µ0 ∂t ∂z ∂z ∂t ∂t On equating components separately, we get. ∂E y ∂E y ∂H x ∂H x or – = –µ0 = µ0 ∂t ∂t ∂z ∂z ∂ H ∂E x y = –µ0 ∂z ∂t ∂Hz and 0= ∂t Similarly, from fourth Maxwell's equation for free space, that is, →

...(8)

...(9)

...(10) ...(11) ...(12)

→

→

∂E ∇ × H = ε0 ∂t

or

∂E y ∂E x ∂Hz ∂H y ∂Ez ∂H x ∂Hz ɵ ∂H y ∂H x – + aɵ y + aɵz aɵ x – – = ε0 aɵ x + aɵ y + az ∂y ∂t ∂t ∂t ∂z ∂z ∂x ∂x ∂y

For uniform plane wave, Hz = 0 and

...(13)

∂ ∂ = =0 ∂x ∂y

Thus eqn. (13) for uniform plane wave reduces to – aɵ x

∂H y ∂z

+ aɵ y

∂E y ∂H x ∂E x ∂Ez = aɵ x ε0 + aɵ y ε0 + aɵz ε0 ∂z ∂x ∂y ∂z

...(14)

On equating the components separately, we get –

and

∂H y ∂z

= ε0

∂H y ∂E x ∂E x or = – ε0 ∂z ∂t ∂t ∂E y

∂H x = ε0 ∂t ∂z ∂Ez 0= ∂t

Differentiating eqn. (11) with respect to z and eqn. (15) w.r.t. t, we get

...(15) ...(16) ...(17)

Electromagnetic W ave P ropagation ∂2 E x ∂z2

∂2 H y

and

∂t ∂z

= –µ0 = – ε0

465

∂2 H y

...(18)

∂z∂t ∂2 E x

...(19)

∂ t2

Substituting the value of ∂2 H y / ∂t∂z from eqn. (19) in eqn. (18), we get ∂2 E x ∂z

2

∂2 E x

= µ0 ε0

...(20)

∂t2

Similarly, differentiating eqn. (10) w.r.t.z and eqn. (16) w.r.t., t, we get ∂2 E y ∂z

2

= µ0

∂2 H x ∂z∂t

...(21)

∂2 E y ∂2 H x = ε0 ∂z∂t ∂ t2

and

...(22)

Substituting the value of ∂2 H x / ∂z ∂t from eqn. (22) in eqn. (21), we get ∂2 E y ∂z 2

= µ0 ε0

∂2 E y

...(23)

∂t2 →

Plane wave differential equations similar to eqns (20) and (23) are obtained for H components as, ∂2 H x ∂z 2

∂2 H y

and

∂z 2

= µ0 ε0 = µ0 ε0

∂2 H x

...(24)

∂t2

∂2 H y

...(25)

∂t2

Thus, eqns (20), (23), (24) and (25) are the uniform plane wave equations for the propagation of y and x →

→

components of E and H along z-axis with constant velocity in free space. →

For simplicity, let us consider the coordinate axis such that y-axis is parallel to the field vector E and x − →

axis parallel to the field vector H. Since E y is quite independent of E x and H x is quite independent of H y . Therefore, we can assume that E x = 0 and H y = 0, Then, and

→

E = aɵ y E y (z, t)

...(26)

→

H = aɵ x H x (z, t)

...(27)

Since the field of our plane electromagnetic wave, in which we are interested, is a simple periodic →

→

function of time. Therefore, in general the time dependence of E and H is e – jω t for single fixed frequency. Hence we can write eqn. (26) as E y ( z , t) = f ( z ) e – jω t

...(28)

Since eqn. (28) is the solution of plane electromagnetic wave represented by eqn. (23), it must satisfy eqn. (23), that is,

E lectromagnetic F ield Theory

466

∂ 2 E y ∂ 2 f – jωt ∂ f – j ωt and e = 2 e ∂z ∂z ∂z2 ∂z and differentiating eqn. (28) w.r.t, t, we get ∂E y

∂E y ∂t ∂2 E y

or

∂ t2

=

= – jωf (z)e – jωt and

∂z2 or

= (– jω)2 f (z)e – jωt

∂2 E y ∂ t2

...(30) in eqn. (23), we get

e – jωt = µ0 ε0ω2 f (z)e – jωt

∂ 2 f ( z) 2

=

ω2

(∵ µ0 ε0 =

f ( z)

2

∂z v clearly eqn. (31), has two solutions and

∂ t2

= ω2 f (z)e – jωt

Substituting the values of ∂2 E y / ∂z2 and ∂ 2 f ( z)

∂2 E y

...(29)

f (z) = E0

y

e jω z / v

f (z) = E0

y

e – jω z / v

1 v2

)

...(31)

...(32)

where E0 y is constant. Hence a general solution of eqn. (23) is E y = E0 y [e

z jω ( t + ) v

+e

z – jω ( t – ) v ]

...(33)

Hence, either term of eqn (33) is a solution or the sum as in eqn. (33) is a solution of uniform plane electromagnetic wave represented by eqn. (23). We can consider first term of eqn. (33) and taking the imaginary part as solution, then we get z ...(34) E y = E0 y sin ω t ± v The positive sign (+) represents a wave travelling to the left and negative (–) sign represents a wave travelling to the right along the z-axis. For a wave travelling in the positive direction of z-axis, eqn. (34) reduces to, E y = E0 y sin ω t – or

z v

E y = E0 y sin(ωt – kz ), where k =

...(35) ω 2π = v λ

...(36)

We can obtain the value of H x by putting the value of E y from eqn. (35) in eqn. (10), we get ∂H x ∂ [ E sin ω (t – z / v)] = µ0 ∂t ∂z 0 y E0 or

y

ω ∂H x cos ω(t – z / v). – = µ0 v ∂t

∂H x ω =– E ∂t µ0 v 0

y

cos ω (t – z / v)

...(37)

Electromagnetic W ave P ropagation

467

Integrating eqn. (37), we get

∫

∂H x ω dt = – cos ω t – ∂t µ0 v ∫ Hx =

ω E0 y vµ0

z dt v

sin ω(t – z / v )

or

z H x = H0 x sin ω t – v

or

H x = H0 x sin(ωt – kz ), where H0 x =

or

Bx = B0 x sin(ωt – kz ), where B0 x =

...(38) E0 y

...(39)

vµ0

E0 y

...(40)

v

Hence eqns. (36) and (40) are the equations for the propagation of the uniform plane electromagnetic waves in free space or in vacuum. E0 µ = 0 = 377 ohm H0 ε0

or

This ratio is called impedance of free space The value of E y / B x is real and depends on the medium in which wave is travelling eqns. (36) and (40) →

represent linearly polarised plane wave travelling along the positive z-axis in which E is every where →

→

parallel to y-axis and B or H is parallel x −axis as shown in fig. 1.

Ey

Z

Fig.1

From eqns (36) and (40) we have Ey Bx

=

E0

y

B0 x

= v or

Ey Bx

=

1 µ0 ε0

1 ∵ v = µ0 ε0

E lectromagnetic F ield Theory

468

P lane Wave Equation

and its Solution in Isotropic Dielectric (or Perfect Dielectric or Lossless) Medium

A non-charged, current free dielectrics are non-conducting media, therefore in non-conducting media other than free space the current density J = 0 ∵ σ = 0, and in homogeneous isotropic medium, the charged density is zero, ρ = 0 because there is no volume distribution of charge. Further in these →

non-conducting media electric displacement D →

→

→

at any point is in the direction of E and magnetic

induction B is in the direction of H. Therefore, →

→

→

→

→

→

D = ε E, B = µ H and J = σ E = 0, ε = ε, µ = µ

The Maxwell’s equations for such media are reduce to →

div E = 0

...(1)

→

div H = 0

...(2) →

→

curl E = − µ

∂H

...(3)

→

∂t →

→

∂E ∂t

curl H = ε

...(4)

Taking curl of eqn. (3), we obtain →

→

∂H ∂ → → curl curl E = − µ ∇ × ( ∇ × H) = −µ ∂t ∂t →

→

→

→

Substituting the value of curl H (or ∇ × H) from eqn. (4), we get →

→

curl curl E = − µ ε →

→

→

∂2 E

...(5)

∂ t2

→ → →

→ → →

Using vector identity, A × ( B × C) = ( A . C) B − ( A . B) C we can expand left hand side of eqn. (5) as →

→

→

curl curl E = grad div E − ∇2 E →

From eqn. (1), div E = 0, therefore →

→

→

∂2 E

curl curl E = − ∇2 E Hence, eqn. (5) becomes ∇2 E = µ ε

→

∂ t2

...(6)

In a similar manner by taking curl of eqn. (4) and using eqn. (2), one can derive the wave equation →

satisfied by H as 2

→

∇ H = µε

→

∂2 H ∂ t2

If eqns. (6) and (7) are compared with general wave equation of the form 2 1 ∂ u ∇2 u = 2 v ∂ t2

...(7)

...(8)

Electromagnetic W ave P ropagation

469

where v is the speed of wave. The comparison of eqns. (6) and (8) gives the velocity of propagation of electromagnetic wave through non-conducting medium as 1 1 v= = µε µ0 µ r ε0 ε r v=

or

∵

c µ r εr

1 = c µ0 ε0

...(9)

where µ r is the relative permeability, ε r the relative permitivity of the medium and c the velocity of light. Since µ r > 1 and ε r > 1; v < c, that is, the speed of electromagnetic waves in an isotropic dielectric is less than the speed of electromagnetic waves in free space. We know that, the refractive index n of the medium is n = c / v or v = c / n

...(10)

Comparison of eqns. (9) and (10) gives n = √ (µ r ε r )

...(11)

For non-magnetic material µ r = 1, therefore or n2 = ε r

n = √ εr

...(12)

Eqn. (12) is the well known Maxwell's relation that has been verified in a number of experiments. If we replace µ ε by 1/v2 in eqns. (6) and (7), we get wave equations as, →

→

∇2 E − →

∇2 H −

and →

1 ∂2 E v2 ∂ t2 2 1 ∂ H

v2 ∂ t2

=0

...(13)

=0

...(14)

→

The field vectors E and H in exponential form may be expressed as, →

→

E = E0 e jω t and H = H0 e jω t

Differentiating the above equations twice with respect to 't', we get. ∂2 E ∂ t2 ∂2 E

or Similarly Substituting the values of

∂t

2 →

∂2 H ∂t ∂2 E ∂t

2

2

=

∂ ∂E ∂ jω t jω t 2 2 = ( jωE0 e ) = j ω E0 e ∂t ∂t ∂t →

= –ω2 E

(∵ J 2 = –1)

...(15)

→

= – ω2 H

...(16)

from eqn. (15) and →

→

∂2 H ∂ t2

from eqn. (16) in eqns. (6) and (7), we get

→

→

∇2 E = – µ ε ω2 E and ∇2 H = – µ ε ω2 H or and

→

→

→

→

∇2 E = – β2 E ∇2 H = –β2 H

...(17) ...(18)

E lectromagnetic F ield Theory

470

where β2 = ω2 µ ε or β = ω µ ε , Expansion of eqns. (17) and (18) give aɵ x and

∂2 E x ∂x

2

∂2 H x

aɵ x

∂x 2

+ aɵ y + aɵ y

∂2 E y 2

∂y

∂2 E y ∂y2

+ aɵz + aɵz

∂2 Ez ∂z2

∂2 Ez ∂z2

= –β2 [aɵ x E x + aɵ y E y + aɵz Ez ] = –β2 [aɵ x H x + aɵ y H y + aɵz Hz ]

Ez = 0 because wave is travelling along z-direction, therefore. aɵ x and

aɵ x

∂2 E x ∂x 2

∂2 H x ∂x 2

+ aɵ y + aɵ y

∂2 E y ∂y2 ∂2 E y ∂y2

= –β2 [aɵ x E x + aɵ y E y ]

...(19)

= –β2 [aɵ x H x + aɵ y H y ]

...(20)

Thus x and y- components of eqns (19) and (20) are ∂2 E x ∂x 2

and

∂2 E y ∂y2 ∂2 H x ∂x 2

and

∂2 H y ∂y2

= –β2 E x

...(21)

= –β2 E y

...(22)

= –β2 H x

...(23)

= –β2 H y

...(24)

Let us consider E x component only. Eqn (21) is the second order differential equation whose general solution is of the form. E x = Ae – jβz + Be jβz

...(25)

Where A and B are constants and β is the phase shift constant. Time variation of eqns. (25) becomes E x ( z , t) = e jω t E x or

E x (z, t) = Ae j(ω t −βz) + Be j(ω t +βz)

...(26)

Eqn. (26) is complex and represents the sum of two travelling waves in opposite directions. Real part of eqn. (26) becomes E x (z, t) = A cos(ω t – βz) + B cos(ω t + βz)

...(27)

Similar equation for H y (z, t) is H y (z, t) = C cos(ω t – βz) + D cos(ω t + βz) The intrinsic wave impedance of uniform plane wave in perfect dielectric is given by η=

or

η=

µ0 µ r Ex µ = = Hy ε ε0 ε r 4 π × 10 –7 8.85 × 10

–12

µr µr = 377 ohm εr εr

...(28)

Electromagnetic W ave P ropagation

471

P lane

Electromagnetic Wave and its Solution in Conducting Media (Lossy dielectric or Partially Conducting Media)

In a conducting medium →

→

J =σE

where σ represents the conductivity of the isotropic and homogeneous medium.Thus, for conducting medium, the Maxwell’s equations reduce to → →

∇. E = 0

...(1)

→ →

∇. H =0

→

→

→

→

...(2) →

∂H ∂t

∇ × E = −µ

→

→

∇ × H = σE + ε

and

...(3) ∂E ∂t

Taking curl of eqn. (3), we obtain →

→

→

∂H ∂ → → ∇ × ( ∇ × E) = − µ ∇ × = −µ ( ∇ × H) ∂t ∂t →

→

...(4)

→

→

Substituting the value of ∇ × H from fourth Maxwell’s eqn. (4) and using vector identity →

→

→

→ → →

→ → →

A × ( B × C) = ( A . C) B − ( A . B) C, we get → → →

→

∂ ∇ ( ∇ . E) − ∇ E = − µ ∂t 2

→ → ∂ E σ E + ε ∂ t

→ →

Putting ∇ . E = 0 from Maxwell’s first eqn. (1), we get →

→

− ∇2 E = − µ σ

→

∂E ∂2 E − µε 2 ∂t ∂t →

→

→

∂E ∂2 E ∇ E =µσ + µε ∂t ∂ t2 2

or

...(5)

→

Similar wave equation for H is obtained by taking curl of eqn. (4) as → → → → → ∂ → → ∇ × ( ∇ × H) = σ( ∇ × E) + ε ( ∇ × E) ∂t

→ →

...(6)

Again using above vector identity and substituting the value of ∇× E from eqn. (3) in eqn. (6), we get

→ →

→

∂H ∂2 H –µ ε 2 ∇ .( ∇ . H) – ∇ H = – σ µ ∂t ∂t

→

→ →

2

→

Using ∇. H = 0 from eqn. (2), we get →

→

∂H ∂2 H – ∇ H = – σµ – µε 2 ∂t ∂t 2

→

E lectromagnetic F ield Theory

472

or

→

→

∂H ∂2 H ∇ H = – σµ + µε 2 ∂t ∂t 2

→

...(7) →

→

Wave equation in terms of displacement vector D is obtained by substituting in eqn. (5), E =

Therefore,

→ ∂2 D ∂t2 ε

→ → ∂ D D ∇ =µσ +µε ε ∂t ε 2

→

D . ε

∂ t2 →

∂D ∂2 D ...(8) or ∇ D = µσ +µε 2 ∂t ∂t The eqns. (5) to (8) are also true for lossy dielectric or partially conducting media in which a electromagnetic wave loses power as it propagates or in other words a lossy dielectric is an imperfect dielectric or imperfect conductor with σ ≠ 0. →

2

→

If we consider the propagation of plane E M wave in z-direction with electric field vector E polarised in →

x-direction and magnetic field vector H in y-direction only, then →

∂ ∂ and =0 = ∂x ∂ y

→

E = E x aɵ x , H = H y aɵ y

Applying above conditions in equations (5) and (7)after expressing in cartesian coordinates, we obtain ∂2 E x ∂z and

2

∂2 H y ∂z

2

=µε

=µε

∂ 2 Ex ∂t

2

+µσ

∂2 H y ∂t

∂ Ex ∂t

+µσ

...(9)

∂ Hy

...(10)

∂t

If electric field E x varies harmonically with time, then E x = E0 e j ω t

∴ and

∂ Ex = j ω E0 e j ω t ∂t ∂2 E x ∂ t2

...(11) or

= ( j ω) ( j ω) E0 e j ω t

∂ Ex = j ω Ex ∂t or

∂2 E x ∂ t2

= − ω2 E x

Substituting the above values of ∂ E x / ∂t and ∂2 E x / ∂ t2 in eqn. (9) we obtain ∂2 E x ∂ z2

= − µ ε ω2 E x + µ σ j ω E x

or

∂2 E x

or

∇2 E x = γ2 E x or ∇2 E = γ2 E

∂ z2

= j ω µ ( σ + j ω ε) E x

(∵ − 1 = j 2) ...(12)

where γ2 = jωµ (σ + jωε), γ is called propagation constant, which is a complex quantity and has real and imaginary parts. Suppose γ =α + jβ where the real part α is called attenuation constant and imaginary part β the phase constant.

Electromagnetic W ave P ropagation

473

Eqn. (12) is a wave equation representing the propagation of uniform plane electromagnetic wave through a conducting (or dielectric) medium of permittivity ε and permeability µ. Let us suppose that the solution of eqn. (12) is of the form E x = E0 e − γ . z or E x = E0 e − (α

+ j β). z

...(13)

j ω µ σ − µ ε ω2

with

γ=

j ω µ (σ + j ω ε) =

or

γ=

ε ω j ω µ σ 1 + j σ

...(14)

Multiplying eqn. (13) by e j ω t and taking the real part, we obtain Ex = E0 e − α z cos (ωt − βz ) This is a uniform plane wave that propagates in the forward z-direction with phase constant β, but which (for positive α) loses amplitude with increasing z according to the factor e − α z . Thus, the general effect of complex value of γ is to give a travelling wave that changes its amplitude with distance. If α is positive, it is called attenuation coefficient. If α is negative, the wave grows in amplitude with distance, and the α is called the gain coefficient. For good conductors σ > > 1 or ωε

ωε 0 or v g < vp

Hence, in normally dispersive media the group velocity is less than the phase velocity. All non-coloured transparent substances exhibit normal dispersion in the visible region. 2.

Anomalously Dispersive Media : In these media, the phase velocity decreases with the increase of wavelength, that is, the ratio of the change in phase velocity to the change in wavelength is negative. Therefore, for anomalously dispersive media, dvp dλ

< 0 or v g > v p

Hence, in anomalously dispersive media group velocity is greater than phase velocity. A conductor is an example of an anomalously dispersive medium. In fact, in this case the terms normal and anomalous are arbitrary, the significance being simply that the two dispersions are different.

E lectromagnetic F ield Theory

490

Relation between V P and V g For a particular frequency at which band width is extremely small ∆ω dω v g = lim = dβ ∆ ω → 0 ∆ β We know that, phase velocity,

vp =

ω β

∴ ω = β vP Substituting the value of ω in eqn. (18), we obtain d (β vP ) vg = dβ vg = v p + β

or

d vp

dβ dβ 2π 2π β β= ∴ =− 2 =− dλ λ λ λ

But

dλ λ Substituting the value of dβ in eqn. (20) from eqn. (21), we get d vp vg = v p − λ dλ dβ = −β

or

...(18)

...(19)

...(20)

...(21)

...(22)

Eqns. (20) and (22) are well known relations between vp and v g . v g can also be obtained in terms of refractive index n as dω , Since vg = dβ d vp

so

dβ

=

d vp dω dvp . = vg dω dβ dω

Substituting this value of dvp / dβ from eqn. (23) in eqn. (20), we obtain dvp v g = vp + βv g dω

...(23)

...(24)

Using vp = c / n and differentiating it with respect to ‘ ω’, we get d vp c dn =− 2 dω n dω Substituting this value of

d vp dω

in eqn. (24), we get c dn v g = vp + βv g − 2 n dω β c v g dn n2 dω

∴

v g = vp –

or

β c dn v g 1 + 2 = vp n dω

But

β=

ω ωn = vp c

(∵ vp = c / n)

Electromagnetic W ave P ropagation ∴ or

491

c ω dn v g 1 + = vp = n dω n c vg = n + ω (dn / dω )

...(25)

This is another form of group velocity. From eqn. (22) it is evident that the group velocity (v g ) exceeds the phase velocity (vp ) when phase velocity decreases with increasing wavelength or when dvp / dλ is negative (A case of anomalously dispersive medium). On the other hand, the group velocity is smaller than phase velocity when vp increases with the increase of wavelength or when dvp /dλ is positive (A case of normally dispersive medium). In interstellar space, that is, in vacuum there is no dispersion, that is, dvp / dλ = 0. From eqn. (22) we have v g = vp That is, in interstellar space, the group velocity is same as the phase velocity. Example 7: Find the group velocity for a plane electromagnetic wave of 1 MHz travelling in a normally dispersive loss less medium which has a phase velocity v p equals to the velocity of light. The phase velocity is a function of λ which is given as, v p = k √ λ, where k is a constant. Solution: The group velocity of a wave is related to the phase velocity as, d vp v g = vp − λ dλ Here ∴ or or or

vp = k √ λ

d (k λ ) λk = vp − dλ 2 λ vp 1 v g = vp − k λ = vp − 2 2 8 vp m/ sec 3 × 10 c vg = = = 2 2 2 vg = 1.5 × 10 8 m/sec v g = vp − λ

Example 8: Determine the phase velocity of a plane electromagnetic wave at frequency 10 GHz in polyethene. For polyethene µ r = 1, ε r = 2 .3 and σ = 2 .56 × 10 −4 mhos/m Solution: For insulator, like polyethene vp = Here ∴

ω 1 1 = = β µε µ0µ r ε0 ε r

µ r = 1, µ0 = 4 π × 10 −7N/A2 , ε r = 2 . 3 and ε0 = 8 .854 × 10 −12 Coulomb2 /N-m2 vp =

1 4 π × 10

−7

× 1 × 8 .854 × 10

−12

× 2 .3

= 1.977 × 10 8 m/sec

E lectromagnetic F ield Theory

492

C onductors and Dielectrics In electromagnetic, materials are classified roughly as conductors and dielectrics. In fact the dividing line between conductors and dielectrics is not sharp, it depends on frequency. A particular material can behave as a conductor in low radio frequency range and as a dielectric in high frequency range. For example, fresh water behaves as a good conductor for frequency, f ≤ 103 sec −1 and as a dielectric for f ≥ 10 −7 sec −1. →

→

Maxwell’s equation relating to magnetic field vector H and the electric displacement vector D is →

→

→

→

∂D ∇×H= J + ∂t

...(1)

→

where J is the current density. →

→

→

J = σ E and D = D0 e j ω t

But ∴

→

→

→

→

→

→

(For harmonically time varying field)

→

→

∇ × H = σ E + jω D →

→

∇ × H = σ E + jω ε E

or

→

(∵ ∂ D / ∂t = jω D) →

(∵ D = ε E) ...(2) →

For a linearly polarized plane electromagnetic wave propagating in the z-direction with E polarized in →

x-direction and H confined to y-direction, we have ∂ ∂ = = 0 and ∂x ∂ y

→

→

E = E x aɵ x , H = H y aɵ y

With these conditions eqn. (2) reduces to the scalar phasor equation as −

∂Hy ∂z

= σ Ex + j ω ε Ex

∂H y → → ∵ ∇× H = aɵ x 0 – ∂z

...(3)

The first term σ E x on the right hand side of eqn. (3) corresponds to the conduction current density while the second term ω ε E x corresponds to the displacement current density in the medium. Therefore, the rate of change of magnetic field intensity H y is the sum of conduction current and displacement current densities. Eqn. (3) may be rearrange as −

∂Hy ∂z

= (σ + j ω ε) E x σ = j ω ε − j E x ω

or where

−

∂Hy

= j ω ε1 E x

...(4)

σ ε1 = ε − j ω

...(5)

∂z

Therefore the medium may behave like a dielectric, conductor or quasi-conductor, depending upon whether ω ε >>, = or > σ, that is, when the displacement current is much greater than conduction current, the medium behaves like an insulator or dielectric. Insulating behaviour of medium further depends on the conductivity σ of the medium. If conductivity σ is approximately equal to zero, the medium behaves as a perfect dielectric or loss-less dielectric. On the other hand if conductivity is non-zero, that is, σ ≠ 0, the medium behaves as a lossy or partially conducting imperfect dielectric. However, if ω ε / σ >> 1, medium generally behaves as dielectric. Case II : When ω ε ~ σ, that is, when the conduction current is approximately equal to the displacement current, the medium is said to be quasi-conductor but not semi-conductor. Case III : ω ε > 1 (say ≥ 100), the medium can be classified as good conductor over entire radio frequency ωε range. σ If > 1 and µ = µ0 µ r , ε = ε0 ε r ωε Maxwell's equations for harmonically or sinusoidally time varying fields are: → →

∇. E = 0

...(2)

→ →

∇. H =0

→

...(3)

→

→

∇ × E = – jµω H

→

→

→

→

...(4) →

∇ × H = (σ + jωε) E

and

....(5)

Taking curl of eqn. (4), we get →

→

→

∇ × ( ∇ × E) = – j µ ω ( ∇ × H)

→

→

Substituting the value of ∇ × H from eqn. (5) and using vector identity, we get → → →

→

→

∇ ( ∇ . E) – ∇2 E = – j µ ω (σ + jω ε) E → →

∇. E =0

But from eqn. (2)

→

→

∴

– ∇2 E = – jµ ω (σ + jω ε) E

or

∇2 E = jµ ω (σ + jωε) E

or

∇2 E = γ E

→

→

→

→

or

→

→

∇2 E – γ E = 0

...(6)

→

Similar equation for H is, →

→

∇2 H – γ H = 0

...(7)

where γ is the propagation constant that can be expressed as γ2 = j ω µ (σ + j ω ε) j ω ε = j ω µ σ 1 + σ Since

ωε σ >>1, ∴ > ω ε0 , therefore silver is a good conductor For good conductor, η2 =

ωµ ∠45 ° σ

∴

η2 =

2 π × 200 × 106 × 4 π × 10 –7 × 1 ∠45 ° . × 106 617

or

η2 = 5.056 × 10 –3

In free space

η1 = η3 =

µ0 = 377Ω ε0

→

We know that

τE = 1

→

| E2 | →

| E1|

=

2 η2 2 × 5.056 × 10 –3 10112 × 10 –3 . = = – 3 η1 + η2 377 + 5.056 × 10 377005 .

× 10 –3 × 10 –3 × 100 10112 10112 . . | E1| = 377005 37705 . .

∴

| E2 | =

or

| E2 |= 2.68 × 10 –3 V / m

→

For good conductor,

α = β = πfµσ = (3 .14 × 200 × 106 × 4 π × 10 –7 × 61. 7 × 106 )

or

α = 2 . 206 × 10 5

∴

→

→

| E3 | =| E2 | e – α x , Here x = 5 µ m = 5 × 10 –6 m →

| E3 | = 2 . 68 × 10 –3 e –(2 .206 ×10 ∴

→

5

×5 ×10 –6 )

E4

= 2 . 68 × 10 –3 e –1.103

| E3 | = 2 . 68 × 10 –3 × 0 . 3319 = 8 . 89 × 10 –4 V /m

µ0 ε0

E lectromagnetic F ield Theory

542

→

τE =

We know that

2

→

| E4 | =

or

| E4 | →

| E3 |

=

2 η3 2 × 377 = η3 + η2 5.056 × 10 –3 + 377

→ 754 × 8 . 89 × 10 754 | E3 | = 377005 377. 005 .

–4

= 1. 78 × 10 –3 V/m

Example 31: A free space -silver interface has Ei0 = 100 V /m on the free space side. The frequency is 15 MHz and for silver medium ε r = µ r = 1 and σ = 617 . MS/m. Determine the magnitudes of reflected and transmitted electric field intensities (Er0 and Et0 ) at the interface. Solution: For perfect dielectric (free space)- conductor (silver) interface, the incident wave is totally reflected, that is, the reflection coefficient, Γ is –1 or Γ = –1 Ei0 = –1 ∴ Er0 = – Ei0 or Er0 = –100 V /m Er0

or The transmission coeffcient,

τ=

Et0 2η2 = Ei0 η1 + η2

As silver medium is good conductor therefore η2 =

2 π × 15 × 106 × 4 π × 10 –7 × 1 (∵ µ = µ0 µ r ) ∠45 ° = σ . × 106 617

ωµ

or

η2 = 1.3848 × 10 –3 Ω

For free space,

η1 =

µ0 = 377Ω ε0

∴

2 × 1. 3848 × 10 –3 Et0 = = 7. 346 × 10 –6 Ei0 377 + 1. 3848 × 10 –3

so

Et0 = 7. 346 × 10 –6 × Ei0 or Et0 = 7. 346 × 10 –6 × 100 V / m

or

Et0 = 7. 346 × 10 –4 V/m

S tanding Wave Ratio When uniform plane electromagnetic waves travelling in a perfect dielectric medium (σ1 = 0, α1 = 0) are ωµ2 incident normally on the interface with perfect conductor (σ2 = β = ∞ and η2 = = 0), The waves get σ2 totally reflected in the same medium and no parts of the waves are transmitted into the second medium. It is due to zero intrinsic impedance (η2 = 0) for perfect conductor and zero transmission coefficient → → η2 − η1 (τ = 0) for field vectors E and H. The reflected waves combine with (Since, Γ = = − 1 for η2 = 0 η2 + η1 and τ =

2 η2 η1 + η2

= 0) incident waves produce standing waves.

Electromagnetic Wave Propagation

543

Let us discuss a different case in which uniform plane waves travelling in a prefect dielectric medium fall normally on the interface with a medium for which intrinsic impedance is non-zero (η2 ≠ 0) and reflection coefficient for electric and magnetic fields lies between −1 and +1 (−1 < Γ < + 1). Under these conditions plane waves are partially reflected from the boundary of the two media and partially transmitted into the second medium. Medium I therefore supports a field that is composed of both a travelling wave and a standing wave. It is customary to describe this field as standing wave, even though a travelling wave is also present. It will be seen that the field does not have zero amplitude at any point for all time, and the extent to which the field is divided between a travelling wave and a true standing wave is represented by the ratio of the maximum amplitude to the minimum amplitude. The ratio of maximum to minimum amplitudes is called standing wave ratio and is expressed as, →

S.W.R, S =

| E|max →

→

=

| E|min

| H|max →

| H|min

Let us consider a case in which the intrinsic impedance of the medium II is greater than the intrinsic impedance of medium I, (that is, η2 > η1) then reflection coefficient, Γ > 0. When the plane wave travels in such a medium, there is transmitted wave in medium II and a standing wave in medium I. However the incident and reflected waves in medium I have amplitudes that are not equal in magnitudes. Suppose the plane waves are propagating along X-axis, then net electric and magnetic fields in medium I are, →

→

→

→

→

E1P = EiP + ErP

...(1)

→

H1P = HiP + HrP

and

→

...(2) →

We know that the electric field E and magnetic field H in phasor form are expressed as →

EiP = E io e − γ 1 x aɵ y

...(3)

→

ErP = E r o e − γ 1 x aɵ y →

→

H1P =

and

...(4)

→

HrP =

E io η1

Ero η1

e − γ 1 x aɵz

...(5)

e γ 1 x aɵz

...(6)

where γ 1 is the propagation constant. →

→

→

Substituting the values of E iP and E r P from eqns. (3) and (4) in eqn. (1) and the values of H1 P and →

H r P from eqns. (5) in (6) in eqn. (2), we get →

E1P = E io e − γ 1 x aɵ y + E r o e γ 1 x aɵ y

and

→

H1 P =

E io η1

e − γ 1 x aɵz +

Ero η1

e γ 1 x aɵz

...(7) ...(8)

E lectromagnetic F ield Theory

544

But reflection coefficient, Γ =

Ero E io

and for dielectric medium σ1 = 0, α1 = 0, thus γ 1 = j β1

Thus eqns. (7) and (8) becomes →

E1P = E io [e − jβ1 x + Γ e jβ1 x ] aɵ y →

H1P =

and

E io η1

...(9)

[e − jβ1 x + Γ e jβ1 x ] aɵz

...(10)

Since e − jθ = cos θ − j sin θ and e jθ = cos θ + j sin θ, eqn. (9) becomes →

E1P = E io [{ cos (β1 x) − j sin (β1 x)} + Γ {cos (β1 x) + j sin (β1 x)}] →

E1P = Ei0 [(1 + Γ ) cos (β1 x) − j sin (β1 x) (1 − Γ )]

or

...(11)

→

| E1P | = E io (1 + Γ )2 cos2 (β1 x ) + (1 − Γ )2 sin2 (β1 x )

or →

Similarly for H →

H1P = →

E io

(1 + Γ )2 cos2 (β1 x ) + (1 − Γ )2 sin2 (β1 x )

η1

→

→

...(12)

→

The magnitudes of E and H, that is, | E1P| or | H1P|. Will be maximum − β1 x max = 0, π , 2 π, ...., nπ

When

or

− β1 x max = nπ

...(13) where n = 0,1, 2, 3,....

Negative sign indicates that the standing waves are in negative (–) x- direction. nπ x max = − ∴ β1 x max = −

or

2π ∵ β1 = ...(14) λ1

nλ1 2

where n = 0,1, 2,.... →

| E1P |max = E io (1 + Γ )

Thus,

...(15)

Similarly for H →

| H1P |max = →

→

E io (1 + Γ )

...(16)

η1 →

→

The magnitudes of E and H, that is | E1P| or | H1P|. will be minimum When

− β1 x min =

π 3π 5π π , , ,....,(2 n + 1) 2 2 2 2

or

π 2 where n = 0,1, 2,....

− β1 x min = (2 n + 1)

Again negative sign signifies that standing wave are in − x direction.

...(17)

Electromagnetic Wave Propagation ∴

x min = −

545

(2 n + 1) π 2 β1

x min = − (2 n + 1)

or

λ1 4

...(18)

where n = 0,1, 2,....

→

| E1P |min = E io (1 − Γ )

Thus,

...(19)

→

Similarly for H →

| H1P |min =

E io (1 − Γ )

...(20)

η1 →

→

→

If η2 < η1, reflection coefficient, Γ < 0. In this situation the locations of | E1|max and | E1|min or | H1|max →

and | H1|min are interchanged. Since transmitted wave in medium II is purely a travelling wave, therefore, there are no maxima and minima in this region.

σ1 = 0

σ2 = 0

→

S=

| E1|max | E1|min

→

=

| H1|max | H1|min

1 + |Γ| = 1 − |Γ|

– 3λ1/2

Standing wave due to reflection at normal incidence between two perfect dielectric media are shown in Fig. 16.

– λ1/2

→

Ei0 (1+Γ) = |Ε1|max

From eqns. (15) and (19) or from eqns. (16) and (20), the standing wave ratio is,

X

0

– λ1/4

→

Ei0 (1 – Γ) = |Ε1|min Fig. 16

Example 32: A uniform plane wave in air partially reflects from the surface of a material whose properties are unknown. Measurement of electric field in the region in front of the interface yield a 1.5 m spacing between maxima with first maximum occuring 0.75 m from the interface. A standing wave ratio of 5 is measured. Determine the intrinsic impedance η of the unknown material. Solution: The spacing between maxima, λ /2 = 15 . m ∴

λ = 15 . × 2 = 3.0 m

Since the first maximum occurs at 0.75 m from the interface (fig. 17), it means that a field minima occurs at the boundary. Thus reflection coefficient, Γ is negative and real. We know that 1 + |Γ| S −1 or SWR, S= Γ= 1 −|Γ| 1+ S Here, ∴ We also know that

S =5 5 −1 4 2 Γ= = = 1+ 5 6 3 Γ=

η2 − η1 η2 + η1

|← λ/2=1.5m —|

3λ /2

λ/2 Medium I η1 Fig. 17

←0.75→

λ/4

Medium II η2

E lectromagnetic F ield Theory

546

2 η1 |Γ| − 1 (η2 − η1) − (η2 + η1) = =− |Γ| + 1 (η2 + η1) + (η2 − η1) 2 η2 η1 [|Γ| + 1]

or

η2 =

or

η2 = 75.4 Ω

[|Γ| − 1]

=

377 Ω (1/3) 377 = Ω 5 (5 /3)

Example 33: Consider a plane interface (z = 0) between free space and glass (σ = 0 , ε r = 4 and µ r = 1). →

Given a plane wave with E = 1cos (wt − β z ) aɵ x V / m incident from free space medium I normal to the glass-medium II. Determine (a) reflection coefficient, (b ) transmission coefficient and (c ) standing wave ratio (S) for medium I. Solution: If η1 and η2 are the intrinsic impedances for media I and II respectively, then Reflection Coefficient,

Γ=

η2 − η1 η2 + η1 µ0

Here

η1 =

and

η2 =

or

η2 = 60 π

∴

Γ=

and Transmission Coefficient.

τ=

ε0

=

4 π × 10 −7 8854 . × 10 −12

= 120 π

µ0 µ r µ0 µ = = ε ε0 ε r ε0

µr 1 = 120 π 4 εr

60 π − 120 π 60 =− = − 0 .333 60 π + 120 π 180 2 η2 η1 + η2

=

2 × 60 π 2 = 180 π 3

τ = 0 .667

or Standing Wave Ratio, ∴

(S) =

1 + |Γ| 1 + 0.333 1333 . = = 1 −|Γ| 1 − 0.333 0.667

S≈2

R eflection of Uniform Plane Electromagnetic Wave by a Perfect Dielectric (or Insulator) for Oblique Incidence

When a plane electromagnetic wave travelling in a medium of dielectric permittivity ε1 be incident →

obliquely, at an angle φ1 upon a plane dielectric interface which is not parallel to the plane containing E →

and H, then a part of the incident wave is refracted at an angle φ2 and moves into the second medium of

Electromagnetic Wave Propagation

547

permittivity ε2 and rest is reflected into the same medium (Fig. 18). During the time the edge B of the incident ray reaches A, the refracted ray reaches C after traversing a distance OC in second medium and reflected ray covers the distance OD in the same medium. If v1 and v2 are the speeds of propagation of the waves in the medium I and II respectively, then, Medium I (ε1, σ1)

incident waves

(90° – φ1) φ1

Interface

φ3

D

B φ 1

Reflected waves

φ3 Interface

φ1

φ2

A

φ2 C

Medium II (ε2, σ2)

Transmitted wave or Refracted waves

Fig. 18

BA = v1t ∴

OC = v2 t

and

BA v1 = OC v2

...(1)

From simple geometry one can show that ∠ BOA = φ1

and

BA = sin φ1 and OA or

BA / OA sin φ1 = OC / OA sin φ2

or

sin φ1 BA = sin φ2 OC

∠ OAC = φ2 (Fig, 18) OC = sin φ2 OA

...(2)

Comparing eqns. (1) and (2), we get sin φ1 v = 1 sin φ2 v2

...(3)

If µ1 and µ2 are the magnetic permeabilities of the media I and II respectively, then the speeds of propagation of the waves v1 and v2 , in media I and II are v1 =

1 ε1µ1

and

Substituting these values of v1 and v2 in eqn. (3), we get

v2 =

1 ε2 µ2

E lectromagnetic F ield Theory

548

sin φ1 1 / ε1µ1 = = sin φ2 1/ ε2µ2

ε2 µ2 ε1 µ1

For a non-magnetic media, µ1 ≈ µ2 ≈ µ0 sin φ1 = sin φ2

ε2 ε1

...(4)

From fig. 18.

AB = OD

...(5)

or

AB OD = OA OA

or

sin φ1 = sin φ3

or

φ1 = φ3

∴

Angle of incidence = Angle of reflection

∴

...(6)

That is, the angle of incidence is equal to the angle of reflection, which is one of the laws of reflection. If n1 and n2 are the refractive indices of medium I and medium II respectively, then

or

n1 =

c v1

and

n2 =

c v2

v1 =

c n1

and

v2 =

c n2

Substituting these values of v1 and v2 in eqn. (3), we obtain sin φ1 n2 = = n sin φ2 n1 1 2

...(7)

Since the index of refraction is constant for each medium for a given wavelength, the ratio of the sine of the angle of incidence to the sine of angle of refraction is a constant and is equal to the refractive index of medium I with respect to, II, which is Snell’s law of refraction. From the above discussion we concluded that, the electromagnetic wave obeys all the laws of reflection and refraction at a surface separating two dielectric media. The energy per unit area per unit time transported by the fields associated with a plane electromagnetic wave is expressed by Poynting →

vector S as →

→

→

→

1 ^ → ( n × E) η

S= E×H

H=

But

...(8) ...(9)

where ^ n is the unit vector along the direction of propagation of electromagnetic wave and η is the intrinsic impedance of the medium. →

Substituting the value of H from eqn. (9) in eqn. (8), we get

Electromagnetic Wave Propagation →

S= =

→

S=

or

549

1→ ^ → E × ( n × E) η 1 →→ ^ →^→ [( E. E) n − ( E. n) E] η E2 ^ n η

→

→

→

→

→→→

→→ →

[∵ A × ( B × C) = ( A.C)B − ( A. B) C]

...(10)

→

∴ Because E .^ n = 0 , the angle between E and ^ n is 90°. Thus, the energy transmitted per square meter S=

E2 η

...(11)

Hence, the energy per square meter associated with the incident wave, reflected wave and transmitted waves are:

and

Si ∝

E2i cos φ1 η1

Sr ∝

E2r cos φ1 η1

St ∝

E2t cos φ2 η2

where η1 and η2 are the intrinsic impedances of the media I and II respectively. According to the law of conservation of energy. Energy associated with the incident wave = Energy associated with reflected wave + Energy associated with refracted wave ∴ or or But

E2i E2 E2 cos φ1 = r cos φ1 + t cos φ2 η1 η1 η2 1= E2r E2i

Er2 Ei2

= 1−

η1 = η2

+

...(12)

Et2 η1 cos φ2 Ei2 η2 cos φ1

η1 E2t cos φ2 η2 E2i cos φ1

µ1 / ε1 µ2 / ε2

For perfect dielectric and non-magnetic media µ1 ~ µ2 ~ µ0 ε2 η1 ∴ = η2 ε1

...(13) ...(14)

...(15)

Substituting this value of η1 / η2 in eqn. (13), we get E2r E2i

=1 −

ε2 E2t cos φ2 ε1 Ei2 cos φ1

...(16)

We shall now derive expressions for the reflection and transmission coefficients in two separate situations; one in which the incident plane wave is linearly polarised with its electric vector (or

E lectromagnetic F ield Theory

550

polarisation vector) perpendicular to the plane of incidence or parallel to the boundary surface, and the other in which the polarisation vector (or electric vector) is parallel to the plane of incidence.

→

Case I: Electric Vector E perpendicular to the Plane of Incidence (Perpendicular Polarisation) In this case of perpendicular or horizontal polarisation, we shall

Medium I (ε1, σ1)

assume the electric vector to lie normal to the plane of incidence →

in fig. 19. Since the electric vector Ei associated with incident wave

Er

Ei

and magnetic vector H parallel to the plane of incidence as shown Hi

φ1

φ1

Hr

is perpendicular to the plane of incidence, the electric vector Er associated with reflected wave and Et associated with transmitted wave are also perpendicular to the plane of incidence. Clearly, the x-component of electric field represents a tangential component which should be continuous across the interface (boundary condition). Therefore, Er E = t Ei Ei

1+

φ2

Et

Ht Medium II (ε2, σ2) Fig. 19

Ei + Er = Et or

Interface

...(17)

Substituting this value of Et / Ei from eqn. (17) in eqn. (16), we get E2r E2i or

or or

=1 −

ε2 E 1 + r Ei ε1

E 1 − 2 = 1 + r Ei Ei

2

E E E 1 + r 1 − r = 1 + r E E Ei i i

2

E2r

1–

Er E = 1 + r Ei Ei

2

cos φ2 cos φ1

ε2 cos φ2 ε1 cos φ1 ε2 cos φ2 ε1 cos φ1 ε2 cos φ2 ε1 cos φ1

√ (ε1) cos φ1 ( Ei − Er ) = √ (ε2 ) cos φ2 ( Ei + Er )

or [√ (ε1) cos φ1 + √ (ε2 ) cos φ2 ] Er = [√ (ε1) cos φ1 − √ (ε2 ) cos φ2 ] Ei Er = Ei

or

ε1 cos φ1 −

ε2 cos φ2

ε1 cos φ1 +

ε2 cos φ2

...(18)

∴ Reflection coefficient or reflectance for horizontal or perpendicular polarisation, that is, Γ⊥ =

Er = Ei

ε1 cos φ1 −

ε2 cos φ2

ε1 cos φ1 +

ε2 cos φ2

...(19)

Electromagnetic Wave Propagation

551

Similarly, the transmission coefficient or transmittance for horizontal or perpendicular polarisation, that is, τ⊥ =

Et E =1 + r Ei Ei τ⊥ = 1 +

∴

τ⊥ =

or

ε1 cos φ − ε2 cos φ2 ε1 cos φ1 +

ε2 cos φ2

2 ε1 cos φ1 ε1 cos φ1 +

...(20)

ε2 cos φ2

From eqn. (4), ε1 sin φ2 = ε2 sin φ1

or

ε1 = ε2

sin φ2 sin φ1

Substituting the above value of √ (ε1) in eqns. (19) and (20), we obtain Γ⊥ =

sin φ2 cos φ1 − sin φ1 cos φ2 sin φ2 cos φ1 + sin φ1 cos φ2

...(21)

and

τ⊥ =

2 sin φ2 cos φ1 sin φ2 cos φ1 + sin φ1 cos φ2

...(22)

or

Γ⊥ =

sin (φ1 − φ2 ) sin (φ1 + φ2 )

...(23)

(∵ sin ( A + B) = sin A cos B + cos A sin B and sin ( A − B) = sin A cos B − cos A sin B) 2 sin φ2 cos φ1 and τ⊥ = sin (φ1 + φ2 )

...(24)

→

Case II : E Parallel to the Plane of Incidence (Parallel Polarisation) Medium I (ε1, σ1)

In the case of parallel or vertical polarisation, we will assume →

the electric vector E to lie in the plane of incidence and the

Ei

→

magnetic vector H perpendicular to the plane of incidence as shown in fig. 20. Since the electric vector Ei associated with the

Hi

Er φ1

φ1

Hr

incident wave lies in the plane of incidence, the electric vectors

Interface

Er and Et associated with the reflected and transmitted waves, respectively, also lie in the plane of incidence. Apply the boundary condition that the tangential component of electric →

field E is continuous across the boundary, that is, Ei cos φ1 − Er cos φ1 = Et cos φ2 ...(25) or

or

E E cos φ2 1− r = t Ei Ei cos φ1 Et E cos φ1 = 1 − r Ei Ei cos φ2

...(26)

φ2

Et

Ht Medium II (ε2, σ2) Fig. 20

E lectromagnetic F ield Theory

552

Substituting the value of Et / Ei from eqn. (26) in eqn. (16), we get E2r E2i

=1 −

ε2 E 1 − r Ei ε1

E = 1 − r Ei E2i

2

E E E 1 − r 1 + r = 1 − r E E Ei i i

2

1−

or

E2r

E E 1 + r = 1 − r Ei Ei

or

2

cos2 φ1 cos φ2 cos2 φ2 cos φ1

ε2 cos φ1 ε1 cos φ2 ε2 cos φ1 ε2 cos φ2 ε2 cos φ1 ε1 cos φ2

√ (ε1) cos φ2 ( Ei + Er ) = ( Ei − Er ) √ (ε2 ) cos φ1 [√ (ε1) cos φ2 + √ (ε2 ) cos φ1] Er = [√ (ε2 ) cos φ1 − √ (ε1) cos φ2 ] Ei or

Er = Ei

ε2 cos φ1 −

ε1 cos φ2

ε1 cos φ2 +

ε2 cos φ1

...(27)

∴ Reflection coefficient or reflectance for vertical polarisation, that is, Er = Ei

τ|| =

ε2 cos φ1 −

ε1 cos φ2

ε1 cos φ2 +

ε2 cos φ1

...(28)

Similarly, the transmission coefficient or transmittance for vertical polarisation, that is, Et E =1 + r Ei Ei

τ|| = ∴

τ|| = 1 +

or

τ|| =

ε2 cos φ1 − ε1 cos φ2 ε1 cos φ2 +

ε2 cos φ1

2 ε2 cos φ1 ε1 cos φ2 +

...(29)

ε2 cos φ1

Again from eqn. (4), ε1 sin φ2 = ε2 sin φ1

or

ε1 = ε2

sin φ2 sin φ1

...(30)

Substituting the above value of √ (ε1) in eqn. (28) from eqn. (30), we obtain Γ|| = =

or

Γ|| =

sin φ1 cos φ1 − sin φ2 cos φ2 sin φ2 cos φ2 + sin φ1 cos φ1 sin 2 φ1 − sin 2 φ2 sin 2 φ1 + sin 2 φ2

2 cos (φ1 + φ2 ) sin (φ1 − φ2 ) 2 sin (φ1 + φ2 ) cos (φ2 − φ1)

(∵ 2 sin A cos A = sin 2 A)

Electromagnetic Wave Propagation

553

A − B A+B A+B A − B cos sin ∵ sin A − sin B = 2 cos and sin A + sin B = 2 sin 2 2 2 2 or

Γ|| =

tan (φ1 − φ2 ) tan (φ1 + φ2 )

...(31)

Similarly, substituting the value of √ (ε1) from eqn. (30) in eqn. (29), we get

or

τ|| =

2 sin φ1 cos φ1 2 sin φ1 cos φ1 = cos φ2 sin φ2 + cos φ1 sin φ1 sin 2 φ1 + sin 2 φ2

τ|| =

2 sin φ1 cos φ1 sin(φ1 + φ2 )cos(φ2 − φ1 )

...(32)

B rewster Angle and Total Internal Reflection Sir David Brewster performed a series of experiments and observed that when a parallel polarised plane →

→

wave (an electromagnetic wave whose electric vector E is parallel and magnetic vector H is perpendicular to the plane of incidence or a case of parallel polarisation) is incident on the interface of two media at a particular angle of incident, no reflected wave is observed. This particular angle of incidence for which no reflected wave (or no reflection) is observed is called Brewster angle or polarising angle and is denoted by φB . Thus, Brewster angle is a particular angle of incidence for which there is no reflection when the incident plane wave is parallel polarised. The law is known as Brewster law. According to this law, for parallel polarisation, the tangent of Brewster angle (φ B ) is equal to the square roots of the ratio of the absolute permittivity of medium '2' to the permittivity of the medium '1'. Obviously, Brewster law is mathematically expressed as, ε tan φ B = 2 ε1 Brewster angle is called Polarising angle because an unpolarised plane wave incident on an interface of →

two media at this angle is reflected as polarised plane wave with electric vector E normal to the plane of incidence. It is further seen that there is no such Brewster angle for perpendicular polarisation because unless ε2 = ε1, there will always be reflection. Proof: Condition for no reflection for an incident parallel polarised wave is obtain by equating the expression for reflection coefficient, that is, ( E r / E i) for parallel polarised wave to zero. That is Er Ei

=

(ε2 ) cos φ1 − (ε1) cos φ2 (ε2 ) cos φ1 + (ε1) cos φ2

or

(ε2 ) cos φ1 − (ε1) cos φ2 = 0

or

(ε2 ) cos φ1 = (ε1) cos φ2

=0

...(1)

E lectromagnetic F ield Theory

554

Square of both sides gives ε2 cos2 φ1 = ε1 cos2 φ2 or

ε2 (1 − sin2 φ1) = ε1 (1 − sin2 φ2 )

...(2)

From Snell's law µ2 ε2 sin φ1 = sin φ2 µ1 ε1 For non-magnetic media µ1 = µ2 = µ0 ε sin φ1 = 2 sin φ2 ε1

∴

ε sin φ1 = 2 sin φ2 ε1

or

...(3)

Where φ1is the angle of incidence. Squaring of eqn. (3) gives ε1 sin2 φ1 ε2

sin2 φ2 =

...(4)

Substituting this value of sin2 φ2 in eqn. (2), we get ε2 (1 − sin2 φ1) = ε1 (1 −

ε1 sin2 φ1) ε2

or

ε12 sin2 φ1 − ε2 sin2 φ1 = ε1 − ε2 ε2

or

ε12 sin2 φ1 − ε22 sin2 φ1 = ε2 (ε1 − ε2 ) ε2 (ε1 − ε2 )

ε2 ε1 + ε2

or

sin2 φ1 =

∴

cos2 φ1 = 1 − sin2 φ1 = 1 −

or

cos2 φ1 =

ε12

− ε2

2

=

ε1 ε1 + ε2

2

cos φ1 or Under this situation ∴

= tan2 φ1 =

ε tan φ1 = 2 ε1 φ1 = θ B tanθ B =

ε2 ε1

ε2 ε1 + ε2 ...(6)

Dividing eqn. (5) by eqn. (6), we get sin2 φ1

...(5)

ε2 ε1

Electromagnetic Wave Propagation

555

T otal Internal Reflection Let us discuss the phenomenon of total internal reflection in case of perpendicular polarisation in which a plane electromagnetic wave in going from optical denser medium to a rarer medium (less denser medium) incident on the interface at an angle equal to or exceeding the critical angle the wave will be totally internally reflected back into the denser medium and there is no transmitted wave in rarer medium. Therefore for total internal reflection to take place, medium I should be more denser than medium II and the angle of incidence should be equal or greater than critical angle φ c . The reflection coefficient, Γ⊥ for perpendicular polarisation is expressed as, Γ⊥ =

Er Ei

=

ε1 cos φ1 − ε2 cos φ2 ε1 cos φ1 + ε2 cos φ2

...(1)

Where φ1 is the angle of incidence, φ2 the angle of reflection, ε1 and ε2 are the permittivities of the denser and rarer media respectively. ε ε cos φ1 − 2 cos φ2 cos φ1 − 2 (1 − sin2 φ2 ) ε1 ε1 or Γ⊥ = ...(2) Γ⊥ = ε2 ε2 2 cos φ1 + cos φ2 cos φ1 + (1 − sin φ2 ) ε1 ε1 According to Snell's law of refraction, if v1 is the velocity of wave in denser medium and v2 that in rarer medium, then v1 sin φ1 ...(3) = sin φ2 v2 v1 =

But

∴

1 µ1 ε1

and

v2 =

1 µ2 ε2

µ2 ε2 sin φ1 = sin φ2 µ1 ε1

...(4)

If the dielectric media is non-magnetic, then µ1 = µ2 = µ0 Therefore, eqn. (3) reduces to, sin φ1 sin φ 2

=

ε2 ε1

or

sin φ 2 =

ε1 sin φ1 ε2

Substituting the value of sin φ 2 from eqn. (5) in eqn. (2), we get cos φ1 −

ε2 ε1

ε 2 1 − 1 sin φ1 ε2

cos φ1 +

ε2 ε1

ε 2 1 − 1 sin φ1 ε2

Γ⊥ =

...(5)

E lectromagnetic F ield Theory

556

or

Γ⊥ =

ε cos φ1 − 2 − sin2 φ1 ε1

...(6)

ε cos φ1 + 2 − sin2 φ1 ε1

ε If ε1 > ε2 and sin2 φ1 ≥ 2 , then reflection coefficient Γ⊥ will be complex but the magnitude|Γ | = 1. ε1 Under these conditions, the incident plane wave is totally internally reflected. The incident angle for which Γ = 1 < 0 ° is called critical angle φ c . Thus,

ε2 ε1

sin φ c =

or φ c = sin−1

Thus, the wave incident at an angle φ c = sin −1

ε2 ε1

...(7)

ε2 or exceeding it, the wave will be totally internally ε1

reflected. Example 34: It is given two dielectric medium one is free space and other has ε2 = µ ε0 and µ2 = µ0 . Determine reflection coefficient for oblique incidence θ = 30 ° for a (i) perpendicular polarisation, (ii) parallel polarisation. [U'Khand, B.Tech IV Sem 2011]

Solution: Reflection coefficient for perpendicular polarisation, that is, Γ⊥ =

Er Ei

=

ε1 cos φ1 − ε2 cos φ2 ε1 cos φ1 + ε2 cos φ2

Where φ1 is incident angle and φ 2 angle of reflection. φ 2 can be obtain from Snell's law as, sin φ1 sin φ 2 ∴

So

ε1

Here φ1 = 30 °, ε1 = ε0 and ε2 = 4 ε0

4 ε0 sin 30 ° = ε0 sin φ 2 sin φ 2 = Γ⊥ =

or

ε2

=

Γ⊥ =

1 4

sin 30 ° 1/2 = 2 2

or

sin φ 2 =

or

φ 2 = sin −1(0.25) = 14.478°

ε0 cos 30 ° − 4 ε0 cos 14.478° ε0 cos 30 ° + 4 ε0 cos 14.478° − 10706 . = − 0 .382 2.8026

Reflection coefficient for parallel polarisation

=

0.8660 − 2 × 0.9683 0.8660 + 2 × 0.9683

Electromagnetic Wave Propagation

Γ|| = =

or

557

ε2 cos φ1 − ε1 cos φ2 ε2 cos φ1 + ε1 cos φ2

=

2 ε0 cos 30 ° − ε0 cos 14.478° 2 ε0 cos 30 ° +

ε0 cos 14.478°

2 × 0.8660 − 0.9683 0.7637 = 2 × 0.8660 + 0.9683 2.7003

Γ|| = 0 .2828

or

Example 35: A Perpendicular polarised wave propagates from medium I (ε r = 8.5, µ r = 1 and σ = 0) to medium II (ε r = 1, µ r = 1) with an angle of incidence 15°. Given E i0 = 10 . µ V / m. Find E r0 , E t0 , H r0 and H t0 . Solution : In case of oblique incidence, the reflection and transmission coefficients for perpendicular polarisation are: Γ⊥ = τ⊥ =

and

Er0 Ei0 E t0 Ei0

=

=

η2 cos φ1 − η1 cos φ2 η2 cos φ1 + η1 cos φ2 2 η2 cos φ1 η1 cos φ2 + η2 cos φ1

ε1 cos φ1 − ε2 cos φ2

=

ε1 cos φ1 + ε2 cos φ2 2 ε1 cos φ1

=

ε1 cos φ1 + ε2 cos φ2

Where φ1 and φ2 are the angle of incidence and angle of refraction respectively and η1and η2 are, η1 =

µ0 µ r µ0 µ = = ε ε0 ε r ε0

µ0 µr = εr ε0

1 = 120 π × 0.343 8.5

η1 = 4116 . π

or

η2 =

and

µ0 µ r µ0 µ = = ε ε0 ε r ε0

or η2 = 120 π

µ ∵ 0 = 120 π ε0

From Snell's law of refraction sin φ1 = sin φ2 ∴

sin φ2 =

ε2 = ε1

sin 15 ° 0.2588 = = 0.7545 = sin 48.98° 0.343 0.343

Thus,

φ2 = 48.98°

So

Γ⊥ =

or

or But

Er0 Ei0 Er0 Ei0

ε0 1 = = 0.343 8.5 ε0 ε r

Er0 Ei0

=

η2 cos φ1 − η1 cos φ2 η2 cos φ1 + η1 cos φ2

=

120 π cos 15 ° − 4116 . π cos 48.98° 120 π cos 15 ° + 4116 . π cos 48.98°

=

120 π × 0.9659 − 4116 . π × 0.6572 115.908π − 27.0503 π = 120 π × 0.9659 + 4116 . π × 0.6572 115.908π + 27.0503 π

=

88.8577 142.9583

E i0 = 10 . µV / m

or

Er0 =

88.8577 E 142.9583 i 0

E lectromagnetic F ield Theory

558

∴

E r0 = 0.6215 × 10 .

We know that,

Hr0 =

E t0 Ei0

Er0

=

η1

0.6215 4116 . π

2 ε1 cos φ1

= τ⊥ =

ε1 cos φ1 + ε2 cos φ2 E t0

or

Ei0

=

or

=

or

E r0 = 0 .6215 µ V / m H r0 = 4.81 × 10 −3 µ A / m 2 η2

η1 cos φ2 + η2 cos φ1

H t0 =

2 × 120 π 4116 . π cos 48.98° + 120 π cos 15 °

240 π 240 = = 1.679 120 π × 0.9659 + 4116 . π × 0.6572 142.9583

E t0 = 1679 . E i 0 = 1679 . × 10 . = 1679 . µV / m

or

=

E t0 η2

=

or E t0 = 1679 . µV / m

1679 . µV / m = 4.45 × 10 −3 µ A / m 120 π

Example 36: A parallel polarised wave propagates from air into dielectric at Brewster angle of 75°, calculate the relativity dielectric constant of the medium. Solution: If ε0 and ε are the permittivities of air and dielectric respectively, then Brewster angle φ is given as tan φ =

ε = tan 75 ° ε0

ε r = 13.91

∴

or

or

ε0 ε r = ε r = 3.73 ε0

relative dielectric constant, ε r = 13.91

Example 37: A parallel polarised wave is incident from air into paraffin. Calculate the Brewster angle assuming ε r = 2 for paraffin. Solution: The Brewster angle φ Β is expressed as, ε φ Β = tan −1 2 Here ε2 = ε0 ε r = 2 ε0 and ε1 = ε0 ε1 or φ B = 54.72 ° φ B = tan −1 2 = tan −1 14142 .

∴

Example 38: Find the critical angle of total internal reflection for glass ε r = 4, polystyrene ε r = 2.52 and polyethylene ε r = 2.25 to air surface. Solution: The wave will totally internally reflected if the wave is incident at an angle critical angle φ c . sin φ c =

Given by Since medium II is air, so For glass: ε r = 4,

∴

ε2 ε1

or

φ c = sin −1

ε 2 = ε0

ε1 = ε 0 ε r = 4 ε 0

and

ε 2 = ε0

ε2 ε1

Electromagnetic Wave Propagation Hence,

φ c = sin −1

ε0 = sin −1 (0.5000) 4 ε0

For polystyrene: ε r = 2 . 52, ∴ ε1 = 2 . 52 ε 0 Hence, For polyethylene: ε r = 2 . 25, Hence,

1.

φ c = sin −1

559

ε 2 = ε0

and

ε0 = sin −1 (0.6299) 2.52 ε0

∴ ε1 = 2 . 25 ε 0

φ c = 30 °

or

and

φ c = 39.12 °

or

ε 2 = ε0

ε0 φ c = sin −1 = sin −1 (0.6667) 2 . 25 ε0

or

φ c = 4181 . °

Derive electromagnetic wave equation in linear medium. On the basis of it deduce the plane wave equation for (a) Non-conducting medium (b) Conducting medium (c) Free space. [U'Khand, B.Tech IV Sem 2010]

2.

What do you mean by lossy dielectric ? Derive wave equation in lossy dielectric. [GBTU, B.Tech III Sem 2011]

3.

Derive the wave equation for conducting medium.

4.

Discuss the wave propagation in conducting media.

[U'Khand, B.Tech IV Sem 2010]

5.

Explain the wave propagation in lossy dielectric.

[U'Khand, B.Tech IV Sem 2011]

6.

Derive the expression for α and β in a conducting medium. [GBTU, B.Tech III Sem 2010, III Sem 2012]

7.

Derive the relation between E and H in uniform plane wave.

8.

Derive the expression for attenuation constant α and phase constant β in a dielectric medium.

→

[GBTU, B-Tech IV Sem 2011, III Sem 2012]

→

[GBTU, B.Tech III Sem 2010]

[GBTU, B.Tech IV Sem 2011]

9.

Derive field expressions for transverse electric waves.

[U'Khand, B.Tech IV Sem 2010]

10. Derive the wave propagation equation in dielectric medium. Find out the propagation constant, attenuation constant and phase shift constant also.

[UPTU, B.Tech IV Sem 2008]

11. Derive the wave equation for electric field and indicate how the propagation constant controls the wave propagation in the medium.

[UPTU, B.Tech IV Sem 2005]

12. Define phase velocity, group velocity, propagation constant and phase shift constant. [GBTU, B.Tech IV Sem 2011]

13. Discuss the following terms : (i) Phase velocity (ii) Group velocity and (iii) Impedance of the medium. [UPTU, B.Tech IV Sem 2005]

E lectromagnetic F ield Theory

560

14. Discuss wave propagation in good dielectrics and good conductors. Show that the angle of characteristic impedance is always 45° for good conductors.

[UPTU,, B.Tech IV Sem 2006]

15. What is skin depth and discuss its significance ?

[GBTU, B-Tech III Sem 2011]

16. What is the depth of penetration of the medium ?

[UPTU, B.Tech IV Sem 2008]

17. Derive an expressions for phase velocity and penetration depth of a plane wave propagating in an imperfect dielectric medium.

[UPTU, B.Tech IV Sem 2006]

18. What is skin effect ? Discuss the depth of penetration of a wave in a conducting medium. [UPTU, B.Tech IV Sem 2005; GBTU, III Sem 2012]

19. Discuss the solution of plane wave equation in conducting and non-conducting media. [UPTU, B.Tech IV Sem 2004]

20. What is understood by polarization of electromagnetic wave ? Explain linear, elliptical and circular polarisation with appropriate figures. Show that a linearly polarised wave can be interpreted as a combination of two circularly polarised waves of equal magnitude and angular velocities rotating in opposite directions.

[UPTU, B.Tech IV Sem 2003]

21. What do you mean by polarisation of a wave ? Show that the linear polarisation is related to circular polarisation of the wave.

[UPTU, B.Tech IV Sem 2005]

22. Explain linear, elliptical and circular polarization with appropriate figures. What do you mean by polarisation ?

[UPTU, B.Tech IV Sem 2003]

23. Explain the poynting vector and poynting theorem.

[U'Khand, B.Tech IV Sem 2009]

24. State and prove poynting theorem. State the physical interpretation of poynting vector. [U'Khand, B.Tech IV Sem 2010]

25. Explain poynting theorem in detail.

[U'Khand, B.Tech IV Sem 2011]

26. Discuss the proof of the poynting theorem and also mention its application. [GBTU, B.Tech IV Sem 2010] 27. State and explain the poynting theorem.

[UPTU, B.Tech IV Sem 2007]

28. What is poynting vector ? Discuss the poynting theorem and explain the physical meaning of each integral involved there in.

[UPTU, B.Tech IV Sem 2005]

29. Derive the complex poynting theorem and explain the transmitter and receiver action. [UPTU, B.Tech IV Sem 2004]

30. The average poynting theorem gives the energy balance in a system. Justify it by obtaining necessary relation. 31. The average poynting vector is given by Pw =

[UPTU, B.Tech IV Sem 2002]

1 η E2 2

[UPTU, B.Tech IV Sem 2002]

Electromagnetic Wave Propagation

561

32. A uniform plane wave is propagating in the +z-direction in a good conductor having conductivity σ S / m. The permittivity and permeability in the conductor are the same as in free space and the electric field is E x0 at z = 0. What power is dissipated in the medium for z > 0 ? Assume σ >> w ε. [GBTU, B.Tech III Sem 2012]

33. Discuss reflection of plane electromagnetic wave incident normally on a perfect dielectric and obtain an expressions for the two reflection coefficients of electric and magnetic fields. [UPTU, B.Tech IV Sem 2006]

34. Derive the formula, S =

1+ Γ 1− Γ

, where S is the standing wave ratio and Γ reflection coefficient. [UPTU, B.Tech IV Sem(old) 2007]

35. Derive expressions for reflection and transmission coefficients for (i) horizontal polarisation and (ii) Vertical polarisation. →

36. Discuss reflection of uniform plane wave by perfect dielectric for oblique incidence (a) when E is →

parallel to plane of incidence and (b) when electric field E is perpendicular to plane of incidence. 37. Derive Brewster angle and E r / E i for perfect insulator.

[U'Khand, B.Tech IV Sem 2009]

38. Explain the Brewster angle.

[U'Khand, B.Tech IV Sem 2011]

39. What is brewster angle ? Derive expression for Brewster angle in terms of dielectric constants of the two media involved.

[UPTU, B.Tech IV Sem 2006]

40. A plane electromagnetic wave is incident at an angle φ1 at the surface discontinuity between two homogeneous isotropic dielectrics with permittivity ε1and ε 2 , ε 2 being the permittivity of the dielectric into which the wave gets reflected at angle φ 2 . If E i, E r and E t are the electric intensities respectively of the incident, reflected and transmitted waves, show that the reflection coefficient for parallel polarisation is given by Er Ei

=

tan (φ1 − φ 2 ) tan (φ1 + φ 2 )

E lectromagnetic F ield Theory

562

1.

If the electric field strength of a plane electromagnetic wave is 1 Volt / m. Calculate the magnitude →

of H for a plane electromagnetic wave. →

2.

A perfect dielectric medium has σ = 0, µ r = 1 and ε r = 4. The magnetic field H associated with an electromagnetic wave is given by →

H = − 0.1 cos (ω t − z) aɵ x + 0.5 sin (ω t − z) aɵ y Amp /m

Find (a) phase constant, (b) angular velocity, (c) the intrinsic impedance and (d) the components of electric field intensity of the wave. 3.

The electric field of a uniform plane electromagnetic wave in free space is given by →

E = 30 sin (6 π × 108 t − 2 π x) aɵ y + 30 cos (6 π × 108 t − 2 π x) aɵz V /m →

Find (a) β, (b) ω, (c) f , (d) λ, (e) vp , (f) η0 and (g) H, where parameters have their usual meaning. →

4.

If the magnetic field intensity H has only aɵz component. This component is expressed as Hz = 3 x cos β + 6 y sin γ →

Find the expression for current density J . 5.

Find the characteristic impedance of the medium whose relative permittivity is 4 and relative permeability is 1.

6.

After which frequency, the earth may be considered as perfect dielectric ?

7.

The constitution parameter of aluminium are given by µ r =1, ε r =1 and σ = 3.54 × 10 7mho /m Find the frequency for which the skin depth of aluminium is 0.01 mm.

8.

Determine the propagation constant γ for a material having µ r =1, ε r = 8 and σ = 0 . 25 p S/m, if the wave frequency is 1.6 MHz.

9.

Determine the following for a damp soil at 1 MHz, ε r =12, µ r =1 and σ = 2 × 10 −12 mho /m. Calculate α , δ, vp and η.

10. A plane wave 200 MHz is travelling in a medium for which σ = 0, µ r = 2, ε r = 4. If the average poynting vector is 5 W/m2 , find E r m s , H r m s , phase velocity and impedance of the medium. 11. Find the skin depth δ at a frequency of 1.6 MHz in aluminium, where σ = 38.2 Ms /m and µ r = 1. Also find the propagation constant and wave velocity.

Electromagnetic Wave Propagation

563

12. Calculate following : (a) depth of penetration δ, (b) propagation constant γ, (c) velocity of propagation v at frequency 1.6 × 106 Hz in aluminium, where conductivity σ = 38.2 × 106 S /m and µ r = 1. 13. For a non-magnetic material for which relative permittivity ε r and conductivity σ are 2.5 and 10 −4 S/m respectively. Calculate (a) loss tangent, (b) attenuation constant, (c) phase constant,

(d) velocity of wave propagation and (e) intrinsic impedance of an electromagnetic wave having a frequency of 3 MHz. Assuming the material to be a good dielectric. 14. At what frequency may earth be considered a perfect dielectric, if σ = 5 × 10 −3 S /m, µ r = 1 and ε r = 8? Calculate the attenuation constant at these frequencies (consider (σ /ωε) ≤ 0.01). 15. The loss tangent for ice at a frequency of 3000 MHz is 0.0009 and ε = 3.20 ε0 . A uniform plane wave with an amplitude of 100 V/m is propagating in it. Find the time average power density at x = 0 and x =10 m. →

16. A travelling electric field E in free space of amplitude 100

ε0, µ0

V/m, strikes a perfect dielectric as shown in adjacent

µr = 1 εr = 20 σ=0

ε0, µ0

figure 21. Find the electric field intensity in medium III. . 17. A normally incident electric field has amplitude E = 10 V/m in free space just outside sea water in which ε r = 80, µ r = 1, σ = 2.5 S /m. For frequency of 30 MHz, at what

Ei Et

Ei Er Medium I

Medium II Medium III |← 10 µm →|

depth the amplitude of E be 1 mV/m.

Fig. 21

→

18. Travelling E waves in free space are normally incident on the interface with a perfect dielectric for which ε r = 3.0. Compute the magnitudes of reflection coefficient and transmission coefficient. 19. The amplitude of E i in free space (medium I) at the boundary with medium II is 1 V/m. If Hr0 = 141 . × 10 −3 A /m, ε r2 = 18.5 and σ2 = 0. Find µ r2 .

20. Given two dielectric media, medium I is free space and medium II has ε2 = 4 ε0 and µ = µ0 . Determine reflection coefficient for oblique incidence φ1 = 30 ° for (a) perpendicular polarisation and (b) parallel polarisation. 21. Determine the critical angle for plane wave passing from glass to air (ε / ε0 = 9). 22. A parallel polarised wave propagates from air into a dielectric ε r2 = 13.93. Find Brewster angle. 23. A parallel polarised wave propagates from air into a dielectric at Brewster angle of 75°. Find ε r . 24. A plane wave travelling in air normally incident on a block of paraffin with ε r = 2.2. Find the standing wave ratio.

E lectromagnetic F ield Theory

564

nswers to Unsolved Numerical Problems 1.

2. β = 1 rad / m, ω = 1.5 × 108 rad / s, 60 π ohm

2.6 m A /m

→

E = 94.12 sin (ω t − z) aɵ x + 18.83 cos (ω t − z) aɵ y

3.

2 π rad / m, 6 π × 108 rad /s, 3 × 108 Hz,1m,

4.

→

J = 6 sin γ aɵ x − 3 cos β aɵ y

3 × 108 m / s, 376.72 ohm 7963 . × 10 −2 cos (6 π × 108 − 2 π x) aɵ y 5.

η = 60 π ohm

6.

7.

7156 . MHz

8.

9. α = β = 2.808 × 10 −6 m −1, δ = 3.56 × 105 m, vp = 2.236 ×10

12

3

m /sand η =140 .425 ×10 >ω L

G > R and ω C >> G. Therefore, for high frequency transmission lines the propagation constant γ and characteristic impedance Z0 may be reduced as follows : For high frequency lines, ω L >> R and ω C >> G γ = √ [(R + j ω L) (G + j ωC)] Neglecting R and G from γ, we get or

γ = √ [( j ω L) ( j ω C)] = j ω √ ( LC) = α + j β

Transmission Lines

`

579

Equating real and imaginary parts, we get α = 0 and β = ω √ (LC) Therefore, velocity of propagation, vp =

ω ω = = β ω LC

1 LC

and characteristic impedance Z0 may be evaluated as

or

Z0 =

R + jω L = G+ jωC

Z0 =

L C

jω L jω C

Infinite Transmission Line If a source is connected to a hypothetical line of infinite length then the voltage and current will travel down the line and reduce to zero at a far distant end (or at an infinite end). Such an infinite line, therefore, have no reflection components of voltage and current. Thus, in case of infinite line reflected component is zero, that is, in the solution of transmission line equation for current, I0− = 0. We have therefore for infinite line, I = I0+ e − γ x

...(1)

where I0 + is constant and γ the propagation constant. If I S is the current at the sending end, that is, at x = 0, then from eqn. (1), we have I s = I0 +

∴

I = Is e – γ x

...(2)

γ = [(R + j ω L) (G + j ω C)]

where

Considering transmission line equations, − or

dI dx

= ( G + j ω C) V

V =−

...(3)

dI 1 ( G + j ω C) d x

...(4)

Differentiating eqn. (2) with respect to, x and then substituting the value of d I /d x in eqn. (4), we get V =− = Is

1 (− γ I S e − γ x ) ( G + j ω C) γ e−γ x ( G + j ω C)

...(5)

(R + j ω L) (G + j ω C)

or

V = Is

or

R + j ω L − γ V = Is e G + j ω C

( G + j ω C) x

e−γ x

...(6)

E lectromagnetic F ield T heory

580

The voltage VS at the sending end of the line is obtained by putting x = 0 in eqn. (6). Therefore, R + j ω L Vs = I s G + j ω C or

R+jωL Vs = = Z0 Is G + j ωC

...(7)

where Z0 is the characteristic impedance of infinite line. In terms of modulus and phase angles, Z0 may be expressed as, 1 /2

(R 2 + ω2 L2 ) ∠ tan −1 (ω L /R ) Z0 = ( G2 + ω2 C2 ) ∠ tan −1 (ω C /G) or

R 2 + ω2 L2 Z0 = 2 2 2 G + ω C

1 /4

∠

1 {tan −1 (ω L / R ) − tan −1 (ω C /G)} 2

√ (R/G)

From eqn. (8), it is clear that, when ω → 0, R2 | Z0 | = 2 G

...(8)

1 /4

=

R G

Z0

√ (L/C)

and when ω → ∞ R2 /ω2 + L2 = | Z0 | = 2 2 2 G /ω + C

L C

Frequency

Fig. 8

Since (R / G) is always greater than L / C, the variation of characteristics impedance with frequency will be in accordance to the fig. 7.

Secondary Line Constants or Coefficients of Transmission Line in Terms of Primary Constants The characteristic impedance Z0 and propagation constant γ of a transmission line are known as secondary constants or coefficients of a transmission line. The propagation constant γ is a complex quantity, the real part α of which accounts for changes in magnitude of the signal at different values of x and is called attenuation constant. The imaginary part β of which accounts for the phase shifts in the signal as it travels down the line and is called phase shift constant. The values of these constants in terms of primary constants (or distributed constant) may be obtained as follows : We know that,

γ = α + j β = [(R + j ω L) (G + j ω C)] = [(R G − ω2 L C) + j ω ( LG + C R )]

or

γ2 = (α + j β)2 = (RG − ω2 L C) + j ω ( LG + C R )

or

α 2 − β2 + 2 j α β = (RG − ω2 L C) + j ω ( LG + C R )

...(1)

...(2)

Equating real and imaginary part of both sides, we get α 2 − β2 = RG − ω2 L C

...(3)

Transmission Lines

`

581

2 αβ = ω ( LG + C R )

...(4)

In terms of modulus and phase angles, eqn. (1), may also be written as γ = √ ((α 2 + β2 ) ∠ [tan −1 (β /α )] = {[ (R 2 + ω2 L2 )∠ tan –1( Lω / R )] [ (G2 + ω2 C2 )∠ tan –1(C ω / G)]}1 /2 ωL ω C 1 = [(R 2 + ω2 L2 ) (G2 + ω2 C2 )]1 /4 × ∠ tan −1 + tan −1 R G 2 | γ| = √ (α 2 + β2 ) = [(R 2 + ω2 L2 ) (G2 + ω2 C2 )]1 /4 or

α 2 + β2 = √ [(R 2 + ω2 L2 ) (G2 + ω2 C2 )]

...(5) ...(6)

Adding eqns. (3) and (6), we get 2 α 2 = R G − ω2 LC + [(R 2 + ω2 L2 ) (G2 + ω2 C2 )] or

α =

1 (RG −ω2 L C) + 2

( R 2 + ω2 L2 ) ( G 2 + ω2 C2 )

...(7)

Subtracting eqns. (3) from eqn. (6), we obtain 2 β2 = √ [(R 2 + ω2 L2 ) (G2 + ω2 C2 )] − (R G − ω2 LC) or

β=

1 2

(R 2 + ω2 L2 ) ( G 2 +ω2 C2 ) − ( R G −ω2 LC)

...(8)

In underground cables, conductors of small diameter are used so that they have greater number of turns for a given overall area of cross-section of the cable. Thus the resistance R per unit length of the cable will be large and since the spacing between the conductors of line is small, the inductance L per unit length of the cable will be small and the capacitance C per unit length of the cable will be large. Since the insulation between the conductors of the of the line is good, the conductance G per unit length will also be very small. Therefore, for an unloaded underground cable, at audio frequencies, R >> ω L

and

ω C >> G

Therefore, eqn. (1) becomes γ = √ [(R ) ( j ω C)] = √ ( j ω C R ) or

γ = √ (ω C R ) ∠45 ° or γ = √ (ω C R ) (cos 45 ° + j sin 45 ° ) γ=

ωCR (1 + j) = α + j β 2

Equating real and imaginary part, we have α =

and

β=

ω CR nepers per unit length 2 ω CR rad per unit length. 2

E lectromagnetic F ield T heory

582

Example 1: For a transmission line the primary line constants are, R = 78 ohm/kilometer, G = 62 µ mhos/kilometer, L = 1. 75 milli henries/kilometer, and C = 0 . 094 µF/kilometer. At a frequency of 1600 Hz, find (a) characteristic impedance, (b) attenuation constant, (c) phase constant, (d) velocity and (e) time taken by a wave to travel 200 kilometer along the line. [U'khand, B.Tech IV Sem. 2010]

Solution: (a) Characteristics impedance Z0 of the line is given by Z0 =

R + jω L

...(1)

G + jω C

Here R = 78 ohm/km, G = 62 µ mhos/km, C = 0 .094 µF/km, L = 1.75 milli henery/km and f = 1600 Hz R + j ω L = 78 + j × 2 π × 1600 × 1.75 × 10 − 3

∴

= 78 + j 17 .58 or

2

2

√ [R + (ω L) ] ∠ tan

−1

(ω L / R ) = √ [(78)2 + (17 .58)2 ] ∠ [tan −1 (17 .58 /78)] = 79 .95 ∠ 12 .70 °

Similarly

[∵ A + jB = √ ( A2 + B2 ) ∠ tan −1 (B / A)]

G + j ω C = 62 × 10 −6 + j × 2 × 3 .14 × 1600 × 0 .094 × 10 −6 = 62 × 10 −6 + j 944 .5 × 10 −6

or √ [G2 + (ω C)2 ] ∠ tan −1 (ω C / G) = 946 .53 × 10 −6 ∠86 .24 ° Substituting numerical values of R + j ω L and G + j ω C in eqn. (1), we get, ∴ or

Z0 =

79 .95 ∠12 .70 ° 946 . 53 × 10 − 6 ∠ 86 . 24 ° =

79 .95 946 . 53 × 10 −6

∠ [12 (12 . 70 ° − 86 . 24 ° )]

Z0 =290.63 ∠ − 36.77 ° ∵ A = A ∠ α = A ∠ (α − β ) and A n B n = A n Bn ∠ (nα + n β ) B B∠β B

(b) and (c)

The propagation constant γ of the line is given by γ = √ [(R + j ω L) (G + j ω C)] 1 /2

ω C ωL = R2 + (ω L)2 ∠ tan−1 G2 + (ω C )2 ∠ tan−1 G R = [(79 .95 ∠ 12 .70 ° ) (946 .53 × 10 −6 ∠ 86 .24 ° ) γ = (79 .95) × 946 .53 × 10 −6 ) ∠ [12 (12 .70 ° + 86 .24 ° )] = 0 .275 ∠ 49 .47° or

γ = 0 .275 cos 49 .47° + j 0 .275 sin 49 .47° = 0 .275 × 0 .650 + j 0 .275 × 0 .760 γ = 0 .179 + j0 .209 = α + j β

Transmission Lines

`

583

∴

Attenuation constant, α = 0 .179 neper/km

and

phase constant, β = 0 . 209 rad/km

(d) The velocity of propagation, v =

ω 2 × 3 .14 × 1600 = β 0 .209

v = 48 . 07 × 10 3 km / sec

or

(e) The time taken by the wave to travel 200 km along the line is t=

200 48 .07 × 103

= 0.00416sec

Example 2: A telephone line has R = 30ohm/km, L = 100mH/km, G = 0 and C = 20 µ F / km at 1kHz obtain (i) the characeterstic impedance of the line, (ii) the propagation constant. [UPTU, B.Tech IV Sem. 2003]

Solution: (i) The charactersitic impedance Z0 of the transmission line is, Z Z0 = = Y

1 /2

R 2 + (ω L)2 ∠ tan –1(ω L / R ) = G + jωC G2 + (ω C)2 ∠ tan –1(ω C / G) R + jωL

Here R = 30ohm /km, L = 100mH/km, G = 0, C = 20µ F / km and f = 1 kHz 1 /2

∴

(30)2 + (2 π × 1 × 103 × 100 × 10 –3 )2 ∠ tan –1(2 π × 1 × 103 × 100 × 10 –3 /30) Z0 = 0 + (2 π × 1 × 103 × 20 × 10 –6 )2 ∠ tan –1(ω C /0 = ∞)

or

900 + 394384 Z0 = –3 125.6 × 10

1 /2

∠

1 180 ° × 1.523 628 . 716 × 103 1 / 2 2 314 . = ∠(43 . 65 °– 45 ° ) 1 125 . 6 ∠ 90 ° 2

= (5005 . 7)1 / 2 ∠ – 1. 35 ° or Characteristic Impedance, Z0 = 70 .751 ∠ – 135 . ° ohm (ii)

Propagation constant, γ = (R + jωL) (G + jωC) 1 /2

or

γ = R 2 + (ωL)2 ∠ tan –1(ωL / R ) G2 + (ωC)2 ∠ tan –1(ω C / G) 1 /2

= (R 2 + ω2 L2 )(G2 + ω2 C2 ) = (628716 . × 125.6 × 10 –3 )1 / 2 ∠

∠

1 ω C ωL tan –1 + tan –1 R G 2

1 (43.65 ° + 45 ° ) 2

= (78 . 967)1 / 2 ∠44 . 32 ° = 8 . 88∠44 . 32 ° or Propagation constant,

γ = 8 . 88 cos 44 . 32 ° + j 8 . 88 sin 44 . 32 ° = 8 . 88 × 0 . 7161 × j8 . 88 × 0 . 6988

or

γ = 6 .359 + j 6 . 205

E lectromagnetic F ield T heory

584

Example 3: A transmission line operating at 500MHz has, Z0 = 80 Ω, α = 0 .04 Np/m, β = 1. 5 rad/m, find the line parameters. [UPTU, B.Tech IV Sem. 2007]

Solution: We have to determine line parameters, that is the values of resistance R, inductance L. Capacitance C and conductance G. In the given problem

Thus,

α = 0.04 Np / m and β = 1.5 rad/m, that is, γ = 0.04 + j1. 5 1.5 γ = (0. 04)2 + (1.5)2 ∠ tan –1 = 2 .2516 tan –1(1.544) 0 .04 180 = 1.5 ∠ × 1.544 = 1.5 ∠ 88.50 ° 3 .14 R + j ωL Z Z = or Z02 = G + j ωC Y Y

We know that,

Z0 =

and

γ2 = (R + jωL) (G + jωC) = ZY or γ2 = ZY

...(1) ....(2)

Substituting the value of Y from eqn. (2) in eqn (1), we get γ2 Z = Z02 or Z 2 = γ2 Z02 or Z = γ Z0 Z

...(3)

Substituting the numerical values of γ and Z0 in eqn. (3)., we get, Z = 80 × 1. 5 ∠88 . 50 ° = 120 ∠88 . 50 ° R + jωL = Z = 120 ∠88 . 50 ° = 120 cos 88 . 50 °+ j 120 sin 88 . 50 ° [∵ A ∠α = Α cos α + jβ sin α ] Thus,

Z = 120 × 0 . 0262 + j 120 × 0 . 9997

∴

R = 3 .144 Ω / m and ω L = 119 .964 119 .964 L= = 0 .382µ H /m 2 × 3 .14 × 500 × 106

= 3 .144 + j119.964 = R + jωL ∴ Similarly, ∴

Y = G + j ωC =

γ2 (1. 5 ∠ 88 . 50) = = 0.0187 ∠88 .50 ° Z 120 ∠88 . 50

G + j ω C = 0 . 0187 cos 88 . 50 °+ j 0 . 0187 sin 88 . 50 ° = 0 . 0187 × 0 . 0262 + j0. 0187 × 0 . 9997 = 4 . 899 × 10 –4 + j 0.01869

Hence,

G = 4 . 899 × 10 –4 mho/m and ω C = 0 . 01869

or

C=

or

C = 5.95 × 10 –5 F / m

0.01869 2 × 3 .14 × 500 × 106

= 5.95 × 10 –5

Example 4: A loss free transmission line has distributed inductance of 1.2 mH/km and a distributed capacitance of 0.05 µF/km. Calculate characteristic impedance Z0 and propagation constant γ of the line.

Transmission Lines

`

585

Solution: Characteristic impedance of a loss-less line, (R = G = 0) Z0 =

L C

where, L = 1.2 × 10 −3 H/ km and C = 0 .05 × 10 −6 F/km Z0 =

∴

1.2 × 10 −3 0 .05 × 10 −6

= 154 .92 ohms

Propagation constant for a loss free transmission line γ = j β = jω √ ( LC) = jω √ (1.2 × 10 −3 × 0 .05 × 10 −6 ) γ = j 7.74 × 10 −6 ω

or

Example 5: A telephone line has R = 30 Ω/km, L=100 mH/km, G=0 and C=20 µ F/km at f =1 kHz, obtain; (i) The characteristic impedance of the line (ii) The propagation constant (iii) The phase velocity [UPTU, B.Tech. IV Sem. 2008]

Solution: (i) The characgterstic impedance, Z0 of the line is

1 /2

Z0 =

–1 2 2 R + j ω L R + (ω L) ∠ tan (ω L / R ) = G + jωC G2 + (ωC)2 ∠ tan –1(ω C / G)

In this problem G = 0, therefore 1 /2

R 2 + (ω L)2 ∠ tan –1(ω L / R ) Z0 = ωC ∠ tan –1(ω C / 0) θ10 =

where

1 /2

180 180 tan –1(ω L / R ) and θ02 = tan –1(ωC / G) π π

ω = 2 π f = 2 × 3 .14 × 1 × 103 = 6 . 28 × 103

Here ∴

R 2 + (ωL)2 or Z0 = ωC

1 /2

R 2 + ω2 L2

1 /2

= (30)2 + (6 .28 × 103 )2 (100 × 10 –3 )2

= (900 + 394 . 384 × 103 )1 /4 = (395284)1 /4 = 25 .074 Similarly

(ω C)1 /2 = (6 .28 × 103 × 20 × 10 –6 )1 /2 = (0.1256)1 /2 = 0 .3544

∴

1 180 tan –1(6 .28 × 103 × 100 × 10 –3 / 30) 25 .074 ∠ × 2 3 .14 Z0 = 1 180 tan –1 ∞ 0 .3544 ∠ × 2 3 .14

or

Z0 = 70 . 75

or

Z0 = 70 . 75 ∠ – 1. 35 °

∠43 .65 ° ∠α = 70 .75 ∠ (43 . 65 °– 45 ° ) ∵ = ∠ α –β ∠β ∠45 °

1 0 θ 2 1 1 ∠ θ02 2 ∠

E lectromagnetic F ield T heory

586

(ii) The propagation constant, γ = (R + jωL)(G + jωC) = (R + jω L) jω C (∵ G = 0) ∴

γ = (25 . 074 ∠43 . 65 ° ) (0 . 3544 ∠45 ° ) = 0. 886 ∠(43 . 65 ° + 45 ° )

or

γ = 0 . 886 ∠ 88 . 65 ° = 0 . 886 cos 88 . 65 °+ j0.886 sin 88.65 ° γ = 0 .886 × 0 .0253 + j 0 .886 × 0 . 9997 γ = 0.0224 + 0. 8857j = α + βj

(iii) Phase velocity, v =

3 ω 6 .28 × 10 = 7. 0904 × 10 3 m/s = β 0 .8857

Example 6: A telephone cable has resistance of 20 ohms a capacitance of 0.05 µF per mile and negligible distributed inductance and conductance. Loading coils each having an inductance of 80 mH and a resistance of 10 ohms are inserted at intervals of 0.7 mile. Calculate attenuation constant α at a frequency of 5000/2π Hz. Solution: According to the given problem, the loading coils are inserted at intervals of 0.7 mile, therefore 10 loading inductance per mile, Ld = 80 mH × = 114 .286 mH 7 and loading resistance per mile, R d = 10 Ω × ∴

10 = 14 .286 ohms 7

Line resistance/mile, R = R c + R d = 20 + 14 .286 = 34 .286 ohms, L = 114 .286 mH, C = 0 .05 µF and G is negligible.

The attenuation constant for telephone cable which is distortion less (or for a low loss line), is given as

or

or

α=

R 2

C G + L 2

L C

α=

R 2

C L

α=

34 .286 0 .05 × 10 −6 2 114 .286 × 10 −3

(∵ G is negligible)

α = 0.0113 neper / mile

Example 7: A loss free transmission line has distributed inductance of 1.8 mH/km and a distributed capacitance of 0.08 µF/km. For a 0.4 km section of the above line, find (i) the frequency at which the line length is equivalent to one wavelength, and (ii) the velocity of propagation. Solution: (i) Line length = 0.4 km. If wavelength equals to line length, that is, λ = 0 .4 km, then frequency, f = v /λ = c /λ, where v is the velocity of propagation which is equal to the velocity of radiation in free space, that is, the velocity of light, c

Transmission Lines

`

∴

f =

3 .0 × 108 0 .4 × 103

587

= 750 kHz

(ii) The velocity of propagation v is expressed as ω and β = ω √ ( LC) [For loss free line] v= β ∴

v=

1 LC

For 0.4 km line length, distributed inductance is 1.8 mH/cm and distributed capacitance is 0.08 µF/km. Therefore, L = 1.8 × 10 −3 × 0 .40 = 0 .720 × 10 −3 H C = 0 .08 × 10 −6 × 0 .40 = 0 .032 × 10 −6 F ∴

v=

1 0 .720 × 10

−3

× 32 × 10 −9

= 208.33 ×10 3 m /sec =208.33 km / sec

C haracteristic Impedance or Surge Impedance The characteristic impedance of a transmission line is expressed as Z0 =

R + jω L G+ jωC

...(1)

where R, L, G and C are the line parameters or the primary constants of the line. The primary constants of the transmission line or line parameters depend on the dimensions of the line and on the properties of the conductors and dielectric materials. Usually the conductivity of a conductor in a line is so high that the effect of series resistance (R ) may be assumed negligible for the purpose of computation of propagation constant. That is, γ = [(R + j ω L) (G + j ω C)]

...(2)

= [ j ω L (G + j ω C)] γ = j ω √ ( LC) (1 + G / j ω C)1 /2

or

...(3)

Earlier, we have derived an expression for propagation constant for a Transverse Electromagnetic (TEM) wave in a medium of permeability µ and permittivity ε, where the conductivity is σ as σ γ = j ω µ ε 1 + j ω ε

1 /2

Comparison of eqn. (3) and (4) gives, G σ = C ε Eqn. (4) is true for a loss-less line.

and

LC = µ ε

...(4)

E lectromagnetic F ield T heory

588

T ransmission Line Terminated in a Load Impedance Z L Let us consider a transmission line terminated in a load impedance Z L as shown in fig. 8. Suppose VS and I S are the sending end voltage and current respectively. In order to find the voltage and current at a distance x from the sending end we begin with the basic line equations as, V = Ae −γ x + Be γ x I =Ce

−γ x

+ De

Is

Ix

Vs Z0

Vx

ZL

...(1)

γx

...(2)

where A, B, C and D are constants having values that depends upon the conditions existing in the line and γ the propagation constant.

x Fig. 8

At a distance x from the sending end the voltage Vx and current I x may be expressed Vx = A e − γ x + Be γ x Ix = Ce Also

−

−γ x

+ De

...(3)

γx

...(4)

dV = (R + j ω L) I x [Well known transmission line equation] dx

...(5)

where R is the resistance per unit length and L the inductance per unit length of the line. Differentiating eqn. (1) with respect to x, we get dV dV = γ A e − γ x − γ Be γ x = − γA e − γ x + γ B e γ x or – dx dx

...(6)

Equating eqns. (5) and (6), we get (R + j ω L) I x = γ A e or

Ix =

−γ x

− γ B eγ x

γ ( Ae − γ x − Beγ x) (R + j ω L)

But Z0 = (R + j ω L) /(G + jω C) and ∴ R + j ω L = Z0 γ where Z0 is the characteristic impedance of the line 1 ∴ Ix = ( Ae − γ x − B e γ x ) Z0 Using hyperbolic functions,

e γ x = cosh γ x + sinh γ x

and

γ = [(R + jωL) (G + jω C)]

...(7) e − γ x = cosh γ x − sinh γ x

In terms of hyperbolic functions eqn. (7), may be written as 1 Ix = [ A (cosh γ x − sinh γ x)] − B (cosh γ x + sinh γ x)] Z0 or

Ix =

1 [( A − B) cosh γ x − ( A + B) sinh γ x] Z0

...(8)

Substituting the value of e − γ x and e γ x in terms of hyperbolic function in eqn. (3), we obtain Vx = A (cosh γ x − sinh γ x) + B (cosh γ x + sinh γ x) or

Vx = ( A + B) cosh γ x − ( A − B) sinh γ x

...(9)

Transmission Lines

589

To find the value of A + B and A − B applying boundary conditions, that is, at x = 0, Vx = Vs , I x = I s , sinh γ x = 0 and cosh γ x = 1 in eqns. (8) and (9), we get Is =

1 ( A − B) Z0

A − B = Z0 I s

or

and Vs = A + B

Substituting these values of A − B and A + B in eqns. (9) and (8), we get Vx = Vs cosh γ x − Is .Z0 sinh γ x

...(10)

1 (Z0 I s cosh γ x − Vs sinh γ x) Z0

and

Ix =

or

Ix = Is cosh γ x −

Vs sinh γ x Z0

...(11)

Eqns. (10) and (11) can be used for any values of Z L and for any distance x.

I

nput Impedance of a Loss-Less Transmission Line Is

The input impedance or sending end impedance of a

IR

transmission line is equal to the impedance offered by the line at its input terminals or at its sending end terminals, and is denoted by Zs . To determine the input impedance Z s ,

Vs , Zs

Z0

ZL

VL

let us consider a transmission line of length l, having propagation constant γ and characteristic impedance Z0 connected to a load Z L as shown in fig. 9. The general voltage and current equations at a

x=l

x=0

x=l

distance x from sending end along the line may be expressed as, Vx = Vs cosh (γ x) − I s Z0 sinh (γ x) and

I x = I s cosh (γ x) −

Fig. 9

...(1)

Vs sinh (γ x) Z0

...(2)

Since the line is terminated at length x = l with load impedance Z L , the voltage VL and current I L at the receiving end ( x = l).

Zs Vs

Z0

ZL

Is

ZS Vs Is Fig. 10: Various Form of Transmission Lines

The voltage and current at the load end are

ZL

E lectromagnetic F ield T heory

590

VL = Vs cosh (γ l) − I s Z0 sinh (γ l) I L = I s cosh (γ l) –

...(3)

Vs sin h (γ l) Z0

...(4)

The voltage across the load impedance Z L may also be expressed as VL = I L Z L

...(5)

Substituting the value of VL and I L from eqns. (3) and (4) in eqn. (5), we get Vs cosh (γ l) − I s Z0 sinh (γ l) = Z L [I s cosh (γ l) −

Vs sinh (γ l)] Z0

or

Vs cosh (γ l) Z0 − I s Z02 sinh (γ l) = I s Z L Z0 cosh (γ l) − Vs Z L sinh (γ l)

or

Vs [Z0 cosh (γ l) + Z L sinh (γ l)] = I s Z0 [Z0 sinh (γ l) + Z L cosh (γ l)]

or

Vs Z0 [Z L cosh(γ l) + Z0 sinh(γ l)] = Z0 cosh(γ l)+ Z L sinh (γ l) Is

But

Vs = Z s , the input impedance of the given transmission line, Is Z cosh(γ l ) + Z0 sinh (γ l ) Zs = Z0 L or Zs = Z0 Z0 cosh (γ l ) + Z L sin h (γ l )

∴

Z L + Z0 tan(γ l) ...(6) Z0 + Z L tan (γ l)

Eqn. (6) may be used to determine the input impedance of any line at any termination. For a loss-less transmission line, in which energy dissipation is zero, both the line resistance (R ) and conductance (G) are zero. Thus, the propagation constant (γ) for a loss-less transmission line reduced to, γ = √ (R + j ω L) (G + j ω C) = j ω √ ( L C) = α + j β Equating real and imaginary parts, we get, ∴

γ = j β, where β = ω √ ( L C) and α = 0

For the input impedance of a loss-less transmission line, substituting the above value of γ in eqn. (6), we get Z cosh ( j β l) + Z0 sinh ( j β l) Z s = Z0 L Z0 cosh ( j β l) + Z L sinh ( j β l) cosh ( j β l) = cos β l and

But

...(7)

sinh ( j β l) = j sin β l

Therefore, eqn. (7) becomes

or

Z cos (β l) + j Z0 sin (β l) Z s = Z0 L Z0 cos (β l) + j Z L sin (β l)

...(8)

Z + j Z0 tan β1 Zs = Z0 L Z0 + j Z L tan β1

...(9)

Eqns. (8) and (9) represent the input impedance of a loss-less transmission line and the quantity β l, the electrical length of the line in degree or radians.

Special Case 1.

Input impedance of a transmission line of length l terminated in its characteristic impedance Z0

Transmission Lines

591

If the line is terminated in its characteristic impedance Z0 , then Z L = Z0 . Therefore, eqn. (6) becomes

Z cosh (γ l) + Z0 sinh (γ l) Z s = Z0 0 Z0 cosh (γ l) + Z0 sinh (γ l)

or

ZS = Z0

...(10)

That is, a line terminated in its characteristic impedance has input impedance equals to its characteristic impedance Z0 . In such transmission lines there is no reflection and thus the line behaves as an infinite line. 2.

Input impedance of a transmission line short circuited at its receiving end. If the line is short circuited at the receiving end, then Z L = 0. Therefore for short circuited line eqn. (6) becomes

0 × cosh (γ l) + Z0 sinh (γ l) Z0 sinh (γ l) (Z s )SC = Z0 = Z0 Z0 cos h (γ l) Z0 cosh (γ l) + 0 × sinh (γ l) or 3.

(Z s ) SC = Z 0 tanh (γ l)

...(11)

Input impedance of a transmission line open circuited at its receiving end. For a line open circuited at its receiving end, that is, Z L = ∞. We divide the numerator and denominator of eqn. (6) by Z L and then put Z L = ∞. Therefore, eqn. (6) becomes

cosh (γ l) + (Z0 / Z L ) sinh (γ l) Z s = Z0 (Z0 / Z L ) cosh (γ l) + sinh (γ l) cosh (γ l) cosh (γ l) + (Z0 /∞) sinh (γ l) = Z0 = Z0 Z l l ( / ) cosh ( γ ) sinh ( γ ) ∞ + 0 sinh (γ l) or

(Z s )OC = Z 0 coth (γ l)

...(12)

The equations (11) and (12) for input impedances of short circuited and open circuited lines are also used to determine characteristic impedance Z0 and propagation constant γ as follows: Multiplying eqn. (11) by eqn. (12), we get

(Z s )sc . (Z s )oc = Z0 2

or

Z0 = √ [( Zs )sc . (Z s )OC ]

...(13)

Dividing eqn. (11) by (12), we get

tanh (γ l) (Z s )SC = = tan2 h (γ l) (Z c )OC cosh (γ l) 4.

or

tanh (γ l) =

(Z s )SC (Z s )OC

...(14)

Input impedance of a quarter-wavelength loss less transmission line: The input impedance of a quarter-wavelength (l = λ /4) loss-less transmission line is determined by eqn. (8) as

2π λ 2π λ Z L cos λ . 4 + j Z0 sin λ . 4 Z s = Z0 Z cos 2 π . λ + j Z sin 2 π . λ L λ 4 λ 4 0 or

0 + jZ0 Z s = Z0 0 + jZ L

or Z s =

Z2 0 ZL

2π ∵ For λ / 4 line, β = λ

E lectromagnetic F ield T heory

592

or

Z0 =

Zs × ZL

Therefore, a quarter wave line transforms an impedance into an admittance. If there is a short circuit at the receiving end of the line, the input impedance of the line becomes infinite. Thus, a quarter wavelength line terminated into an open circuit line acts like a short circut as viewed from the input terminals. Similarly a quarter wavelength line terminated into a short circuit behaves as an opencircuit at it input terminals. 5.

Input impedance of a Half-wavelength loss-less transmission line The input impedance of a one-half wavelength long (l = λ /2) loss-less line is determined by eqn. (8) as,

2π λ 2π λ Z L cos λ . 2 + j Z0 sin λ . 2 Z s = Z0 Z cos 2 π . λ + j Z sin 2 π . λ L 0 λ 2 λ 2 or

Z cos π + jZ0 sin π Z = Z0 L Z s = Z0 L Z0 Z0 cos π + jZ L sin π

or Z s = Z L

It means that the load impedance is replicated at every one-half wavelength in transmission line.

S hort-Circuited and Open Circuited Loss-Less Lines The basic equations for voltage and current at any point along a transmission line may be expressed as V = Vs cosh (γ x) − (I s Z0 ) sinh (γ x) and

I = I s cosh (γ x) − (Vs / Z0 ) sinh (γ x)

...(1) ...(2)

where Vs and I s are the voltage and current at the sending end, Z0 characteristic impedance, and γ the propagation constant of the line. For a loss-less line the propagation constant is imaginary and characteristic impedance is a pure resistance, and are given as γ = j β and Z0 =

L = R0 C

...(3)

Therefore the basic line equations for loss-less line becomes,

and

V = Vs cosh ( j β x) − I s R0 sinh ( j β x)

...(4)

V I = I s cosh ( j β x) − s sinh ( j β x) R0

...(5)

But

cosh ( j β x) = cos β x

and

sinh ( j β x) = j sin β x (Well known relations)

Therefore, eqns. (4) and (5) take the form V = Vs cos β x − j Is R 0 sin β x and

V I =Is cos β x − j s sin β x R0

...(6) ...(7)

Transmission Lines

593

1. Short Circuit Termination (Z L = 0) If the loss-less line of length l ( x = l) is short circuited at the receiving end then the voltage at the receiving end will be zero, that is, V = VL = 0 and current would be equal to I l (say). Therefore, eqns. (6) and (7) become, O = Vs cos β l − j I s R0 sin β l

...(8)

and

V I l = I s cos β l − j s sin β l, where R 0 =√ ( L / C) R0

...(9)

Eqn. (8) gives

V I s = − j s cot β l R0

...(10)

Substituting the value of I s from eqn. (10) in eqns. (6) and (7), we obtain current and voltage at any point along the line as, V V = Vs cos β x − s cot β l R0 .sin β x R0 or

V=Vs (cos β x −cot β l . sin β x)

...(11)

and

V V I = − j s cot β l. cos β x − j s sin β x R R0 0

or

V I = − j s [cot β l . cos β x +sin βx] R0

...(12)

From eqn. (11) it is clear that the voltage along the line will be zero at points where Vs (cos β x − cot β l. sin β x) = 0 or

cos β x = cot β l sin β x

or

tan β x = tan β l

...(13)

V

For a short circuited transmission line

I Current node

under consideration tan β l is constant. Therefore, tan β x will satisfy eqn. (13) at

Voltage node

a number of points. These values occur when β x = [(2 π / λ ) x] is increased by π , 2 π , 3 π , ..., that is, when the distance ( x) is increased by λ /2, λ, 3λ /2, ... .

l = λ /2

l = λ /2

ZL = 0

Short circuited transmission line

Therefore, the transmission line which

Fig. 11

is short circuited at its receiving end has a number of points at which the voltage is zero. The points at which the voltage is zero are called voltage nodes. Similarly, from eqn. (12) current nodes or the points at which current is zero, will occur at points, where cot β l . cos β x + sin β x = 0 or

tan β x = − cot β l

...(14)

E lectromagnetic F ield T heory

594

The current nodes also separated by a distance λ /2 and occur mid way between the voltage nodes as shown by dotted curve in fig. 11. Hence, as a result of reflection at the receiving end of the line, stationary waves are set up along the line as shown in fig. 11. From eqn. (10) input impedance of short circuited loss free transmission line is Vs = Z s = j R0 tan β l or Z s = j ( L / C) tan β l Is

...(15)

The value of tan β l can vary from − ∞ to + ∞, therefore, the input impedance of a short circuited loss-less transmission line can either be purely inductive or purely capacitive as tan β l can be either positive or negative depending upon the values of β l. Therefore, line behaves as a pure reactance when viewed from the sending end. The reactance is inductive when the length of the line is less then λ /4 (or β l = π /2) or between λ /2 and 3 λ /4 (or β l between π and 3 π /2) etc., and capacitive when the length is between λ /4 and λ /2 (or β l between π /2 and π), 3 λ /4 and λ etc. Fig. 12 shows the variation of Z s = R0 tan β l with β l. Hence, at very high frequencies (300 MHz to 3 GHz) the transmission line can be used as circuit element.

Zs = R0 tan βl

For short circuited line Inductive

0

π /2

π

3π /2

2π

5π /2

3π

βl

βl

Capacitive Fig. 12

For a very short length of a line β l R0

V(x)

V(x) oc

I(x)

I(x) sc

x

2λ 7λ/4 3λ/2 5λ/4

λ

sc oc

3λ/4 λ/2

λ/4

0

Standing waves on short circuited (sc) and open circuited (oc) loss less lines

Fig. 15 (a)

Fig. 15 (b)

Reflection Coefficients Let us suppose that a transmission line of characteristic impedance Z0 is connected to another transmission line of characteristic impedance Z L as shown in

Transmitted wave (V2,I2)

Incident wave (V, I) Reflected wave (V1, I1)

Fig. 16. Let V and I are the voltage and current

+V

Z0

ZL

associated with the incident wave. In steady state, if V1 and I1 are the voltage and current associated with the reflected wave at the load end, and V2 and I2 that are associated with the transmitted wave (Fig. 16), then I2 = I + I1

Voltage or current

15(a), and that for open circuited (oc) and short circuited (sc) loss-less lines are shown in fig. 15(b).

...(1)

–V st

1 line

2nd line Fig. 16

ZL

E lectromagnetic F ield T heory

598

V2 = V + V1

and

I=

where

...(2)

V V , I =− 1 Z0 Z0 1

and I2 =

V2 ZL

...(3)

The reflected current I1 is taken as negative if reflected voltage is considered as positive because the circulation of current I1 is in opposite direction with respect to I. Substituting the values of I, I1 and I2 from eqn. (3) in eqn. (1), we get V − V1 V2 V V V = − 1 or 2 = − Z L Z0 Z0 ZL Z0

...(4)

Substituting the value of V2 from eqn. (2) in eqn. (4), we obtain V + V1 ZL

=

V − V1 Z0

or V Z0 + V1Z0 = Z LV − Z LV1

V (Z0 − Z L ) = − V1 (Z0 + Z L ) or

or

V1 Z L − Z0 = = ΓV V Z L + Z0

...(5)

where ΓV is the voltage reflection coefficient at the load and defined as the ratio of the voltage of the reflected wave to the voltage of the incident wave at the load. Similarly, current reflection coefficient at the load is obtained by substituting the values of V, V1 and V2 from eqn. (3) in terms of current components in eqn. (2) as V = I Z0 , V1 = – I1 Z0 and V2 = I2 Z L Therefore, eqn. (2) takes the form I2 Z L = I Z0 − I1 Z0

...(6)

Putting I2 = I + I1 from eqn. (1) in eqn. (6). We obtain (I + I1) Z L = Z0 I − Z0 I1 I1 (Z L + Z0 ) = I (Z0 − Z L ) or or

I1 Z0 − Z L = = ΓI I Z0 + Z L

Z − Z0 = − ΓV ΓI = − L Z L + Z0

...(7)

where ΓI is the current reflection coefficient at the load and defined as the ratio of the current of the reflected wave to the current of the incident wave at the load. It is clear from eqn. (7) that the current reflection coefficient ΓI at a point on the line is the negative voltage reflection coefficient ΓV at that point. Similarly, the transmission coefficient (τ L ) is defined as the ratio of the voltage of the transmitted wave to the voltage of the incident wave at the load, that is, τ= From eqn. (2)

V2 V

V2 V = 1+ 1 V V

Transmission Lines

But

599

V1 = Γ, reflection coefficient V

∴

τ = 1+ Γ

or Transmission coefficient = 1 + reflection coefficient. The reflection and transmission coefficients have both magnitudes and phases and therefore are vector quantities. Cases : (i) If the line is terminated in its characteristic impedance Z0 , that is, Z L = Z0 , then from eqn. (5) we have

Γ=

ZL − ZL =0 ZL + ZL

Hence, there is no reflected wave. (ii) If the line is short circuited at the receiving end, then Z L = 0. Then from eqn. (5) 0 − Z0 Γ= = −1 Z0 Since Γ is negative the polarity of reflected wave is reversed. Hence, for a line short circuited at the receiving end ( ZL = 0 ) the magnitude of reflected wave is equal to incident wave but in opposite direction. (iii) If the line is open circuited, that is, Z L = ∞, then from eqn. (5), we have 1 − (Z0 / Z L ) 1 − 0 Γ= = =1 1 + (Z0 / Z L ) 1 + 0 That is, for an open circuited line the reflected wave is identical to the incident wave in magnitude and sign. (iv) If Z L > Z0 , that is when Z L = 3 Z0 (say), then from eqn. (5) 3 Z0 − Z0 1 Γ= = 3 Z0 + Z0 2 Since Γ is positive, reflected wave will be of the same polarity with amplitude half of the incident wave, and if Z L < Z0 , that is, when Z L = Z0 /3 (say), then from eqn. (5) (Z /3) − Z0 1 Γ= 0 =− (Z0 / 3) + Z0 2 That is, the reflected wave travels in opposite direction to that of incident wave with half of the magnitude of incident wave. Therefore, the maximum reflection occurs when the line is either short circuited or open circuited. In fact the reflection coefficient on a transmission line varies with the distance from the load. The reflection coefficient at a point distant d from the load is defined as the ratio of the reflected component to the incident component at a point distant d from the load. Thus,

Γd =

or

Γd =

Reflected voltage or current component at a point on the line Incident voltage or current component at that point on the line (Z L − Z0 ) e − γ d (Z L + Z0 ) e γ d

or

Z − Z0 –2 γ d e Γd = L ZL + Z0

E lectromagnetic F ield T heory

600

At the load end, d = 0. Therefore Γ =

Z L − Z0 Z L + Z0

Relation between input Impedance Z s and Reflection Coefficient Γ The input impedance of a transmission line of length l, propagation constant γ, characteristic impedance Z0 and terminating impedance Z L is expressed as Z cosh (γ l) + Z0 sinh (γ l) Z s = Z0 L Z0 cosh (γ l) + Z L sinh (γ l)

...(1)

Substituting the values of cosh (γ l) and sinh (γ l) in exponential form as cosh (γ l) =

eγ l + e−γ l 2

ZL Z s = Z0 Z 0

and sinh (γ l) =

in eqn. (1), we get

e−γ l (Z L − Z0 ) 2 e−γ l (Z L − Z0 ) 2

e−γ l (Z L − Z0 ) 2 eγ l (Z L + Z0 ) 2 e−γ l (Z L − Z0 ) 2 eγ l (Z L + Z0 ) 2

or

−2 γ l 1 + e Z s = Z0 1 − e − 2 γ l

or

1+ Γ e − 2 γ l Zs = Z0 1 − Γ e − 2 γ l

where Γ =

2

eγ l −e− γ l eγ l + e−γ l + Z0 2 2 eγ l −e− γ l eγ l + e−γ l + ZL 2 2

eγ l (Z L + Z0 ) + = Z0 2γ l e 2 (Z L + Z0 ) − 1 + = Z0 1 −

eγ l − e−γ l

eγ l (Z L + Z0 ) 2 γl e (Z L + Z0 ) 2

Z L − Z0 Z L + Z0 Z L − Z0 Z L + Z0

Z L − Z0 is the reflection coefficient Z L + Z0

Eqn. (2) relates the reflection coefficient Γ and input impedance Z s of a transmission line.

...(2)

Transmission Lines

601

S tanding Wave Ratio (SWR) When a transmission line is terminated with a terminating impedance other than its characteristic impedance, a reflected wave is generated

Relative Amplitude

on the line and interference occurs between the incident and reflected wave Vmax

as a result of which standing waves of current and voltage are set up along the line. The resultant voltage or current at any point along such a line will be the

Vmin

V Vmin

Position along line

phasor sum of incident and reflected wave

Load

Fig. 17

components. At certain points reflected and incident waves interfere in such a way that the resultant voltage (or current) become maximum. On the other hand at certain other point they interfere in such a way that the resultant voltage (or current) become minimum at that points on the line. Thus, the reflected wave in conjunction with the incident wave creates nodes and antinodes in the voltage or current along the line at fixed locations. The ratio of maximum to the minimum magnitudes of voltage on the Voltage Standing Wave Ratio (VSWR) S. Thus, from fig. 17. Standing wave ratio,

S =

Maximum magnitude of voltage |Vmax | = Minimum magnitude of voltage |Vmin|

Similarly, the ratio of maximum to the minimum magnitudes of current on the current standing wave is defined as Current Standing Wave Ratio (CSWR) S and is expressed as S =

Maximum magnitude of current | I max | = Minimum magnitude of current | I min|

Thus, the Standing Wave Ratio (SWR) is a measure of mismatching between the terminating load and transmission line. Higher the value of standing wave ratio (SWR) greater the mismatch between load and transmission line. A high standing wave ratio on a line is undesirable because it results in a large power loss and noise. Hence, a low value of SWR is always appreciable except when the transmission line is used as a phase reactance or turned circuit. For perfectly matched transmission line (Vmax = Vmin ) the value of SWR is unity (S = 1) because the reflection coefficient is zero and for a mismatched line S is always greater than unity (S > 1). For a short circuited and open circuited loss-less transmission line SWR is infinity (S = ∞) because the magnitude of the reflection coefficient is each case is unity.

Relation between Standing Wave Ratio (SWR) S and Reflection Coefficient Γ It is well known that for an incorrectly terminated transmission line the voltage maxima occur at points on the line at which incident and reflected waves are in phase so that their components are added up directly. On the other hand voltage minima occur at point on which incident and reflected waves are in

E lectromagnetic F ield T heory

602

out of phase and tend to neutralize each other. If Vi and Vr the r.m.s, values of incident and reflected voltages respectively as measured from generator end, then

and

|Vmax | = |Vi| + |Vr |

...(1)

|Vmin| = |Vi | − | Vr |

...(2)

Thus, the voltage standing wave ratio S in terms of voltage maxima and minima is given by VSWR (S) =

|Vmax | | Vi | + | Vr | = | Vmin | | Vi | − | Vr |

...(3)

| Vr | | Vi | 1 + | Γ | S= = | Vr | 1 − | Γ | 1− | Vi | 1+

...(4)

where | Γ | or (| Vr |/| Vi |) is the voltage reflection coefficient. Thus, Voltage Standing Wave Ratio (VSWR) may be calculated if reflection coefficient Γ is known. Eqn. (4) clearly indicates that when reflection coefficient tends to zero, standing wave ratio tends to unity, whereas when reflection coefficient is unity, standing wave ratio tends to infinity. From eqn. (4) S [1 − | Γ |] = 1 + | Γ | or S − S| Γ | = 1 + | Γ | or

S| Γ | + | Γ | = S − 1

∴

|Γ | =

or | Γ |[S + 1] = S − 1

S −1 S+1

...(5)

Eqn. (5) gives the relation VSWR and reflection coefficient | Γ |. But

|Γ | =

Z L − Z0 Z L + Z0

.

..(6)

where Z L is the load impedance and Z0 the characteristic impedance of the transmission line. Substituting the value of | Γ | from eqn. (6) in eqn. (4), we get

or

S=

1 + (Z L − Z0 ) /(Z L + Z0 ) 1 − (Z L − Z0 ) / (Z L + Z0 )

S=

Z L + Z0 + Z L − Z0 Z L = Z L + Z0 − Z L + Z0 Z0

or S =

ZL Z0

...(7)

Thus, the standing wave ratio SWR may be calculated either by knowing the value of reflection coefficient Γ or by knowing load impedance ( ZL ) and characteristic impedance ( Z0 ). It is well known that for a loss-less line, at points where maxima occur in voltage standing wave, the minima will certainly be occur in current standing wave and vice-versa. Thus, the voltage maximum positions and current minimum positions are same or voltage maximum and a current minimum occur at the same point. Therefore,

Transmission Lines

I min =

603

Vmin Z0

=

| Vi | − | Vr |

...(8)

Z0

At this position impedance of line is purely resistive and has maximum value. It is given by Z max =

| Vmax | | Vi | + | Vr | = | I min | | Vi | − | Vr | Z0

...(9)

| V | + | Vr | = Z0 i | Vi | − | Vr |

or

But, ∴

Z max

1 + |Γ | 1 −| Γ |

| Vr | 1 + | V | i = Z0 1 − | Vr | |V | i

1 + | Γ | or Z max = Z0 1 − | Γ |

| Vr | ∵ = | Γ | | V | i

...(10)

=S

S max = Z0 × S

...(11)

Thus, the maximum impedance of a line is the product of Voltage Standing Wave Ratio (VSWR)S and characteristic impedance Z0 of the line. The positions at which voltage is minimum and current is maximum, impedance of line is minimum. Therefore, Z min = But

I max =

∴

Z min =

Vmin I max | Vi | + | Vr | Z0

...(12) and Vmin = | Vi | − | Vr |

| Vi | − | Vr | | Vi | + | Vr | Z0

| V | − | Vr | = Z0 i = Z0 | Vi | + | Vr |

or

| Vr | 1 − | V | i 1 + | Vr | | Vi |

1 − | Γ | | Vr | Z min = Z0 = | Γ | ∵ 1 + | Γ | | Vi | = Z0

Again,

...(13)

1 1 + | Γ | 1 − | Γ |

1 + |Γ | =S 1 −| Γ |

...(14)

E lectromagnetic F ield T heory

604

∴

Z min =

Z0 S

...(15)

Hence, the minimum impedance of the line is equal to the product of reciprocal of Voltage Standing Wave Ratio (VSWR) S and characteristic impedance of the line Z0 .

Relations for Reflection and Transmission Coefficients and Voltage Standing Wave Ratio of a Line Reflection coeffcient for Voltage,

ΓV =

Z L – Z0 Z L + Z0

Reflection coefficient for current,

ΓI =

Z0 – Z L = –ΓV Z0 + Z L

Transmission coefficeint for Voltage, τV =

2ZL = 1 + ΓV Z0 + Z L

Transmission coefficient for current, τ I =

2 Z0 = 1 + ΓI Z0 + Z L

Voltage standing wave ratio, VSWR

=

1 + |ΓV | 1 + |ΓI| = 1–|ΓV | 1–|ΓI|

Magnitude of Γ interms of VSWR, |ΓV | =|ΓI| =

VSWR –1 VSWR +1

Power Transmitted along a Low-Loss Line The computation of power transmitted by a transmission line is comfortable at the point of maximum or minimum of voltage where the impedance is purely resistive. At a point of maximum voltage and minimum current, we have Vmax = Z max or Vmax = Z max . I min I min But

Z max = Z0 × S,

∴

Vmax = I min Z0 . S or I min =

Vmax Z0 S

So the power transmitted on the line may be written as P = | Vmax | × | I min | = | Vmax | × or

P=

| Vmax | Z0 S

|Vmax | 2 Z0 S

where S is the voltage standing wave ratio and Z0 the characteristic impedance of the line. Example 8: A 10 km long transmission line has the following constants : Characteristic impedance Z0 = 100 ohms, attenuation constant α = 0 .1 neper/km and phase constant β = 0 . 05 rad/km. Find the received current and voltage when 20 mA current is sent over the line from one end and the other end is short circuited.

Transmission Lines

605

Solution: The equation of current at the receiving end of a transmission line of length x, characteristic impedance Z0 and propagation constant γ is given by I x = I s cosh γ x −

Vs sinh γ x Z0

...(1)

If Z s is the input impedance of the line then: Vs = I s Z s For a short circuited line, Z s = Z0 tanh γ x, therefore Vs = I s Z0 tanh γ x or

Vs = I s tanh γ x Z0

...(2)

Substituting the value of Vs / Z0 from eqn. (2) in eqn. (1), we get I x = I s cosh γ x − I s tanh γ x.sinh γ x or

I x = Is

or

Ix =

(cosh2 γ x − sinh2 γ x) cosh γ x

Is cosh γ x

(∵ cosh2 γ x − sinh2 γ x = 1)

Here I s = 20 mA, α = 0 .1 neper/km, β = 0 . 05 rad/km; γ = α + j β = 0 .1 + j 0 . 05 and x = 10 km ∴

Ix =

20 20 = cosh [(0 .1 + j 0 .05) 10] cosh (1 + j 0 .5)

We know that, cosh ( A + B) = cosh A cosh B + sinh A sinh B cosh ( jB) = cos B ∴ Hence

and

sinh (B j) = j sin B

cosh ( A + jB) = cosh A cos B + j sinh A sin B cosh (1 + 0 .5 j) = cosh (1) cos (0 .5) + j sinh (1) sin (0 .5) 0 .5 × 180 ° 0 .5 × 180 ° = cosh (1) cos + j sinh (1) sin π π = cosh (1) cos (28 .66 ° ) + j sinh (1) sin (28 .66 ° ) = 1.543 × 0 .8775 + j 1.175 × 0 .479 = 1. 354 + j 0 .562 0 .562 = (1.354)2 + (0 .562)2 ∠ tan −1 1.354 = √ (2 .149) ∠ [tan −1 (0 .415)]

∴

= 1.466 ∠ 22 .54 ° 20 Ix = = 13.64 ∠ −22.54 ° mA 1.466 ∠ 22 .54 °

Since, the line is short circuited at the receiving end, the receiving voltage is zero. Example 9: A loss-less transmission line has a characteristic impedance of 200 ohm and is quarterwavelength long. What is the voltage at the open circuited receiving end if the sending end is connected to a generator which has 40 ohm internal impedance and a voltage of 5 volt..

E lectromagnetic F ield T heory

606

Solution: For a transmission line of lengths l, the sending end voltage, VS and receiving end voltage VL are related as, VS = e γ l or VL = VS e – γ l VL For loss- less line,

α = 0, β =

2π and γ = jβ λ

VL = VS e – jβl = VS (cos βl – j sin βl)

∴

For quarter wavelength line, l = λ / 4, ∴ β l =

(∵ e – jθ = cos θ – j sin θ)

2π λ π . = λ 4 2

π π VL = VS cos – j sin = – j VS 2 2

∴

....(1)

In an open circuited receiving end no current flows, That is, Here

IL = 0 Z0 = 200 ohm, Vg = 5 volt and R g = 40 ohm

Thus the sending end current I S is IS =

Vg Z0 + Z g

=

5 5 = = 0.0208 Amp 200 + 40 240

and

VS = Z0 I S = 200 × 0.0208 = 416 . volt

From eqn. (1),

VL = – jVS = – j4.16 volt.

Example 10: What length of line is needed and how should it be terminated to act as a capacitance of 2pF at 500 mc/sec ? The line impedance is 300 ohms. Solution: Input impedance of the line open circuited at its receiving end is expressed as (Z s )oc = Z0 coth (γ l) For a loss-less line γ = j β, Therefore, (Z s ) oc = − j Z0 cot β l or |(Z s )oc | = Z0 cot β l

[∵ coth ( j β) = − j cot β l]

If l is the length of the line at which the line should be terminated to act as a capacitance, then Z s = Capacitive reactance = ∴ or or

or

1 1 = = 159.23 ohms 2 π f C 2 × 3 .14 × (500 × 106 ) × 2 × 10 −12

2 π 159.23 = Z0 cot β l = 300 cot l λ 300 2 π 159 .23 2 π or tan cot l = l = λ λ 159 .23 300 2π λ l = tan −1 (1.88) or l = × 61.99 ° λ 2π l=

61.99 λ 360

but λ =

10 c 3 .0 × 10 = cm = 60 cm f 500 × 106

(∴ β = 2 π / λ )

Transmission Lines

∴

l=

607

61.99 360

× 60 = 10.33cm

For a short circuited loss-less line, the input impedance is given by |(Z s )sc| = Z0 tan β l If l is the length of the line at which the line should be terminated to act as a capacitance, then 1 2 π = 159 .23 ohms = Z0 tan λ 2π f C

(Z s )sc = ∴ ∴ or

l

2 π 159 .23 = 0 .53 tan l = λ 300 2π l = tan −1(0 .53) = 27 .92 ° λ l=

27 .92 27 .92 λ= × 60 = 4.653 cm 2π 360

Example 11: The short circuited and open circuited impedance of a transmission line of length 100 miles are : ( Zs )sc = 700 ∠30 ° and

( Zs )oc = 500 ∠ − 40 °

Find the characteristic constants of the line. Solution : For a short circuited loss-less line, the input impedance is given by (Z s )sc = Z0 tanh (γ l)

...(1)

Similarly, the input impedance of an open circuited line is given by (Z s )oc = Z0 coth (γ l)

...(2)

Multiplying eqns. (1) and (2), we get Z02 = (Z s )sc (Z s )oc or Z0 = [(Z s )sc . (Z s )oc ] ∴

Characteristic impedance, Z0 =

...(3)

[(700 ∠ 30 ° ) (500 ∠ − 40 ° )]

30 ° 40 ° = (700 × 500)1 /2 ∠ − 2 2 or

Z0 = 591. 608 ∠ −5 °

...(4)

Dividing eqn. (1) by (2), we get (Z s )sc (Z s )sc = tanh2 γ l or tanh γ l = (Z s )oc (Z s )oc 1 /2

∴

700 ∠ 30 ° tanh γ l = 500 ∠ − 40 °

=

700 .∠ 500

30 ° 40 ° = 1.183 ∠ 35 ° − – 2 2

tanh γ l = 1.183 cos 35 ° + j 1.183 sin 35 ° = 1.183 × 0 .819 + j 1.183 × 0 .5736 or Further,

tanh γ l = 0 .969 + j 0 .678 tanh γ l =

2γ l

−1

2γ l

+1

e e

∴ e2 γ l =

...(5) 1 + tanh γ l But γ l = (α + j β) l 1 − tanh γ l

E lectromagnetic F ield T heory

608

∴

e2 (α

+ j β)l

=

1 + 0 .969 + j 0 .678 1.969 + j 0 .678 = 1 − 0 .969 − j 0 .678 0 .031 − j 0 .678

[From eqn. (5)]

0 .678 (1.969)2 + (0 .678)2 ∠ tan −1 2 .082 ∠ 19 .00 ° 2 .082 1.969 = = = ∠ (19 .00 ° + 87 .38° ) ∠ − . 0 679 87 . 38 ° 0 .679 . 0 678 −1 2 2 (0 .031) + (0 .678) ∠ tan − 0 .031 or

e2 (α

∴

+ j β) l

= 3 .066 ∠ 106 .38° or e2 α l ∠2 β l = 3 .066 ∠106 .38°

e2 α l = 3 .066 and 2 β l = 106 .38° e2 α l = 3 .066 or 2 α l = log e 3 .066 α=

1.120 1.120 = 0.0056 neper / mile = 2l 2 × 100

or

β=

106 .38 106 .38 × π 106 .38 × 3 .14 rad = = 2 × 100 200 × 180 200 × 180

or

β =0.00928 rad / mile

or Similarly, 2 β l = 106 .38°

Example 12: Calculate the characteristic impedance of line when its input impedance is measured to be 286∠ 40° ohm with the receiving end open, and it is 1520 ∠ 16 ° ohm when the receiving end is shorted. Solution: If Z oc is the input impedance of the line with receiving end open and Z sc when the receiving end is short circuited, then the characteristic impedance Z0 of the line is given by, Z0 = Z oc × Z sc Here

Z0 C = 286 ∠40 ° and Z sc = 1520 ∠ 16 °

∴

40 ° 16 ° Z0 = 286 ∠40 °×1520 ∠16 ° = 286 × 1520 ∠ + 2 2

or

Z0 = 659.33 ∠28 ° ohm

Example 13: A load having impedance of ZL = 400 ohm is to be connected to a line of R0 = 144 ohm by a quarter wave matching transformer. find characteristic impedance Z0 of the matching transformer. Solution: The characteristic impedance Z0 of the quarter wave matching transformer is given by Z0 = Z L × Z S = Z L × R0 Here Z L = 400ohm and

Z S = R0 = 144 ohm

∴

Z0 = 400 × 144 = 240 ohm

Example 14: A 25 meter long loss- less transmission line is terminated with a load having an equivalent impedance of 40+ j 30 ohm at 10 MHz. The per unit length inductance and capacitance of the line are 310.4nH/m and 38.28 pF/m, respectively. Calculate the input impedance at the sending and of the line. Solution: The input impedance at the sending end of the loss-less transmission line is,

Transmission Lines

609

Z + jZ0 tan βl Z S = Z0 L Z0 + jZ L tan βl The characteristic impedance, Z0 and the phase constant β of the line are; Z0 =

310 .4 × 10 –9 L = = 90 .05 ≈ 90 ohm C 38 . 28 × 10 –12

β = ω LC = 2 π × 10 × 106 × (310 .4 × 10 –9 × 38.28 × 10 –12 ) or

β = 6 .28 × 107 × 34 .47 × 10 –10 = 0 .2165 rad

Here l = 25 meter, Z l = 40 + j30 ohm 180 ° In degree, β = 0 .2165 × = 0 .2165 × 57. 32 ° = 12 .4 ° π or Here ∴

β l = 12 .4 °×25 = 310 ° or

β l = 2 π – 310 ° = 50 °

Z L = 40 + j30 (40 + j30) + j90 tan 50 ° 40 + j30 + j90 × 11918 . Z S = 90 = 90 + + j j ° + + × 90 ( 40 30 ) tan 50 90 ( 40 j 30 ) j 11918 . 40 + j30 + j107. 26 40 + j137. 26 = 90 = 90 + 90 47 . 67 j – 35 . 75 54 . 25 + j47. 67

or

Z S = 90 Z S = 90 ×

or

(40)2 + (137. 26)2 ∠ tan –1(137. 26 / 40) (54 . 25)2 + (47. 67)2 ∠ tan –1(47. 67 / 54 . 25) 142 . 97 ∠ tan –1(3 . 4315) = 178 .17(73 . 76 °–41. 30 ° ) 72 . 22 ∠ tan –1(0 . 8787)

ZS = 178 .17 ∠ 32 . 46 °

Example 15: A loss-less transmission line of 200 ohms characteristic impedance is connected to a load of 300 ohms. Determined the voltage reflection coefficient and standing wave ratio (SWR). Solution: The voltage reflection coefficient, Γ =

Z L − Z0 Z L + Z0

where Z L is the load impedance and Z0 the characteristic of the line. Here, ∴

Z0 = 200 Ω, and Z L = 300 Ω Γ=

Standing wave ratio (SWR)S =

300 − 200 100 1 = = = 0.2 300 + 200 500 5 1 + | Γ | 1 + 0 .2 1.2 =1.5 = = 1 −| Γ | 1 − 0 .2 0 .8

Example 16: A transmission line with a characteristic impedance of 400 ohms is terminated in a purely resistive load. While making standing wave ratio (SWR) measurements, the meter reads a maximum voltage of 8.0 µV and a minimum voltage of 4 µV. What should be the load resistance ?

E lectromagnetic F ield T heory

610

Solution: Standing wave ratio (SWR) is given as, 1 + |Γ | Z − Z0 , where | Γ | = L S= 1 −| Γ | Z L + Z0 Z L − Z0 Z L + Z0 Z + Z0 + Z L − Z0 Z L = L = S= Z L − Z0 Z L + Z0 − Z L + Z0 Z0 1− Z L + Z0 1+

∴

Also Here, ∴

| Vmax | | Vmin |

=

| Vmax | ZL or Z L = Z0 Z0 | Vmin |

| Vmax | = 8 .0 µV , | Vmin | = 4 .0 µV , 400 × 8 .0 ZL = = 800 ohms 4 .0

and

Z0 = 400 Ω

Example 17: What would be the reflection coefficient if the standing wave ratio is 3 ? What would be standing wave ratio for zero reflection coefficient. Solution: Standing wave ratio, S =

1 + |Γ | S −1 or | Γ | = 1 −| Γ | S +1

Here

S =3 ∴ Γ =

For zero reflection, Γ = 0

S=

3 −1 2 = = 0.5 3 +1 4

1+ 0 Z =1 = L 1−0 Z0

That is, Z L = Z0 . It means that the line is perfectly matched. Example 18: Find the voltage reflection coefficient for a line with characteristic impedance equal to 250Ω.This transmission line is terminated in the load impedance of 120 + j 180 . Solution: We know that, the reflection coefficient, Γ = Here ∴

Z L − Z0 Z L + Z0

Z L = 120 + j 180 and Z0 = 250 Ω 120 + j 180 − 250 − 130 + j 180 (18 j − 13) (− 18 j + 37) Γ= = = × 120 + j 180 + 250 370 + j 180 (18 j + 37) (− 18 j + 37) = =

(18)2 + 37 × 18 j + 13 × 18 j − 37 × 13 (18)2 + (37)2

=

324 + 666 j + 234 j − 481 324 + 1369

900 j − 157 1693

Γ = 0.53j − 0.093 Example 19: An open wire transmission line with Z0 = 300 Ω is terminated by ZL = 600 Ω . Find the reflection coefficient and standing wave ratio. [U'khand, B.Tech IV Sem. 2011]

Transmission Lines

611

Solution: The reflection coefficient ΓL is given by ΓL = Standing wave ratio (SWR) =

Z L – Z0 600 – 300 300 = = = 0.33 Z L + Z0 600 + 300 900 1 + |ΓL| 1 + 0 .33 1.33 = = 1–|ΓL| 1 – 0 .33 0 .67

S = 19 . Example 20: A transmission line of Z0 = 50 Ω is terminated by RL = ZL = 100 Ω. Find SWR, Zmin , and Zmax . Solution: The standing wave ratio, SWR =

Z L 100 = =2 Z0 50

[U'khand, B.Tech IV Sem. 2009]

At the point of maximum voltage, the current is minimum Z max = Z0 ∴

1+|ΓL| = Z0 S 1–|ΓL|

Z max = 50 × 2 = 100 ohm

At the point of minimum voltage, the current is maximum Z min =

Vmin I max

Vmax = V+ e – αx (1 + | S|) Vmin = V+ e – αx (1 – | S|) I max =

or

– αx Vmax V+ e = (1 + | S|) Z0 Z0

Z min =

Vmin V+ e – αx (1–| S|) Z0 = I max V+ e – αx (1+| S|)

Z min =

Z0 Z 50 = 0 = = 25 ohm (1 + | S|) / (1–| S|) S 2

Example 21: A transmission line with a characteristic resistive impedance 30ohm is connected to a 200 ohm resistive load to a 50 volt D.C. Source with 10 ohm internal resistance. Calculate the voltage reflection coefficient at the load and at the source. Solution: Voltage reflection coefficient at the load is ΓL =

Z L – Z0 R L – Z0 = Z L + Z0 R L + Z0

ΓL =

200 – 30 170 = = 0.74 200 + 30 230

Similarly, the voltage reflection coefficient at the source is,

E lectromagnetic F ield T heory

612

R S – Z0 10 – 30 20 1 = =– = – =– 0.5 R S + Z0 10 + 30 40 2

ΓS =

Example 22: A radio frequency line of characteristic impedance 600 ohm is terminated is an impedance 400 + j 200 ohm. find the voltage standing wave ratio. Solution: Voltage standing wave ratio (VSWR) in terms of reflection coefficient Γ is expressed as, 1 + |ΓL| 1–|ΓL|

VSWR, S =

and reflection coefficient, ΓL =

Z L – Z0 Z L + Z0

Here

Z L = 400 + j200 and Z0 = 600

∴

ΓL =

or

–(400 + j200) + 600 200 – j200 1– j = = 400 + j200 + 600 1000 + j200 5 + j

(1 – j) (5 – j) 1 × 5 + 1 – 5 j – j 4 – 6 j or ΓL = 0 .1538 – 0 .2307 j = = (5 + j) (5 – j) 25 + 1 26 0 .2307 |ΓL| = (0 .1538)2 + (0 .2307)2 ∠ tan –1 – 0 .1538 ΓL =

= 0 . 2772 ∠ – tan –1(1.5) or |ΓL|=0.2772 ∠ –56 .4 ° Hence

VSWR,S =

1 + |ΓL| 1 + 0 .2772 1.2772 or S = 1. 767 = = 1 –|ΓL| 1 – 0 .2772 0 .7228

Example 23: In a transmission line having a characteristic impedance of 50 ∠ 0 ° ohms, the load impedance is a resistance of 100 ohm shunted by a capacitance of 160 µµF. Calculate the magnitude and phase of the reflection coefficient at 200 kc/sec. Solution: The situation of the problem is shown in Fig. 18. If Z L is the terminating impedance, then R + (1/ j ω C) 1 1 1 = + = Z L R (1/ j ω C) R (1/ j ω C) or

ZL =

R (1/ j ω C) R = R + (1 / j ω C ) 1 + j ω C R

The reflection coefficient Γ is given by, Γ=

Z0 − Z L Z − R /(1 + j ω C R ) = 0 Z0 + Z L Z0 + R /(1 + j ω C R )

=

(1 + j ω C R ) Z0 − R (1 + j ω C R ) Z0 + R

=

Z0 − R + j ω C R Z0 Z0 + R + j ω C R Z0

Load end Transmission line Z0=50Ω

R=100Ω

C

=160 µµF

Fig. 18

ZL

Transmission Lines

|Γ | =

or

|Γ | =

∴ Here

613

(Z0 − R )2 + (ω C R Z0 )2 2

ω C R Z0 −1 ω C R Z0 ∠ tan −1 − tan (Z0 − R ) (Z0 + R )

2

(Z0 + R ) + (ω C R Z0 ) (Z0 − R )2 + (ω CR Z0 )2

ω C R Z0 −1 ω C R Z0 and phase = tan −1 − tan (Z0 + R ) + (ω CR Z0 ) (Z0 − R ) (Z0 + R ) 2

2

f = 200 kc/sec = 200 × 103 c /sec, ω = 2 π f = 2 × 3 .14 × 200 × 103 , C =160µµF = 160 × 10 −12 F,

R = 100 Ω and Z0 = 50 Ω ω C R Z0 = 2 × 3 .14 × 200 × 103 × 160 × 10 −12 × 100 × 50 = 1.005

∴

Γ=

Thus,

|Γ | =

and ∴

(50 − 100) + j 1.005 − 50 + j 1.005 = (50 + 100) + j 1.005 150 + j 1.005 (50 − 100)2 + (1.005)2 2

2

(50 + 100) + (1.005)

=

50 .01 150 .003

≈

1 3

magnitude of Γ, | Γ | = 0.33

and

1.005 1 1.005 −1 −1 − tan −1 phase = tan −1 − tan = tan − 50 150 − 50 1 1 = π − tan −1 − tan −1 150 50

1 150

[∵ tan −1 (− x) = π − tan −1 ( x)]

= (180 ° − 1.146 ° − 0 .382 ° ) or ∴

phase = 178.47 ° Γ = 0.33 ∠178.47 °

Example 24: A loss-less transmission line has characteristic impedance of 50 ohms and is terminated in a load Zl . The generator connected to the line supplies 500 mW power. The measured voltage standing wave ratio (VSWR) on the line is 1.4. Calculate maximum and minimum value of voltage and current on the line. Solution: On account of mismatch standing waves are exist on the line, that is, voltage maxima and current minima occurs simultaneously. Impedance of the line at these points is maximum. Therefore, | Vmax | 1 + |Γ | = Z max = Z0 = Z0 . S | I min | 1 −| Γ | Power transmitted by the line, P =| Vmax |.| I min | but ∴

| I min | =

| Vmax | |V | , ∴ P =| Vmax | max Z max Z max

| Vmax |2 = P . Z0 . S

or

or P =

| Vmax | = (P. Z0 . S)

Here P = 500 mW = 500 × 10 −3 W = 0 .5 W, Z0 = 50 Ω and S = 1.4 ∴ Similarly,

| Vmax | = (0 .5 × 50 × 1.4)= 5. 92 volt P =| Vmin |.| I max |

| Vmax |2 |Vmax|2 = Z max Z0 . S

E lectromagnetic F ield T heory

614

But

| I max | = S, ∴ | I max | = S | I min | | I min |

∴

P = | Vmin | S | I min| = | Vmin |2 S

or

P =| Vmin |2 S .

or

| Vmin | = | I min | =

and

P. Z0 S

1 Z0

=

| I min | | Vmin |

or |Vmin|2 =

0 .5 × 50 1.4

P . Z0 S

= 4. 225 volt

0 .5 P = = 0.084 amp S |Vmin| 1.4 × 4 .225

| I max | = S.| I min | = 1.4 × 0 .084 = 0.117 amp

Hence, |Vmax | = 5 . 92 volt, | I max | = 0 .117 amp, | Vmin | = 4 . 225 volt and |I min | = 0 . 084 amp. Example 25: The characteristic impedance of a transmission line is 728 ∠ – 15.5° ohm and its propagation constant is 0.00875 + j 0.0291 per km. The line is 200 km long and is terminated in an impedance of 400 ∠ 45° ohm. find the current through the load if the sending end voltage is 1 Volt. Solution: The current through the load Z L is given by, VS = VL e γ l or VL = VS e – γ l VL VS e – γ l = ZL ZL

∴

IL =

Here

Z0 = 728 ∠ – 15 .5 ° ; Z L = 400 ∠45 ° , VS = 1 Volt

and

γ = 0.00875 + j0.0291 and l = 200 km

Hence,

VL = 1 × e –(0.00875 + j0.0291) × 200

or

VL = e – (1.75 + j5.82)

or

VL = e –1.75 e – j5.82 = 01737 . e – j5.82 5.82 rad =

∴

5.82 × 180 = 333.63 ° = 2 π –333.63° = 26.37° π

VL = 01737 . e – j26.37° = 0 .1737 (cos 26 .37°– j sin 26 .37° ) (∵ e – jθ = cos θ – j sin θ) = 01737 . (0.8809 – j 0.4439) = 01530 . – j0.0771

or

0.0771 |VL| = (01530 . )2 + (0.0771)2 tan –1 01530 .

or

|VL| = 0 .1713 tan –1 (–0 .5039) = 0.1713 ∠ –26.73 ° Volt ∠ – 26.73 ° . |VL| 01713 = = 0.00043 ∠ (–26.73 °–45° ) | Z L| 400 ∠45 °

Thus

IL =

or

IL = 0.00043 ∠ –71.73 ° Amp

Transmission Lines

615

Example 26: An ideal loss-less transmission line of characteristic impedance 60 ohm is connected to a resistive load. If the standing wave ratio on the line is 4, calculate (i) the value of load impedance ZL and (ii), the reflection coefficient at the load. Solution: The reflection coefficient at the load is related with standing wave ratios as ΓL =

S –1 4 –1 3 = = = 0.6 S +1 4 +1 5

In terms of load Z L and characteristic impedance Z0 . the reflection coefficient ΓL is given by ΓL =

Z L – Z0 Z L + Z0

∴

Z L – Z0 +1 Z – Z0 + Z L + Z0 ΓL + 1 Z L + Z0 Z = L = =– L ΓL – 1 Z L – Z0 Z L – Z0 – Z L – Z0 Z0 –1 Z L + Z0

or

Γ + 1 . 0.6 + 1 60 × 16 Z L = – Z0 L = –60 = 0.6 – 1 0.4 ΓL – 1

or

Z L = 240 ohm

Example 27: The characteristic impedance of a transmission line is terminated by reactance. Characteristic impedance and reactance terminated are 600 Ω and j 150 Ω respectively. Determine the input impedance of a section 0.25 m long at a frequency of 300 MHz. Solution : The input impedance of l meter long section of a transmission line is given by Z cos (β l) + j Z0 sin (β l) Z s = Z0 L Z0 cos (β l) + j Z L sin (β l) Here Z0 = 600 Ω, Z L = j 150 Ω, l = 0 .25m and β=

300 × 106 2π f = 2π = 2 π × = 2 π rad/sec v λ 3 × 108

(∵ Velocity of propagation will be equal to that of radiation in free space, that is, c = 3 × 108 m/sec) ∴

j 150 cos (2 π × 0 .25) + j 600 sin (2 π × 0 .25) Z s = 600 600 cos (2 π × 0 .25) + j . j 150 sin (2 π × 0 .25) 150 × 0 + 600 × 1 150 cos (π /2) + 600 sin (π /2) = 600 j = 600 j ) 600 cos ( π / 2 ) − 150 sin ( π / 2 600 × 0 − 150 × 1

or

600 Z s = 600 j = – j 2400 Ω − 150

E lectromagnetic F ield T heory

616

Example 28: A loss-less line of 200 ohm characteristics impedance connects a 100 KHz generator to a 250 ohm load. The load power is 200 mW. Calculate (a) voltage reflection coefficient, (b) standing wave ratio (SWR) (c) positions and values of voltage maxima and current maxima and (d) positions and values of voltage minima and current minima. Solution: (a) Voltage reflection coefficient, Γ = Here, ∴

Z L = 250 ohm and Z0 = 200 Ω Γ=

(b) Standing wave ratio (SWR), S = or

Z L − Z0 Z L + Z0

S=

250 − 200 50 1 = = = 0.111 250 + 200 450 9 1 + | Γ | 1 + 0 .111 = 1 − | Γ | 1 − 0 .111 1.111 = 1.25 0 .889

(c) and (d) Since the load impedance is resistive and greater than the characteristic impedance Z0 , voltage maximum lies at the load end, and current maximum at a distance of 0.25 λ from it. SubsequentlyVmax and I max points are situated at a distance of 0.5 λ from these points. Voltage maxima and current maxima points are also current minima and voltage minima points respectively. Hence, load power,

P = 200 mW = 200 × 10 –3 watt

∴

P=

or

| Vmax |2 |Vmax |2 or = P or | Vmax | = P. Z L Z0 S ZL

(∵ S = Z L / Z0 )

|Vmax| = (200 × 10 −3 × 250) = 7 . 07 volt | I min | = | Vmin | =

| Vmax | 7 .07 = 0.028 amp = ZL 250

|Vmax| 7 .07 = 5.65 volt = S 1.25

| I max | = S.| I min | = 1.25 × 0 .028 = 0.035 amp Example 29: A loss-less transmission line of characteristic impedance 60 ∠ 0° ohms and half wavelength long is left open circuited at the far end. The r.m.s., value of the open circuited voltage is 15 volt. Calculate the r.m.s., value of voltage and current at a distance of one-eighth wavelength away from the open circuit end. Solution: The voltage at a distance l from an open circuit end is expressed as V = VL cos β l + j I L sin β l The characteristic impedance of a loss-less transmission line of half wavelength long, Z0 = 60 ∠0 °. Since the line is open circuited, therefore I L = 0 Here

VL = 15 volts, l = λ /8

Transmission Lines

∴

βl =

617

2π 2π λ π l= . = λ λ 8 4

Thus, the voltage at a distance λ /8 from the open circuit end is given by V = 15 cos (π /4) + j (0) sin (π /4) = 15 × (1/√ 2) = 10.61 volt (r. m. s. ) The current at a distance λ /8 from the open circuit end is given by I = I L cos β l + j

VL sin β l Z0

or

15 15 1 π π sin = j × = j 0 .1768 I = (0) cos + j 4 4 60 ∠0 ° 60 2

or

I =0.1768 ∠ 90 ° amp. (r. m. s. )

(∵ j = ∠90 ° )

Example 30: The short circuit and open circuit impedance of 10 km long open wire transmission line are ZSC = 2930 ∠ 26 ° and Z0 C = 260 ∠ –32° at a frequency of 1 kHz. Calculate the characteristic impedance and phase velocity. [GBTU, B.Tech IV Sem. 2010, III Sem 2012]

Solution: The short circuit impedance of the line is given by (Z s )sc = Z0 tanh (γ l)

...(1)

Similarly, the open circuit impedance of the line is expressed as, (Z s )oc = Z0 coth (γ l)

...(2)

Multiplication of eqns. (1) and (2) give Z0 = (Z s )sc .(Z s )oc where Z0 is the characteristic impedance impedance ∴

26 ° 32 ° Z0 = (2930 ∠26 ° ) (260 ∠ – 32 ° ) = 2930 × 260 ∠ – 2 2

or

Z0 = 872.81 ∠ – 3 °

Dividing eqn (1) by (2). we get

( Z s )sc ( Z s )oc ∴

= tan2 h (γ l) or tan h (γ l) =

tanh(γ l) =

(Z s )sc (Z s )oc

° ° 2930 ∠26 ° 2930 26 32 = ∠ – – 2 260 ∠ – 32 ° 260 2

tanh(γ l) = 3 .357∠29 ° tanh(γ l) = 3 . 357 cos 29 °+ j 3 . 357 sin 29 ° = 3 .357 × 0 .8746 + j3 .357 × 0.4848 or

tanh(γ l) = 2.9360 + j16275 .

In exponential form tanh (γ l) is expressed as,

E lectromagnetic F ield T heory

618

tanh (γ l) =

eγ l – e– γ l eγ l + e– γ l

=

e2 γ l – 1 e2 γ l + 1

Applying compodendo and dividendo, we get

or

e2 γ l =

. 1 + tanh(γ l) 1 + 2.9360 + j 16275 = 1 – tanh (γ l) 1 – 2.9360 – j162 . 75

e2 γ l =

. 3.9360 + j 16275 –19360 . – j 16275 .

e2 (α + jβ) l

. 16275 ∠ tan –1 3.9360 + j 16275 . (3.9360)2 + (16275 . )2 3.9360 =– = . 19360 . . + j16275 (19360 . )2 + (16275 . )2 ∠ tan –1 16275 19360 . 4 .2592 ∠ tan –1 (0.4135) = 1684 . ∠(22.5 °– 40.06 ° ) 2 .5292 ∠ tan –1(0.8406)

or

e2(α + jβ)l =

or

e2 (α + jβ)l = 1.684 ∠ – 17. 56 ° e2 α l ∠2βl = 1.684 ∠ – 17. 56 °

∴

e2 α l = 1.684 and 2 β l = 17. 56 ° 17. 56 ° 17. 56 × π 17.56 × 3.14 = rad= 4 2l 2 × 104 × 180 2 × 10 × 180

Thus,

β=

or

β = 1.53 × 10 –5 rad /m

We know that

vp =

3 ω 2 πf 2 × 3 .14 × 1 × 10 = = = 4.104 × 10 8 m/s 5 – β β 1.53 × 10

Example 31: A RF milliameter (100 mA range, 1 Ω resistance) is used as a voltmeter by feeding the RF voltage to be measured through a loss-less λ /4 line of 100 ohm characteristics impedance. If the meter reads full scale, find the voltage being measured. Solution: Since a RF milliameter of 1 Ω resistance is connected to the loss-less λ /4 line of 100 ohm characteristic impedance, the load to the λ /4 line is the RF milliameter having R = 1 Ω ∴ Input impedance Z s of the voltage source is expressed as, 2

Zs =

Z0 R

=

(100)2 = 104 ohm 1

We know that

| Vs | = VL

and

|VL| = IR = 1 × 100 × 10 −3 = 0 .1 volt

∴

| Vs | =

Zs Z0

104 102

or | Vs | = | VL |.

× 0 .1 = 1 volt

Zs Z0

Transmission Lines

I

619

mpedance Matching

Transmission lines are used to transmit power from a source to a load. Maximum power is transmitted by the line to the load with minimum losses when no reflected wave is present or when the load impedance is equal to the characteristic impedance of the line, that is, when the load is perfectly matched. Usually this condition is not achieved. Under exceptional cases, when the load impedance is a resistance, the load impedance is exactly equal to the characteristic impedance of the line. Therefore, for the transfer of power with maximum efficiency from generator to the load it is essential to provide means for matching the actual load impedance to the characteristic impedance of the line. Again it is often desired that the line impedance be independent of the distance to the load. There are various techniques of impedance matching. Two such techniques which are widely used are as follows : 1.

Impedance matching with quarter wave transformer.

2.

Impedance matching by stub lines.

Impedance Matching with Quarter Wave Transformer Sections of transmission lines that are exactly a quarter wavelength long has unique impedance transforming properties and are frequently used in radio communication system. Since a quarter wave line behaves as an impedance transformer, it can be used as an impedance matching device. consider a short section of a loss-less transmission line of length λ /4. Input impedance of such a line may be obtained with the help of an expression for input impedance of a loss-less line of length l as,

or

Z cos (β l) + j Z0 sin (β l) Z s = Z0 L Z0 cos (β l) + j Z L sin (β l)

...(1)

Z + j Z0 tan β l Z s = Z0 L Z0 + j Z L tan β l

...(2)

where Z0 is the characteristic impedance of the line with phase constant β and Z L is the load impedance connected at the load end. 2π λ

For a quarter wave line,

λ = 4 l and β =

Therefore, substituting

l = λ /4 and β = 2 π / λ in eqn. (1), we get 2 π λ 2 π λ . + j Z0 sin . Z L cos λ 4 λ 4 Z s = Z0 2 π λ 2 π λ + j Z sin . Z cos . L 0 λ 4 λ 4

or

Z cos (π /2) + j Z0 sin (π /2) 0 + j Z0 Z s = Z0 L = Z0 Z0 cos (π /2) + j Z L sin (π /2) 0 + j Z L

or

Zs =

Z20 or Z0 = Zs . Z L ZL

...(3)

620

Therefore, by inserting a line of length λ /4 (quarter wave

E lectromagnetic F ield T heory Main line λ/4 transformer

line) of characteristic impedance Z0 = √ (Z s . Z L ) prior to a mismatched load of impedance Z L in a main transmission line of input impedance Z s , then a mismatched load Z L can be matched properly as shown

ZS

in Fig. 19. It is also clear from eqn. (3) [Z S = Z02 / Z L ) that a quarter wave transmission line transforms a load impedance Z L connected to the main line, which is

ZL

ZS=Z0 λ/4

Load matching by λ/4 transformer Fig. 19

smaller than Z0 , into a value Z s that is larger than Z0 and vice versa. Therefore, a quarter wave line transforms an impedance of small magnitude into one of larger magnitude and vice-versa. Thus, a loss-less quarter wave line acts as an impedance changing device, that is whatever be the terminating impedance, an inverse impedance will always appear at the input terminal. On account of step up or step down impedance transforming property of loss-less quarter wave transmission line, it is also called impedance inverting transformer. In a transmission line properly matched with a quarter wave impedance transformer (or a quarter wave line) standing waves will exist between transformer and the load but no standing wave on the main line will exist. Eqn. (3) also indicates that the input impedance Z s is inversely proportional to the load impedance Z L . That is, when the load impedance is resistive, the effect of insertion of the transformer is thus to transform this resistive load into another resistive load Z s , which is inversely proportional to Z L . Again when Z L is a capacitive reactance, then the impedance transforming action of the line causes Z s to be an inductive reactance having a magnitude inversely proportional to the capacitive reactance of Z L . Similarly, if Z L is inductive reactance, then the impedance transforming action causes Z s to be capacitive reactance. The main drawback of quarter transformer is that, it is frequency sensitive device. It means that impedance matching will be perfect only at a single frequency. At other frequencies there will be a mismatching, resulting in standing waves.

Impedance Matching by Stubs At very high and microwave frequencies, when it is not possible to control the terminating impedance of the transmission line, impedance matching is achieved by connecting a stub line in shunt with the main transmission line. The distance of the point of attachment of the stub to the main line from the load end and the length of the stub are so chosen that the reflected wave produced by the stub line at the junction of main line and stub becomes equal and opposite in polarity to the reflected wave on transmission line at the same junction so that both will cancel out and there will be no reflected wave on the transmission line towards generator. Thus, although a reflected wave is present in the length between the point of contact and load end due to reflection from load impedance Z L , but there will be no reflected wave on generator side of the stub line.

Transmission Lines

621

In fact a small section of loss free stub line introduces a reactance at the point of attachment, the value of which depends upon the distance of stub from the load end and upon its length. This reactance created by stub neutralizes the reactance of main transmission line at the point of attachment and provides impedance matching. Thus, the principle on which the stub matching technique is based consists of finding a point nearest to the load end at which the real part of the line admittance (reciprocal of the impedance) is equal to the characteristic admittance (reciprocal of characteristic impedance Z0 ) of the main line. The imaginary part of the line admittance is then cancelled by connecting in parallel with main line a short circuit stub of appropriate length which produces an input susceptance which is equal in magnitude but opposite in sign to the imaginary part of the line admittance to the right of the stub. Thus, the total line admittance at the point of attachment will be equal to the real part of the line admittance. Hence, the line impedance seen from the left of the stub into the junction of the line and stub is therefore, equal to the characteristic impedance of the line. In this way the impedance matching is achieved. To understand the principle of stub matching in the simplest way, let us consider a ultra high frequency line terminated in a resistance Z L = R L

Z0

which is different from the characteristic

Y0

Y0 + jB –j B

resistance Z0 = R0 of the line. Let a short circuited loss free stub line of length lt be connected at a point distant l from the load end. At a point distant l = λ /4 from the load end the

ZL

l Short circuited stub

input impedance is purely resistive and is

lt Impedance matching by single stub

expressed as Rs =

R02 RL

Fig. 20

If R L < R0 then R s > R0 and if R L > R0 then R s < R0 . Therefore, some where between l = 0 and l = λ /4 the resistive component of input impedance will be R0 . However, there will also be a reactive component of input impedance at such point and if this is matched by an equal and opposite reactance then only resistive component will remain and the line will be perfectly matched. Now, let us consider a line terminated in a load impedance Z L which is different from its characteristic impedance Z0 as shown in fig. 20. At a point distant l from the load end where the stub is connected, the input impedance of such a transmission line is given as

or

Z cosh γ l + Z0 sin h γ l Z s = Z0 L Z0 cosh γ l + Z L sinh γ l

...(1)

Z + Z0 tanh (γ l) Z s = Z0 L Z0 + Z L tanh (γ l)

...(2)

E lectromagnetic F ield T heory

622

Since the stubs are generally connected in parallel with main line it is easier to deal with the problem in terms of admittances rather than in terms of impedance. The input admittance of the transmission line may be expressed as

or

where Y0 =

Ys =

1 1 Z0 + Z L tanh (γ l) = Z s Z0 Z L + Z0 tanh (γ l)

Ys =

1 (1/Y0 ) + (1/Y L ) tanh (γ l) Z0 (1/Y L ) + (1/Y0 ) tanh (γ l)

1 1 1 and Y s = are the characteristic, load and input admittances of the line ,Y = Z0 R Z L Zs

respectively. Therefore, in terms of admittances eqn. (2) may be transformed as Y + Y0 tanh (γ l) Y s = Y0 L Y0 + Y L tanh (γ l)

...(3)

For ultra high frequency line, attenuation constant α is zero, therefore, propagation constant would be j β. Hence, putting γ = j β in eqn. (3) and on simplification, we get Y + j Y0 tan (β l) Y s = Y0 L Y0 + j Y L tan (β l)

(∵ tanh x = j tan x) ...(4)

To express eqn. (4) in terms of conductance GS and susceptance Bs and to separate real and imaginary part multiplying and dividing eqn. (4) by the complex conjugate of Y0 + jY L tan β l, that is, by Y0 − j Y L tan β l, we get Y s = Gs + j Bs = Y0

= Y0

or

or

(Y L + j Y0 tan β l) (Y0 − j Y L tan β l ) (Y0 + j Y L tan β l) (Y0 − j Y L tan β l )

[Y LY0 − j Y L2 tan β l + j Y02 tan β l + Y L Y0 tan2 β l ] Y02 + Y L2 tan2 β l

Y Y (1 + tan2 β l) + j tan β l (Y 2 − Y L2 ) 0 Y s = Y0 L 0 Y02 + Y L2 tan2 β l

Ys = Y0

Y Y02 L (1 + tan2 β l) + j tan β l (1 − Y L2 /Y02 ) Y 0 Y02 [1 + (Y L2 /Y02 ) tan2 β l]

...(5)

Expressing eqn. (5) in terms of normalised admittances, we get or

Ys =

Y L (1 + tan2 β l) (1 + Y L2 tan2 β l)

+

j tan β l (1 − Y L2 ) (1 + Y L2 tan2 β l )

...(6)

where Y s (= Y s /Y0 ) and Y L (= Y L /Y0 ) are the normalised input admittance and normalised load admittance respectively. If Gs and Bs are the normalised conductance and susceptance respectively, then

Transmission Lines

623

1 + tan2 β l tan β l (1 − Y L2 ) Y s = Gs + j Bs = Y L + j 2 2 1 + Y L2 tan2 β l 1 + Y L tan β l

...(7)

Equating real and imaginary parts, we get ∴

Gs =

and

Bs =

Y L [1 + tan2 β l]

...(8)

1 + Y L2 tan2 β l tan β l (1 − Y L2 )

...(9)

1 + Y L2 tan2 β l

The reflected wave will be absent if the input admittance is equal to the characteristic admittance. That is, Y s = Y0 or Y s = 1 Therefore, for no reflection Y s = Gs + j Bs = 1 or

Y s = Gs =1 and Bs = 0

From eqn. (8) it is clear that Gs is the function of the line length ‘ l’ and thus by proper selection of this length impedance matching is achieved. If ls is the distance of the point of attachment of the stub from the load end at which there is no reflection, then from eqn. (8), we have Gs = 1 = or

Y L [1 + tan2 β ls ]

(∵ l = ls )

1 + Y L2 tan2 β ls

...(10)

Y L [1 + tan2 β ls ] = 1 + Y L2 tan2 β ls Y L + Y L tan2 β ls = 1 + Y L2 tan2 β ls or tan2 β ls [Y L − Y L2 ] = [1 − Y L ] tan2 β ls . Y L [1 − Y L ] = [1 − Y L ] or YL ∴ tan β ls = Y0

Since

YL =

or

β ls = tan −1

Y0 YL

Y0 YL

or ls =

Y L tan2 β ls = 1 or tan β ls = or tan β ls =

1 tan−1 β

1 YL

Y0 YL

Y0 YL

...(11)

...(12)

...(13)

The distance of point of attachment of the stub from the load end in term of admittance may be expressed as ls =

ZL 1 tan−1 Z0 β

...(14)

If the characteristic and load impedances are resistive, that is , Z L = R L and Z0 = R0 then ls =

λ tan−1 2π

RL R0

...(15)

E lectromagnetic F ield T heory

624

Eqn. (15) gives the position of the point where the resistive parts of line impedance and the characteristic impedance are equal. To find the length of the stub lt we use eqn. (9) which represents the susceptance Bs of the line at the junction of the main line and stub as Bs =

tan β ls (1 − Y L2 )

(∵ l = ls )

1 + Y L2 tan2 β ls

...(16)

Substituting the value of tan β ls from eqn. (11) in eqn. (16), we get Bs =

or

Bs =

( 1/Y L ) (1 − Y L2 ) 1 + Y L2 ( 1 / Y L )2

or =

1 YL

1 − Y L2 1 + Y L

1 (1 − Y L ) YL

...(17)

Substituting Y L = Y L /Y0 and Bs = Bs /Y0 , in eqn. (17) we get Y0 Y Bs = 1 − L Y0 YL Y0

∴

Bs =

Y0 (Y0 −YL ) YL

...(18)

This is the susceptance added by the transmission line of length ls at the point of attachment of the stub with main line. If this susceptance is neutralize, then the line will be properly matched. This is achieved by inserting a short circuited stub of appropriate length at this junction such that the susceptance offered by the stub is equal and opposite to the susceptance offered by the line so that the total susceptance at the point of attachment becomes zero. If lt be the length of the stub, then the impedance offered by the stub at the point of attachment is given by Z t = j Z0 tan (β lt ) The equivalent admittance is written as Y t =

...(19) 1 1 = Z t j Z0 tan ( β lt )

or

Y t = − j Y0 cot ( β lt )

But

Y t = Gt + j Bt = − jY0 cot ( β lt )

Separating real and imaginary parts, we get ∴

Gt = 0 and Bt = − Y0 cot ( β lt )

...(20)

According to the condition of impedance matching, the sum of the susceptances offered by transmission line (Bs ) and by the stub (Bt ) at the point of attachment must be zero, that is, Bs + Bt = 0

...(21)

Substituting the values of Bs from eqn. (18) and Bt from eqn. (20) in eqn. (21), we obtain. Y0 (Y − Y L ) − Y0 cot (β lt ) = 0 YL 0

...(22)

Transmission Lines

Y0 cot ( β lt ) =

or

625

Y0Y L Y0 (Y − Y L ) or tan ( β lt ) = Y0 − Y L YL 0

Y0Y L Y0Y L 1 or lt = tan −1 β lt = tan −1 Y − Y Y Y − β 0 0 L L

or

lt =

or

Y0 YL λ tan−1 2π Y0 − YL

...(23)

...(24)

In terms of impedance eqn. (24) may be written as lt =

Z0 Z L λ tan−1 2π Z L − Z0

...(25)

Eqn. (25) represents the length of the loss-less short circuited stub required for proper impedance matching. If Z L = R L and Z0 = R0 then eqn. (25) reduces to lt =

R L R0 λ tan −1 2π R L − R0

...(26)

The position and the length of stub may also be conveniently computed by a circular chart commonly known as Smith Chart.

Disadvantage of Using Single Stub for Impedance Matching There are two main drawbacks in using single stub as a line matching device, which are as follows: 1.

The value of terminating impedance that can be transferred by this technique is limited. The position and the length of the stub change with the change of terminating impedance. Therefore, it is essential to adjust the location and length of the stub each time for using different terminating impedances. The adjustment of location and length of the stub is easier in open wire transmission line but quite tedious for coaxial transmission line.

2.

The single stub technique is applicable only for a fixed frequency. For variable frequencies the position of stub has to be varied. It is a frequency sensitive or narrow band device.

T he Position (l s ) and Length (l t ) of the Stub in Terms of Reflection Coefficient

The input impedance Z s of a transmission line in terms of characteristic impedance Z0 and load impedance Z L may be expressed as, Z cosh (γ l) + Z0 sinh (γ l) Z s = Z0 L Z0 cosh (γ l) + Z L sinh (γ l) Substituting,

cosh (γ l) =

eγ l + e− γ l 2

and

sinh (γ l) =

...(1)

eγ l − e− γ l 2

in eqn. (1), we obtain

E lectromagnetic F ield T heory

626

ZL Z s = Z0 Z 0

eγ l + e−γ l eγ l + Z0 2 eγ l eγ l + e−γ l + ZL 2

− e− γ l 2 − γ l −e 2

e γ l (Z + Z ) + e − γ l (Z − Z ) 0 L 0 = Z0 γ l L −γ l + + − e Z Z e Z Z ( ) ( L L ) 0 0 or

But

∴

Zs =

Z0 e γ l(Z L + Z0 ) 1 + e − 2 γ l × (Z L − Z0 ) /(Z L + Z0 ) e γ l (Z L + Z0 ) 1 − e − 2 γ l × (Z L − Z0 ) /(Z L + Z0 )

Z L − Z0 = Γ, the reflection coefficient Z L + Z0 1 + Γ e − 2 γ l Zs = Z0 −2 γ l 1 − Γ e

...(2)

For a loss-less high frequency transmission line, α = 0, that is, γ = j β.The reflection coefficient is a complex quantity, therefore, we can write, Γ =| Γ | e j φ

...(3)

where | Γ | is the magnitude of Γ and φ is the phase angle of reflection coefficient. Thus,

Γ e −2 γ l = | Γ | e j φ . e − 2 jβ l = | Γ | e j(φ − 2 β l)

Substituting the value of Γ e

−2 γ l

=| Γ | e

j (φ − 2 β l )

...(4)

in eqn. (2), we get

1 + | Γ | e j (φ − 2 β l) Z s = Z0 1 –| Γ | e j(φ − 2 β l)

...(5)

Eqn. (5) may be expressed in terms of admittance as, j (φ − 2 β l) 1 1 1 − | Γ | e or = Z s Z0 1 + | Γ | e j(φ − 2 β l) 1 1 In terms of admittance, = Y0 . Therefore, = Y s and Z0 Zs 1 − |Γ | e j(φ − 2 β l) Ys = Y0 1 + | Γ | e j ( φ − 2 β l )

...(6)

In terms of conductance and susceptance, input admittance may be expressed as Y s = Gs + j Bs If the characteristic impedance is purely resistive, that is, Z0 = R0 then 1 G0 = = Y0 Z0 Substituting Y0 = G0 in eqn. (6), we get 1 − | Γ | e j(φ − 2 β l) Y s = G0 1 + | Γ | e j(φ − 2 β l)

...(7)

...(8)

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627

Expanding e j(φ − 2 β l) in terms of cos and sin functions, we get 1 − | Γ | cos (φ − 2 β l) − j | Γ |sin(φ − 2 β l) Y s = G0 1 + | Γ | cos (φ − 2 β l) + j | Γ |sin (φ − 2 β l)

or

...(9)

To separate real and imaginary parts, multiplying and dividing eqn. (9) by 1 + | Γ | cos (φ − 2 β l) − j | Γ |sin (φ − 2 β l) as 1 − | Γ | cos (φ − 2 β l) − j | Γ |sin (φ − 2 β l) Y s = G0 1 + | Γ | cos (φ − 2 β l) + j | Γ |sin (φ − 2 β l) 1 + | Γ | cos (φ − 2 β l) − j | Γ |sin (φ − 2 β l) × 1 + | Γ | cos (φ − 2 β l) − j | Γ |sin (φ − 2 β l) [1 − j | Γ |sin (φ − 2 β l)] 2 − [| Γ | cos (φ − 2 β l)] 2 = G0 [1 + | Γ | cos (φ − 2 β l)] 2 + [| Γ |sin (φ − 2 β l)] 2 2 2 2 2 1 − | Γ | sin (φ − 2 β l) − 2 j | Γ |sin (φ − 2 β l) − | Γ | cos (φ − 2 β l) = G0 1 + | Γ |2 cos2 (φ − 2 β l) + 2 | Γ | cos (φ − 2 β l) + | Γ |2 sin2 (φ − 2 β l) 2 1 − | Γ | − 2 j | Γ |sin (φ − 2 β l) = G0 2 1 + | Γ | + 2 | Γ | cos (φ − 2 β l)

or

Ys =

G0 (1 − | Γ |2 ) 1 + | Γ |2 + 2 | Γ |cos (φ − 2 β l)

But

Y s = Gs + j Bs

Therefore,

Gs =

and

Bs =

+ j

− 2 G0 | Γ |sin (φ − 2 β l) 1 + | Γ |2 + 2 | Γ |cos (φ − 2 β l)

G0 (1 − | Γ |2 ) 1 + | Γ |2 + 2 | Γ |cos (φ − 2 β l) − 2 G0 | Γ |sin (φ − 2 β l) 1 + | Γ |2 + 2 | Γ |cos (φ − 2 β l)

...(10)

...(11) ...(12)

According to the first part of the condition of impedance matching, the resistive component of input impedance at a point of attachment of a stub distant ls from the load end must be equal to the resistive component of characteristic impedance of the line. That is, Gs =

Gs 1 1 so that =1 = G0 Z0 R0

1 ∵ G0 = ...(13) R0

Therefore, from eqn. (11), we obtain 1 − | Γ |2 Gs =1 = G0 1 + | Γ |2 + 2 | Γ | cos (φ − 2 β ls ) or or or But

...(14)

1 + | Γ |2 + 2 | Γ | cos (φ − 2 β ls ) = 1 − | Γ |2 cos(φ − 2 β ls ) = − | Γ | φ − 2 β ls = cos −1 (− | Γ |) cos −1 (− | ρ | = − π + cos −1)| Γ |

...(15)

E lectromagnetic F ield T heory

628

∵ cos θ = − 1 when θ = π and cos θ = 1, when θ = 0 ∴ π = cos −1 (− 1) and 0 = cos−1 (+ 1) −1 −1 so cos (− 1) = cos (+ 1) + π = π + 0 = π or − π −1 −1 Similarly , cos (− | Γ |) = − π + cos | Γ | ∴

φ − 2 β ls = cos −1 | Γ | − π

...(16)

2 β ls = φ + π − cos −1 | Γ |

or

ls =

or But

φ + π − cos 2β

−1

...(17)

|Γ |

β=

2π , therefore, λ

ls =

λ [φ + π − cos −1 | Γ |] 4π

...(18)

...(19)

Eqn. (19) represents the distance of the point of attachment of stub on the line from the load end or position of stub on the line. The find the length of the stub lt we use eqn. (12) which represents the susceptance Bs of the line at the point of attachment of the stub with the main transmission line as, Bs =

− 2 G0 | Γ |sin (φ − 2 β ls )

(∵ l = ls ) ...(20)

1 + | Γ |2 + 2 |Γ | cos (φ − 2 β ls )

Substituting the value of ls from eqn. (18) or the value of φ − 2 β ls from eqn. (16) in eqn. (20), we get, Bs =

...(21)

1 + | Γ |2 + 2| Γ | cos (cos −1 | Γ | − π)

Bs = G0

or

− 2 G0 | Γ |sin (cos −1 | Γ | − π )

+ 2 | Γ |sin (π − cos −1 | Γ |) 1 + | Γ |2 + 2 | Γ | cos (π − cos −1 | Γ |) [∵ sin (θ − π ) = − sin (π − θ) and cos (θ − π ) = cos (π − θ)]

Bs = G0

∴

Now, putting ∴ and

2 | Γ |sin (cos −1 | Γ |)

cos −1 | Γ | = θ, then cos θ = | Γ | sin (cos −1 | Γ |) = sin θ = √ (1 − cos 2 θ) = cos (cos

−1

...(22)

1 + | Γ |2 − 2 | Γ | cos (cos −1 | Γ |)

(∵ sin (π − θ) = sin θ and cos (π − θ) = − cos θ) (1 − | Γ |2 )

...(23)

| Γ |) = cos θ = | Γ |

...(24)

Substituting the values of sin (cos −1| Γ |) and cos (cos −1| Γ |) from eqns. (23) and (24) in eqn. (22), we obtain Bs = G0

+ 2 | Γ | (1 − | Γ |2 ) 1 + | Γ |2 − 2 | Γ |2

or Bs =

+ 2 G0 | Γ | (1 − | Γ |2 ) 1 − | Γ |2

...(25)

Transmission Lines

or

629

Bs =

+ 2 G0 | Γ |

...(26)

1 − | Γ |2

Now according to the second part of the condition of impedance matching, the reactive part of the input impedance, at the point of attachment of the stub, offered by the main line must be equal and opposite to the reactive impedance of the stub line so that the total input impedance at the point of attachment is zero. Hence the stub line must offers only reactive impedance. Such line whose input impedance is purely reactive are short circuited or open circuited lines. Let us connect a short circuited line. Assuming lt to be the length of the stub, the impedance offered by the stub line at the point of attachment is given by Z t = j R0 tan β lt Therefore, the admittance Y t is Yt =

1 1 = = − j G0 cot β lt Z t j R0 tan β lt

But,

Y t = Gt + j Bt = − j G0 cot (β lt )

∴

Bt = − G0 cot (β lt )

...(27)

...(28)

At the point of attachment, the sum of the line susceptance Bs and the stub susceptance Bt must be zero that is, Bs + Bt = 0 or Bs = − Bt

...(29)

Substituting the values of Bs from eqn. (26) in eqn. (29), we get Bt = −

2 G0 | Γ |

...(30)

1 − | Γ |2

Equating eqns. (28) and (30), we get 2 G0 | Γ | – = − G0 cot (β lt ) 1 − | Γ |2 or

or

cot ( β lt ) =

2|Γ | 1 − | Γ |2

...(31)

or

tan ( β lt ) =

1 − | Γ |2 2|Γ |

1 − | Γ |2 ( β lt ) = tan −1 2|Γ |

or

lt =

1 − | Γ |2 1 tan −1 2|Γ | β

or

lt =

λ tan−1 2π

1 − | Γ |2 2 |Γ |

2π ∵ β = ...(32) λ

Eqn. (32) gives the length of the stub in terms of reflection coefficient. To find the position and length of the stub in terms of standing wave ratio, we put |Γ | =

S −1 (S being the standing wave ratio) in eqn. (32) we get, S +1

E lectromagnetic F ield T heory

630

lt =

λ tan −1 2π

[1 − (S − 1)2 /(S + 1)2 ]1 /2 2 (S − 1) / (S + 1)

or

lt =

λ tan −1 2π

[(S + 1)2 − (S − 1)2 ]1 /2 2 (S − 1)

or

lt =

4S λ tan −1 2π 2 (S − 1)

or

lt =

S λ tan−1 2π S − 1

...(33)

Eqn. (33) gives the length of stub in terms of standing wave ratio. The position of stub may also be expressed in terms of standing wave ratio by replacing | Γ | in terms of SWR, S in eqn. (19) as ls =

λ [ φ + π − cos −1 [(S − 1) / (S + 1)] 4π

...(34)

Impedance matching by single stub is suitable for open parallel wire lines and is unsuitable for coaxial lines where the insertion of stub at the exact required point is quite difficult.

M erits of a Short Circuited Stub Over an Open Circuited Stub Short circuited stubs are preferred over open circuited stubs due to following reasons: 1.

A short circuited loss-less stub line can be constructed easily as compared to open circuited stub line.

2.

At the open end of an open circuited line, the insulation cannot be maintained, that is, it cannot be confirmed that stub is actually open circuited.

3.

Normally open circuited stub line radiates some energy at high frequencies while short circuited line is sealed at the load end and therefore stops all field propagation.

I

mpedance Matching by Double Stubs

To overcome the difficulty of insertion of a stubs at an exact required position on a main transmission line and to achieve impedance matching at frequencies in the Ultra High Frequency (UHF) range double stubs technique of impedance matching is employed. At ultra high frequencies parallel wire line with single stub is not applicable. Further, double stubs arrangement of impedance matching is suitable particularly for coaxial transmission lines, as it avoids the mechanical problems involved in moving the position of a shunting stub along a coaxial line. In a double stubs impedance matching technique, two stubs are connected at arbitrary positions on the main transmission line and lengths of stubs are adjusted to provide perfect matching. The spacing between two stubs is kept in between λ /4 and 3 λ /8. Half wavelength (λ /2 ) spacing between the stubs is avoided as it makes the two stubs parallel which provides only one effective adjustment. Two

Transmission Lines

631

stubs cannot be kept very close to each other as it also creates the same problem. The 3 λ /8 spacing between the two stubs is an optimum spacing. In the typical case where spacing is made one-eighth wavelength (λ /8), the matching can be obtained only if the conductance component of the impedance at the stub nearest the load and looking towards the load is less than 2 /Z0 . In Fig. 21 double stubs impedance matching system in cases of open wire and coaxial lines are shown. The operation of double stubs arrangement is based on the fact that it introduces a reflected wave which is adjusted to produce a reflection equal in magnitude and opposite in phase to the reflected wave produced by the load impedance. The phase of reflection so introduced by the stubs is controlled by moving the stubs along the line while keeping the spacing between them constant.

ls1

ls 2 2

1

ZL 2

lt2

1

lt 2

lt 1

Zs To generator

l t1

ls 2

Double stubs arrangements

Open wire line (a)

ZL

To load

ls 1

Coaxial line (b)

Fig. 21

To understand the process of impedance matching by double stub consider an arrangement depicted in fig. 21 (a) in which the double stub matching in open wire lines is shown. Suppose the first stub of length lt1 be located on the line at a point (1, 1) distant ls1 from the load end. The normalized input admittance of the loss-less transmission line is represented as, Y + j tan (β l) Ys = L 1 + j Y L tan (β l)

(∵ For loss less line, γ = j β) ...(1)

For the first stub at point (1, 1), Y s = Y s1 and l = ls1 , therefore, Y L + j tan ( β ls ) 1 Y s1 = 1 + j Y L tan ( β ls1 ) To separate real and imaginary part multiplying and dividing eqn. (2) by 1 − j Y L tan (β ls1 ) , we get Ys = 1

[ Y L + j tan ( β ls1 )] [1 − j Y L tan ( β ls1 )] [1 + j Y L tan ( β ls1 )] [1 − j Y L tan ( β ls1 ]

...(2)

E lectromagnetic F ield T heory

632

=

Y L + j tan ( β ls1 ) − j Y L2 tan ( β ls1 ) + Y L tan2 ( β ls1 )

=

Y L (1 + tan2 ( βls1 ) + j tan ( β ls1 ) [1 − Y L2 ]

1 + Y L2 tan2 ( β ls1 ) 1 + Y L2 tan2 ( β ls1 ) Y L sec2 ( β ls1 )

or

Y s1 =

But

Y s1 = Gs1 + j Bs1

1 + Y L2 tan2 ( βls1 )

+ j

( 1 − Y L2 ) tan (β ls1 ) 1 + Y L2 tan2 ( β ls1 )

...(3) ...(4)

where Gs1 is the normalized conductance and Bs1 the normalized susceptance offered by the first stub. ∴

Gs1 =

and

Bs1 =

Y L sec2 (βls1 ) 1 + Y L2 tan2 (β ls1 ) (1 − Y L2 ) tan (β ls1 ) 1 + Y L2 tan2 (β ls1 )

...(5)

...(6)

The normalized input admittance is Ys =

Ys Y0

...(7)

When the stub of susceptance B2 is added to the point (1, 1), the new admittance will be Y s2 = Gs1 (= Gs2 ) + j Bs2

...(8)

The addition of stub at (1, 1) changes only the susceptance part but the conductance part remains unchanged, that is, Gs1 = Gs2 .The value of the admittance Y s2 should be such that the admittance Y at the point (2, 2) is equal to 1 + j B2 . The length of the stub at (2, 2) should be adjusted in such a manner that the new value of Y s becomes unity to effect the matching. For the smooth function of the line the input 2

admittance Y s of the line looking towards the load end at (2, 2) must be unity. That is, Ys =

Ys = 1 or Y s = Gs G0

Per unit admittance of the location of point (2, 2) is Ys =

Ys = 1 + j B2 G0

Usually two stubs are kept at fixed points on the line with optimum separation of about 3 λ /8. The first stub at a point (1, 1) nearest to the load end is adjusted in such a manner that the conductance or real part of admittance at points (2, 2) on the line becomes equal to the characteristic conductance of the main transmission line, in the absence of second stub at (2, 2). The second stub at (2, 2) is then adjusted in such a way that the positive susceptance at the point (2, 2) must be neutralized by a negative susceptance provided by the stub at (2, 2). In this way the impedance matching is achieved. With the help of double stub matching technique Voltage Standing Wave Ratio (VSWR) can be reduce-below 1.2. The arbitrary values of positions of stubs that is, ls1 and ls2 cannot be calculated analytically. The values of ls1 and ls2 can be determined with the help of Smith chart. The length of the stubs as obtained by Smith chart are in the range of lt1 = 0 .348 λ and

lt2 = 0 .110 λ

Transmission Lines

633

A pplications of Smith Chart Smith chart or transmission line chart represents various properties of transmission lines in a graphical manner. The most important and useful Smith charts are those which can be used to evaluate various impedance relations that exist along a loss-less transmission line for different load conditions. Smith chart shows very simply and directly the Standing Wave Radio (SWR) corresponding to a given load impedance and also the line impedance at any desired point, giving the standing wave ratio and the impedance at any other point on the line. The construction of Smith chart is based on two sets of perpendicular circles. One set indicates the ratio R / Z0 , where R is the resistive component of the line impedance, and Z0 the characteristic impedance which is also resistive for loss-less line. The second set of circles gives the ratio j X /Z0 , where X is the reactive component of the line impedance. Before discussing any more about Smith chart we will first determine the equations of two circles. These equations are based on the input impedance equation for a loss-less transmission line which can be expressed in terms of reflection coefficient Γ as, 1 + Γ e −2 jβ l Z s = Z0 1 − Γe −2 jβ l Reflection coefficient being complex may be expressed as

(∵ For a loss-less line γ = j β)...(1)

Γ =| Γ | e j φ

...(2)

where | Γ | is the magnitude and φ the phase angle of the reflection coefficient. Substituting the value of Γ from eqn. (2) in eqn. (1), we get 1 + | Γ | e j φ . e −2 jβ l Z s = Z0 1 − | Γ | e j φ . e − 2 jβ l or

1 + | Γ | e j(φ − 2 β l) Z s = Z0 1 − | Γ | e j(φ − 2 β l)

or

Z s 1 + | Γ | ∠ (φ − 2 β l) = Z0 1 − | Γ | ∠ (φ − 2 β l)

...(3)

where Z s / Z0 is normalized input impedance. Therefore, the normalized input impedance Z s may also be expressed as 1 + | Γ | ∠ (φ − 2 β l) ...(4) Zs = 1 − | Γ | ∠ (φ − 2 β l) In terms of normalized resistive component R s and reactive component X s the normalized input impedance may be expressed as Z s = Rs + j Xs ∴

Rs + j Xs =

1 + | Γ | ∠ (φ − 2 β l) 1 − | Γ | ∠ (φ − 2 β l)

In term of standing wave ratio S, the reflection coefficient may be expressed as S −1 |Γ | = S +1 Expressing eqn. (6) in term of standing wave ratio S, we get

...(5) ...(6)

...(7)

E lectromagnetic F ield T heory

634

Rs + J Xs =

1 + [(S − 1) /(S + 1)] ∠ (φ − 2 β l) 1 − [S − 1] / (S + 1)] ∠ (φ − 2 β l)

S − 1 S − 1 ∠ (φ − 2 β l) = 1 + ∠ (φ − 2 β l) (R s + j X s ) 1 − S + 1 S + 1 S − 1 S − 1 ∠ (φ − 2 β l) − ∠ (φ − 2 β l) = 1 R s + j X s − (R s + j X s ) S + 1 S + 1

or or

S − 1 ∠ (φ − 2 β l) [R s + j X s + 1] = 1 − R s − j X s − S + 1

or

...(8)

S − 1 j(φ − 2 β l) e [R s + j X s + 1] = Rs + j Xs − 1 S + 1

or

Expanding e j(φ − 2 β l) in terms of cos and sin, we get S − 1 {cos (φ − 2 β l) + j sin (φ − 2 β l)} = R s + j X s − 1 or [R s + j X s + 1] S + 1 Equating real and imaginary parts of both sides, we get S − 1 S − 1 cos (φ − 2 β l) − X s sin (φ − 2 β l) = R s − 1 [R s + 1] S + 1 S + 1 S − 1 S − 1 sin (φ − 2 β l) + X s cos (φ − 2 β l) = X s (R s + 1) S + 1 S + 1

and

...(9)

...(10) ...(11)

Squaring and adding eqns. (10) and (11), and after clearing of fractions, we get, S2 + 1 + X s2 = − 1 R s2 − R s S S2 + 1 + X s2 = − 1 R s2 − 2 R s 2S

or

...(12)

2

S2 + 1 on both sides of eqn. (12), we get Adding 2S S2 + 1 S2 + 1 + R 2s − 2 R s 2S 2S 2

2

S2 + 1 + X 2s = − 1 + 2S

2

2

or

S2 + 1 S2 + 1 + X s2 = R s − 2S 2 S

or

S2 + 1 S4 + 1 + 2 S2 − 4 S2 + X s2 = R s − 4 S2 2 S

or

S2 + 1 S4 − 2 S2 + 1 + X s2 = R s − 4 S2 2 S

or

S2+ 1 R s – 2 S

−1

2

2

2

2 2 2 S − 1 + Xs = 2 S

...(13)

Transmission Lines

Eqn. (13) is of the form,

635

( x − c)2 + y2 = r 2

...(14)

S2 − 1 S2 + 1 with centres shifted by c = units from This is the equation of a circle of radius r = 2S 2 S the origin on the positive x = (R s ) axis. According to eqn. (13), the centre of family of circles and their radii are obtained as,

and

c=

S2 + 1 S + (1/S) = 2S 2

...(15)

r=

S2 − 1 S − (1/ S) = 2S 2

...(16)

1.5

Family of circles obtained from eqn. (13) is shown in fig. 22. For minimum value of S, that is, for S = 1, c = 1 [eqn. (15)] and r = 0 [eqn. (16)], from eqn. (14), we get ( x − 1)2 + y2 = 0.

1.0 XS

Thus, circle lies at (1, 0) point. For S > 1 all the circles surrounded the circle (1, 0). For maximum value of S, that is, for S = ∞ (for open or short

3=S 2.5 2

0.5 0

1.5 M

1

N 2

–0.5

circuited lines) or when S → ∞, the circle becomes X s axis

–1.0

(Fig. 22). Between S = 0 and S = ∞, as S increases radius of S

–1.5

circle increases with its centre shifted more and more toward

3

RS

Fig. 22

right. It is important to mention here that for any circle the intercept near the origin is at 1/S, whereas the intercept far from origin is at a distance S on the X s axis. As an example for S = 2, the intercept near the origin is at OM(= 12 = 1/ S) while the intercept at a point far from the origin is at ON (= 2 = S) on the X s axis as shown in fig. 22.

When normalized input impedance Z s (= Z s / Z0 ) lies on R s axis, then at a point remote from R s = S, X s = 0, we have Zs =

1 + |Γ | Zs =S= 1 −| Γ | Z0

...(17)

It is possible only when, φ − 2 β l = 0. Therefore from eqn. (3), we have 1 + | Γ | Z s = Z0 . S = Z0 1 − | Γ |

...(18)

That is, input impedance of line is purely resistive at a point R s = S, X s = 0 on the S circle. At a point near to the origin, R s = 1/ S. Z s will assumed to be a minimum value. Zs =

1 −| Γ | 1 1 = = S 1 + |Γ | 1 + |Γ | 1 −| Γ |

...(19)

We can obtain this value of normalized input impedance only when φ − 2 β l is replaced by π in equation (3), that is,

E lectromagnetic F ield T heory

636

Zs =

jπ 1 −| Γ | Z s 1 −| Γ | e = = j π Z0 1 + | Γ | e 1 + |Γ |

Now for resistive load

Γ =| Γ | so that φ = 0

∴

φ − 2 β l = π or βl = −

or

0 −2βl = π

π 2

Hence, on arriving at Z s =1/S point, β l has undergone a change of − π /2. Such a change can be represented by considering β l scale on the S-circles. The β l scale increases along clockwise direction, that is, in the direction of increase of negative angle. Eqn. (8) may be rewritten as, S − 1 (R s + j X s + 1) ∠ (φ − 2 β l) = R s + j X s − 1 S + 1 or

S − 1 R + j Xs − 1 ∠ (φ − 2 β l) = s S + 1 Rs + j Xs + 1

...(20)

For equating real and imaginary part multiplying and dividing R.H.S. of eqn. (20) by R s + 1 − j X s , that is, S − 1 (R − 1 + j X s ) (R + 1 − j X s ) ∠ (φ − 2 β l) = s × s S + 1 (R s + 1 + j X s ) (R s + 1 − j X s ) =

(R s − 1) (R s + 1) − j X s (R s − 1) + j X s (R s + 1) + X 2s (R s + 1)2 + X 2s 2

=

or

R s − 1 + 2 j X s + X 2s 2

(R s + 1)2 + X s

2 2 S − 1 j (φ − 2 β l) Rs − 1 + Xs 2 Xs e = + j 2 2 2 S + 1 (R s + 1) + X s (R s + 1)2 + X s

...(21)

To start β l scale at 0° on the abscissa (or R s- axis) angle φ is made zero: Substituting φ = 0 in eqn. (21), we get S − 1 −2 e S + 1

or

jβ l

2

=

2

Rs − 1 + Xs 2

(R s + 1) +

2 Xs

+ j

2 Xs 2

(R s + 1)2 + X s

...(22)

2 2 S − 1 2 Xs Rs − 1 + Xs [cos (−2 β l) + j sin (− 2 β l)] = + j 2 2 2 S + 1 (R s + 1)2 + X s (R s + 1) + X s

Separating real and imaginary parts, we have 2 2 S − 1 Rs −1 + Xs cos (2 β l) = 2 S + 1 (R + 1)2 + X s

s

...(23)

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637

S − 1 2 Xs sin (2 β l) = − 2 S + 1 (R s + 1)2 + X s

...(24)

Dividing eqn. (24) by eqn. (23), we get − tan (2 β l) =

or

2

2

Rs + Xs +

2 Xs R s2

2

2 Xs

−1 +

2

or R s − 1 + X s +

2 Xs =0 tan (2 β l)

2 Xs =1 tan (2 β l)

...(25)

Adding 1/tan2 (2 β l) on both sides of eqn. (25), we get 2

2

Rs + Xs +

2 Xs 1 + tan (2 β l) tan (2 β l)

2

1 =1 + tan (2 β l)

2

2

1 1 2 Rs + Xs + =1 + 2 tan (2 β l) tan (2 β l) 2

or

tan2 (2 β l) + 1 sec2 (2 β l) 1 R s2 + X s + = = tan (2 β l) tan2 (2 β l) tan2 (2 β l)

or

1 R s2 + Xs + tan (2 β l)

2

=

1 2

sin (2 β l)

...(26)

Eqn. (26) is again the equation of circle of the form ( x + C)2 + y = r 2 Thus, in this case

C=

...(27)

1 1 and r = tan (2 β l) sin (2 β l)

A family of a such circles are drawn in fig. 23. 1.5

XS

150°

1.0

170°

120° 0.5 0

180°

O

90°

2

5

RS

–0.5 –1.0

40°

10°

30°

20°

–1.5

Fig. 23

All β l circles pass through R s = 1, X s = 0 points. The superposition of β l circles and S circles gives a scale of β l angles and a circle diagram.

...(28)

E lectromagnetic F ield T heory

638

The Smith chart is a modified form of circle diagram and is shown in fig. 24.

Fig. 24 Smith Chart

B asis of Smith Chart To understand the basis of Smith chart, let us start from eqn. (20) as S − 1 R + j Xs − 1 ∠ (φ − 2 β l) = s S + 1 Rs + j Xs + 1 | Γ | ∠ (φ − 2 β l) =

Rs − 1 + j Xs Rs + 1 + j Xs

[∵ (S − 1) /(S + 1) = | Γ |]

...(29)

Introducing new variable U + jV such that U + jV =

Rs −1 + j Xs Rs + 1 + j Xs

...(30)

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639

For separating real and imaginary parts multiply and divide R.H.S. of eqn. (30) by R s + 1 − j X s and on rearranging, we get

2

U + jV =

R s2 − 1 + X s + j 2 X s

...(31)

2

[R s + 1]2 + X s

Equating real and imaginary part, we get 2

U = V=

and

R s2 − 1 + X s

...(32)

2

[R s + 1]2 + X s 2 Xs

...(33)

2

[R s + 1]2 + X s

Eliminating X s from eqn. (32) and then R s from eqn. (33), we obtain 2

Rs 1 + V 2 = U − R s + 1 R s + 1 and

1 (U − 1)2 + V − Xs

2

1 = Xs

2

...(34)

2

...(35)

Eqns. (34) and (35) are also equations of circles. Eqn. (34) is the equation of family of R s circles with centres at

Rs on U-axis and radii = 1/(R s + 1) and eqn. (35) gives a family of constant X s circles with Rs + 1

centres at 1 + ( j / X s ) and radii = 1/ X s . These families of circles

are shown in fig. 24. Commercially

available Smith chart is shown in fig. 24. Smith chart is used to determine the input impedance, load impedance, reflection coefficient and standing wave ratio on the transmission line. It is also used to find the positions of voltage minima and maxima from the termination end of the line. Example 32: What should be the length and characteristic impedance of a quarter wave transformer to match a 500 ohm transmission line to a 72 ohm dipole antenna at 200 MHz/sec ? Solution: We know that a quarter wave transformer matches the input impedance, Z s and output impedance Z L , provided its characteristics impedance Z0 is the geometrical mean of the two impedances. That is, Z0 = √ (Z s . Z L ) = √ (500 × 72) =189.74 Ω The length of the quarter wave transformer is given by l=

∴

c 1 λ 1 v = . = . (∵ for a loss free line velocity of propagation, v = c) f 4 4 4 f l=

3 .0 × 1010 200 × 10 6

×

1 = 37.5cm 4

Example 33: A generator with internal impedance 25 + j15 is to be matched to a load impedance (70 − j 42 ) using a low loss transmission line section at 3 MHz. What should be the length and characteristic impedance of the line ?

E lectromagnetic F ield T heory

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Solution: A quarter wave low loss line section is used to provide matching between the load impedance and the internal impedance of the generator, provided its characteristic impedance is, Z0 = (Z S . Z L ) = [(25 + j 15) (70 − j 42)] = [(25 × 70 − j 42 × 25 + j 15 × 70 + 42 × 15)] = [(1750 − j 1050 + j 1050 + 630)] or

Z0 = (2380) =48.78 ohm

The length of the line,

l=

or

l=

λ 1 v c 1 = = . (∵ For a loss free line, propagation velocity, v = c) 4 4 f f 4 3 × 108 3 × 106 × 4

= 25 meter

Example 34: A loss-less line of 100 ohm characteristic impedance connects a 100 KHz generator to a 140 ohm load. A properly terminated λ /4 section of this line is connected across the load to provide matching. What should be termination on this λ /4 section ? Solution: Since λ /4 section of the loss-less line is connected in parallel to 140 ohm load, the combination should offer a resistance of 100 ohm. Therefore, if the resistance offered by this section is R, then

But or

1 1 1 = + R L R 140

or R L =

R L = 100 ohm,

∴

R=

140 × R R + 140

100 (R + 140) = 140 R

140 × 100 = 350 Ω 40

For a λ /4 section

Z0 = (Z s . Z L )

Here

Z0 = 100 Ω and Z s = 350 ohm

∴

ZL =

2

Z0

Zs

=

100 × 100 = 28 .57ohm 350

Example 35: An ultra high frequency loss-less transmission line working at 1GHz is connected to an unmatched load producing a voltage reflection coefficient of 0.5 ∠ 30 °. Calculate the lengths and the position of a single stub to match the line Solution: The distance of the point of attachment of the stub with main line from the load end is given by lS =

λ [φ + π – cos –1|ΓL |] 4π

Given that f = 1 GHz = 109 Hz, ΓL = 0 .5, φ = 30 ° λ=

30 × π π c 3 × 108 = 0 .3 m and φ = = rad = 180 6 f 1 × 109

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641

∴

ls =

λ π λ π + π – 60° + π – cos –1(0 .5) = 4 π 6 4 π 6

or

ls =

0 .3 π 0 .3 5 π π +π– = 4 π 6 3 4 π 6

∴

ls =

1.5 24

= 0 .0625 m or ls = 6.25cm

The length lt of the stub is given by, lt =

1 –|ΓL|2 λ tan –1 2π 2|ΓL|

∴

lt =

1 – (0 .5)2 0 .3 0 .3 tan –1(0 .8660) tan –1 = 2π 2 × 0 .5 2π

or

lt =

or

lt = 0 .0341m or lt = 3.41 cm

0 .3 2π

(40 .9 ° ) or lt =

0 .3 2π

× 40 .9 ×

π 180

Example 36: A 100 ohm line with air dielectric is terminated by a load impedance of 75+ j40 ohm and is excited at 1GHz by a matched generator. Find the position of a single matching stub of 100 ohm impledance on the line, and determine the length of the stub. [UPTU, B.Tech IV Sem. 2007]

Solution: The position of a single matching stub attached with main line or the distance of the point of attachment from the load end is given by ls =

λ [φ + π – cos –1|ΓL|] 4π

where ΓL is the reflection coefficient. If Z L is the load impedance and Z0 is the characteristic impedance, then ΓL =

Z L – Z0 Z L + Z0

Here

Z L = 75 + 40 j ohm & Z0 = 100 ohm

∴

ΓL =

75 + 40 j – 100 –25 + 40 j = 75 + 40 j + 100 175 + 40 j

ΓL =

(–25 + 40 j) (175 – 40 j) –25 × 175 + 25 × 40 j + 175 × 40 j – (40 j)2 = (175 + 40 j) (175 – 40 j) [(175)2 – (40 j)2 ]

or

ΓL =

–4375 + 8000 j + 1600 –2775 + 8000 j = 30625 + 1600 32225

or

ΓL = –0 .0861 + 0 .2482 j

or

0.2482 ΓL = (0 .0861)2 + (0.2482)2 ∠ tan –1 – 0.0861

E lectromagnetic F ield T heory

642

or

ΓL = 0 .2627∠ tan –1(–2 . 8827) [ ∴ tan(– φ) = – tan φ and – tan φ = tan (180 – φ)]

But ∴ ∴

–1

φ = tan (2 .8827) or 180 – φ = tan –1 (2 .8827) 180 – φ = 70.9 ° or φ = 180 °–70 .9 ° = 109.1° ΓL = 0 . 2627∠109 .1° or |ΓL| = 0 . 2627

Hence

ls =

λ [φ + π – cos –1|ΓL|] 4π

Here

λ=

c 3 × 108 π π or φ = = 0.3 m and φ = 109.1°× = f 1 × 109 180 165 .

∴

ls =

0 .3 π + π – cos –1(0.2627) 4 π 165 .

=

0 .3 4π

π π 0 .3 + π – 74 .8° × = 1 . 65 180 4π

π π + π– 1 . 65 2 . 40

0 .3 0.3 π × [0.6060 + 1 – 0 .4167] = × 11893 . 4π 4

or

ls =

or

ls = 0 .0892m

The length of the stub,

lt =

1–|Γ |2 λ L tan –1 2π 2|ΓL|

∴

lt =

1 – (0 .2627)2 0 .3 tan –1 2π 2 × 0 .2627

or

lt =

0 .3 0 .3 π (61.4 ° ) or lt = × 61.4 °× 2π 2π 180

or

lt =

0.3 × 61.4 = 0.05117m 2 × 180

or

lt = 5.117 cm

or ls = 8.92 cm 0 .3 = tan –1(1.8364) 2π

Example 37: An aerial having an input impedance of 100∠0 ohm is to be matched, at a frequency of 100 MC/sec to a loss free open wire feeder having a surge impedance of 600 ∠0 ohms by means of a loss-free short-circuited stub. Assuming that the stub has the same characteristic impedance as the feeder, find the point of attachment and the length of the stub. Solution: The distance of the point of attachment of a short circuited stub with the main line from load end is given by 1 /2 1 /2 Z R λ λ ls = tan −1 L tan −1 = 2π 2π Z0 R0 Here

R = 100 ohm and R0 = 600 ohm.

Since the stub and the feeder are loss free, the velocity of propagation will be equal to that of radiation in free space. That is,

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643

λ= λ=

v c = (∵ the velocity of propagation for a loss-less line, v = c) f f 3 × 108 100 × 106

= 3 meter

and

ls =

3 100 3 = tan −1 tan −1(0 .41) 2π 600 2 π

or

ls =

3 3 π rad = 0.186 meter × 22 .29 ° = × 22 .29 × 2π 2π 180

The length of the stub, is given by,

lt =

R R0 Z L Z0 1 λ or lt = tan −1 tan −1 β 2π Z L − Z0 R − R0

or

lt =

100 × 600 244 .95 3 3 = tan −1 tan −1 2π − 500 100 − 600 2 π

3 tan −1(− 0 .49) 2π θ = tan −1 (− 0 .49) or (180 ° − θ ° ) = tan −1(0 .49)

lt = But,

[∵ tan (–θ) = – tan θ and – tan θ = tan(180 – θ] ∴ Hence, or

180 − θ = 26 .10 ° or θ = 180 ° − 26 .10 ° = 153 .9 ° 3 π 3 rad lt = × 153 .9 ° = × 153 .9 × 2π 180 2π lt = 1.282 meters

L osses and Distortions in Transmission Lines In the preceding articles we have emphasized loss- less transmission lines having perfect conductors and dielectrics, but in actual practice neither a conductor is perfect nor a dielectric is pure insulator. Imperfect materials develop losses along a transmission line when wave is propagated. the imperfections are basically due to the finite resistance of the conductor and due to the finite conductance of the dielectric material separating the conductors. Thus, actual transmission lines have always some power losses because of imperfect conductors and dielectric consequently, the existence of line losses can not be ruled out. These losses increase with the increase of the frequency of operation. Some of the important losses taking place in the line are: Copper losses, dielectric losses, and induction and radiation losses.

Copper Losses In radio frequency transmission lines, whenever current flows through the conductors, due to finite resistance of the conductor, some energy is dissipated as heat. In addition to d.c. resistance, there is an a.c. resistance of the line conductors which increases with the increase of operating frequency. In fact in calculating the resistance and inductance of a conductor, we assume that the current is distributed uniformly over its cross-section. It is true for thin conductors operating at low frequency. In case of high frequencies when rate of change of current becomes very large, the current distribution becomes non uniform. At high frequencies the induced emf or induced current developed in the

E lectromagnetic F ield T heory

644

conductor is largest in the centre. The induced current is always in a direction to reduce the original current and thus it forces the current to confine itself near the outer surface of the conductor. Thus, inner region of the conductor becomes practically void of any current. Thus the effective cross-sectional area decreases as frequency increases. Since the resistance is inversely proportional to the cross-sectional area, the resistance of the conductor will increase as the frequency is increased. Also, since power loss increases as resistance increases, power losses increase with the increase in frequency because of skin effect. These copper losses can be minimized and conductivity can be increased in the radio frequency transmission line by plating the line with silver. Since silver is better conductor than copper, most of the current will flow through the silver layer.

Dielectric Losses Due to non-zero conductivity (σ ≠ 0) of the dielectric material between the conductors of the line, there is always power loss in the form of heat. The heat produced is dissipated into the surrounding medium. Thus, the part of the power given to the transmission line is waste in heating the dielectric. The dielectric losses can be reduced by using rubber and polyethylene as dielectric between the conductors of the transmission line as they consumed less power in dielectric heating.

Radiation and Induction Losses High frequency currents flowing through an open wire line produce an electromagnetic field surrounding the conductor. A part of this electromagnetic field is radiated in the free space representing a power loss due to radiation as the wave propagates from one end of the line to the other. Radiation and Induction losses are similar in the sense that both are caused by the fields surrounding the conductors. Induction losses are due to the electromagnetic field about a conductor cuts through any near by metallic object and a current is induced in that object. As a result power is dissipated in the object and lost. Radiation losses occurs because some magnetic lines of force about a conductor do not return to the conductor when the cycle alternates. these lines of force are projected into the space as radiation and this results in power losses.

Various Distortions in Lines Due to mismatch in transmission lines there are various types of losses due to which the energy given to the transmission line, may become dissipated before reaching the load. Distortion due to mismatch in transmission line gives attenuation loss, reflection loss, transmission loss, return loss and insertion loss.

Attenuation Loss Attenuation loss in a transmission line occured when the part of incident energy is absorbed during the propagation of a signal from sending end to the load end, that is, some energy is lost in the system. Attenuation loss is a measure of power loss due to absorption of the signal in the transmission line and is mathematically expressed in decible as, Energy absorbed Ei – Er Attenuation loss (dB) = 10 log10 = 10 log10 Energy transmitted Et where Ei is the incident energy, Er is the reflected energy from load and Et is the transmitted energy to the load.

Transmission Lines

645

Reflection Loss When power, from a generator, is applied at the input end of a transmission line terminated by its characteristic impedance (Z L = Z0 ) all the input power is absorbed by the load at far end and no part of the power is reflected, that is, there exists no reflected wave. When the line is terminated by an impedance other than characteristic impedance (Z L ≠ Z0 ) then only a part of the power is absorbed at the load end and remaining power will be reflected back to generator end of the line reducing the available energy at the load. In this way reflected waves are set up in the line which propagates from load end to the input end. These reflected waves, like to incident waves, cause losses on the line. Greater the difference between the load impedance (Z L ) and the characteristic impedance (Z0 ), more stronger will be reflected wave and consequently, a lagaer loss due to reflection. Thus, the reflection loss is a measure of the power loss due to reflection of the signal due to impedance mismatch of the transmission line and given as,

Input energy Ei Reflection loss (dB) = 10 log10 = 10 log10 Lost energy Ei – Er Where Ei is the incident energy Er is the reflected energy and Ei – Er is the energy lost in the line.

Transmission Loss Transmission loss is a measure of the power loss during the propagation of a signal through the transmission line and is given as,

E Transmission loss (dB) = 10 log10 i Et where Ei is the incident energy and Et is the transmitted energy to the load.

Return Loss Return loss is the loss of the signal power resulting from the reflection caused at a discontinuity in a transmission line. When a signal travelling along a transmission line encounters a discontinuity, part of the signal is reflected and rest continues towards the next discontinuity if exist, where another portion is reflected and the remaining portion is continues and so forth. The reflected portions are re-reflected at discontinuities, tending perhapes to reinforce the original signal or not depending on the phase, and so on. The attenuation of reflected light or the return loss measurement describes the ratio of the power in the reflected wave to the power in the incident wave in units of decible. That is,

P Return loss, RL (dB) = 10 log10 i Pr P RL' (dB) = 10 log10 r Pi In term of reflection coefficient, Γ =

Vr Z L – Z S = Vi Z L + Z S

where Z L and Z S are the load impedance and input impedances of the transmission line respectively.

RL(dB) = –20 log10|Γ | or

RL (dB) = Pi (dBm ) – Pr (dBm )

The standard output for return loss is positive, so a large return loss value mean that the power in reflected wave is small compare to the power in the incident wave and indicate a better impedance match.

E lectromagnetic F ield T heory

646

Insertion Loss Insertion loss is the loss of signal power resulting from the insertion of a device in a transmission line. Insertion loss is a measure of the energy loss through a transmission line as compared to direct transmission of energy without the transmission line, and is expressed as,

E Insertion Loss, IL(dB) = 10 log10 1 E2 where E1 is the energy received by the load when connected directly to the source without the transmission line is inserted between the source and the load, keeping the input energy constant. The insertion loss is due to mismatch losses at the input and output plus the attenuation loss in the transmission line.

Application of Transmission Lines In many electrical circuits, the length of the wires connecting different components is ignored. That is, the voltage on the wire at a given time be assumed same at all points. However, when the voltage changes in time interval comparable to the time it takes for the signal to travel down the wire, the length become important and the wire is treated as transmission line. In other words, the length of the wire is important when the signal includes the frequency components with corresponding wavelengths comparable to or less than the length of the wire. Electrical transmission lines are used to convey electromagnetic energy and high frequency signals over long or short distances with minimum power loss. The energy may be in the form of light, heat, mechanical work or information-speech, music, pictures, data etc. Transmission lines are used to connect a transmitter to an antenna; TV or radio aerial to the receiver; to interconnect components of a sterio system; computer in a network; hydroelectric generating plant to a sub station several hundred kilometers away etc. Transmission lines are also used as pulse generators. By charging the transmission line and then discharging it into a resistive load, a rectangular pulse equal in length to twice the electrical length of the line can be obtained, although with half the voltage. A Blumlein transmission line is a related pulse forming device that overcomes this shortcoming. These are some times used as the pulse energy sources for radar-radar transmitters and other related devices. When an open-circuited or short circuited transmission line is wired parallel with a line used to transfer signals from one place to another, then it will function as a filter. The method for making stubs is to take an open-circuited length of transmission line wired in parallel with the feeder delivering signals from an aerial. By cutting the free end of the transmission line, a minimum in the strength of the signal observed at a receiver can be found. At this juncture, the stub filter will reject this frequency and the odd harmonics, but if the free end of the stub is short circuited then the stub will become a filter rejecting the even harmonics.

Transmission Lines

647

W aveguides A system of parallel conductors, in which the conductor separation is small compared with the wavelength of the propagating wave and guides low frequency (upto 300 MHz) Type Transverse Electromagnetic (TEM) wave is called transmission line. For the propagation of the various other possible modes, like TE and TM waves through parallel planes a certain minimum separation (in wavelength) between the conductors is required. They must have large cross-sectional dimensions of the guiding system. A system of conductor in which the separation of conductors is of the order of wavelength and which propagates TE or TM waves is called waveguides. Transmission line has at least two separate conductors between which voltage can exist, whereas waveguide involve only one conductor, for example a hollow rectangular or circular cylinder within which wave propagates. Thus, waveguides are rectangular or circular metallic tubes inside which an electromagnetic wave is propagated and is confined by the tube. Waveguides are not capable of transmitting TEM mode, but capable of transmitting TE and TM modes. At high frequencies where the wavelengths are of the order of meters or millimeters the only most suitable way of generating and transmitting electromagnetic radiation involves metallic structures with dimensions comparable to the wavelength involved. Any system of conductors or insulators through which electromagnetic waves can propagate, are known as waveguides. All transmission lines are infact waveguides, because they guide the electromagnetic waves from place to place. Similar to a coaxial line in which an electromagnetic wave, travelling along it is confined to the space between the conductors, in waveguides too, wave is confined to move in the direction of metallic wave guides. Though any arrangement of conductors and insulators that is used to carry electromagnetic fields to a desired destination may be termed as waveguide, but the term is more commonly reserved for hollow metallic pipes or tubes of suitable shapes that are used to carry energy in the upper range of ultra high frequency band. Waveguides most commonly employed are of rectangular cross-section, though some times wave guides have circular or other cross-sections are also used. In fact guiding system, waveguides have many different form that depend on the requirement, and on the frequency of the waves to be transmitted. Usually a metal pipe with constant rectangular or circular cross-sections are used as waveguides because they are easiest to manufacture and most amenable to mathematical treatments. The simplest form is the parallel-plate guide [Fig. 25(a)] The other forms are the hollow pipe rectangular waveguides [Fig. 25(b)] and hollow pipe cylindrical wave guides (Fig. 25(c)]. Dielectric slab waveguides [Fig. 25(d)] and optical fibre waveguides [Fig. 25(e)] are suitable at optical frequencies. Depending upon the use and the frequency of the wave to be transmitted, each of these possesses certain advantages over the others.

E lectromagnetic F ield T heory

648

d x

b

b y

Parallel Pla t e Waveguide (a)

a Rectangular Waveguide (b)

Cylindrica l Waveguide (c)

µ

2

µ2 µ2 µ2

µ2

µ1

µ1 a b

Symmetrical Dielectric Slab Waveguide (d)

Optical Fibre Waveguide (e)

Fig. 25

imilarities and Dissimilarities between Transmission Lines and S Waveguides The behaviour of a guiding energy transfer device or waveguide is similar in many respects to the behaviour of a transmission line. Waves travelling along a waveguide have a phase velocity and are attenuated like a transmission line. The waves are attenuated in the waveguides due to non-availability of a waveguide of perfectly conducting walls and dielectric within the guide. A wave reaches the end of a waveguide is reflected unless the load is carefully adjusted to absorb the wave; also an irregularity in a waveguide produces reflection just as does an irregularity in a transmission line. Further, reflected waves can be eliminated by the use of impedance matching by stub tuners, exactly as with a transmission line. Finally, when both incident and reflected waves are simultaneously present in a waveguide, the result is a standing wave pattern as in transmission line.

Transmission Lines

649

In the waveguide propagation the electric and magnetic fields of TE and TM waves are equivalent with those of current and voltages on a suitably loaded transmission line. The concept of a waveguide as an equivalent transmission line with a certain characteristic impedance and propagation constant, helps a lot in solving several problems by means of well-known transmission line theory. In some other respects, the wave propagation along the transmission lines and through the waveguides have some significant differences. Transmission lines are used at low frequencies, whereas waveguides are used at ultrahigh frequencies, that is, microwaves because they are relatively much less lossy in comparison to transmission lines. Transmission lines consists of two or more conductors, whereas a waveguide may consist of one or more →

conductors, or no conductor at all. In the transmission lines, the electromagnetic waves have both the E →

and the H vectors transverse to the direction of propagation and are generally referred to as TEM (Transverse Electromagnetic) waves. However, in the waveguides the propagating waves are either in TE mode (transverse electric mode, in which the electric field is everywhere transverse to the axis of the guide and has no component any where in the direction of the guide axis. The associated magnetic field does, however, have a component in the direction of the axis. This TE mode sometimes also called H-mode) or in TM mode (transverse magnetic mode, the magnetic field being everywhere transverse to the guide axis while at some places the electric field has components in the axial direction. This TM mode sometimes also called E-mode). The most striking difference between transmission line and waveguide is that a particular mode will propagate down a waveguide with low attenuation only if the wavelength of the wave is less than some critical value determined by the dimensions and geometry of the waveguide. If the wavelength of the propagating wave is greater than this critical cut-off value of wavelength the transmission is no longer possible. At this cut-off wavelength, the waves in the waveguide die out very rapidly in amplitude even when the walls of the guide are of material having infinite conductivity. The TEM mode, however, has no cut-off; it will be supported to any frequency.

W aveguide Modes The waves can travel in a waveguide in different field configurations, each such configuration is term as mode. In fact, any actual configuration of electric and magnetic fields existing in a waveguide is the result of a series of modes that are superimposed upon one another. If the phase, magnitude and position along the axis of each individual mode is properly chosen, then the sum of the fields of individual modes is equal to any actual electric and magnetic fields that can be present. Thus, modes in waveguides are similar to the harmonics of a periodic wave, because a periodic wave of arbitrary shape is the outcome of sum of series of properly chosen harmonic components.

650

E lectromagnetic F ield T heory

In one type of mode, the electric field is everywhere transverse to the direction of propagation (axis of the guide) and has no component along the guide axis while the magnetic field will have a component along the direction of propagation (or along the guide axis). In this mode magnetic field may have the components transverse to as well as along the direction of propagation. Such modes in waveguide terminology, are called Transverse Electric or TE-modes or H-modes. In the other type, the magnetic field is everywhere transverse to the guide axis and has no component along the guide axis or along the direction of propagation, while the electric field has components in the direction of propagation or the guide axis as well as transverse to the guide axis. It should be remembered that the transverse component of electric field is also normal to the transverse component of magnetic field. This type of mode is called Transverse-Magnetic or TM-modes or E-modes. Modes are labelled by double subscripts as TEmn or TMmn depending upon whether the electric field (or magnetic field) is transverse to the direction of propagation. In this system of nomenclature the first subscript, m denotes the number of half wavelength of intensity of electric field or the number of half period variation of electric (or magnetic) field in the transverse plane in the direction of broader side (along x-axis). The second subscript, n denote, the number of half-period variation of the same field in the direction of short side (along y-axis). The possible modes that can pass through commonly known waveguides are; TE10 , TE01, TE11, TM11, TE21 and TM21. Modes in circular waveguides are designed in the same manner as other waveguides, but the letters ‘m’ and ‘n’ have different nomenclature. The letter ‘m’ denotes the full-wave intensity variations around the circumference, while ‘n’ denotes the number a half-wave intensity variations radially from the centre to the walls. A dominant mode that a waveguide can transmit, is one which has the lowest cut-off frequency of all possible mode. The critical wavelength or frequency at which transmission is no longer possible through a waveguide is called cut-off wavelength or cut-off frequency. For receiving or setting up of TEmn and TMmn different arrangement of antenna system is needed.

Cut-off Wavelength or Frequency Waveguides are those transmitting system that will not transmit low frequencies but will transmit extremely high frequencies. Therefore, there must be some intermediate frequency at which there is a transition from one condition to the other. Thus, there is a cut-off frequency or wavelength associated with each guided mode. If the operating frequency is below the cut-off frequency, the mode will not propagate through the waveguide. If the operating frequency is above cut-off, the mode propagates. The critical wavelength or frequency at which transmission is no longer possible is called the cut-off wavelength or frequency. Hence, at frequencies higher than cut-off, the mode propagates without attenuation, while at frequencies lower than cut-off, the mode is attenuated. In fact, if the frequency of the wave to be transmitted is less than a certain limiting value or cut-off value, the wave is simply reflected backwards and forwards across the guides and makes no forward motion.

Transmission Lines

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Q uestion Bank 1.

Derive transmission line voltage and current equations. Discuss the concept of distortionless and loss-less line.

[GBTU, B.Tech III Sem. 2011]

2.

Derive transmission line equations.

3.

What are the advantages of transmission lines? Discuss various types of transmission lines.

[U'Khand, B.Tech IV Sem. 2011]

[U'Khand, B.Tech IV Sem. 2010]

4.

Derive transmission line differential equations. Derive the conditions of loss less transmission from it. [UPTU, B.Tech IV Sem. 2007]

5.

What is meant by distortionless line? Compare the advantage and disadvantage of coaxial cables and two wire transmission line.

6.

[GBTU, B.Tech IV Sem. 2010]

Using the general current and voltage equations for a transmission line. Obtain an expression for the input impedance of a lossless transmission line when the line is terminated by load impedance Z L . [GBTU, B.Tech IV Sem. 2011]

7.

Relate short circuit, open circuit and characteristic impedance of transmission line. [GBTU, B.Tech III Sem. 2012]

8.

Describe the wave characteristic of finite transmission line.

9.

Describe the primary and secondary transmission line parameters. What is a distortion less transmission

[GBTU, B.Tech IV Sem. 2010]

line? Derive the condition to be satisfied by distortion less transmission.

[UPTU. B.Tech IV Sem. 2006]

10. Derive the expression for the input impedance of a lossy transmission line terminated in a mismatched load.

[UPTU, B.Tech IV Sem. 2006]

11. Derive expressions for sending end-voltage and current along a transmission line in terms of receiving end quantities for a loss less line.

[UPTU, B.Tech IV Sem. 2003]

12. Explain the term standing wave ratio related to transmission line. What will be the value of input impedance when output impedances are (i), short circuited, (ii) open-cricuited, (iii) characteristic impedance.

[GBTU, B.Tech IV Sem. 2011]

13. Derive input impedance of transmission line. Define standing wave ratio. [U'Khand, B.Tech IV Sem. 2011] 14. Derive, input impedance SWR and power of a transmission line of length l terminated by load impedance Z L .

[UPTU, B.Tech IV Sem. 2008]

15. Discuss input and characteristic impedance of a transmission line. 16. Derive, transmission coefficient = 1 + reflection coefficient 17. Define reflection loss, transmission loss and return loss.

[UPTU, B.Tech IV Sem. 2004] [U'Khand, B.Tech IV Sem. 2009]

[GBTU, B.Tech III Sem. 2010; III Sem. 2012]

18. Define and explain the meaning of the term ''standing wave ratio'' What is the formula for it if the load is purely resistive? What is the value of standing wave ratio for a purely reactive load? [UPTU, B.Tech IV Sem. 2006]

E lectromagnetic F ield T heory

652

19. VSWR and reflection indicate the matching condition between two networks. Justify it. [UPTU, B.Tech IV Sem. 2002]

20. Discuss transmission line equation and its characteristic.

[UPTU, B.Tech IV Sem. 2002]

21. Discuss the distortion less transmission line.

[UPTU, B.Tech IV Sem. 2004]

22. What are the techniques used for impedance matching on transmission line? Discuss one technique in detail.

[UPTU, B.Tech IV Sem. 2003]

23. Discuss impedance matching and λ /4 transformer.

[UPTU, B.Tech IV Sem. 2004]

24. Explain the double stub method for impedance matching on a transmission line. What are the advantages of double stubs over the single stub?

[UPTU, B.Tech IV Sem. 2007]

25. Discuss the structure of Smith chart. How it is used for the measurement of impedances and VSWR? [GBTU, B.Tech III Sem. 2012]

26. Explain the role of Smith chart in measurement of various parameters in transmission line. [GBTU, B.Tech III Sem. 2011]

27. What is Smith chart and why it is useful in making transmission line calculation? [UPTU, B.Tech IV Sem. 2007; GBTU, IV Sem. 2010]

28. What are the applications of Smith chart in matching?

[UPTU, B.Tech IV Sem. 2002]

29. Discuss various distortions in lines.

[UPTU, B.Tech IV Sem. 2002]

30. Write short notes on the following: (a) quarter wave impedance inverting transformer, (b) single stub matching, and (c) double stub matching. 31. Show that a transmission line will be distortion less if CR = LG. 32. Show that when the series resistance R and the shunt capacitance C of a transmission line are small but not negligible, the attenuation constant may be written as

α=

R 2

C G L + L 2 C

Transmission Lines

653

nsolved Numerical Problems 1.

Find the characteristic impedance, propagation constant and velocity of propagation for a transmission line having the following parameters.

R = 84 ohm/km, G = 10 −6 mho/km, H = 0 .01 henry/km, C = 0 .061 µF/km and frequency = 1000 Hz. 2.

Calculate Z and Y at ω = 5 × 103 rad/sec for a transmission line which has following constants :

R = 10 .4 Ω, L = 3 . 666 mH, C = 0 . 00835 µF and G = 0 . 08 µ mho. 3.

A pair of high frequency parallel wire transmission line has distributed capacitance and inductance of 0.04 µF and 9.8 mH per mile respectively. What is the surge impedance (Z0 ) of the line ?

4.

Determine the primary constants and the phase velocity on the transmission line. The measurement on a transmission line at 1 KHz gave the following results.

Z0 = 710 ∠ – 16 ° ohm, α = 0 .01 neper/m and β = 0 .035 rad/m. 5.

The primary constants of a cable are R = 80 ohm per mile, L = 2 mH/mile, G = 0 .3 × 10 −6 mho/mile, and C = 0 .07 µF/mile. Calculate the characteristics constants of the line at 500 Hz.

6.

At 1200 cycles the characteristic impedance and propagation constant of an open wire transmission line are :

Z0 = 650 − j 150 and γ = 0 .005 + j 0 .007. Calculate the distributed parameters of the line. 7.

What would be the reflection coefficient if the standing wave radio (SWR) is 4. What would be SWR for zero reflection coefficient.

8.

Find the voltage reflection coefficient for a transmission line with characteristic impedance equal to 300 ohm. This transmission line is terminated in a load impedance of 100 + j 200.

9.

A transmission line of characteristic impedance of 150 ohm is terminated by a resistor of 200 ohm. Calculate, (a) Voltage standing wave ratio of the transmission line. (b) Impedances at the voltage maximum and minimum positions.

10. A loss-less transmission line of characteristic impedance 50 ∠0 ° ohm and half wavelength long is left open circuited at the far end. The r.m.s value of the open circuited voltage is 10 volt. Determine the r.m.s value of voltage and current to a distance of one-eighth wavelength away from the open circuit. 11. Find out the characteristic impedance of the transmission line, when the far end of a finite length transmission line is short circuited the impedance measured at the sending end is 4.61 Ω resistive and when that is open circuited, that input impedance becomes 1390 Ω resistive. 12. An open wire transmission line whose characteristic impedance Z0 = 600 Ω is terminated by a resistive load of 900 Ω. Find out the SWR and give arrangement for single stub matching. 13. A transmission line of characteristic impedance 50 Ω is terminated by a resistor 100 Ω. Determine (a) VSWR, and (b) impedance at the voltage maximum and minimum positions.

E lectromagnetic F ield T heory

654

14. A ultra high frequency (UHF) loss-less transmission line working at 16 Hz is connected to an unmatched load producing a voltage reflection coefficient is 0.5 ∠30 °. Find out the length and the position of a single stub to match the transmission line. 15. Determine the input impedance of a short circuited transmission line of length λ /8 and having the characteristic impedance Z0 = 600 Ω. 16. A 30 m long transmission line of impedance 50 Ω is operating at 2 MHz. The line is terminated with a load of Z l = 60 + j 40 ohm. The velocity of propagation on the line is 1. 8 × 108 m/sec. Find (a) the reflection coefficient, (b) the standing wave ratio, (c) the input impedance of the line. 17. Determine the position and length of a short circuited stub designed to match a 200 ohm load to a transmission line having characteristic impedance Z0 = 300 Ω. 18. Calculate the characteristic impedance of quarter wave transformer if a 120 Ω load is to be matched to a 75 Ω line.

nswers to Unsolved Numerical Problems 1.

Z0 = 523 . 16 ∠ − 26 . 53 ° ohm,

−6 2. 21. 05 ∠ 60 . 4 ° , 41. 77 × 10 ∠ 89 . 9 °

v = 34 . 9 × 106 m/sec,

4.

C = 8 .1528 nF/m, vp = 1. 794 × 105 m/sec

γ = 0 . 09 + j 0 .18 per km 3.

495 ohm

5.

Z0 = 605 ∠ − 42 . 8° , α = 0 . 0903 neper/mile, β = 0 . 0977 rad/mile, λ = 64 . 4 miles,

L = 3 . 487 mH/meter, G = 0 .1177 µmho/m,

6.

R = 4 . 8 ohms/mile, L = 0 . 398 mH/mile G = 6 . 75 µ mhos/mile, C = 0 . 00143 µF/mile

8. 0 . 6 − 0 . 6 j

v = 322000 mile/sec.

10. 7 . 071 volts, 0 .141 ∠ 90 °

7.

ρ = 0 . 6, S = 1

12.

S = 1. 5, ls = 0 .14 λ, lt = 0 .188 λ

9.

S = 1. 33, Z max = 200 Ω, Z min = 112 .5 Ω

14.

ls = 6 . 25 cm, lt = 3 . 41 cm

11. 80 . 04 Ω 13.

S = 2, Z max = 100 Ω, Z min = 25 Ω

15. j 600 ohms

16. 0 . 3523 ∠ 56 ° , 2 . 088, 24 + j 135 . Ω 17. ls = 10 . 89 λ cm, lt = 31.16 λ cm, 18. 95 ohms ❍❍❍