Electromagnetic field theory 9781259006319, 125900631X

Table of contents :
Title
Contents
1 Vector Analysis
1.1 Introduction
1.2 Scalar and Vector Quantities
1.3 Fields
1.4 Properties of Vectors
1.5 Vector Algebra
1.6 Coordinate Systems
1.7 General Curvilinear Coordinates
1.8 Differential Elements (Lengths, Areas and Volumes) in Different Coordinate Systems
1.9 Vector Calculus
1.10 Gauss’ Divergence Theorem
1.11 Stokes’ Theorem
1.12 Classifications of Vector Fields
1.13 Helmholtz Theorem
Summary
Important Formulae
Exercises
Review Questions
Multiple Choice Questions
Answers
2 Electrostatics
2.1 Introduction
2.2 Electric Charge
2.3 Coulomb’s Law
2.4 Principle of Superposition of Charges
2.5 Electric Field and Field Intensity (E)
2.6 Different Charge Densities
2.7 Electric Fields due to Continuous Charge Distribution
2.8 Electric Flux (Displacement) (y ) and Flux Density (Displacement Density) (D)
2.9 Electric Field Lines (Lines of Force) and Flux Lines
2.10 Gauss’ Law
2.11 Applications of Gauss’ Law
2.12 Electric Potential (V)
2.13 Principle of Superposition of Electrostatic Fields
2.14 Potential Gradient and Relation between E and V
2.15 Equipotential Surfaces
2.16 Electric Dipole
2.17 Multipole (or Far Field) Expansion of Electric Potential
2.18 Energy Stored in Electrostatic Fields
2.19 Poisson’s and Laplace’s Equations
2.20 Uniqueness Theorem
2.21 General Procedure for Solving Poisson’s and Laplace’s Equations
2.22 Capacitor and Capacitance
2.23 Method of Images
2.24 Electric Boundary Conditions
2.25 Dirac Delta Representation in Electrostatic Fields
Summary
Important Formulae
Exercises
Review Questions
Multiple Choice Questions
Answers
3 Magnetostatics
3.1 Introduction
Part I: Behaviour of Different Electrical Materials in Electric Field
3.2 Electric Current and Current Density
3.3 Classification of Electrical Materials
3.4 Behaviour of Conductors in Electric Field
3.5 Behaviour of Dielectric Materials in Electric Field
3.6 Behaviour of Semiconductor Materials in Electric Field
Part II: Magnetostatic Field
3.7 Introduction
3.8 Biot–Savart Law
3.9 Magnetic Field Intensity (H )
3.10 Magnetic Flux (f) and Flux Density (B)
3.11 Gauss’ Law of Magnetostatic Interpretation of Divergence of Magnetic Field— Maxwell’s Equations
3.12 Ampere’s Circuital Law—Interpretation of Curl of Magnetic Field
3.13 Applications of Ampere’s Law
3.14 Magnetic Potentials
3.15 Forces due to Magnetic Fields
3.16 Magnetic Torque
3.17 Magnetic Dipole
3.18 Magnetic Materials
3.19 Magnetic Boundary Conditions
3.20 Self Inductance and Mutual Inductance
3.21 Magnetic Energy (Energy Stored in Magnetic Fields)
Summary
Important Formulae
Exercises
Review Questions
Multiple Choice Questions
Answers
4 Electromagnetic Fields
4.1 Introduction
4.2 Faraday’s Law of Induction for Time-Varying Fields
4.3 Induced EMF for Time Varying Fields
4.4 Inconsistency in Ampere’s Law for Time-Varying Fields
4.5 Maxwell’s Equations for Time-Varying Fields
4.6 Time-varying Potentials for EM Fields
Summary
Important Formulae
Exercises
Review Questions
Multiple Choice Questions
Answers
5 Propagation, Reflection, Refraction and Polarisation of Electromagnetic Waves
5.1 Introduction
5.2 Three-Dimensional Wave Equations (Helmholtz Equation)
5.3 Properties of Electromagnetic Waves
5.4 Standing Electromagnetic Waves
5.5 Phase Velocity and Group Velocity of Electromagnetic Waves
5.6 Intrinsic Impedance
5.7 Propagation of Uniform Plane Waves through different Media
5.8 Polarisation
5.9 Reflection and Refraction of Plane Electromagnetic Waves at the Interface between Two Dielectrics
5.10 Reflection and Refraction of Plane Electromagnetic Waves by Perfect Conductor
5.11 Poynting Theorem and Poynting Vector
5.12 Energy Flux in a Plane Electromagnetic Wave
5.13 Radiation Pressure
Summary
Important Formulae
Exercises
Review Questions
Multiple Choice Questions
Answers
6 Transmission Lines
6.1 Introduction
6.2 Types of Transmission Lines
6.3 Transmission Line Modes
6.4 Transmission Line Parameters
6.5 Transmission Line Equations (Telegrapher’s Equations)
6.6 Characteristic Impedance of Transmission Line
6.7 Input Impedance of Transmission Line
6.8 Reflection Coefficients of Transmission Line
6.9 Standing Waves and Standing Wave Ratio (S) of Transmission Line
6.10 Input Impedance as a Function of Position along a Transmission Line
6.11 Losses in Transmission Lines
6.12 Smith Chart
6.13 Load Matching Techniques in a Transmission Line
Summary
Important Formulae
Exercises
Review Questions
Multiple Choice Questions
Answers
7 Waveguides
7.1 Introduction
7.2 Parallel Planes Waveguides between Two Infinite Parallel Conducting Planes
7.3 Rectangular Waveguides
7.4 Circular or Cylindrical Waveguides
7.5 Power Transmission in Waveguides
7.6 Power Losses and Attenuation in Waveguides
7.7 Cavity Resonator or Resonant Cavities
7.8 Dielectric Slab Waveguides
7.9 Transmission Line Analogy for Waveguides
7.10 Applications of Waveguide
Summary
Exercises
Review Questions
Multiple Choice Questions
Answers
Typical Short Answer Type Questions with Answers
Bibliography
Index

Citation preview

Electromagnetic Field Theory

About the Authors Shankar Prasad Ghosh received his BE (Hons) in Electrical Engineering from National Institute of Technology, Durgapur and Master of Electrical Engineering degree from Jadavpur University, with specialisation in High Voltage Engineering. Prof. Ghosh joined College of Engineering & Management, Kolaghat as Lecturer in 2002. He is presently working as Assistant Professor in the Department of Electrical Engineering at College of Engineering & Management, Kolaghat. He has authored and co-authored several books, all of which have been published by Tata McGraw-Hill. He has also published several papers in national and international conferences. He is currently pursuing his PhD. His areas of interest include electromagnetics, high voltage applications, and power systems among others.

Lipika Datta received her BE (Hons) in Electrical Engineering from National Institute of Technology, Durgapur, and ME in Computer Science and Engineering from West Bengal University of Technology with specialisation in Embedded Systems. She joined College of Engineering & Management, Kolaghat as Lecturer in 2006. Prof. Datta is presently working as Assistant Professor in the Department of Computer Science and Engineering at College of Engineering & Management, Kolaghat. She is also pursuing her PhD at Calcutta University. Her areas of interest include embedded systems, operating systems, and computer architecture, among others.

Electromagnetic Field Theory

Shankar Prasad Ghosh Assistant Professor Department of Electrical Engineering College of Engineering & Management, Kolaghat Purba Medinipur, West Bengal

Lipika Datta Assistant Professor Department of Computer Science & Engineering and Information Technology College of Engineering & Management, Kolaghat, Purba Medinipur, West Bengal

Tata McGraw Hill Education Private Limited NEW DELHI McGraw-Hill Offices New Delhi New York St Louis San Francisco Auckland Bogotá Caracas Kuala Lumpur Lisbon London Madrid Mexico City Milan Montreal San Juan Santiago Singapore Sydney Tokyo Toronto

Published by Tata McGraw Hill Education Private Limited, 7 West Patel Nagar, New Delhi 110 008. Electromagnetic Field Theory Copyright © 2012 by Tata McGraw Hill Education Private Limited No part of this publication may be reproduced or distributed in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise or stored in a database or retrieval system without the prior written permission of the author. The program listings (if any) may be entered, stored and executed in a computer system, but they may not be reproduced for publication. This edition can be exported from India only by the publishers, Tata McGraw Hill Education Private Limited, ISBN (13): 978-1-25-900631-9 ISBN (10): 1-25-900631-X Vice President and Managing Director —McGraw-Hill Education: Ajay Shukla Head—Higher Education Publishing and Marketing: Vibha Mahajan Publishing Manager—SEM & Tech Ed.: Shalini Jha Editorial Executive: Koyel Ghosh Copy Editor: Preyoshi Kundu Sr. Production Manager: Satinder S Baveja Asst. Manager Production: Anjali Razdan Marketing Manager—Higher Ed.: Vijay Sarathi Sr. Product Specialist—SEM & Tech VOC.: Tina Jajoriya Graphic Designer—Cover: Meenu Raghav General Manager—Production: Rajender P Ghansela Production Manager: Reji Kumar Information contained in this work has been obtained by Tata McGraw-Hill, from sources believed to be reliable. However, neither Tata McGraw-Hill nor its authors guarantee the accuracy or completeness of any information published herein, and neither Tata McGraw-Hill nor its authors shall be responsible for any errors, omissions, or damages arising out of use of this information. This work is published with the understanding that Tata McGrawHill and its authors are supplying information but are not attempting to render engineering or other professional services. If such services are required, the assistance of an appropriate professional should be sought. Typeset at The Composers, 260, C.A. Apt., Paschim Vihar, New Delhi 110 063, and printed at Pushp Print Services, B-39/12 A, Gali No.-1, Arjun Mohalla, Moujpur, Delhi – 110 053 Cover Printer: SDR Printers RAXBCRCHDXDQY

CONTENTS

Preface

1

ix

Vector Analysis

1

Introduction 1 Scalar and Vector Quantities 1 Fields 1 Properties of Vectors 2 Vector Algebra 3 Coordinate Systems 12 General Curvilinear Coordinates 25 Differential Elements (Lengths, Areas and Volumes) in Different Coordinate Systems 28 1.9 Vector Calculus 30 1.10 Gauss’ Divergence Theorem 66 1.11 Stokes’ Theorem 76 1.12 Classifications of Vector Fields 80 1.13 Helmholtz Theorem 82 Summary 85 Important Formulae 89 Exercises 90 Review Questions 93 Multiple Choice Questions 94 Answers 95 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8

2

Electrostatics 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8

Introduction 96 Electric Charge 96 Coulomb’s Law 97 Principle of Superposition of Charges 98 Electric Field and Field Intensity (E ) 104 Different Charge Densities 109 Electric Fields due to Continuous Charge Distribution 110 Electric Flux (Displacement) (y ) and Flux Density (Displacement Density) ( D ) 123

96

vi Contents

2.9 Electric Field Lines (Lines of Force) and Flux Lines 124 2.10 Gauss’ Law 126 2.11 Applications of Gauss’ Law 129 2.12 Electric Potential (V) 139 2.13 Principle of Superposition of Electrostatic Fields 147 2.14 Potential Gradient and Relation between E and V 150 2.15 Equipotential Surfaces 152 2.16 Electric Dipole 153 2.17 Multipole (or Far Field) Expansion of Electric Potential 160 2.18 Energy Stored in Electrostatic Fields 161 2.19 Poisson’s and Laplace’s Equations 165 2.20 Uniqueness Theorem 168 2.21 General Procedure for Solving Poisson’s and Laplace’s Equations 169 2.22 Capacitor and Capacitance 176 2.23 Method of Images 188 2.24 Electric Boundary Conditions 201 2.25 Dirac Delta Representation in Electrostatic Fields 207 Summary 209 Important Formulae 213 Exercises 214 Review Questions 219 Multiple Choice Questions 222 Answers 227

3

Magnetostatics 3.1 Part 3.2 3.3 3.4 3.5 3.6 Part 3.7 3.8 3.9 3.10 3.11 3.12 3.13 3.14 3.15 3.16 3.17 3.18

Introduction 228 I: Behaviour of Different Electrical Materials in Electric Field 228 Electric Current and Current Density 228 Classification of Electrical Materials 240 Behaviour of Conductors in Electric Field 240 Behaviour of Dielectric Materials in Electric Field 258 Behaviour of Semiconductor Materials in Electric Field 266 II: Magnetostatic Field 266 Introduction 266 Biot–Savart Law 266 Magnetic Field Intensity (H ) 275 Magnetic Flux (f ) and Flux Density ( B ) 275 Gauss’ Law of Magnetostatic Interpretation of Divergence of Magnetic Field— Maxwell’s Equations 276 Ampere’s Circuital Law—Interpretation of Curl of Magnetic Field 278 Applications of Ampere’s Law 279 Magnetic Potentials 296 Forces due to Magnetic Fields 309 Magnetic Torque 315 Magnetic Dipole 317 Magnetic Materials 319

228

Contents

3.19 Magnetic Boundary Conditions 326 3.20 Self Inductance and Mutual Inductance 329 3.21 Magnetic Energy (Energy Stored in Magnetic Fields) Summary 345 Important Formulae 351 Exercises 352 Review Questions 356 Multiple Choice Questions 358 Answers 361

vii

341

4 Electromagnetic Fields

362

4.1 Introduction 362 4.2 Faraday’s Law of Induction for Time-Varying Fields 362 4.3 Induced EMF for Time Varying Fields 367 4.4 Inconsistency in Ampere’s Law for Time-Varying Fields 376 4.5 Maxwell’s Equations for Time-Varying Fields 377 4.6 Time-varying Potentials for EM Fields 386 Summary 390 Important Formulae 392 Exercises 392 Review Questions 393 Multiple Choice Questions 395 Answers 397

5

Propagation, Reflection, Refraction and Polarisation of Electromagnetic Waves Introduction 398 Three-Dimensional Wave Equations (Helmholtz Equation) 398 Properties of Electromagnetic Waves 403 Standing Electromagnetic Waves 408 Phase Velocity and Group Velocity of Electromagnetic Waves 411 Intrinsic Impedance 414 Propagation of Uniform Plane Waves through different Media 415 Polarisation 435 Reflection and Refraction of Plane Electromagnetic Waves at the Interface between Two Dielectrics 439 5.10 Reflection and Refraction of Plane Electromagnetic Waves by Perfect Conductor 453 5.11 Poynting Theorem and Poynting Vector 460 5.12 Energy Flux in a Plane Electromagnetic Wave 465 5.13 Radiation Pressure 479 Summary 480 Important Formulae 485 Exercises 485 Review Questions 487 Multiple Choice Questions 491 Answers 495 5.1 5.2 5.3 5.4 5.5 5.6 5.7 5.8 5.9

398

viii

Contents

6 Transmission Lines

496

6.1 Introduction 496 6.2 Types of Transmission Lines 496 6.3 Transmission Line Modes 496 6.4 Transmission Line Parameters 497 6.5 Transmission Line Equations (Telegrapher’s Equations) 498 6.6 Characteristic Impedance of Transmission Line 500 6.7 Input Impedance of Transmission Line 507 6.8 Reflection Coefficients of Transmission Line 515 6.9 Standing Waves and Standing Wave Ratio (S) of Transmission Line 517 6.10 Input Impedance as a Function of Position along a Transmission Line 521 6.11 Losses in Transmission Lines 528 6.12 Smith Chart 529 6.13 Load Matching Techniques in a Transmission Line 543 Summary 561 Important Formulae 563 Exercises 564 Review Questions 565 Multiple Choice Questions 567 Answers 568

7

Waveguides 7.1 Introduction 569 7.2 Parallel Planes Waveguides between Two Infinite Parallel Conducting Planes 7.3 Rectangular Waveguides 595 7.4 Circular or Cylindrical Waveguides 623 7.5 Power Transmission in Waveguides 637 7.6 Power Losses and Attenuation in Waveguides 643 7.7 Cavity Resonator or Resonant Cavities 656 7.8 Dielectric Slab Waveguides 665 7.9 Transmission Line Analogy for Waveguides 677 7.10 Applications of Waveguide 681 Summary 682 Exercises 686 Review Questions 688 Multiple Choice Questions 689 Answers 691 Typical Short Answer Type Questions with Answers Bibliography Index

569 570

692 722 725

PREFACE

Electromagnetic Field Theory is the study of characteristics of electric, magnetic and combined fields. The history of electromagnetism dates back over several thousand years, and in the twentyfirst century, its applications are far reaching. Electromagnetism manifests as both electric fields and magnetic fields. In fact, all the forces involved in interactions between atoms can be explained by electromagnetic force, together with how these particles carry momentum by their movement. Use of any electric or magnetic or electromagnetic device involves the presence of electromagnetism. Its increased usefulness in science and engineering has made it gain an important position in various areas of technology or physical research. Electromagnetic Field Theory is designed for undergraduate and postgraduate students studying electromagnetic field theory. The highlight of the book is its lucid and easy-to-understand language, with in-depth coverage of all the important topics sequentially which are supported by numerous illustrations. The book begins with a discussion on experimental laws and gradually synthesises them in the form of Maxwell’s equations in an adoptive approach. The latter part of the book takes the axiomatic way of presentation, starting with Maxwell’s equations, identifying each with the appropriate experimental law, and then specialising the general equations to static and time-varying situations for analysis. Being one of the extreme events of the human intellect, electromagnetic theory is, therefore, the foundation of the technologies of electrical and computer engineering. The study of electromagnetism and electromagnetic field theory thus becomes imperative for all branches of engineering dealing with electricity and electronics and related applications.

Salient Features of the Book Simple and lucid language In-depth discussion on vector algebra and coordinate systems to build strong fundamentals for the course Comprehensive coverage of topics like electric and magnetic fields, wave propagation, wave guides and antenna Exhaustive treatment of electrostatics and electromagnetic waves and their applications Complexities of subject overcome through easy explanation, illustrative examples and simplified derivations Excellent pedagogy with several hundreds of solved examples, review questions, exercises and multiple-choice-questions. Solved Examples: 300 Exercises: 157

x Preface

Multiple Choice Questions: 189 Review Questions: 130 Illustrative examples are interspersed throughout the book at appropriate locations. Most questions have been selected carefully from different university question papers and competitive examinations. With so many years of teaching experience, we have found that such illustrations permit a level of understanding otherwise unattainable. As an aid to both, the instructor and the student, multiple-choice questions, review questions and the exercise problems provided at the end of each chapter progress from easy to hard levels.

Structure of the Book The book is organized in seven chapters with the first chapter beginning with a discussion on vector which is the most basic requirement in the study of electromagnetism. In the subsequent chapters, the effects of non-varying electric charges, uniformly varying electric charges and varying electric charges with accelerating or decelerating velocity, have been discussed in detail. The concluding chapters cover some of the practical applications, such as transmission lines, waveguides. Chapter 1 gives an introduction to vector analysis with an emphasis on scalar and vector fields, properties of vectors, vector algebra, coordinate systems, general curvilinear coordinates, differential elements, vector calculus, Gauss’ divergence theorem, Stokes’ theorem and classifications of vector fields. Since the book has been written with a vector approach of the fundamental quantities, and the knowledge of vector is indispensable for the study of electromagnetism, this chapter serves as a foundation for the study of electromagnetic theory. Chapter 2 elaborates on the basic physics behind the interaction of static electric charges with the nature and associated phenomena. It explains electric fields and discusses electric charge, Coulomb’s law, superposition of charges and electric fields, electric flux displacement and flux density, electric potential, potential gradient, equipotential surfaces, electric dipoles, Poisson’s and Laplace’s equations, uniqueness theorem, capacitance, method of images, electric boundary conditions and Dirac–Delta representations among others. Chapter 3 provides the basis for understanding the physics behind the interaction of static magnetic charges, in the form of either magnets or moving electric charges, with the nature and associated phenomena. This chapter is divided into two parts. Part I deals with behavior of different electrical materials in an electric field and Part II deals with magnetostatic field. The Biot– Savart law and Ampere’s law are discussed in this part. Chapters 2 and 3 together cover the concept of electrostatics to the core. The following two chapters enlighten the concepts of electromagnetic fields and wave propagation. The functions of different practical electromagnetic devices and their accessories can be understood after completing these chapters. Both these chapters also open up the scope of different research activities related to electromagnetic wave phenomena. Chapter 4 discusses electromagnetic fields with special emphasis on time-varying fields. The topics taken up are Faraday’s law of induction, induced emf, and inconsistencies in Ampere’s law, Maxwell’s equations and potentials for electromagnetic fields. Chapter 5 is on electromagnetic wave propagation, reflection, refraction and polarisation of electromagnetic waves. It also covers Helmholtz equation, properties of electromagnetic waves, standing electromagnetic waves, intrinsic impedance, propagation of uniform plane waves through different media, Poynting theorem and Poynting vector.

Preface xi

Chapter 6 covers transmission lines, their types, modes, parameters, equations, characteristic and input impedance, and the Smith chart. The concluding chapter, Chapter 7, explains waveguides and their properties and types—rectangular, circular-power losses and attenuation, cavity resonator and resonant cavities, dielectric slab waveguides, transmission line analogy, and finally, the applications of waveguides. Both these chapters describe two different applications of electromagnetic field theory— Transmission lines and waveguides, which provide the base for understanding the functionaries of many modern electromagnetic devices. In addition, each chapter contains a summary and a list of important formulae for quick review. A large number of solved examples, short-answer-type question-answers, exercise problems, objectivetype questions and review questions taken from different university question papers and other competitive examinations have also been included. These features make it an indispensable source of study of electromagnetic theory for students and practicing engineers alike. The book will help them improve their problem solving capability and also will guide them prepare for different competitive examinations.

Web Supplements The following additional information is available at http://www.mhhe.com/ghosh/eft1 Chapter on Antennas and Wave Propagation Additional Exercises

Acknowledgements We are grateful to Prof. S N Bhadra, Professor, Department of Electrical Engineering, College of Engineering and Management, Kolaghat and Ex-Professor, IIT Kharagpur, Prof. C. K. Roy, Professor, Department of Electrical Engineering, College of Engineering and Management, Kolaghat and ExProfessor, Jadavpur University and Prof. A. K. Chakraborty, Department of Electrical Engineering, NIT Agartala for their constant inspiration and encouragement in the field of academics. We would also like to express our gratitude to the following reviewers for reviewing the book and providing constructive suggestions for the improvement of the book. Amit Kant Pandit Anirban Neogi Shishir K Balita P G Polgawande M. Karthikeyan C Murali Preeta Sharan Amit Kant Pandit Sanghamitra Chatterjee Abhijit Lahiri K T Mathew S C Raghavendra

Mata Vaishno Devi University, Jammu and Kashmir Bengal Institute of Technology, West Bengal Silicon Institute of Technology, Odisha Shri Shivaji Vidya Prasarak Sanstha’s Bapusaheb Shivaji Rao Deore College of Engineering, Maharashtra Paavai Engineering College, Tamilnadu Ramachandra College of Engineering, Andhra Pradesh Vemana Institute of Technology, Karnataka University College of Engineering, Rajasthan Technical University, Rajasthan Camellia Institute of Technology, Kolkata Calcutta Institute of Engineering and Management, Kolkata Cochin University of Science and Technology, Kerala People’s Education Society School of Engineering, Karnataka

xii

Preface

We are also thankful to the editorial and production staff of Tata McGraw Hill Education Private Limited for taking interest in publishing this edition. Last but not the least, we acknowledge the support offered by our family members, our daughter- Adrita, in particular, without whom this work would not have been successful. S P Ghosh L Datta

Feedback Criticism and suggestions for improvement shall be gratefully acknowledged. Readers may contact the authors at [email protected] and [email protected].

Publisher’s Note Don’t forget to write to us! We are always open to new ideas (the best ideas come from you!). You may send your comments to [email protected] (don’t forget to mention the title and author name in the subject line). Piracy-related issues may be reported as well!

Visual Walkthrough

xiii

VISUAL WALKTHROUGH Learning Objectives This chapter deals with the following topics: ■ ■ ■ ■ ■

Sources of electrostatics Basic laws of electrostatics To acquire knowledge of fundamental quantities of electrostatics Boundary conditions in electrostatics Concepts of capacitance

Learning Objectives offer an overview of chapter ideas. Each chapter opens with a list of objectives that will be discussed and explained throughout the chapter.

4.1 INTRODUCTION

Introduction at the beginning of each chapter lays a foundation for the topics that have been explained in detail in the succeeding pages.

In the previous chapters, we have studied different concepts of electrostatic and magnetostatic fields. In general, electrostatic fields are produced by stationary charges and magnetostatic fields are produced by motion of electric charges with uniform velocity (i.e. steady currents). However, if the current is time-varying, the field produced is also time-varying and is known as electromagnetic fields or waves. In this chapter, we will discuss the concepts of electromagnetic fields and contribution of Maxwell to the laws of electromagnetism.

4.2 FARADAY’S LAW OF INDUCTION FOR TIME-VARYING FIELDS English physicist Michael Faraday and American scientist Joseph Henry independently and simultaneously, in 1831, observed experimentally that any change in the magnetic environment of a coil of wire will cause a voltage (emf) to be induced in the coil. If the circuit is a closed one, this emf will cause flow of current. This phenomenon is known as electromagnetic induction. The results of Faraday and Henry’s experiment led to two laws: 1. Neumann’s Law: When a magnetic field linked with a coil or circuit is changed in any manner, the emf induced in the circuit is proportional to the rate of change of the flux-linkage with the circuit. 2. Lenz’s Law: The direction of the induced emf is such that it will oppose the change of flux producing it.

Example 4.1 An infinite straight wire carries a current I is placed to the left of a rectangular loop of wire with width w and length l, as shown in Fig. 4.2. (a) Determine the magnetic flux through the rectangular loop due to the current I. (b) Suppose that the current is a function of time with I(t) = a + bt, where a and b are positive constants. What is the induced emf in the loop and the direction of the induced current? Solution (a) Using Ampere’s law,

Fig. 4.2

Ú B ◊ dS = m0 I enc the magnetic field due to a current-carrying wire at a distance r away is B=

m0 I 2p r

Rectangular loop near a wire

Illustrated Examples have been carefully selected from various university question papers and competitive examinations. These examples will help students achieve a level of understanding which is not possible through theory alone.

xiv

Visual Walkthrough

NOTE Vector Representation of Surface Differential surface area is defined as a vector whose magnitude corresponds to the area of the surface and whose direction is perpendicular to the surface.

Notes offer additional information on the topics discussed in the book.

\

dS = dSan

For closed surface, the outward normal direction is taken as positive direction. For an open surface, the normal created by positive periphery according to the right-hand cork screw rule is taken to be positive. If the surface is not a plane, then the surface is subdivided into smaller elements that are considered to be plane and a vector surface is considered for each of these elemental surfaces. Vector addition of these elemental vector surfaces gives the total surface.

Summary Faraday’s law states that the emf induced in a closed circuit is proportional to the rate of change of the magnetic flux-linkage and the direction of the current flow in the closed circuit is such that it opposes the change of the flux. df E=dt Different forms of Faraday’s law are: —¥E =-

Differential Form: Integral Form:

∂B ∂t

Summary at the end of each chapter act as an overview of all the topics discussed in that particular chapter.

d Ú E ◊ d l = - dt Ú B ◊ dS C S

Induced emf for different cases are as follows: Stationary loop in time varying magnetic field (transformer emf) ES = - Ú S

∂B ◊ dS ∂t

Moving loop in static magnetic field (motional emf) E m = Ú Em ◊ d l = Ú (v ¥ B) ◊ d l C

C

Important Formulae Faraday’s law Integral form of Faraday’s law

E=-

df dt

d Ú E ◊ d l = - dt Ú B ◊ dS

C

Differential form of Faraday’s law

List of Important Formulae after Summary will help students review all the important formulae in a short period of time!

Transformer EMF

S

—¥E =ES = - Ú S

∂B ∂t

∂B ◊ dS ∂t

Motional EMF

E m = Ú (v ¥ B ) ◊ d l

EMF induced in Faraday disc generator

1 E m = w Ba 2 2

C

Modified Ampere’s circuital law Inhomogeneous wave equation

—¥H = J + —2V - me

∂D ∂t

r ∂ 2V ∂2 A = - and —2 A - me 2 = - m J e ∂t 2 ∂t

Visual Walkthrough

xv

Exercises [NOTE: * marked problems are important university problems] Easy 1. A conductor 1 cm in length is parallel to the z-axis and rotates at radius of 25 cm at 1200 r.p.m. [ 157.08 mV] Find the induced voltage, if the radial field is given by B = 0.5ar T. 2. A square conducting loop with sides 25 cm long is located in a magnetic field of 1 A/m varying at a frequency of 5 MHz. The field is perpendicular to the plane of the loop. What voltage will be read on a voltmeter connected in series with one side of the loop? [2.54 V] 3. A square loop of wire 25 cm has a voltmeter (of infinite impedance) connected in series with one side. Determine the voltage indicated by the meter when the loop is placed in an alternating field, the maximum intensity of which is 1 A/m. The plane of the loop is perpendicular to the magnetic field, the frequency is 10 MHz. [4.93 V]

Chapter-end Exercises have been divided into three levels as easy, medium and hard which will help students assess their level of understanding

Review Questions

Review Questions include theoretical questions to allow students to confirm their mastery of topics and concepts

[NOTE: * marked questions are important university questions.] 1. State and explain Faraday’s law of electromagnetic induction. ∂D ∂B . and (ii) — ¥ H = J + ∂t ∂t 3. Show that the electric field E induced by a time-varying magnetic field B is given by the ∂B . expression — ¥ E = ∂t

*2. From the fundamental principle, establish the relation (i) — ¥ E = -

Multiple Choice Questions 1. Maxwell’s equations in differential form from Ampere’s law are obtained from (a) M. M. F. area (b) Electric potential area (c) Magnetic flux volume (d) Electric current area. 2. Maxwell’s equations are not completely symmetrical because (a) isolated magnetic charges do not exist. (b) it is difficult to get curl of a vector in spherical coordinates. (c) — ◊ D is always zero. (d) — ¥ H does not exist in free space. 3. A circular disc of 1 m radius rotates at 150 radians per second anticlockwise on f-r plane with flux density B = 10 az tesla. The voltage induced between two stationary brushes connected at the centre and at the circumference will be (a) 0.15 V (b) 0.075 V (c) 1.0 V (d) 0.85 V 4. A conductor 4 m long lies along the y-axis with a current 10.0 A in the ay direction, the force on the conductor if the field in the region be B = 0.05ax T is az N (b) 2.0az N (c) 2.0ay N (d) 2.0ax N 5 A conducting rod of length L revolves about its mid point at uniform angular speed w in a uniform

Set of Multiple Choice Questions at the end of each chapter serves as an exercise for students for quick understanding and analysis

1 VECTOR ANALYSIS

Learning Objectives This chapter deals with the following topics: ■ ■

1.1

Vector algebra and calculus Different laws of vector

INTRODUCTION

Electromagnetics is the branch of physics or electrical engineering in which the electric and magnetic phenomena are studied. The basic knowledge for analysing the performance of any electrical network is the knowledge of circuit theory. However, the approach of circuit theory is a simplified approximation of a more exact field theory. Field theory is more difficult than circuit theory because of the larger number of variables involved. For most electromagnetic field problems, there are three space variables and thus, the solutions become complex. This can be overcome by the use of vector analysis. Thus, knowledge of vector analysis is an essential prerequisite to the study of electromagnetic field theory. The use of vector analysis in the study of electromagnetic field theory results in less time for solutions.

1.2

SCALAR AND VECTOR QUANTITIES

A quantity that has only magnitude is said to be a scalar quantity. Examples of scalar quantities are time, mass, distance, temperature, work, electric potential, etc. Scalar quantities are represented by italic letters, e.g., A, B, a, b, and F. A quantity that has both magnitude and direction is called a vector quantity. Examples of vector quantities are force, velocity, displacement, electric field intensity, etc. Vector quantities are represented by a letter with an arrow on the top, such as A and B or with a bold letter, such as F, a, B.

1.3

FIELDS

If at each point of a region, there is a value of some physical function, the region is called a field. Fields may be classified as scalar fields and vector fields.

2 Electromagnetic Field Theory

1.3.1

Scalar Fields

If the value of the physical function at each point is a scalar quantity, then the field is known as a scalar field. Some examples of scalar fields are: temperature distribution in a building, sound intensity in a theatre, height of the surface of the earth above sea level and electric potential in a region. A scalar field independent of time is called a stationary or steady-state scalar field.

1.3.2

Vector Fields

If the value of the physical function at each point is a vector quantity, then the field is known as a vector field. Some examples of vector fields are: gravitational force on a body in space, wind velocity in the atmosphere and the force on a charge body placed in an electric field. A time-independent vector is called a stationary vector field.

1.4

PROPERTIES OF VECTORS

We will consider the following essential properties that enable us to represent physical quantities as vectors: 1. Vectors can exist at any point in space. 2. Vectors have both the direction and the magnitude. 3. Any two vectors that have the same direction and magnitude are equal no matter where they are located in space; this is called vector equality. 4. Unit vector: A vector A has both magnitude and direction. The magnitude of A is a scalar written as A or | A |. A unit vector a A along A is defined as a vector whose magnitude is unity and its direction is along A. In general, any vector can be represented by its magnitude and its direction as follows. A = Aa A = | A | a A

(1.1)

where A or | A | represents the magnitude of the vector and a A direction of the vector A. This a A is called a unit vector. From Eq. (1.1), the unit vector is given as aA =

A A = A | A|

(1.2)

Note that, a A = 1 In Cartesian coordinates, the unit vectors along the three axes, e.g., x-axis, y-axis and z-axis, are represented as (ax , a y , az ) or as (i , j , k ). 5. Component vectors: Any vector A in Cartesian coordinates may be represented as (Ax, Ay, Az) or A = Ax ax + Ay a y + Az az where Ax, Ay, Az are called the component vectors in x, y and z directions, respectively. Using the Pythagorean Theorem, the magnitude of A is given as A = | A| =

Ax2 + Ay2 + Az2

(1.3)

Vector Analysis

3

The unit vector along A is given as aA =

Ax ax + Ay a y + Az az A A = = A | A| Ax2 + Ay2 + Az2

(1.4)

6. Vector decomposition: Choosing a coordinate system with an origin and axes, we can decompose any vector into component vectors along each coordinate axis. In Fig. 1.1, we choose Cartesian coordinates. A vector at P can be decomposed into the vector sum A = Ax + Ay + Az where, Ax is the x-component vector pointing in the positive or negative x-direction, and Ay is the y-component vector pointing in the positive or negative y-direction and Az is the z-component vector pointing in the positive or negative Fig. 1.1 Vector decomposition z-direction (Fig. 1.1). 7. Direction angles and direction cosines of a vector: The direction cosines of a vector are merely the cosines of the angles that the vector makes with the x, y, and z axes, respectively. We label these angles a (angle with the x-axis), b (angle with the y-axis), and g (angle with the z-axis). Given a vector (Ax, Ay, Az) in three-space, the direction cosines of this vector are given as l = cos a = m = cos b = n = cos g =

Ax Ax2

+ Ay2 + Az2 Ay

Ax2 + Ay2 + Az2 Az Ax2

+ Ay2 + Az2

(1.5a)

(1.5b) (1.5c)

Here, the direction angles a, b, g are the angles that the vector makes with the positive x-, y- and z-axes, respectively. In formulas, it is usually the direction cosines that occur, rather than the direction angles. We have cos 2 a + cos 2 b + cos 2 g = 1

Direction cosines define the orientation of a vector in three dimensions.

1.5

VECTOR ALGEBRA

We consider the addition, subtraction and multiplication of vectors.

1.5.1

Vector Addition and Subtraction

Let A and B be two vectors given as A = Ax ax + Ay a y + Az az B = Bx ax + B y a y + Bz az

(1.5d)

4 Electromagnetic Field Theory

We define a new vector, C = A + B , the vector addition of A and B and is given as C = ( A + B ) = ( Ax ax + Ay a y + Az az ) + ( Bx ax + B y a y + Bz a z ) = ( Ax + Bx )ax + ( Ay + B y )a y + ( Az + Bz )az Similarly, we define a new vector, D = A - B , the vector subtraction of A and B and is given as D = ( A - B) = ( Ax ax + Ay a y + Az az ) - ( Bx a x + B y a y + Bz a z ) = ( Ax - Bx )ax + ( Ay - By )a y + ( Az - Bz )az Physically, vector subtraction ( A - B) is the addition of vector A and vector B after reversing the direction of vector B. Graphically, vector addition and subtraction are obtained by the triangle or parallelogram rules as explained below.

Triangular Rule of Vector Addition An arrow is drawn that represents the vector A. The tail of the arrow that represents the vector B is placed at the tip of the arrow for A as shown in Fig. 1.2 (a). The arrow that starts at the tail of A and goes to the tip of B is defined to be the vector addition, C = A + B . This is the triangle rule of vector addition.

Fig. 1.2

(a) Vector addition triangle rule, (b) vector addition parallelogram rule, (c) vector subtraction triangle rule, and (d) vector subtraction parallelogram rule

Parallelogram Rule of Vector Addition The vectors A and B can be drawn with their tails at the same point. The two vectors form the sides of a parallelogram. The diagonal of the parallelogram corresponds to the vector C = A + B , as shown in Fig. 1.2 (b). This is the parallelogram rule of vector addition. Vector addition and subtraction satisfies the following properties: 1. Commutivity: The order of adding vectors does not matter. A+B=B+ A

Vector Analysis

5

2. Associativity: When adding three vectors, it does not matter which two we start with A + ( B + C ) = ( A + B) + C 3. Identity element for vector addition: There is a unique vector, 0, that acts as an identity element for vector addition. This means that for all vectors A, A+0=0+ A= A 4. Inverse element for vector addition: For every vector A, there is a unique inverse vector ( -1) A ∫ - A A + ( - A) = 0

such that

This means that the vector - A has the same magnitude as A, i.e., | A| = | - A| = A ; but they point in opposite directions. 5. Distributive law for vector addition: Vector addition satisfies a distributive law for multiplication by a number. Let c be a real number. Then c( A + B) = cA + cB

1.5.2

Vector Multiplication or Product

When two vectors A and B are multiplied, the result may be a scalar or a vector depending on how they are multiplied. There are two types of vector multiplication: 1. Scalar or Dot Product, and 2. Vector or Cross Product.

1.5.3

Scalar or Dot Product ( A · B )

Definition The scalar or dot product of two vectors A and B, written as A ◊ B, is defined as (Fig. 1.3) A ◊ B = AB cos q AB

(1.6)

where qAB is the smaller angle between A and B, A = | A| and B = | B | represent the magnitude of A and B , respectively.

Fig. 1.3

Dot product of two vectors

The dot product can be positive, zero, or negative, depending on the value of cos qAB. The result of A ◊ B is always a scalar quantity. Dot product is geometrically defined as the product of magnitude of B and the projection of A onto B or vice versa. This is illustrated in Fig. 1.4 (a) and (b).

6 Electromagnetic Field Theory

Fig. 1.4

Projection of vectors and the dot product

Properties of Dot Product 1. The first property is that the dot product is commutative A◊ B = B ◊ A 2. The second property involves the dot product between a vector cA which is a scalar and a vector B cA ◊ B = c( A ◊ B) 3. The third property involves the dot product between the sum of two vectors A and B with a vector C ( A + B) ◊ C = A ◊ C + B ◊ C This shows that the dot product is distributive. 4. Since the dot product is commutative, similar relations are given, e.g., A ◊ cB = c( A ◊ B ) C ◊ ( A + B) = C ◊ A + C ◊ B

Vector Decomposition and Dot Product We now develop an algebraic expression for the dot product in terms of components. We choose a Cartesian coordinate system with the two vectors having component vector as A = Ax ax + Ay a y + Az az B = Bx ax + B y a y + Bz az \

A ◊ B = ( Ax ax + Ay a y + Az az ) ◊ ( Bx ax + By a y + Bz az ) = Ax Bx (ax ◊ ax ) + Ay By (a y ◊ a y ) + Az Bz (az ◊ az ) + Ax B y (ax ◊ a y ) + Ax Bz (ax ◊ az ) + Ay Bx (a y ◊ ax ) + Ay Bz (a y ◊ az ) + Az Bx (a z ◊ ax ) + Az B y (a z ◊ a y )

Now, ax ◊ ax = a y ◊ a y = az ◊ az = 1 cos(0∞) = 1 and ax ◊ a y = a y ◊ az = az ◊ ax = a y ◊ ax = az ◊ a y = ax ◊ az = 1 cos (90∞) = 0 Hence, we get A ◊ B = ( Ax Bx + Ay By + Az Bz )

(1.7)

Vector Analysis

7

Application of Dot Product 1. The dot product is to find the work done by a force F for a displacement of D, given as W = F ◊ D = FD cos q 2. It is used to find out the line integral of a vector over a path, e.g., to calculate the electric potential between two points in an electric field. Fig. 1.5 Work done by a force 3. It is used to find out the surface integral over a surface, e.g., to calculate the total charge enclosed by a surface placed in an electric field.

Example 1.1

Given the two vectors A = - 7 ax + 12a y + 3az and B = 4ax - 2a y + 16az

Find the dot product and the angle between the two vectors.

Solution The dot product between the two vectors is given as A ◊ B = ( Ax Bx + Ay By + Az Bz ) = ( - 7) ¥ 4 + 12 ¥ ( - 2) + 3 ¥ 16 = - 4 Since the dot product is negative, it is expected that the angle between the two vectors will be greater than 90°. Here, | A| = | B| =

( - 7)2 + 122 + 32

=

202

42 + ( - 2) 2 + 162 =

276

Hence, the angle between the two vectors is given as, Ê A◊ B ˆ Ê q = cos -1 Á = cos -1 ˜ Á Ë | A| | B | ¯ Ë

1.5.4

-4 ˆ = 90.97∞ ˜ 202 ¥ 276 ¯

Vector or Cross Product ( A × B )

Definition The cross product of two vectors A and B , written as ( A ¥ B) , is defined as A ¥ B = AB sin q AB an

(1.8)

where an is the unit vector normal to the plane containing A and B The vector multiplication is called cross product due to the cross sign. It is also termed as vector product because the result is a vector. The direction of the cross product is obtained from a common rule, called right-hand rule.

Right-hand Rule for Direction of Cross Product

The direction of an is taken as the direction of right thumb when the fingers of the right hand rotate from A to B [Fig. 1.6 (a)]. Alternatively, the direction of an is taken as that of the advance of right-handed screw as A is turned into B [Fig. 1.6 (b)].

8 Electromagnetic Field Theory

Fig. 1.6

(a) Right-hand rule, and (b) Right-hand cork-screw rule

We can give a geometric interpretation to the magnitude of the cross product by writing the definition as | A ¥ B | = A( B sin q ) The vectors A and B form a parallelogram. The area of the parallelogram equals the height times the base, which is the magnitude of the cross product. In Fig. 1.7, two different representations of the height and base of a parallelogram are illustrated. As depicted in Fig. 1.7 (a), the term B sin q is the projection of the vector B in the direction perpendicular to the vector A.

Fig. 1.7

Projection of vectors and the cross product

We could also write the magnitude of the cross product as | A ¥ B | = ( A sin q ) B Now the term A sin q is the projection of the vector A in the direction perpendicular to the vector B as shown in Fig. 1.7(b).

Properties of Cross Product 1. The cross product is anti-commutative since changing the order of the vector’s cross product changes the direction of the cross product vector by the right-hand rule: A¥ B = -B ¥ A 2. The cross product between a vector cA where c is a scalar and a vector B is cA ¥ B = c( A ¥ B ) A ¥ cB = c( A ¥ B) Similarly, 3. The cross product is not associative. A ¥ ( B ¥ C ) π ( A ¥ B) ¥ C

Vector Analysis

9

4. The cross product between the sum of two vectors A and B with a vector C is, ( A + B) ¥ C = A ¥ C + B ¥ C A ¥ (B + C) = A ¥ B + A ¥ C

Similarly,

This shows that the cross product is distributive.

Vector Decomposition and Cross Product

We now develop an algebraic expression for the cross product in terms of components. We choose a Cartesian coordinate system with the two vectors having component vector as A = Ax ax + Ay a y + Az az B = Bx ax + B y a y + Bz az \

A ¥ B = ( Ax ax + Ay a y + Az az ) ¥ ( Bx ax + B y a y + Bz az ) = Ax Bx (ax ¥ ax ) + Ay B y (a y ¥ a y ) + Az Bz (az ¥ az ) + Ax By (ax ¥ a y ) + Ax Bz (ax ¥ az ) + Ay Bx (a y ¥ ax ) + Ay Bz (a y ¥ az ) + Az Bx (az ¥ ax ) + Az By (az ¥ a y )

Now, for right-handed coordinate system ax ¥ ax = a y ¥ a y = az ¥ az = 0 and, ax ¥ a y = a z = - a y ¥ a x

a y ¥ az = ax = - az ¥ a y

az ¥ ax = a y = - ax ¥ az

This is illustrated in Fig. 1.8 (a) and (b).

Fig. 1.8 results

(a) Moving clockwise leads to positive results, and (b) Moving counterclockwise leads to negative

\

A ¥ B = ( Ay Bz - Az By ) ax + ( Az Bx - Ax Bz ) a y + ( Ax By - Ay Bx ) az ax

ay

az

= Ax

Ay

Az

Bx

By

Bz

Hence, we get ax

ay

az

A ¥ B = Ax

Ay

Az

Bx

By

Bz

(1.9)

10

Electromagnetic Field Theory

Example 1.2

Given the two vectors A = 8ax + 3a y - 10az and B = -15ax + 6a y + 17 az

Find the cross product between the two vectors and the unit vector normal to the plane containing the two vectors A and B.

Solution The cross product between the two vectors is given as ax

ay

az

A ¥ B = Ax

Ay

Az = ( Ay Bz - Az By ) ax + ( Az Bx - Ax Bz ) a y + ( Ax B y - Ay Bx ) az

Bx

By

Bz

= [3 ¥ 17 - ( -10) ¥ 6] ax + [( -10) ¥ ( -15) - 8 ¥ 17] a y + [8 ¥ 6 - 3 ¥ ( -15)] az = 111ax + 14a y + 93az The unit vector normal to the plane containing the vectors A and B is given as an =

111ax + 14a y + 93az

A¥ B = | A| | B |

82 + 32 + ( -10) 2 = 0.36ax + 0.04a y + 0.30az

( -15) 2 + 62 + 17 2

=

111ax + 14a y + 93az 308.46

Two vectors are represented by A = 2i + 2 j , B = 3i + 4 j - 2k . Find the dot and cross-products and the angle between the vectors. Show that A ¥ B is at right angle to A.

Example 1.3

Solution Here, A = 2i + 2 j , B = 3i + 4 j - 2k \

A ◊ B = 2 ¥ 3 + 2 ¥ 4 + 0 ¥ ( - 2) = 14

\

Now,

ax

ay

az

ax

ay

az

A ¥ B = Ax

Ay

Bx

By

Az = 2 3 Bz

2 4

0 = - 4a x + 4a y + 2a z -2

B=

32 + 42 + ( - 2) 2 = 5.39

A=

22 + 22 + 0 = 2.83;

A ◊ B = AB cos q fi

Ê A◊ Bˆ 14 Ê ˆ = 23.2∞ q = cos -1 ÁË ˜ = cos -1 Á AB ¯ Ë 2.83 ¥ 5.39 ˜¯

For A ¥ B to be at right angle to A, ( A ¥ B) ◊ A should be zero. ( A ¥ B) ◊ A = 2 ¥ ( - 4) + 2 ¥ 4 + 0 ¥ 2 = 0

(Proved)

Applications of Cross Product 1. To find the torque about a point P which can be described mathematically by the cross product of a vector from P to where the force acts, and the force vector. 2. To find the force experienced by a current carrying conductor placed in a magnetic field.

Vector Analysis

1.5.5

11

Triple Products

Multiplication of three vectors A, B and C is called vector triple product. The product of three vectors is classified into two categories: 1. Scalar triple product, and 2. Vector triple product.

Scalar Triple Product

For the three vectors A, B and C, scalar triple product is defined as A ◊ ( B ¥ C ) = B ◊ (C ¥ A) = C ◊ ( A ¥ B)

Since the result is a scalar quantity, this is known as scalar triple product. If the components of three vectors A, B and C are given as A = ( Ax , Ay , Az ), B = ( Bx , B y , Bz ), C = (C x , C y , C z ) , respectively, then the scalar triple product is obtained by the determinant of a 3 ¥ 3 matrix given as

Bx

By

Bz

Similarly, B ◊ (C ¥ A) = C x

Cy

Cz

Ax

Ay

Az

Cx

Cy

Cz

and C ◊ ( A ¥ B) = Ax

Ay

Az

Bx

By

Bz

Ax

Ay

Az

A ◊ ( B ¥ C ) = Bx

By

Bz

Cx

Cy

Cz

We know from a determinant theorem that if any two columns or rows of a determinant are interchanged, its value remains the same, but the sign changes. Hence, we can write Bx

By

Bz

Bx

By

Bz

Ax

Ay

Az

Ax

Ay

Az

B ◊ (C ¥ A) = C x

Cy

C z = - Ax

Ay

Az = ( -) ( -) Bx

By

Bz = Bx

By

Bz

Ax

Ay

Az

Cx

Cy

Cz

Cx

Cy

Cz

Cx

Cy

Cz

Cx

Cy

Cz

Bx

By

Bz

Ax

Ay

Az

Ax

Ay

Az

C ◊ ( A ¥ B) = Ax

Ay

Az = - Ax

Ay

Az = ( -) ( -) Bx

By

Bz = Bx

By

Bz

Bx

By

Bz

Cy

Cz

Cy

Cz

Cy

Cz

Similarly,

Cx

Cx

Cx

Therefore, we can write that A ◊ ( B ¥ C ) = B ◊ (C ¥ A) = C ◊ ( A ¥ B )

(1.10)

12

Electromagnetic Field Theory

Example 1.4

If the components of three vectors

A, B and C are given as, A = ( Ax , Ay , Az ), B = ( Bx , By , Bz ), C = (C x , C y , C z ) respectively, then show that A ◊ ( B ¥ C ) is the volume of a parallelepiped of sides A, B and C .

Fig. 1.9

Scalar triple product

Solution Here, B ¥ C = BC sin q an = Area of the rectangle with side B and C and perpendicular to the plane containing the vectors B and C = Pan = P \

A ◊ ( B ¥ C ) = A ◊ P = AP cos f = Product of the component of A normal to BC -plane and the area of the parallelogram with sides B and C = Volume of the parallelepiped of sides A, B and C

Vector Triple Product

For the three vectors A, B and C, vector triple product is defined as A ¥ ( B ¥ C ) = B( A ◊ C ) - C ( A ◊ B)

(1.11)

Since the result is a vector quantity, this is known as vector triple product. It may be noted that A ¥ ( B ¥ C ) is in the plane containing B and C and is perpendicular to A. Associative law does not hold good for vector triple product, i.e., A ¥ ( B ¥ C ) π ( A ¥ B) ¥ C Rather, ( A ¥ B) ¥ C = - C ¥ ( A ¥ B) = - [ A(C ◊ B) - B(C ◊ A)] = B(C ◊ A) - A(C ◊ B) The concept of vector triple product is used in deriving wave equations from Maxwell’s equations.

1.6

COORDINATE SYSTEMS

Coordinate system is defined as a system to describe uniquely the spatial variation of a quantity at all points in space. All coordinate systems can be broadly classified into two categories: 1. Orthogonal coordinate systems: Three coordinate axes are perpendicular to each other. 2. Non-orthogonal coordinate systems: Coordinate axes are not perpendicular to each other. From another point of view, the coordinate systems are of two types: 1. Right-handed coordinate systems: These systems follow the right-hand cork-screw rule. This means that if one rotates from the first coordinate axis towards the second coordinate axis, a right-hand screw will advance in the positive direction of the third coordinate axis.

Vector Analysis

13

2. Left-handed coordinate systems: These systems follow the opposite movement of a right-handed screw. Here, we will discuss the most useful three right-handed orthogonal coordinate systems; namely, 1. Cartesian or rectangular coordinates, 2. Circular or cylindrical coordinates, and 3. Spherical coordinates.

1.6.1

Cartesian or Rectangular Coordinates (x, y, z)

A point P in Cartesian coordinates is represented as P(x, y, z). The ranges of coordinate variables are x y z

(1.12)

From Fig. 1.10 (b), it is understood that any point in rectangular coordinates is the intersection of three planes (i) constant x-plane (ii) constant y-plane and (iii) constant z-plane, which are mutually perpendicular.

Fig. 1.10

(a) Cartesian coordinates, and (b) Constant x, y, z planes

A vector A in Cartesian coordinate system is written as A = Ax ax + Ay a y + Az az where, ax , a y , az are the unit vectors along the x, y and z directions, respectively. From the definitions of dot product, we see that ax ◊ ax = a y ◊ a y = az ◊ az = 1 ax ◊ a y = a y ◊ az = az ◊ ax = 0

(1.13)

(1.14)

From the definitions of cross product, we see that ax ¥ ax = a y ¥ a y = az ¥ az = 0 ax ¥ a y = az ; a y ¥ az = ax ; az ¥ ax = a y

(1.15)

14

Electromagnetic Field Theory

1.6.2

Cylindrical or Circular Coordinates (r, f, z)

A point P in cylindrical coordinates is represented as P(r, f, z). Here r = radius of the cylinder passing through P = radial distance from the z-axis f = angle measured from the x-axis in the xy-plane, known as azimuthal angle z = same as in Cartesian coordinates The ranges of coordinate variables are 0 r 0 f

2p

(1.16)

z From Fig. 1.11 (b), it is understood that any point in cylindrical coordinates is an intersection of three planes viz. (i) constant ‘r’ plane (a circular cylinder) (ii) constant f plane (semi-infinite plane with its edge along the z-axis (iii) constant z-plane (parallel to xy-plane).

Fig. 1.11

(a) Cylindrical coordinates, and (b) Constant r, f, z planes

A vector A in cylindrical coordinate system is written as A = Ar ar + Af af + Az az

(1.17)

where ar , af , az are the unit vectors along the r, f and z directions, respectively. From the definitions of dot product, we see that ar ◊ ar = af ◊ af = az ◊ az = 1 ar ◊ af = af ◊ az = az ◊ ar = 0

(1.18)

From the definitions of cross product, we see that ar ¥ ar = af ¥ af = az ¥ az = 0 ar ¥ af = az ; af ¥ az = ar ; az ¥ ar = af

(1.19)

Vector Analysis

15

Relations between Cartesian (x, y, z) and Cylindrical (r, f, z) Coordinates

The relationship between Cartesian (x, y, z) and cylindrical (r, f, z) coordinates are obtained from Fig. 1.11(a) and are written as r=

x2 + y 2

Ê yˆ f = tan -1 Á ˜ , Ë x¯

x = r cos f

y = r sin f

z=z

(1.20)

and z=z

(1.21)

The relationships between the unit vectors are obtained from Fig. 1.12 and are given as

Fig. 1.12

Unit vector transformation between Cartesian and cylindrical coordinates

ax = cos f ar - sin f af a y = sin f ar + cos f af az = az

(1.22)

ar = cos f ax + sin f a y af = - sin f ax + cos f a y az = az

(1.23)

and

The relationships between the component vectors (Ax, Ay, Az) and (Ar, Af, Az) are obtained by using Eqs. (1.22) and (1.23) and then rearranging the terms. This is given as A = ( Ax cos f + Ay sin f ) ar + ( - Ax sin f + Ay cos f ) af + Az az = ( Ar cos f - Af sin f ) ax + ( Ar sin f + Af cos f ) a y + Az az

(1.24)

Thus, the relationships between the component vectors can be written in matrix forms as È Ar ˘ È cos f sin f 0 ˘ È Ax ˘ Í ˙ Í ˙Í ˙ Í Af ˙ = Í - sin f cos f 0 ˙ Í Ay ˙ ÍA ˙ Í 0 0 1 ˙˚ ÍÎ Az ˙˚ Î z˚ Î

and

È Ax ˘ È cos f Í ˙ Í Í Ay ˙ = Í sin f ÍA ˙ Í 0 Î z˚ Î

- sin f 0 ˘ È Ar ˘ Í ˙ cos f 0 ˙˙ Í Af ˙ 0 1 ˙˚ ÍÎ Az ˙˚

(1.25)

16

Electromagnetic Field Theory

1.6.3

Spherical or Polar Coordinates (r, q, f)

A point P in spherical coordinates is represented as P(r, q, f). Here, r = distance of the point from the origin = radius of a sphere centered at the origin and passing through the point P, q = angle between the z-axis and the position vector P, known as colatitudes, and f = angle measured from the x-axis in the xy-plane, known as azimuthal angle (same as in cylindrical coordinates). The ranges of coordinate variables are 0 r 0 q p 0 f 2p

(1.26)

From Fig. 1.13 (b), it is understood that any point in spherical coordinates is an intersection of three planes, viz. (i) constant ‘r’ plane (a sphere with its centre at the origin), (ii) constant q-plane (circular cone with z-axis as its axis and the origin at its vertex), and (iii) constant f-plane (semi-infinite plane as in cylindrical coordinates).

Fig. 1.13 (a) Spherical coordinates, (b) Constant r, q, f planes, and (c) Point P and unit vectors in spherical coordinates

Vector Analysis

17

A vector A in spherical coordinate system is written as A = Ar ar + Aq aq + Af af

(1.27)

where ar , aq , af are the unit vectors along the r, q and f directions, respectively. From the definitions of dot product, we see that ar ◊ ar = aq ◊ aq = af ◊ af = 1 ar ◊ aq = aq ◊ af = af ◊ ar = 0

(1.28)

From the definitions of cross product, we see that ar ¥ ar = aq ¥ aq = af ¥ af = 0 ar ¥ aq = af ; aq ¥ af = ar ; af ¥ ar = aq

(1.29)

Relations between Cartesian (x, y, z) and Spherical (r, q, f) Coordinates

The relationships between Cartesian (x, y, z) and spherical (r, q, f) coordinates can also be obtained from Fig. 1.13 (a) and can be written as r=

x2 + y 2 + z 2

Ê q = tan -1 ÁË

x2 + y 2 ˆ ˜¯ , z

Ê yˆ f = tan -1 Ë ¯ x

(1.30)

and x = r sin q cos f

y = r sin q sin f

z = r cos q

(1.31)

The relationships between the unit vectors are obtained from Fig. 1.14 and are given as ax = sin q cos f ar + cos q cos f aq - sin f af a y = sin q sin f ar + cos q sin f aq + cos f af az = cos q ar - sin q aq

(1.32)

ar = sin q cos f ax + sin q sin f a y + cos q az aq = cos q cos f ax + cos q sin f a y - sin q az af = - sin f ax + cos f a y

(1.33)

and

The relationships between the component vectors (Ax, Ay, Az) and (Ar, Aq, Af) can be obtained by using Eqs. (1.32) and (1.33) and then rearranging the terms. This is written in matrix form as È Ar ˘ È sin q cos f sin q sin f Í ˙ Í Í Aq ˙ = Í - cos q cos f cos q sin f Í A ˙ Í - sin f cos f Î f˚ Î

Fig. 1.14

Unit vector transformation for Cartesian and spherical coordinates

cos q ˘ È Ax ˘ Í ˙ - sin q ˙˙ Í Ay ˙ 0 ˙˚ ÍÎ Az ˙˚

(1.34)

18

Electromagnetic Field Theory

and È Ax ˘ Èsin q cos f cos q cos f Í ˙ Í Í Ay ˙ = Í sin q sin f cos q sin f Í ˙ Í - sin q Î Az ˚ Î cos q

- sin f ˘ È Ar ˘ Í ˙ cos f ˙˙ Í Aq ˙ 0 ˚˙ ÎÍ Af ˚˙

(1.35)

Relations between Cylindrical (r, f, z) and Spherical (r, q, f) Coordinates

The relationships between cylindrical (r, f, z) and spherical (r, q, f) coordinates are obtained from Fig. 1.13 (a) and are written as r=

r2 + z2

r q = tan -1 Ê ˆ , Ë z¯

f =f

(1.36)

and r = r sin q

f =f

z = r cos q

(1.37)

The relationships between the unit vectors are obtained from Fig. 1.15 and are given as ar = sin q ar + cos q az aq = cos q cos f ar - sin q az af = af

(1.38)

ar = sin q ar + cos q cos f aq af = af az = cos q ar - sin q aq

(1.39)

and

Fig. 1.15

The relationships between the component vectors (Ar, Aq, Af and (Ar, Af, Az) can be obtained by using Eqs. (1.38) and (1.39) and then rearranging the terms. This is written in matrix form as

Unit vector transformation for cylindrical and spherical coordinates

È Ar ˘ È sin q Í ˙ Í Í Aq ˙ = Í cos q ÍA ˙ Í 0 Î f˚ Î

0 cos q ˘ È Ar ˘ Í ˙ 0 - sin q ˙˙ Í Af ˙ 1 0 ˙˚ ÍÎ Az ˙˚

(1.40)

È Ar ˘ È sin q Í ˙ Í Í Af ˙ = Í 0 Í A ˙ Í cos q Î z˚ Î

cos q 0 - sin q

0 ˘ È Ar ˘ Í ˙ 1 ˙˙ Í Aq ˙ 0 ˙˚ ÍÎ Af ˙˚

(1.41)

and

Vector Analysis

*Example 1.5

Convert the point P(1, 3, 5) from Cartesian to cylindrical and spherical

coordinates.

Solution At point P: x = 1, y = 3, z = 5 Conversion from Cartesian to cylindrical coordinates: Hence, r=

x 2 + y 2 = 12 + 32 = 10 = 3.162

()

Ê yˆ 3 = 71.565∞ f = tan -1 Ë ¯ = tan -1 x 1 z =5 Conversion from Cartesian to spherical coordinates: Here, r=

x 2 + y 2 + z 2 = 12 + 32 + 52 =

Ê q = tan -1 ÁË

35 = 5.916

Ê 12 + 32 ˆ x2 + y 2 ˆ ˜¯ = tan -1 ÁË ˜¯ = 32.311∞ z 5

()

Ê yˆ 3 = 71.565∞ f = tan -1 Ë ¯ = tan -1 x 1 Therefore, the point P is written as P (1, 3, 5) = P(3.162, 71.565∞, 5) = P(5.916, 32.311∞, 71.565∞)

*Example 1.6

19

Transform the vector x2 + y 2

A=

x +y +z 2

2

2

ax -

yz x + y2 + z2 2

az

to cylindrical and spherical coordinates.

Solution For A, the components are given as Ax =

x2 + y 2 x +y +z 2

2

2

Ay = 0

Az = -

yz x + y2 + z2 2

Transformation from Cartesian to cylindrical coordinates: From Eq. (1.25), È x2 + y 2 Í È Ar ˘ È cos f sin f 0 ˘ È Ax ˘ È cos f sin f 0 ˘ Í x 2 + y 2 + z 2 Í ˙ Í ˙Í ˙ Í ˙Í 0 Í Af ˙ = Í - sin f cos f 0 ˙ Í Ay ˙ = Í - sin f cos f 0 ˙ Í Í ÍA ˙ Í 0 Í ˙ yz 0 1 ˙˚ Î Az ˚ ÍÎ 0 0 1 ˙˚ Í Î z˚ Î 2 Í x + y2 + z2 Î

˘ ˙ ˙ ˙ ˙ ˙ ˙ ˙ ˚

20

Electromagnetic Field Theory

\

x2 + y 2

Ar = Af = Az = -

But,

x = r cos f

y = r sin f

cos f

x2 + y 2 + z 2 x2 + y 2

x2 + y 2 + z 2 yz

sin f

x2 + y 2 + z 2

\ x2 + y 2 = r

Substituting these relations, we get \

r cos f

Ar = Af = Az = -

r2 + z2 r sin f r2 + z2 rz sin f r2 + z2

Hence, the vector is expressed in cylindrical coordinates as A=

r cos f r +z 2

2

ar -

r sin f r +z 2

A=

2

af -

rz sin f

r r + z2 2

r +z 2

2

az =

r r + z2 2

(cos f ar - sin f af - z sin f az )

(cos f ar - sin f af - z sin f az )

Transformation from Cartesian to spherical coordinates: From Eq. (1.34), È Ar ˘ È sin q cos f sin q sin f Í ˙ Í Í Aq ˙ = Í cos q cos f cos q sin f Í A ˙ Í - sin f cos f Î f˚ Î È sin q cos f sin q sin f = ÍÍ cos q cos f cos q sin f ÍÎ - sin f cos f

cos q ˘ È Ax ˘ Í ˙ - sin q ˙˙ Í Ay ˙ 0 ˙˚ ÍÎ Az ˙˚ È x2 + y 2 Í 2 2 2 cos q ˘ Í x + y + z Í ˙ - sin q ˙ Í 0 Í yz 0 ˙˚ Í Í x2 + y 2 + z 2 Î

˘ ˙ ˙ ˙ ˙ ˙ ˙ ˙ ˚

Vector Analysis

Ar =

\

Aq =

x2 + y 2 x + y +z 2

2

x2 + y 2 x +y +z

Af = But, x = r sin q cos f

2

2

2

2

sin q cos f cos q cos f +

x2 + y 2 x2 + y 2 + z 2

y = r sin q sin f

\

yz x + y2 + z2 2

yz x + y2 + z2 2

cos q

sin q

sin f z = r cos q

x 2 + y 2 = r sin q

and

x2 + y 2 + z 2 = r

Substituting these relations, we get r sin 2 q cos f r sin q sin fr cos 2 q r r 2 2 = (sin q cos f - r sin q cos q sin f ) r sin q cos q cos f r sin 2 q sin fr cos q Aq = + r r = (sin q cos q cos f + r sin 2 q cos q sin f ) r sin q sin f Af = = - sin q sin f r

\

Ar =

Hence, the vector is expressed in spherical coordinates as A = (sin 2 q cos f - r sin q cos 2 q sin f ) ar

(

)

+ sin q cos q cos f + r sin 2 q cos q sin f aq - sin q sin f af = sin q (sin q cos f - r cos q sin f ) ar + sin q cos q (cos f + r sin q sin f ) aq - sin q sin f af 2

A = sin q (sin q cos f - r cos 2 q sin f ) ar + sin q cos q (cos f + r sin q sin f ) aq - sin q sin f af

Example 1.7

Express the following vectors in Cartesian coordinates:

(a) A = rz sin f ar + 3r cos f af + r cos f sin f az (b) B = r 2 ar + sin q af

Solution (a) A = rz sin f ar + 3r cos f af + r cos f sin f az Here, Ar = rz sin f, Af = 3r cos f, Az = r cos f sin f È Ax ˘ È cos f Í ˙ Í Í Ay ˙ = Í sin f ÍA ˙ Í 0 Î z˚ Î

- sin f 0 ˘ È Ar ˘ È cos f Í ˙ cos f 0 ˙˙ Í Af ˙ = ÍÍ sin f 0 1 ˚˙ ÎÍ Az ˚˙ ÎÍ 0

- sin f 0 ˘ È rz sin f ˘ cos f 0 ˙˙ ÍÍ 3r cos f ˙˙ 0 1 ˚˙ ÎÍ r cos f sin f ˚˙

21

22

Electromagnetic Field Theory

Ax = rz sin f cos f – 3r sin f cos f Ay = rz sin2 f + 3r cos2 f Az = r cos f sin f But r =

x

x 2 + y 2 , cos f =

x +y 2

2

y

, sin f =

x + y2 2

Substituting these values Ax =

x2 + y 2 z

xy xy - 3 x2 + y 2 2 = x2 + y 2 x + y2

Ay =

x2 + y 2 z

y2 x2 + 3 x2 + y 2 2 = 2 x +y x + y2

Az =

x2 + y 2

xyz x +y 2

y2 z

2

xy = x + y2

2

x2 + y 2

3 xy

-

x2 + y 2 3x 2

+

x2 + y 2

xy

2

x + y2 2

Hence, the vector in Cartesian coordinates is written as 1 A= [( xyz - 3 xy ) ax + ( y 2 z + 3 x 2 ) a y + xyaz ] 2 2 x +y (b) B = r 2 ar + sin q af Here, Br = r2, Bq = 0, Bf = sin q È Bx ˘ Èsin q cos f cos q cos f Í ˙ Í Í By ˙ = Í sin q sin f cos q sin f Í ˙ Í - sin q Î Bz ˚ Î cos q

- sin f ˘ È r 2 ˘ Í ˙ cos f ˙˙ Í 0 ˙ 0 ˚˙ ÎÍsin q ˚˙

Bx = r 2 sin q cos f - sin q sin f By = r 2 sin q sin f + sin q cos f Bz = r 2 cos q But r=

x 2 + y 2 + z 2 , sin q =

sin f =

y x + y2 2

, cos f =

x2 + y 2 x +y +z x 2

2

, cos q =

2

z x + y2 + z2 2

x2 + y 2

Substituting these values Bx = ( x 2 + y 2 + z 2)

x2 + y 2

x

x +y +z x +y y = x x2 + y 2 + z 2 2 x + y2 + z2 1 [ x( x 2 + y 2 + z 2 ) - y ] = 2 2 2 x +y +z 2

2

2

2

2

-

x2 + y 2 x +y +z 2

2

y 2

x + y2 2

,

Vector Analysis

x2 + y 2

By = ( x 2 + y 2 + z 2 )

y

+

x2 + y 2

x +y +z x +y x +y +z x = y x2 + y 2 + z 2 + x2 + y 2 + z 2 1 [ y ( x 2 + y 2 + z 2 ) + x] = 2 2 2 x +y +z z 2 Bz = ( x + y 2 + z 2 ) = z x2 + y 2 + z 2 2 2 2 x +y +z 2

2

2

2

2

2

2

23

x 2

x + y2 2

Hence, the vector in Cartesian coordinates is written as B=

1 x +y +z 2

2

*Example 1.8

2

[{x( x 2 + y 2 + z 2 ) - y} ax + { y ( x 2 + y 2 + z 2 ) + x} a y + z ( x 2 + y 2 + z 2 ) az ]

Express the field E = 2 xyzax - 3( x + y + z )az , in cylindrical coordinates, and

calculate | E | at the point P (r = 2, f = 60°, z = 3).

Solution In the cylindrical system, È Er ˘ È cos f sin f 0 ˘ È Ex ˘ È cos f sin f 0 ˘ È 2 xyz ˘ Í ˙ Í Í ˙ Í ˙ ˙ Í ˙ 0 Í Ef ˙ = Í - sin f cos f 0 ˙ Í E y ˙ = Í - sin f cos f 0 ˙ Í ˙ ÍE ˙ Í 0 ˙˚ Í E ˙ ÍÎ 0 ˙˚ ÍÎ -3( x + y + z ) ˙˚ 0 1 0 1 Î Î z˚ Î z˚ Er = 2 xyz cos f Ef = –2 xyz sin f Ez = –3(x + y + z) But,

x = r cos f, y = r sin f,

z=z

Substituting these values Er = 2r cos f r sin f z cos f = 2r 2 z sin f cos 2 f Ef = - 2r cos f r sin f z sin f = - 2r 2 z sin 2 f cos f Ez = - 3(r cos f + r sin f + z ) Hence, the vector in Cartesian coordinates is written as E = r 2 z sin 2f cos f ar - r 2 z sin 2f sin f af - 3(r cos f + r sin f + z ) az At P (r = 2, f = 60°, z = 3), the vector is given as 3 1 ¥ = 3 3 = 5.196 2 4 3 1 2 2 Ef = - 2 ¥ 2 ¥ 3 sin 60∞ cos 60∞ = - 24 ¥ ¥ = - 9 4 2 Ê ˆ 1 3 Ez = - 3(2 cos 60∞ + 2 sin 60∞ + 3) = - 3 Á 2 ¥ + 2 ¥ + 3˜ = - 3(4 + 2 2 Ë ¯ Er = 2 ¥ 22 ¥ 3 sin 60∞ cos 2 60∞ = 24

3) = - 17.196

24

Electromagnetic Field Theory

Hence, at P, E = 5.196ar - 9af - 17.196az \

|E| =

Example 1.9

(5.196) 2 - 92 - (17.196) 2 = 20.092

Given the vector field in ‘mixed’ coordinate variables as V =

Ê x cos f 2 yz x2 ˆ a x + 2 a y + Á1 - 2 ˜ a z r Ë r r ¯

Convert the vector completely in spherical coordinates.

Solution Here, Vx =

x cos f ; r

\ \

r= Vx =

x2 + y 2 , x2 ; x2 + y 2

Ê x2 ˆ Vz = Á1 - 2 ˜ Ë r ¯

2 yz ; r2

x = r cos f,

But and

Vy =

y = r sin f,

cos f = Vy =

x x +y 2

2 yz ; 2 x + y2

2

,

z=z sin f =

Vz = 1 -

y x + y2 2

y2 x2 = x2 + y 2 x2 + y 2

Ê 2 yz ˆ Ê y2 ˆ Ê x2 ˆ V =Á 2 a a + + x y Á ˜ Á ˜ az Ë x + y 2 ¯˜ Ë x2 + y 2 ¯ Ë x2 + y 2 ¯

This is the vector in purely Cartesian coordinates. Now, by relations of Eq. (1.34), we get ÈVr ˘ È sin q cos f sin q sin f Í ˙ Í ÍVq ˙ = Í- cos q cos f cos q sin f ÍV ˙ Í - sin f cos f Î f˚ Î

È sin q cos f sin q sin f = ÍÍ- cos q cos f cos q sin f ÍÎ - sin f cos f

\

cos q ˘ ÈVx ˘ Í ˙ - sin q ˙˙ ÍVy ˙ 0 ˙˚ ÍÎVz ˙˚ ÈÊ x 2 ˆ ˘ Í ˙ ÍÁË x 2 + y 2 ˜¯ ˙ cos q ˘ Í ˙ Ê 2 yz ˆ ˙ - sin q ˙˙ ÍÁ 2 ÍË x + y 2 ˜¯ ˙ 0 ˙˚ Í ˙ ÍÊ y 2 ˆ ˙ ÍÁ 2 2 ˜˙ ÍÎË x + y ¯ ˙˚

Ê x2 ˆ Ê 2 yz ˆ Ê y2 ˆ Vr = Á 2 sin q cos f + Á 2 sin q sin f + Á 2 cos q 2˜ 2˜ Ëx + y ¯ Ëx + y ¯ Ë x + y 2 ¯˜ Ê - x2 ˆ Ê 2 yz ˆ Ê y2 ˆ Vq = Á 2 cos cos cos sin q f + q f ÁË x 2 + y 2 ˜¯ ÁË x 2 + y 2 ˜¯ sin q Ë x + y 2 ˜¯ Ê - x2 ˆ Ê 2 yz ˆ sin f + Á 2 cos f Vf = Á 2 2˜ Ëx + y ¯ Ë x + y 2 ˜¯

Vector Analysis

25

However, x = r sin q cos f; y = r sin q sin f; z = r cos q Therefore, the vector components in spherical coordinate system are given as \

1 ( r 2 sin 3 q cos3 q + 3r 2 sin 2 q sin 2 f cos q ) r sin 2 q = (sin q cos3 q + 3sin 2 f cos q ) 1 Vq = 2 2 ( - r 2 sin 2 q cos q cos3 f + 2r 2 sin q cos 2 q sin 2 f - r 2 sin 3 q sin 2 f ) r sin q Ê cos 2 q sin 2 f ˆ = Á2 - cos q cos3 f - sin q sin 2 f ˜ sin q Ë ¯ Vr =

2

Ê cos q cos f ˆ - cos 2 f sin f ˜ Vj = Á 2 sin q Ë ¯

1.7

GENERAL CURVILINEAR COORDINATES

Curvilinear Coordinate System

Curvilinear coordinates are a coordinate system for the Euclidean space based on some transformation that converts the standard Cartesian coordinate system to a coordinate system, with the same number of coordinates in which the coordinate lines are curved. The name curvilinear coordinates, given by the French mathematician Lame, derives from the fact that the coordinate surfaces of the curvilinear systems are curved. Let f1(x, y, z), f2(x, y, z), f3(x, y, z) be three independent, unambiguous and smooth functions, x, y, z be three independent space variables in the Cartesian coordinate system, and u1, u2, u3 be three constant parameters.

We set the equations u1 = f1 ( x, y, z )

u 2 = f 2 ( x, y , z )

u3 = f3 ( x, y, z )

By this equation, three surfaces are defined that can be labelled by these parameters as shown in Fig. 1.16. The common intersection of the surfaces u1 = constant 1, u2 = constant 2, u3 = constant 3 defines one point in the space to which a set of three unique numbers (u1, u2, u3) can be assigned. These numbers are called curvilinear coordinates of that point as illustrated in Fig. 1.16.

Position Vector in Curvilinear System We have the set of equations u1 = f1 ( x, y, z )

u 2 = f 2 ( x, y , z )

u3 = f3 ( x, y, z )

These equations can be solved and the solution can be written in the form x = f1¢(u1 , u2 , u3 ) y = f 2¢ (u1 , u2 , u3 ) z = f1¢(u1 , u2 , u3 )

(1.42)

This defines the position of the point A in the Cartesian system (x, y, z) using the curvilinear coordinates (u1, u2, u3).

26

Electromagnetic Field Theory

Fig. 1.16

To the definition of the curvilinear coordinates of a point A(u1,u2,u3) in the space

Here, r = xax + ya y + za z = x(u1 , u2 , u3) a x + y (u1 , u2 , u3) a y + z (u1 , u2 , u3) az

(1.43)

is the position vector of the point A and ax , a y , az are the unit vectors along the coordinate axes of the Cartesian system (see Fig. 1.16).

Base Vectors in Curvilinear System An elementary displacement of the point A in the space can be described by the differential formula dr =

∂r ∂r ∂r du + du + du ∂u1 1 ∂u2 2 ∂u3 3

d r = du1a1¢ + du2 a2¢ + du3 a3¢

or

(1.44) (1.45)

Here, a1¢, a2¢ , a3¢ are called the base vectors of the general curvilinear coordinate system at point A(u1, u2, u3). It is to be noted that the absolute values of a1¢, a2¢ , a3¢ are not equal to 1; they are generally not unit vectors. If a1¢ ^ a2¢ ^ a3¢ , we have the orthogonal curvilinear coordinate system.

Metric Coefficients The relation of Eq. (1.45) can be rewritten into the form d r = h1 (u1 , u2 , u3 ) du1a1 + h2 (u1 , u2 , u3 ) du2 a2 + h3 (u1 , u2 , u3 ) du3 a3 where we have put a1¢ = h1a1 ;

a1¢ = h2 a2 ;

a3¢ = h3 a3

(1.46)

where a1 , a2 , a3 are now the units vectors of the same directions as vectors a1¢, a2¢ , a3¢. The functions h1, h2, h3 are usually called the metric coefficients. The physical meaning of these coefficients can be understood when defining the length elements along the particular directions given by vectors a1 , a2 , a3 in the local curvilinear coordinate system at point A(r ) corresponding with the elementary displacements of A(u1, u2, u3) by du1, du2, du3.

Vector Analysis

27

Differential Lengths in Curvilinear Coordinates The elementary displacement d s from point A(r ) to A(r + d r ) can be described by

\

d s = d r = h1du1a1 + h2 du2 a2 + h3 du3 a3

(1.47a)

ds 2 = h12 du12 + h22 du22 + h32 du32

(1.47b)

It is seen that the products h1du1, h2du2, h3du3 represent the lengths of projections of the elementary displacement d s onto the vectors a1 , a2 , a3 , respectively. Consequently, the change of the curvilinear coordinate dui can be transformed into corresponding displacement in space by multiplying it by hi corresponding metric coefficient as shown in Fig. 1.17.

Fig. 1.17

To the definition of the elementary displacement ds , surface dS and volume dV, respectively

Therefore, the metric coefficients can be represented in terms of the elementary lengths as h1 =

ds1 du1

h2 =

ds2 du2

h3 =

ds3 du3

(1.48)

Differential Areas in Curvilinear Coordinates

Elementary coordinate surface can be defined corresponding with the elementary changes of a couple of coordinates. It is explicitly defined by the relation (1.49) dSi = d s j ¥ d sk Using expressions for elementary displacements d s j , d sk from Eq. (1.47 a), we can write dS = h2 h3 du2 du3 a1 + h1h3 du1du3 a2 + h1h2 du1du2 a3

(1.50)

In Eq. (1.50), the relations a1 ¥ a2 = a3 , a2 ¥ a3 = a1 , a3 ¥ a1 = a2 , has been used. According to Eq. (1.50), a general elementary surface dS is composed of the three elementary surfaces dS1 , dS2 , dS3 oriented along the unit vectors a1 , a2 , a3 , see Fig. 1.17.

Differential Volume in Curvilinear Coordinates

Elementary volume element dV can be

described by the relation dV = dsi ◊ dSi

(1.51)

dV = h1h2 h3 du1du2 du3

(1.52)

or using metric coefficients, we obtain

28

1.8

Electromagnetic Field Theory

DIFFERENTIAL ELEMENTS (LENGTHS, AREAS AND VOLUMES) IN DIFFERENT COORDINATE SYSTEMS

Now, we will consider the generalised curvilinear coordinates for the three different coordinate systems and determine the differential elements in these three coordinate systems. To obtain the differential elements in length, area and volume, we consider the following figures:

Fig. 1.18

(a) Differential elements in Cartesian coordinates, (b) Differential elements in cylindrical coordinates, and (c) Differential elements in spherical coordinates

Cartesian Coordinate System In this system as shown in Fig. 1.18 (a), the differential components of the general arc ds are identical with the differentials of the coordinates. \ u1 = s1 = x, u2 = s2 = y, u3 = s3 = z Hence, the metric coefficients are given as h1 =

ds1 =1 du1

h2 =

ds2 =1 du2

h3 =

Differential displacement is given as ds = dxax + dya y + dzaz

ds3 =1 du3

Vector Analysis

29

Differential normal area is given as dS = dydzax + dxdza y + dxdya z Differential volume is given as dV = dxdydz

NOTE Vector Representation of Surface Differential surface area is defined as a vector whose magnitude corresponds to the area of the surface and whose direction is perpendicular to the surface.

\

dS = dSan

For closed surface, the outward normal direction is taken as positive direction. For an open surface, the normal created by positive periphery according to the right-hand cork screw rule is taken to be positive. If the surface is not a plane, then the surface is subdivided into smaller elements that are considered to be plane and a vector surface is considered for each of these elemental surfaces. Vector addition of these elemental vector surfaces gives the total surface.

Cylindrical Coordinate System In this system as shown in Fig. 1.18 (b), the differential components are given as \

u1 = r,

u2 = f,

u3 = z

The differential displacements are obtained from Fig. 1.18 (b) as ds1 = dr

ds2 = rdf

ds3 = dz

Hence, the metric coefficients are given as h1 =

ds1 dr = =1 du1 dr

h2 =

ds2 rd f = =r du2 df

h3 =

ds3 dz = =1 du3 dz

Differential displacement is given as d s = drar + rd f af + dzaz Differential normal area is given as dS = rd f dzar + drdzaf + rdrd f az Differential volume is given as dV = rdrd f dz Spherical Coordinate System In this system as shown in Fig. 1.18 (c), the differential components are given as \

u1 = r, u2 = q , u3 = f

The differential displacements are obtained from Fig. 1.18 (c) as ds1 = d r ds2 = rdq

ds3 = r sin q d f

30

Electromagnetic Field Theory

Hence, the metric coefficients are given as h1 =

ds1 d r = =1 du1 d r

h2 =

ds2 rdq = =r du2 dq

h3 =

ds3 r sin q d f = = r sin q du3 df

Differential displacement is given as d s = d r ar + r dq aq + r sin q d f af Differential normal area is given as dS = r 2 sin q dq d f ar + r sin q d r d f aq + rd r dq af Differential volume is given as dV = r 2 sin q d r dq d f These relations are given in Table 1.1. Table 1.1 Differential elements in different coordinate systems Differential Elements

Cartesian Coordinates

Cylindrical Coordinates

Spherical Coordinates

Length

d s = dxax + dya y + dza z

d s = drar + rdf af + dza z

d s = d r ar + rdq aq + r sin q df af

Area

dS = dydzax + dxdza y + dxdyaz

dS = rdf dzar + drdzaf + rdrdf az

dS = r 2 sin q dq df ar + r sin q d r df aq + rd r dq af

Volume

dV = dxdydz

dV = rdrdfdz

dV = r2 sin q dr dq df

1.9

VECTOR CALCULUS

We now discuss the vector calculus, i.e., integrations and differentiations of vector.

1.9.1

Vector Integration or Vector Integrals

The integration of a vector may be obtained in three ways—line integral, surface integral and volume integral as discussed below.

Line Integral The line integral of a vector is the integral of the dot product of the vector and the differential length vector tangential to a specified path. For vector F and a path l, the line integral is given by b

Ú F ◊ dl = Ú | F | cos q dl l

(1.53)

a

We consider a vector F and a specified path a–b. As the magnitude of the vector varies from point to point, the path is divided into a number of small line segments d l1 , d l2 , º with vector magnitudes F1 , F2 ,º Thus, the total work done by the vector F from a to b is given as

Vector Analysis

31

n

W = F1 ◊ dl1 + F2 ◊ dl2 + F3 ◊ dl3 + º = Â Fi ◊ dli i =1

Fig. 1.19

(a) Line integral of a vector, and (b) Path of integration of vector field F

If the lengths of the segments tend to zero, this work done can be written as line integral b

W = Ú F ◊ dl a

If the path of integration is a closed curve, such as abca, the line integral becomes W = Ú F ◊ dl l

If the line-integration of a vector along a closed path is zero, i.e., Ú F ◊ d l = 0 , then the vector is l known as conservative or lamellar vector. The concept of line integral is used to calculate the electric potential for a given electric field intensity or to calculate the total current enclosed by a closed path for a given magnetic field intensity.

*Example 1.10 Find the line integral of the vector F = ( x 2 - y 2 ) ax + 2 xya y around a square of side a which has a corner at the origin, one side on the x axis and the other side on the y axis.

Solution

Fig. 1.20

Line integral of a vector

32

Here,

Electromagnetic Field Theory B

C

D

A

A

B

C

D

Ú F ◊ dl = Ú F ◊ dl + Ú F ◊ dl + Ú F ◊ dl + Ú F ◊ dl

L

Now, Along AB, y = z = 0, dy = dz = 0, d l = dxax B

a

a

A

x=0

x=0

3

a 2 2 Ú F ◊ d l = Ú ( x ax ) ◊ (dxax ) = Ú x dx = 3

\

Along BC: x = a, z = 0, dx = dz = 0, d l = dya y C

a

a

B

y =0

y =0

2 2 3 Ú F ◊ d l = Ú [(a - y ) ax + 2aya y ] ◊ (dya y ) = Ú 2aydy = a

\

Along CD: y = a, z = 0, dy = dz = 0, d l = dxax \

D

0

C

x=a

Ú F ◊ dl =

Ú

[( x 2 - a 2 ) ax + 2 xaa y ] ◊ (dxax ) =

0

È x3 2 3 2 2 2 ˘ Ú ( x - a )dx = ÍÎ 3 - a x ˙˚ = 3 a a x=a 0

Along DA: x = z = 0, dx = dz = 0, d l = dya y A

0

D

y=a

Ú F ◊ d l = Ú 0 ◊ (dya y ) = 0

\

Therefore, total line integral of the vector is given as 3

a 2 3 3 3 Ú F ◊ d l = 3 + a + 3 a = 2a L

*Example 1.11 Compute the line integral of F = 6i + yz 2 j + (3 y + z )k along triangular path shown in Fig. 1.21.

Fig. 1.21

Triangular path of Example 1.11

Vector Analysis

33

Solution Here, F = 6i + yz 2 j + (3 y + z )k d l = dxi + dyj + dzk \

F ◊ d l = {6i + yz 2 j + (3 y + z )k } ◊ (dxi + dyj + dzk ) = 6dx + yz 2 dy + (3 y + z )dz

The closed line integral is given as B

C

A

A

B

C

Ú F ◊ dl = Ú F ◊ dl + Ú F ◊ dl + Ú F ◊ dl

L

Along path AB, dx = dz = 0, x = z = 0; y varies from 0 to 1. \

B

È1

˘

A

Î0

˚z =0

2 Ú F ◊ d l = Í Ú yz dy ˙

=0

Along path BC, dx = 0, x = 0; y varies from 1 to 0 and z varies from 0 to 2. Also, for this path, the equation relating y and z is obtained as y z + =1 fi 1 2 \

z = 2(1 - y )

0 2 0 2 È Ê 2 - zˆ ˘ 2 2 Ú F ◊ d l = Ú Ú yz dy + (3 y + z )dz = Ú y[2(1 - y )] dy + Ú ÎÍ3 Ë 2 ¯ + z ˚˙ dz B y =1 z = 0 y =1 z =0

C

=

0

2

( )

z 3 2 Ú 4( y - 2 y + y )dy + Ú 3 - 2 dz y =1 z =0 0

2

È y 4 2 y3 y 2 ˘ È z2 ˘ = 4Í + + Í3z - ˙ ˙ 3 2 ˚1 Î 4 ˚0 Î 4 1 2 1 = 4 ÈÍ - + - ˘˙ + (6 - 1) Î 4 3 2˚ 14 = 3

Along path CA, dx = dy = 0, x = y = 0; z varies from 2 to 0. \

A

È0

˘

C

Î2

˚ y =0

Ú F ◊ d l = Í Ú (3 y + z )dz ˙

= -2

By addition, we get the closed line integral as 14 8 Ú F ◊ dl = 0 + 3 - 2 = 3

L

*Example 1.12 For a given vector, F = xyax - 2 xa y , evaluÚ F ◊ d l over the path shown in Fig. 1.22.

ate the line integral

Solution Here, F = xyax – 2 xa y , d l = dxax + dya y + dzaz , d S = dxdya z

Fig. 1.22

Path of Example 1.12

34

Electromagnetic Field Theory

\

F ◊ d l = ( xyax - 2 xa y ) ◊ (dxa x + dya y + dza z ) = ( xydx - 2 xdy )

Along the path OA, y = 0, x varies from 0 to 3. A

Ú F ◊ dl = 0

\

O

Along the path AB, x varies from 3 to 0 and y varies from 0 to 3. Also, along this path, the equation relating x and y is obtained as x2 + y2 = 9 B

\

0

3

Ú F ◊ d l = Ú Ú ( xydx - 2 xdy ) x =3 y =0 0

A

=

3

2 2 Ú x( 9 - x )dx - Ú 2( 9 - y )dy

x =3

y =0

3

0

3 y˘ È 1 = - (9 - x 2 ) 2 - Í y 9 - y 2 + 9 sin -1 ˙ 3 3 Î ˚0 3 p = -9 1 + 2

( )

Along the path BO, x = 0, y varies from 3 to 0. A

\

Ú F ◊ dl = 0

By addition, we get

( )

p Ú F ◊ dl = -9 1 + 2

L

Also, — ¥ F =

\

O

ax

ay

az

∂ ∂x xy

∂ ∂y -2x

∂ = - ( x + 2)az ∂z 0 3

3

3

3

Ú (— ¥ F ) ◊ d S = Ú Ú [ -( x + 2)az ] ◊ (dxdyaz ) = - Ú Ú ( x + 2)dxdy

S

x=0 y =0

=-

3

Ú

y =0

=-

x=0 y =0

9 - y2

3

Ú ( x + 2)dxdy = - Ú

x=0

y =0

Ê x2 ˆ + 2 x ˜¯ ÁË 2

9 - y2

dy 0

3 Ê 9 - y2 È9 y3 2ˆ -1 y ˘ 2 + 2 9 = y dy Á ˜ Ú Ë 2 ÍÎ 2 y - 6 + y 9 - y + 9sin 3 ˙˚ ¯ 0 y =0 3

( p2 )

= -9 1 +

Example 1.13 Calculate the circulation of A = r cos f ar + z sin f az around the edge L of the wedge defined by 0 £ r £ 2, 0 £ f £ 60°, z = 0 as shown in Fig. 1.23.

Vector Analysis

Fig. 1.23

Path of Example 1.13

Solution Here, A = r cos f ar + z sin f az dl = drar + rd f af + dzaz \

A ◊ d l = (r cos f ar + z sin f a z ) ◊ (drar + rd f af + dza z ) = (r cos f dr + z sin f dz )

Since the path is on the xy plane, dz = 0. A ◊ dl = r cos f dr

\

A

B

O

O

A

B

Ú A◊ dl = Ú A◊ dl + Ú A◊ dl + Ú A◊ dl

L

Along the path OA, r varies from 0 to 2. \

A

È2

˘

O

Î0

˚f = 0

Ú A ◊ d l = Í Ú r cos f dr ˙

=2

Along the path AB, r is constant and so the integration is zero. B

Ú A◊ dl = 0

\

A

Along the path BO, r varies from 2 to 0. \

O

È0

˘

B

Î2

˚f = 60∞

Ú A ◊ d l = Í Ú r cos f dr ˙

By addition, we get the closed line integral as,

=

1 ¥ ( - 2) = -1 2

Ú A ◊ d l = 2 + 0 - 1 = 1.

L

Example 1.14

Using the concept of line integral, find the periphery of a circle of radius a.

Solution In cylindrical coordinates, the differential length is, d l = adf af By line integral, the periphery is obtained as 2p

2p

2p

f =0

f =0

f =0

Ú af ◊ d l = Ú (af ) ◊ (adf af ) = Ú adf = 2p a

35

36

Electromagnetic Field Theory

Surface Integral

For a vector F , continuous in a region containing a smooth surface S, the surface integral or the flux of F through S is defined as, y = Ú F ◊ d S = Ú F ◊ an dS = Ú | F | cos q dS S

S

(1.54)

S

where, an is the unit normal vector to the surface S. We consider a surface S. We divide the surface into infinitesimal surfaces d S1 , d S2 ,º which are treated as the vector quantities. Let F1 , F2 ,º be the vector magnitudes at the elemental surfaces, respectively. Thus, the sum of the scalar products F1 ◊ d S1 , F2 ◊ d S 2 ,º is written as F1 ◊ d S1 + F2 ◊ d S2 + F3 ◊ d S3 +

Fig. 1.24

Surface integral

n

= Â Fi ◊ d Si i =1

If the elemental surfaces areas tend to zero, this can be written as surface-integral y = Ú F ◊dS S

If the surface is a closed surface, the surface integral is written as y = Ú F ◊dS S

For a closed surface, the surface integral,

Ú F ◊ d S is referred to as the net outward flux of F from

S

the surface. If the surface integral of a vector over a closed surface is zero, i.e.,

Ú F ◊ d S = 0 , then the

S

vector is known as solenoidal vector. The concept of surface integral is necessary to calculate the flux or current from the flux or current density over a surface.

Example 1.15

Using the concept of surface integral, find the surface area of a sphere of radius

a.

Solution In spherical coordinates, the differential surface in the perpendicular direction is given as d S = a 2 sin q dq d f ar By surface integral, the surface area of the sphere is obtained as S = Ú ar ◊ d S = S

p

2p

2 2 p 2 Ú Ú a sin q dq df = a ¥ 2p ¥ ( - cos q )0 = 4p a

q =0f =0

*Example 1.16 Use the cylindrical coordinate system to find the area of a curved surface on the right circular cylinder having radius = 3 m and height = 6 m and 30° £ f £ 120°. Solution Here, the differential surface is given as d S = rd f dzar

Vector Analysis

37

Taking the surface integral, the area of the curved surface is obtained as S = Ú ar ◊ d S = S

2p /3

Ú

6

=3¥6¥

Ú rdf dz

f = p /6 z = 0

r =3

( 23p - p6 ) = 9p

m2

*Example 1.17 Use the spherical coordinate system to find the area of the strip a £ q £ b on the spherical shell of radius ‘a’. What results when a = 0 and b = p? Solution For a fixed radius of a, the elemental surface is d S = r 2 sin q dq d f ar

r=a

= a 2 sin q dq d f ar

Hence, the area of the strip is given as b

S = Ú d S ◊ ar = S

2p

2 Ú Ú a sin q dq df

q =a f = 0

= 2p ¥ a ¥ ( - cos q )ab = 2p a 2 (cos a - cos b ) 2

For a = 0 and b = p, the area is S = 2pa2 (cos 0 – cos p) = 4pa2 This is the area of a sphere of radius a. Given rs = (x2 + xy); calculate 2 Ú rsdS over the region y £ x , 0 < x < 1.

Example 1.18

Fig. 1.25

Arrangement of Example 1.17

Solution x2

x2

x2

1

1 Ê 2 Ê 4 x5 ˆ Ê x5 x 6 ˆ xy 2 ˆ 17 2 Ú rs dS = Ú Ú ( x + xy ) dxdy = Ú ÁË x y + 2 ˜¯ 0 dx = Ú ÁË x + 2 ˜¯ 0 dx = ÁË 5 + 12 ˜¯ 0 = 60 x=0 y =0 x=0 x=0 1

1

k1 + az k2 z; evaluate the scalar surface integral r surface of a closed cylinder about the z-axis specified by z = ±3 and r = 2.

Example 1.19

Given F = ar

Ú F ◊ d S over the s

Solution The cylinder has three surfaces as follows.

Ú F ◊dS = s

Ú

Circular 1

F ◊dS +

Ú

F ◊dS +

Circular 2

Ú F ◊dS

Curved

For the upper circular surface z = 3, an = az

Ú

Circular 1

F ◊dS =

2

2p

Ê ˆ 1 Ú Ú Ë ar r + az k2 z ¯ ◊ (rdrdf az ) r =0f =0 k

= z =3

2

2p

Ú Ú 3k2 rdrdf = 12p k2

r =0f =0

38

Electromagnetic Field Theory

For the bottom circular surface z = - 3, an = - az F ◊dS =

Ú

Circular 1

2

2p

Ê ˆ 1 Ú Ú Ë ar r + az k2 z ¯ ◊ ( - rdrdf az ) r =0f =0 k

= 12p k2 z = -3

For the curved surface r = 2, an = ar F ◊dS =

Ú

Circular 1

2p

3

Ê ˆ 1 Ú Ë ar r + az k2 z ¯ ◊ (rdf ar ) f = 0 z = -3 k

Ú

= r =2

2p

Ú

3

Ú k1df dz = 12p k1

f = 0 z = -3

By addition, total surface area of the closed cylinder is given as

Ú F ◊ d S = 12p (k1 + 2 k2 ) s

Example 1.20

Evaluate Ú r ◊ an dS where S is a closed surface. S

Solution We know, r = xax + ya y + zaz \

Ú r ◊ an dS = Ú — ◊ rdv = Ú 3dv = 3V {

S

V

— ◊ r = 3, see Example 36 (b)}

V

where V is the volume enclosed by the surface S.

Volume Integral

The volume integral of a scalar quantity F over a volume V is written as U = Ú Fdv

(1.55)

V

The concept of volume integral is necessary to calculate the charge or mass of an object, which are distributed in the volume.

Example 1.21

Using the concept of volume integral, find the volume of a sphere of radius a.

Solution In spherical coordinates, the differential volume is given as dv = r 2 sin q d rdq d f where 0£ r£a 0£q £p 0 £ f £ 2p By volume integral, the volume of the sphere is obtained as v = Ú dv = v

a

p

2p

a

3

a 4 3 2 p 2 Ú Ú Ú r sin q d r dq df = 2p ¥ ( - cos q )0 ¥ Ú r d r = 4p ¥ 3 = 3 p a r =0q =0f =0 r =0

Vector Analysis

Example 1.22

39

Obtain the expression for the volume of a sphere of radius a using the concept

of volume integral.

Solution Here, the differential volume in spherical coordinates is given as dv = r 2 sin q d rd q d f where a£r£0 0£q £p 0 £ f £ 2p The volume of the sphere is obtained by the volume integral as given. V = Ú r 2 sin q d rdq d f = v

1.9.2

2p

p

a

3

a 4 3 2 p Ú Ú Ú r sin q d rdq d f = 3 ¥ 2p ¥ ( - cos q )0 = 3 p a

f =0q =0 r =0

Vector Differentiations

In order to understand vector differentiation, we introduce an operator known as del operator or differential vector operator.

Differential Vector Operator (—) or Del Operator

The differential vector operator (—) or

Del or Nabla, in Cartesian coordinates, is defined as —=

∂ ∂ ∂ a + a + a ∂x x ∂y y ∂z z

(1.56)

This Del is merely a vector operator but not a vector quantity. When it operates on a scalar function, a vector is created. Since a vector, in general, is a function of the space and time both, the del operator is a vector space function operator. It is defined in terms of partial derivatives with respect to space. In Eq. (1.56), del has been expressed in Cartesian coordinates. However, this operator can be expressed in general curvilinear coordinates as —=

∂ ∂ ∂ a + a + a ∂s1 1 ∂s2 2 ∂s3 3

(1.57)

But, we know that, dsi = hidui; replacing this in Eq. (1.57), we get —=

1 ∂ 1 ∂ 1 ∂ a + a + a h1 ∂u1 1 h2 ∂u2 2 h3 ∂u3 3

(1.58)

Substituting the values of hi in different coordinate systems, we obtain the relation of del in three different coordinate systems as —=

∂ ∂ ∂ a + a + a ∂x x ∂y y ∂z z

(Cartesian coordinates)

=

1 ∂ ∂ ∂ a + a + a ∂r r r ∂f f ∂z z

=

1 ∂ 1 ∂ ∂ a + a + a ∂r r r ∂q q r sin q ∂f j

(Cylindrical coordinates) (Spherical coordinates)

(1.59a) (1.59b) (1.59c)

40

Electromagnetic Field Theory

Corresponding to three different vector multiplications, there are three possible operations of —. These operations are: 1. Gradient of a scalar F, written as, —F; 2. Divergence of a vector A, written as, — ◊ A ; 3. Curl of a vector A, written as, — ¥ A ; and

Gradient of a Scalar Definition The gradient of a scalar function is both the magnitude and the direction of the maximum space rate of change of that function. Mathematical Expression of Gradient

We consider a scalar function F. A mathematical expression for the gradient can be obtained by evaluating the difference in the field dF between the points P1 and P2. Here, F1 and F2 are the contours on which F is constant.

Fig. 1.26

Gradient of a scalar quantity

∂F ∂F ∂F dx + dy + dz ∂x ∂y ∂z ∂F ∂F ∂F ˆ = ÊÁ a + a + a ◊ (dxax + dya y + dzaz ) Ë ∂x x ∂y y ∂z z ˜¯

dF =

= G ◊ dl ∂F ∂F ∂F ˆ where G = ÊÁ a + a + a Ë ∂x x ∂y y ∂z z ˜¯ and, d l = dxax + dya y + dza z = differential length = differential displacement from P1 to P2 \

dF = G ◊ d l = Gdl cos q

\

dF = G cos q ; dl

For maximum \

dF dl

= max

where q is the angle between G and d l

( dFdl ), q = 0°, i.e., when d l is in the direction of G. dF dF is the normal derivative. = G; where, dn dn

Thus, by definition of gradient, we have the mathematical expression of gradient in Cartesian coordinates given as Grad F = —F =

∂F ∂F ∂F a + a + a ∂x x ∂y y ∂z z

(1.60)

Physical Interpretation The gradient of a scalar quantity is the maximum space rate of change of the function. For example, we consider a room in which the temperature is given by a scalar field T, so at any point (x, y, z) the temperature is T(x, y, z) (assuming that the temperature does not change with time). Then, at any arbitrary point in the room, the gradient of T indicates the direction in which the temperature

Vector Analysis

41

rises most rapidly. The magnitude of the gradient will determine how fast the temperature rises in that direction.

Gradient in General Curvilinear Coordinates

Let F(u1, u2, u3) be a scalar function. The u1 component of the gradient of F, by definition, is the maximum space rate of change of the function, i.e., (Grad F )1 = Lim

du1 Æ 0

F (G ) - F (O) 1 ∂F = h1du1 h1 ∂u1

Similarly, considering directions 2 and 3 and so, the resultant expression of the gradient of F is given as —F =

1 ∂F 1 ∂F 1 ∂F a + a + a h1 ∂u1 1 h2 ∂u2 2 h3 ∂u3 3

(1.61) Fig. 1.27

General Curvilinear coordinates

Substituting the values of hi from Section 1.8, we get the relation of gradient in three different coordinate systems as ∂F ∂F ∂F a + a + a ∂x x ∂y y ∂z z

(Cartesian Coordinates)

(1.62a)

=

1 ∂F ∂F ∂F a + a + a ∂r r r ∂f f ∂z z

(Cylindrical Coordinates)

(1.62b)

=

1 ∂F 1 ∂F ∂F a + a + a (Spherical Coordinates ) ∂r r r ∂q q r sin q ∂f f

—F =

(1.62c)

Properties of Gradient 1. The magnitude of the gradient of a scalar function is the maximum rate of change of the function per unit distance. 2. The direction of the gradient of a scalar function is in the direction in which the function changes most rapidly. 3. The gradient of a scalar function at any point is always perpendicular to the surface that passes through the point and over which the function is constant (points a and b in Fig. 1.26). 4. The projection of the gradient of a scalar function (say, —S) in the direction of a unit vector a, i.e., —S ◊ a is known as the directional derivative of the function S along unit vector a.

Example 1.23

Find the gradient of the following scalar fields:

(a) F = x y + e (b) V = rz sin f + z2 cos2 f + r2 (c) S = cos q sin f ln r + r2 f. 2

z

Solution (a) The gradient in Cartesian coordinates is given as —F =

∂F ∂F ∂F a + a + a = 2 xax + x 2 a y + e z az ∂x x ∂y y ∂z z

42

Electromagnetic Field Theory

(b) The gradient in cylindrical coordinates is given as 1 ∂V ∂V ∂V a + a + a ∂r r r ∂f f ∂z z 1 = ( z sin f + 2r )ar + (rz cos f - z 2 2 cos f sin f )af + (r sin f + 2 z cos 2 f )az r Ê ˆ z2 sin 2f ˜¯ af + (r sin f + 2 z cos 2 f )az = ( z sin f + 2r )ar + ÁË z cos f r

—V =

(c) The gradient in spherical coordinates is given as 1 ∂S 1 ∂S ∂S a + a + a ∂r r r ∂q q r sin q ∂f f Ê cos q sin f ˆ 1 1 (cos q cos f ln r + r 2 )af =Á + 2 rf ˜ ar + sin q sin f ln r aq + r r r sin q Ë ¯ sin q sin f Ê cos q sin f ˆ Ê cot q ˆ ln r aq + Á cos f ln r + r cosec q ˜ af =Á + 2 rf ˜ ar + r r Ë ¯ Ë r ¯

—S =

Example 1.24

Find the gradient of the following scalar fields:

(a) V = 4xz + 3 yz (b) V = 2r(1 + z2)cos f (c) V = r2 cos q cos f 2

Solution (a) V = 4xz2 + 3 yz —V =

∂ ∂ ∂ (4 xz 2 + 3 yz )ax + (4 xz 2 + 3 yz )a y + (4 xz 2 + 3 yz )a z = 4 z 2 a x + 3za y + (8 xz + 3 y )a z ∂x ∂y ∂z

(b) V = 2r(1 + z2)cos f —V =

∂V ∂V 1 ∂V a + a + a = 2(1 + z 2 ) cos f ar - 2(1 + z 2 ) sin f af + 4rz cos f az ∂r r r ∂f f ∂z z

(c) V = r2 cos q cos f \

∂V ∂V 1 ∂V 1 a + a + a ∂r r r ∂q q r sin q ∂f f = 2 r cos q cos f ar - r sin q cos f aq + r cot q sin f af

—V =

*Example 1.25 Find the rate at which the scalar function V = r2 sin 2f in cylindrical coordinates

(

increases in the direction of the vector A = ar + af at the point 2,

)

p ,0 . 4

Or, Find the gradient of the scalar function V = r2 sin 2f and the directional derivative of the function in the p direction (ar + af ) at the point 2, , 0 . 4

(

)

Vector Analysis

43

Solution The gradient in cylindrical coordinates is given as —V =

∂V ∂V 1 ∂V a + a + a = 2 r sin 2f ar + 2 r cos 2f af = 2 r (sin 2f ar + cos 2f af ) ∂r r r ∂f f ∂z z

The direction derivative is given as —V ◊ a A = —V ◊

(

At 2,

)

Ê ar + af ˆ A = (2 r sin 2f ar + 2 r cos 2f af ) ◊ Á ˜= Ë | A| 2 ¯

p , 0 , the directional derivative is given as 4 p —V ◊ a A = 2 ¥ 2 sin + 2

2 ¥ 2 cos

2 r sin 2f +

2 r cos 2f

p =2 2 2

(r )

*Example 1.26 Find — 1 , where r = xax + ya y + zaz Or

()

1 1 Show that for a moving field point and a fixed source point, Grad = - ÊÁ 2 ˆ˜ ar , where r is the Ër ¯ r distance between the source point and the field point.

Solution Here, r = xax + ya y + zaz r=

\ \

—

()

x2 + y 2 + z 2

1 ∂ Ê = r ∂x Á Ë ==-

ˆa + ∂ Ê x ∂y Á x + y + z ˜¯ Ë 1

2

2

2

ˆa + ∂ Ê y ∂z Á x + y + z ˜¯ Ë 1

2

2

2

ˆa z x + y + z ˜¯ 1

2

2

2

2y 1 2x 1 1 2z a -a -a 2 ( x 2 + y 2 + z 2 )3/2 x 2 ( x 2 + y 2 + z 2 )3/2 y 2 ( x 2 + y 2 + z 2 )3/2 z xax + ya y + zaz ( x 2 + y 2 + z 2 )3/2

r { r = rar } r3 1 = - ÊÁ 2 ˆ˜ ar Ër ¯ =-

This problem can be solved easily in cylindrical coordinates as follows. — \

( 1r ) = ∂∂r ( 1r ) a = - ÊÁË r1 ˆ˜¯ a 1 r 1 —( ) = = - ÊÁ ˆ˜ a Ër ¯ r r r

3

r

2

2

r

Divergence of a Vector Definition Divergence of a vector at any point is defined as the limit of its surface integral per unit volume as the volume enclosed by the surface around the point shrinks to zero.

44

Electromagnetic Field Theory

Ê Ú F ◊dS ˆ Ê Ú F ◊ an dS ˆ Á ˜ Á ˜ S S div F = — ◊ F = Lim Ë ¯ = Lim ¯ v v vÆ0 vÆ0 Ë

\

(1.63)

where v is the volume of an arbitrarily shaped region in space that includes the point, S is the surface of that volume, and the integral is a surface integral with an being the outward normal to that surface. Figures 1.28 (a), (b) and (c) show three cases of positive, negative and zero divergence.

Fig. 1.28

(a) Positive divergence, (b) Negative divergence, and (c) Zero divergence

Mathematical Expression of Divergence We consider a hypothetical infinitesimal cubical box oriented along the coordinate axes around an infinitesimal region of space. We consider a vector V at a point P(x, y, z). Let, V1, V2, and V3 be the components of V along the three coordinate axes. In order to compute the surface integral, we see that there are six surfaces to this box, and the net content leaving the box is therefore, simply the sum of differences in the values of the vector field along the three sets of parallel surfaces of the box. The component vectors are given as follows. Ê 1 ∂V1 ˆ Along x-direction: at front surface, ËV1 + Dx ¯ 2 ∂x Ê 1 ∂V1 ˆ at back surface, ËV1 Dx ¯ 2 ∂x Ê 1 ∂V2 ˆ Along y-direction: at left surface, ÁV2 Dy ˜ 2 ∂y Ë ¯ ∂ V Ê 1 2 ˆ at back surface, ÁV2 + Dy ˜ 2 ∂y Ë ¯ ∂ V Ê ˆ 1 3 Along z-direction: at top surface, ËV3 + Dz ¯ 2 ∂z

Fig. 1.29

To derive expression for divergence in Cartesian coordinates

Ê 1 ∂V3 ˆ at bottom surface, ËV3 Dz ¯ 2 ∂z Therefore, the net outward flux of the vector is ∂V Ê Ê 1 ∂V1 ˆ 1 ∂V1 ˆ Along x-direction: ËV1 + Dx ¯ Dy Dz - Ë V1 Dx¯ Dy Dz = 1 DxDy Dz 2 ∂x 2 ∂x ∂x

Vector Analysis

45

∂V Ê Ê 1 ∂V2 ˆ 1 ∂V2 ˆ Along y-direction: Á V2 + Dy ˜ DxDz - Á V2 Dy ˜ DxDz = 2 DxDy Dz ∂y 2 ∂y 2 ∂y Ë ¯ Ë ¯ ∂V Ê Ê 1 ∂V3 ˆ 1 ∂V3 ˆ Along z-direction: ËV3 + Dz ¯ DxDy - ËV3 Dz ¯ DxDy = 3 DxDy Dz 2 ∂z 2 ∂z ∂z The total net outward flow, considering all thee directions, is ∂V

∂V

∂V

∂V

∂V

∂V

Ê 1 Ê 1 3ˆ 3ˆ 2 2 Ú F ◊ d S = ÁË ∂x + ∂y + ∂z ˜¯ DxDy Dz = ÁË ∂x + ∂y + ∂z ˜¯ Dv S where Dv = DxDyDz is the infinitesimal volume of the cube. Hence, the total net outward flow per unit volume is given as

Ú F ◊dS

S

Dv

∂V ˆ ∂V Ê ∂V =Á 1 + 2 + 3˜ ∂y ∂z ¯ Ë ∂x

By definition, this is the divergence of the vector. \

∂V ˆ ∂V Ê ∂V div F = — ◊ F = Á 1 + 2 + 3 ˜ ∂y ∂z ¯ Ë ∂x

(1.64)

Physical Interpretation The physical significance of the divergence of a vector field is the rate at which the density of a vector exits a given region of space. In other words, divergence of a vector field at a given point is an operator that measures the magnitude of the source or sink, in terms of a signed scalar. More technically, the divergence represents the volume density of the outward flux of a vector field from an infinitesimal volume around a given point. For example, we consider air as it is heated or cooled. The relevant vector field for this example is the velocity of the moving air at a point. If air is heated in a region, it will expand in all directions such that the velocity field points outward from that region. Therefore, the divergence of the velocity field in that region would have a positive value, as the region is a source. If the air cools and contracts, the divergence is negative and the region is called a sink.

Divergence in General Curvilinear Coordinates

We consider the general curvilinear

coordinates as shown in Fig. 1.29. We consider the vector F = F1a1 + F2 a2 + F3a3 In order to find a general expression for divergence of the vector, we need to calculate the closed surface integral of the vector per unit volume for the elemental volume as shown in Fig. 1.27(See page 41). The contributions to the closed surface integral Ê Ú F ◊ dS ˆ through the six surfaces are given as ËS ¯ follows. 1. Through surface OABC: Taken in the direction of the outward normal is

Ú F ◊ dS = - F1h2 h3du2 du3

OABC

46

Electromagnetic Field Theory

∂ Ú F ◊ dS = F1h2 h3du2 du3 + ∂u ( F1h2 h3 )du1du2 du3

2. Through surface GFED:

1

GFED

Ú F ◊ dS = - F2 h1h3du1du3

3. Through surface OCDG:

OCDG

4. Through surface ABEF:

∂ Ú F ◊ dS = F2 h1h3du1du3 + ∂u ( F2 h1h3 )du1du2 du3 2

ABEF

Ú F ◊ dS = - F3h1h2 du1du2

5. Through surface OAFG:

OAFG

∂ Ú F ◊ dS = F3h1h2 du1du2 + ∂u ( F3h1h2 )du1du2 du3

6. Through surface BCDE:

3

BCDE

Thus, the closed surface integral is given as ∂ ∂ È ∂ ˘ Ú F ◊ dS = Í ∂u ( F1h2 h3 ) + ∂u ( F2 h1h3 ) + ∂u ( F3h1h2 ) ˙ du1du2 du3 Î

S

1

2

3

˚

Thus, the surface integral per unit volume is obtained as

Ú F ◊ dS

S

dv

Ú F ◊ dS

1 È ∂ ( F h h ) + ∂ ( F h h ) + ∂ ( F h h )˘ du du du = h1du1h2 du2 h3 du3 h1du1h2 du2 h3 du3 ÍÎ ∂u1 1 2 3 ∂u 2 2 1 3 ∂u3 3 1 2 ˙˚ 1 2 3 1 È ∂ ∂ ∂ (F h h ) + (F h h ) + ( F h h )˘ = h1h2 h3 ÍÎ ∂u1 1 2 3 ∂u 2 2 1 3 ∂u3 3 1 2 ˙˚

=

S

By definition, this is the divergence of the vector. —◊ F =

1 È ∂ ∂ ∂ (F h h ) + (F h h ) + (F h h )˘ ∂u 2 2 1 3 ∂u3 3 1 2 ˚˙ h1h2 h3 ÎÍ ∂u1 1 2 3

(1.65)

Substituting the values of hi from Section 1.8, we get the relation of divergence in three different coordinate systems as ∂Fx ∂Fy ∂Fz (Cartesian coordinates) + + ∂x ∂y ∂z 1 ∂ 1 ∂Ff ∂Fz (rFr ) + (Cylindrical coordinates) = + r ∂r r ∂f ∂z 1 ∂ 1 1 ∂Ff ∂ ( r 2 Fr ) + ( Fq sin q ) + (Spherical coordinates) = 2 sin sin ∂ ∂ r r q q r q ∂f r

—◊F =

(1.66a) (1.66b) (1.66c)

Properties of Divergence 1. The result of the divergence of a vector field is a scalar. 2. Divergence of a scalar field has no meaning. 3. Divergence may be positive, negative or zero. A vector field with constant zero divergence is called solenoidal; in this case, no net flow can occur across any closed surface. For example, for

Vector Analysis

47

an incompressible fluid, if V denotes the quantity of the fluid at any point, then — ◊ V = 0 , i.e., an incompressible fluid cannot diverge from, nor converge towards a point. On the other hand, when the valve of a steam boiler is opened, there is a net outward flow of steam at each elemental volume. So, there exists a positive divergence. When an evacuated bulb is broken, there is a transient negative divergence in the space that was inside the bulb before breaking it.

Example 1.27

Find the divergence of the vector field, F = 2 xyax + za y + yz 2 az , at point (2, –1, 3).

Solution The divergence of the vector is given as —◊ F =

∂Fx ∂Fy ∂Fz ∂ ∂ ∂ + + = (2 xy ) + ( z ) + ( yz 2 ) = 2 y + 2 yz ∂x ∂y ∂z ∂x ∂y ∂z

At point, (2, –1, 3), the divergence if obtained as —◊ F

Example 1.28

(2, -1, 3)

= 2 y + 2 yz = 2 ¥ ( -1) + 2 ¥ ( -1) ¥ 3 = - 8

If A = x 2 zi - 2 y 3 z 2 j + xy 2 zk , find div A at the point (1, –1, 1).

Solution —◊ A= \

∂ 2 ∂ ∂ ( x z) + (- 2 y 3 z 2 ) + ( xy 2 z ) = 2 xz - 6 y 2 z 2 + xy 2 ∂x ∂y ∂z

—◊ A

Example 1.29

(1, –1,1)

= 2 – 6 +1= -3

The electric field at a point P, expressed in cylindrical coordinate system is

given by E = 16r 2 sin f ar + 3r 2 cos f af Find the value of divergence of the field, if the location of the point P is given by (1, 2, 3) m in Cartesian coordinate system.

Solution Here, x = 1, y = 2, z = 3 \

r=

x2 + y 2 =

()

Ê yˆ 2 = 63.43∞, z = 3 5, f = tan -1 Ë ¯ = tan -1 x 1

For the electric field, E = 16r 2 sin f ar + 3r 2 cos f af , the divergence in cylindrical coordinates is given as ∂ 1 ∂ 1 ∂Ef ∂Ez 1 ∂ 1 ∂ + = (rEr ) + (16r 3 sin f ) + (3r 2 cos f ) + (0) r ∂r r ∂f ∂z r ∂r r ∂f ∂z = 48r sin f - 3r sin f = 45r sin f

—◊ E =

Hence, the divergence of the field at point P is given as — ◊ E = 45r sin f = 45 ¥

5 sin (63.43∞) = 90

48

Electromagnetic Field Theory

*Example 1.30 An electric field at point P, expressed in cylindrical coordinate system is given by E = 6r 2 sin f ar + 2r 2 cos f af Find the value of the divergence of the field if the location of the point P is given by (1, 1, 1) in Cartesian coordinate system.

Solution The divergence in cylindrical coordinate system is given as 1 ∂ 1 ∂Ef ∂Ez 1 ∂ 1 ∂ + = (rEr ) + (6r 3 sin f ) + (2r 2 cos f ) + 0 r ∂r r ∂f ∂z r ∂r r ∂f = 18r sin f - 2r sin f = 16r sin f

—◊ E =

From the relation between the Cartesian and cylindrical coordinates, we have r=

x2 + y 2

sin f =

— ◊ E = 16r sin f = 16 ¥

\

y x + y2 2

x2 + y 2 ¥

y x + y2 2

= 16 y

Hence, the divergence at point (1, 1, 1) is \

— ◊ E = 16 y = 16 ¥ 1 = 16

Example 1.31

Determine the divergence of the vector field given as V = r cos q ar -

1 sin q aq + 2 r 2 sin q af r

Solution The divergence in spherical coordinate system is given as 1 r2 1 = 2 r

— ◊V =

1 1 ∂Vf ∂ 2 ∂ ( r Vr ) + (Vq sin q ) + r sin q ∂q r sin q ∂f ∂r

1 1 ∂ 3 ∂ Ê 1 ∂ ( r cos q ) + (2 r 2 sin q ) - sin 2 q ˆ˜ + r sin q ∂q ÁË r ∂r ¯ r sin q ∂f 1 = 3 cos q - 2 2 sin q cos q + 0 r sin q Ê 2 ˆ = Á 3 - 2 ˜ cos q r Ë ¯

Curl of a Vector Definition The curl of a vector field, denoted as curl F or — ¥ F , is defined as the vector field having magnitude equal to the maximum circulation at each point and to be oriented perpendicularly to this plane of circulation for each point. Mathematically, it is defined as the limit of the ratio of the integral of the cross product of the vector with outward drawn normal over a closed surface, to the volume enclosed by the surface, as the volume tends to zero.

Vector Analysis

Ê Ú F ¥ dS ˆ Ê Ú F ¥ an dS ˆ ÁS ˜ Á ˜ S Curl F = Lim Ë ¯ = Lim ¯ v v vÆ0 vÆ0 Ë

49

(1.67)

In other words, the component of curl of a vector in the direction of the unit vector an is the ratio of the line integral of the vector around a closed contour, to the area enclosed by the contour, as the area tends to zero. Ê Ú F ◊ dl Curl F = Lim ÁÁ l DS D S Æ0 Ë

\

ˆ ˜a ˜¯ n

(1.68)

where, the direction of the contour is obtained from right hand cork-screw rule.

Mathematical Expression of Curl In order to find an expression for the curl of a vector, we consider an elemental area in the yz-plane as shown in Fig. 1.30. We define a vector F at the centre of the area P(x, y, z). The closed line integral of F around the path abcd is, a

b

c

d

d

a

b

c

a

Ê

Ú F ◊ dl = Ú F ◊ dl + Ú F ◊ dl + Ú F ◊ dl + Ú F ◊ dl l

Now, ∂F D z ˆ

y Ú F ◊ d l = ÁË Fy - ∂z 2 ˜¯ Dy d

∂F Dy ˆ ˜¯ D z

b

Ê z Ú F ◊ d l = ÁË Fz + ∂y 2 a c

Ê

Fig. 1.30

∂F D z ˆ

To derive expression for curl in Cartesian coordinates

y Ú F ◊ d l = - ÁË Fy + ∂z 2 ˜¯ Dy b d

∂F Dy ˆ ˜¯ D z

Ê z Ú F ◊ d l = - ÁË Fz - ∂y 2 c

Summing up, we get Ê ∂F

∂F ˆ

y z Ú F ◊ d l = ÁË ∂y - ∂z ˜¯ Dy D z l

Therefore, the x-component of the curl (since the area is considered in the yz-plane) of the vector is given as Curl x F = Lim

D S Æ0

Ú F ◊ dl l

DS

Ê Ú F ◊ dl = Lim ÁÁ l D S Æ0 Ë Dy D z

ˆ ˜ = Ê ∂Fz - ∂Fy ˆ ∂z ¯˜ ¯˜ ËÁ ∂y

Similarly, considering the area in the xy and xz planes, we will get the other two components of the curl and can be written as

50

Electromagnetic Field Theory

∂F ˆ Ê ∂F Curl y F = Ë x - z ¯ ∂z ∂x Ê ∂Fy ∂Fx ˆ Curl z F = Á ∂y ˜¯ Ë ∂x Thus, the curl of the vector, considering all three directions, is given as ∂Fy ˆ Ê ∂F Ê ∂Fy ∂Fx ˆ ∂F ˆ Ê ∂F Curl F = — ¥ F = Á z ax + Ë x - z ¯ a y + Á a ˜ ∂ ∂ ∂ ∂ ∂y ˜¯ z y z z x Ë ¯ Ë ∂x

(1.69)

In matrix form, this can be written as

—¥F =

ax

ay

az

∂ ∂x Fx

∂ ∂y Fy

∂ ∂z Fz

(1.70)

Physical Interpretation The physical significance of the curl of a vector at any point is that it provides a measure of the amount of rotation or angular momentum of the vector around the point. We consider a stream on the surface of which floats a leaf, in the xy-plane.

Fig. 1.31

(a) Rotation of a floating leaf, and (b) Interpretation of curl

If the velocity at the surface is only in y-direction and is uniform over the surface, there will be no circulation of the leaf. But, if there are vertices or eddies in the stream, there will be rotational movement of the leaf. The rate of rotation or angular velocity at any point is a measure of the curl of the velocity of the stream at that point. In this case, the rotation is about the z-axis and the curl of velocity vector V in the z-direction is written as (— ¥ V ) z . A positive value of (— ¥ V ) z implies a rotation from x to y, i.e., anticlockwise. It is seen that For positive value of

∂V y , rotation is anticlockwise. ∂x

For negative value of

∂Vx , rotation is clockwise. ∂y

Vector Analysis

Fig. 1.32

51

Interpretation of positive and negative velocity gradients

The rate of rotation about the z-axis is therefore, proportional to the difference between these two quantities, i.e., Ê ∂Vy ∂Vx ˆ (— ¥ V ) z = Á ∂y ˜¯ Ë ∂x Now, considering any point within the fluid, there may be rotations about the x and y axes, too. Thus, ∂V y ˆ Ê ∂V (— ¥ V ) x = Á z ∂z ˜¯ Ë ∂y ∂V ˆ Ê ∂V (— ¥ V ) y = Ë x - z ¯ ∂z ∂x A rotation about any axis can be expressed as the sum of the component rotations about the x, y and z axes. Since rotations have both magnitude and direction, the vector sum gives the resultant rotation as ∂V y ˆ Ê ∂V Ê ∂V ax + ÁË x Curl V = — ¥ V = Á z ˜ ∂z ¯ ∂z Ë ∂y ax Ê ∂Vy ∂Vx ˆ ∂ +Á a = ∂y ˜¯ z ∂x Ë ∂x Vx

∂Vz ˆ ˜a ∂x ¯ y a y az

-

∂ ∂y Vy

∂ ∂z Vz

Curl in General Curvilinear Coordinates We consider the general curvilinear coordinates as shown in Fig. 1.27 (see page 41). We consider the vector F = F1a1 + F2 a2 + F3a3 In order to find a general expression for curl of the vector, we need to calculate the closed line integral of the vector per unit area for the elemental volume as shown in Fig. 1.27(See page 41).

52

Electromagnetic Field Theory

The a1 component of curl of F is obtained by taking the line integral through the path OABCO. The contributions to the closed line integral Ê Ú F ◊ d l ˆ through the four paths are given as follows. Ël ¯ A

B

C

O

O

A

B

C

Ú F ◊ dl = Ú F ◊ dl + Ú F ◊ dl + Ú F ◊ dl + Ú F ◊ dl l

∂ ∂ = [ F2 h2 du2 ] + ÈÍ F3 h3 du3 + ( F3 h3 )du3 du2 ˘˙ - ÈÍ F2 h2 du2 + ( F2 h2 )du2 du3 ˘˙ - [ F3 h3 du3 ] ∂ u ∂ u 2 3 Î ˚ Î ˚ ∂ ∂ È ˘ =Í ( F3 h3 ) ( F h ) du du ∂u3 2 2 ˙˚ 2 3 Î ∂u 2 By definition of curl, this equals the a1 component of curl of F , i.e., (— ¥ F )1 multiplied by the area of face OABC. ∂ ∂ (— ¥ F )1 h2 h3du2 du3 = ÈÍ ( F3h3 ) ( F2 h2 ) ˘˙ du2 du3 ∂ ∂ u u 3 Î 2 ˚

\

(— ¥ F )1 =

\

1 È ∂ ∂ (F h ) (F h )˘ ∂u3 2 2 ˙˚ h2 h3 ÍÎ ∂u2 3 3

By cyclic change of indices, we get the other two components of the curl and are written as (— ¥ F ) 2 =

1 È ∂ ∂ (F h ) (F h )˘ ∂u1 3 3 ˙˚ h3h1 ÍÎ ∂u3 1 1

(— ¥ F )3 =

1 È ∂ ∂ (F h ) (F h )˘ ∂u2 1 1 ˙˚ h1h2 ÍÎ ∂u1 2 2

Hence, the expression for curl is given as —¥F =

1 È ∂ ∂ 1 È ∂ ∂ (F h ) ( F h )˘ a + (F h ) ( F h )˘ a h2 h3 ÍÎ ∂u2 3 3 ∂u3 2 2 ˙˚ 1 h3 h1 ÍÎ ∂u3 1 1 ∂u1 3 3 ˙˚ 2 +

1 È ∂ ∂ (F h ) ( F h )˘ a h1h2 ÍÎ ∂u1 2 2 ∂u2 1 1 ˙˚ 3

or in matrix form h1a1 1 ˆ ∂ — ¥ F = ÊÁ Ë h1h2 h3 ˜¯ ∂u1 F1h1

h2 a2 ∂ ∂u 2

h3a3 ∂ ∂u3

F2 h2

F3h3

(1.71)

Substituting the values of hi from Section 1.8, we get the relation of curl in three different coordinate systems as ∂Fy ˆ Ê ∂F Ê ∂Fy ∂Fx ˆ ∂F ˆ Ê ∂F —¥F =Á z ax + ÁË x - z ˜¯ a y + Á a ˜ ∂z ¯ ∂z ∂x ∂y ˜¯ z Ë ∂y Ë ∂x

(Cartesian coordinates)

(1.72a)

Vector Analysis

53

()

È 1 ∂Az ∂Af ˘ ∂Ar ˘ È ∂Ar ∂Az ˘ 1 È∂ —¥ A=Í ˙ ar + Í ∂z - ∂r ˙ af + r Í ∂r (rAf ) - ∂f ˙ az r z ∂ f ∂ Î ˚ Î ˚ Î ˚ (Cylindrical coordinates) (1.72b) ∂Vr ˘ ˘ ∂V ˘ ∂ 1 ˆÈ ∂ 1 È 1 ∂Vr 1È ∂ — ¥ V = ÊÁ (sin q Vf ) - q ˙ ar + Í ( rVf ) ˙ aq + Í ( rVq ) a ˜ Í ∂f ˚ ∂r ∂q ˙˚ f r Î sin q ∂f r Î ∂r Ë r sin q ¯ Î ∂q ˚ (Spherical coordinates) (1.72c) Or in matrix form as

—¥F =

ax

ay

az

∂ ∂x Fx

∂ ∂y Fy

∂ ∂z Fz

ar

raf

az

1 ∂ = ÊË ˆ¯ r ∂r Fr

∂ ∂f rFf

∂ ∂z Fz

(Cartesian coordinates)

(1.73a)

(Cylindrical coordinates)

(1.73b)

ar

r aq

r sin q af

1 ˆ ∂ = ÊÁ 2 Ë r sin q ˜¯ ∂r Fr

∂ ∂q r Fq

∂ (Spherical coordinates) ∂f r sin q Ff

(1.73c)

Properties of Curl 1. The result of the curl of a vector field is another vector field. 2. Curl of a scalar field has no meaning. 3. If the value of curl of a vector field is zero, then the vector field is said to be irrotational or conservative field. Electrostatic field is one of such fields.

Example 1.32

Determine the curl of the following vector fields:

(a) F = x 2 yax + y 2 za y - 2 xzaz (b) A = r 2 sin f ar + r cos 2 f af + z tan f az (c) V =

sin f cos f ar - 2 af r2 r

Solution (a) F = x 2 yax + y 2 za y - 2 xzaz

54

Electromagnetic Field Theory

The curl in Cartesian coordinate system is given as

—¥F =

ax

ay

az

ax

∂ ∂x Fx

∂ ∂y Fy

∂ ∂ = ∂z ∂x Fz x2 y

ay ∂ ∂y y2 z

az ∂ = - y 2 ax + 2 za y - x 2 az ∂z - 2 xz

(b) A = r 2 sin f ar + r cos 2 f af + z tan f az The curl in cylindrical coordinate system is given as

—¥ A=

ar

raf

az

1 ∂ r ∂r Ar

∂ ∂f rAf

∂ ∂z Az

()

()

È 1 ∂Az ∂Af ˘ ∂Ar ˘ È ∂Ar ∂Az ˘ 1 È∂ =Í ˙ ar + Í ∂z - ∂r ˙ af + r Í ∂r rAf - ∂f ˙ az r z ∂ f ∂ Î ˚ Î ˚ Î ˚ 1 1 = ( z sec 2 f - 0)ar + (0 - 0)af + (2r cos 2 f - r 2 cos f )az r r 1 = ÎÈ z sec 2 f ar + (2r cos 2 f - r 2 cos f )az ˚˘ r

()

(c) V =

( )

sin f cos f ar - 2 af 2 r r

The curl in spherical coordinate system is given as ar

r aq

r sin q af

1 ˆ ∂ — ¥ V = ÊÁ 2 Ë r sin q ˜¯ ∂r Vr

∂ ∂q rVq

∂ ∂f r sin q Vf

˘ ∂V ˘ ∂ 1 ˆÈ ∂ 1 È 1 ∂Vr = ÊÁ (sin q Vf ) - q ˙ ar + Í ( rVf ) ˙ aq Í ˜ ∂f ˚ ∂r r Î sin q ∂f Ë r sin q ¯ Î ∂q ˚ + 1 ˆÈ ∂ = ÊÁ Í Ë r sin q ˜¯ ÎÍ ∂q

1 r

∂Vr ˘ È ∂ Í ( rVq ) ˙a ∂q ˚ f Î ∂r

˘ cos f ˆ Ê 1 È 1 ∂ Ê sin f ˆ ∂ Ê cos f ˆ ˘ ÁË sin q r 2 ˜¯ - 0˙ ar + r Í sin q ∂f ÁË r 2 ˜¯ - ∂r ÁË r r 2 ˜¯ ˙ aq ÍÎ ˚˙ ˚˙ +

1 r

È ∂ Ê sin f ˆ ˘ Í0 - ∂q Á 2 ˜ ˙ af Ë r ¯ ˙˚ ÍÎ

cos f ˆ 1 ˆ Ê cos q cos f ˆ 1 Ê cos f = ÊÁ ar + Á 2 + 2 ˜ aq ˜ Á ˜ 2 r Ë r sin q Ë r sin q ¯ Ë r r ¯ ¯

Vector Analysis

ˆ 1 1 Ê cos f — ¥ V = ÊÁ 3 cot q cos f ˆ˜ ar + 3 Á + cos f ˜ aq sin q ¯ r Ë Ër ¯

\

Example 1.33

Determine the divergence and curl of the following vector fields:

(a) f = yzax + 4 xya y + yaz (b) A = r 2 zar + r 3af + 3rz 2 az (c) F =

1 cos q ar + r sin q cos f aq + cos q af r2

Solution (a) f = yzax + 4 xya y + yaz —◊f =

\

\

∂ ∂ ∂ ( yz ) + (4 xy ) + ( y ) = 0 + 4 x + 0 = 4 x ∂x ∂y ∂z

—¥f =

ax

ay

az

∂ ∂x yz

∂ ∂y 4 xy

∂ = ax + ya y + (4 y - z )az ∂z y

(b) A = r 2 zar + r 3af + 3rz 2 az \

—◊ A =

∂ 1 ∂ 1 ∂Af ∂Az 1 ∂ 3 1 ∂ 3 + = (rAr ) + (r z ) + (r ) + (3rz 2 ) = 3rz + 0 + 6rz = 9rz ∂z ∂z r ∂r r ∂f r ∂r r ∂f

( 1r ) a

af

∂ ∂r Fr

∂ ∂f rFf

r

\

(c) F = \

—¥ A=

( 1r ) a ( 1r ) a z

∂ ∂z Fz

r

=

∂ ∂r r2z

af ∂ ∂f r4

( 1r ) a

z

∂ = 0 + (r 2 - 3 z 2 )af + 4r 2 az ∂z 3rz 2

1 cos q ar + r sin q cos f aq + cos q af r2 —◊F = =

1 ∂ 1 1 ∂Ff ∂ ( r 2 Fr ) + ( Fq sin q ) + 2 ∂r r sin q ∂q r sin q ∂f r 1 ∂ 1 1 ∂ ∂ (cos q ) + ( r sin 2 q cos f ) + (cos q ) 2 ∂r sin sin r q ∂ q r q ∂ f r

=0+

2 sin q cos q cos f + 0 sin q

= 2 cos q cos f

55

56

Electromagnetic Field Theory

\

ar

r aq

r sin q af

∂ ∂r

∂ ∂q

∂ ∂f

1 cos q r2

r 2 sin q cos f

r sin q cos q

1 ˆ — ¥ F = ÊÁ 2 Ë r sin q ˜¯

Ê cos 2q ˆ 1 1 =Á + sin f ˜ ar - cos q aq + ÁÊ 2 cos f + 3 ˜ˆ sin q af r Ë r sin q ¯ r ¯ Ë

Example 1.34

If A = xz 3i - 2 x 2 yzj + 2 yz 4 k , find Curl A at the point (1, –1, 1).

Solution ax —¥ A=

∂ ∂x xz 3

ay

az

∂ ∂ = (2 z 4 + 2 x 2 y )ax + (3xz 2 - 0)a y + ( - 4 xyz - 0)az ∂y ∂z - 2 x 2 yz 2 yz 4

= (2 z 4 + 2 x 2 y )ax + 3 xz 2 a y - 4 xyzaz \

—◊ A

(1, -1,1)

= (2 - 2)ax + 3a y + 4az = 3a y + 4az

*Example 1.35 (a) For a vector field A, show explicitly that — ◊ — ¥ A = 0 , i.e., the divergence of the curl of any vector field is zero. (b) For a scalar field V, show that — ¥ —V = 0, i.e., the curl of the gradient of any scalar field is zero. Solution

(a) Let, A = Ax ax + Ay a y + Az az

\

—¥ A=

ax

ay

az

∂ ∂x Ax

∂ ∂y Ay

∂Ay ˆ Ê ∂A Ê ∂Ay ∂Ax ˆ ∂A ˆ Ê ∂A ∂ =Á z ax + ÁË x - z ˜¯ a y + Á a ˜ ∂z Ë ∂y ∂z ¯ ∂z ∂x ∂y ˜¯ z Ë ∂x Az

= Gx a x + G y a y + Gz a z \

(Let)

∂G x ∂G y ∂G z + + ∂x ∂y ∂z ∂ A Ê ˆ y ∂ ∂Az ∂ Ê ∂Ax ∂Az ˆ ∂ Ê ∂Ay ∂Ax ˆ = + Á ˜+ Á ˜ ∂x Ë ∂y ∂z ¯ ∂y Ë ∂z ∂x ¯ ∂z ÁË ∂x ∂y ˜¯

—◊— ¥ A=

Ê ∂2 A ∂2 Ay ˆ Ê ∂2 Ax ∂2 Az ˆ Ê ∂2 Ay ∂2 Ax ˆ z =Á + + Ë ∂x∂y ∂x∂z ¯˜ ËÁ ∂y∂z ∂y∂x ¯˜ ËÁ ∂z∂x ∂z∂y ¯˜ Ï ∂2 Az ∂2 Az ¸ =0Ì = and so on ˝ Ó ∂ x∂ y ∂ y ∂ x ˛ \ — ◊ — ¥ A == 0

Vector Analysis

(b) —V =

\

∂V ∂V ∂V + + ∂x ∂y ∂z ax

ay

az

∂ ∂x ∂V ∂x =0

∂ ∂y ∂V ∂y

Ê ∂ 2V Ê ∂ 2V Ê ∂ 2V ∂ ∂ 2V ˆ ∂ 2V ˆ ∂ 2V ˆ ax + ËÁ a =Á ˜¯ a y + Á ˜ ∂z Ë ∂y∂z ∂z ∂y ¯ ∂z∂x ∂x∂z Ë ∂x∂y ∂y∂x ¯˜ z ∂V ∂z

— ¥ —V =

\ — ¥ —V = 0

NOTE These two vector identities are known as null identities.

Example 1.36 *(a) Find the divergence and curl of the vector field: E =

( rx ) a , where r = xa + ya + za . x

x

y

z

*(b) Prove that — ◊ r = 3 and — ¥ r = 0 where, r = xax + ya y + zaz .

Solution

(a) Here, r = xax + ya y + zaz \

r= E=

x2 + y 2 + z 2

( rx ) a = x

x

x2 + y 2 + z 2 - x \

—◊E =

=

\

∂ Ê ∂x Á Ë

ˆ= 2 2 2 ˜ x +y +z ¯

ax

x + y2 + z2 2

x

1 2

x +y +z 2

2

2x x + y2 + z2 2

2

=

x2 + y 2 + z 2 - x2 ( x 2 + y 2 + z 2 )3/2

y2 + z2 y2 + z2 = ( x 2 + y 2 + z 2 )3/2 r3

—¥E=

ax

ay

az

∂ ∂x x

∂ ∂y

∂ ∂ = Ê ∂z ∂z Á Ë 0

x2 + y 2 + z 2 =-

0

ˆa - ∂ Ê y ∂y Á x + y + z ˜¯ Ë x

2

2

2

x2 y 1 1 x2 z ay + a 2 2 2 3/2 2 2 (x + y + z ) 2 ( x + y 2 + z 2 )3/2 z

ˆa z x + y + z ˜¯ x

2

2

2

57

58

Electromagnetic Field Theory

xy xz ay + 2 az ( x 2 + y 2 + z 2 )3/2 ( x + y 2 + z 2 )3/2 xy xz = - 3 a y + 3 az r r =-

(b) Here, r = xax + ya y + zaz

—¥r =

\

\

ax

ay

az

∂ ∂x x

∂ ∂y y

Ê ∂z ∂y ˆ Ê ∂y ∂x ˆ ∂ ∂x ∂z = a + a + a =0 ∂z ÁË ∂y ∂z ˜¯ x ∂z ∂x y ÁË ∂x ∂y ˜¯ z z

(

—◊r = 3

Solution

)

—¥r =0

and

1 Curl v , where w is a constant vector. 2 Here, r = xax + ya y + zaz and let, w = w x ax + w y a y + w z az If v = w ¥ r , prove that w =

Example 1.37

\

∂ ∂ ∂ ( x) + ( y) + ( z) = 1 + 1 + 1 = 3 ∂x ∂y ∂z

—◊r =

\

ax

ay

az

Curl v = — ¥ v = — ¥ (w ¥ r ) = — ¥ w x

wy

wz

x

y

z

= — ¥ [(w y z - w z y )ax + (w z x - w x z )a y + (w x y - w y x)az ] ax =

ay

az

∂ ∂ ∂ ∂x ∂y ∂z (w y z - w z y ) (w z x - w x z ) (w x y - w y x)

∂ ∂ ∂ ˘ È∂ ˘ = ÈÍ (w x y - w y x) (w x - w x z ) ˙ ax + Í (w y z - w z y ) (w y - w y x) ˙ a y ∂z z ∂x x Î ∂z ˚ Î ∂y ˚ ∂ ∂ + ÈÍ (w z x - w x z ) (w z - w z y ) ˘˙ az ∂y y Î ∂x ˚ = 2w x ax + 2w y a y + 2w z a z = 2(w x ax + w y a y + w z az ) = 2w \ w=

1 Curl v 2

Vector Analysis

59

Laplacian (—2) Operator The Laplacian operator (—2) can operate both on scalar as well as vector field.

Laplacian (—2) of a Scalar

The Laplacian operator (—2) of a scalar field is the divergence of the gradient of the scalar field upon which the operator operates. Practically, it is a single operator, which is the composite of gradient and divergence operators. The Laplacian of a scalar field is also a scalar field. The Laplacian of a scalar field F in Cartesian coordinate system is written as ∂ ∂ ∂ ˆ Ê ∂F ∂F ∂F ˆ ∂ 2 F ∂ 2 F ∂ 2 F —2 F = — ◊ —F = ÁÊ ax + ay + az ˜ ◊ Á ax + ay + a = + 2 + 2 (1.74) ∂y ∂z ¯ Ë ∂x ∂y ∂z z ¯˜ ∂x 2 Ë ∂x ∂y ∂z Similarly, the expression of Laplacian of a scalar in other two coordinate systems can be obtained. The gradient of the scalar field F in curvilinear coordinate system is given as —F = where, A1 =

1 ∂F 1 ∂F 1 ∂F a + a + a = A (say) = A1a1 + A2 a2 + A3 a3 h1 ∂u1 1 h2 ∂u2 2 h3 ∂u3 3

1 ∂F ; h1 ∂u1

A2 =

1 ∂F ; h2 ∂u2

A3 =

1 ∂F h3 ∂u3

Divergence of this vector in curvilinear coordinate system is given as — ◊ A = — ◊ —F = — 2 F =

1 È ∂ ∂ ∂ (A h h ) + (A h h ) + ( A h h )˘ h1h2 h3 ÍÎ ∂u1 1 2 3 ∂u 2 2 1 3 ∂u3 3 1 2 ˙˚

Substituting the values of A1, A2 and A3, we have —2 F =

1 h1h2 h3

È ∂ Ê h2 h3 ∂F ˆ ∂ Í ∂u ÁË h ∂u ˜¯ + ∂u 1 1 1 2 Î

Ê h1h3 ∂F ˆ ∂ ÁË h ∂u ˜¯ + ∂u 2 2 3

Ê h1h2 ∂F ˆ ˘ ÁË h ∂u ˜¯ ˙ 3 3 ˚

(1.75)

Substituting the values of hi from Section 1.8, we get the relation of Laplacian of scalar field in three different coordinate systems as ∂2 F ∂2 F ∂2 F + 2 + 2 ∂x 2 ∂y ∂z

—2 F =

( )

(Cartesian coordinates)

=

1 ∂ ∂F 1 ∂2 F ∂2 F r + 2 + 2 r ∂r ∂r r ∂f 2 ∂z

=

1 ∂ Ê 2 ∂F ˆ 1 ∂ ∂F 1 ∂2 F sin q + 2 2 Á r ∂r ¯˜ + 2 2 ∂r Ë ∂q r r sin q ∂q r sin q ∂f 2

(Cylindrical coordinates)

(

)

(1.76a) (1.76b) (Spherical coordinates) (1.76c)

NOTE If the Laplacian of a scalar field V is zero in a region, i.e., —2V = 0, then the scalar field V is said to be harmonic (containing sine or cosine terms) in that region and the equation —2V = 0 is known as Laplace’s equation. We will learn about Laplace’s equation in more detail in Chapter 2.

60

Electromagnetic Field Theory

Laplacian (—2) of Products of Scalar Fields If a scalar field is represented as the product of two other scalar functions, then the Laplacian of that scalar field in Cartesian coordinate system is written as —2 (uv) = (—2u )v + u (—2v) + 2(—u ) ◊ (—v)

(1.77)

Proof: By definition of scalar Laplacian in Cartesian coordinate system —2 (uv) =

∂2 ∂2 ∂2 (uv) + 2 (uv) + 2 (uv) 2 ∂x ∂y ∂z

Now, ∂2 ∂ È∂ ˘ = ∂ È ∂u v + u ∂v ˘ = ∂ 2 u v + ∂u ∂v + ∂u ∂v + u ∂ 2 v ( uv ) = ( uv ) ∂x ÎÍ ∂x ∂x ˚˙ ∂x 2 ∂x ∂x ∂x ∂x ˚˙ ∂x ÎÍ ∂x ∂x 2 ∂x 2 2 2 ∂ u ∂u ∂v ∂ v = 2v+2 +u 2 ∂x ∂x ∂x ∂x Similarly, ∂2 ∂ 2u ∂u ∂v ∂ 2v (uv) = 2 v + 2 +u 2 2 ∂y ∂y ∂y ∂y ∂y ∂2 ∂ 2u ∂u ∂v ∂ 2v (uv) = 2 v + 2 +u 2 2 ∂z ∂z ∂z ∂z ∂z By summation, we get ∂ 2u ∂u ∂v ∂ 2 v ∂2u ∂u ∂v ∂ 2 v ∂ 2u ∂u ∂v ∂2v v+2 +u 2 + 2 v+2 +u 2 + 2 v+2 +u 2 2 ∂x ∂x ∂y ∂y ∂z ∂z ∂x ∂x ∂y ∂y ∂z ∂z 2 2 2 2 2 2 Ê ∂ u ∂ u ∂ uˆ Ê ∂ v ∂ v ∂ vˆ = Á 2 + 2 + 2 ˜ v + uÁ 2 + 2 + 2 ˜ ∂y ∂z ¯ ∂y ∂z ¯ Ë ∂x Ë ∂x ∂u ∂u ∂u ˆ Ê ∂v ∂v ∂v ˆ Ê a + a ◊ a + a + a + 2 Á ax + ∂y y ∂z z ˜¯ ÁË ∂x x ∂y y ∂z z ˜¯ Ë ∂x

—2 (uv) =

= (—2 u )v + u (—2 v) + 2(—u ) ◊ (—v)

Laplacian (—2) of a Vector

The Laplacian of a vector is defined as the gradient of divergence of the vector minus the curl of curl of the vector; i.e., — 2 F = —( — ◊ F ) - — ¥ — ¥ F

(1.78)

In terms of general curvilinear coordinate system, Laplacian of a vector is written as ∂ Ê h1h3 ∂ ˆ ∂ Ê h1h2 ∂ ˆ ˘ 1 È ∂ Ê h2 h3 ∂ ˆ + + F h1h2 h3 ÍÎ ∂u1 ÁË h1 ∂u1 ˜¯ ∂u2 ÁË h2 ∂u2 ˜¯ ∂u3 ÁË h3 ∂u3 ˜¯ ˙˚ ∂ Ê h1h3 ∂F ˆ ∂ Ê h1h2 ∂F ˆ ˘ 1 È ∂ Ê h2 h3 ∂F ˆ = + + Í Á ˜ Á ˜ h1h2 h3 Î ∂u1 Ë h1 ∂u1 ¯ ∂u2 Ë h2 ∂u2 ¯ ∂u3 ÁË h3 ∂u3 ˜¯ ˙˚

—2 F =

È ∂ Ï h2 h3 ∂ ¸ ¸˘ ∂ Ï h1h3 ∂ Í ∂u Ì h ∂u ( F1a1 + F2 a2 + F3a3 ) ˝ + ∂u Ì h ∂u ( F1a1 + F2 a2 + F3a3 ) ˝˙ 1 Í 1Ó 1 1 2 Ó 2 2 ˛ ˛˙ (1.79) = h1h2 h3 Í ¸˙ ∂ Ï h1h2 ∂ + ( F a + F2 a2 + F3a3 ) ˝ ˙ Í ∂u3 ÌÓ h3 ∂u3 1 1 ˛˚ Î

Vector Analysis

61

This is the general expression for vector Laplacian in curvilinear coordinate system.

Derivation of Vector Laplacian in Cartesian Coordinate System

In Cartesian coordinate

system u1 = x, u2 = y, u3 = z h1 = h2 = h3 = 1 From Eq. (1.76),

} { {

{

}

È∂ ∂ ∂ ∂ ( Fx ax + Fy a y + Fz az ) + ( F a + Fy a y + Fz az ) Í ∂ x ∂ x ∂ y ∂y x x 2 — F =Í ∂ ∂ Í + ( F a + Fy a y + Fz az ) ∂z ∂z x x ÎÍ

}

˘ ˙ ˙ ˙ ˚˙

Now ∂a y ˆ Ê ∂Fz ∂a ˆ Ê ∂Fy ∂a ˆ Ê ∂F ∂ + Fy + Fz z ¯ ( Fx ax + Fy a y + Fz az ) = Ë ax x + Fx x ¯ + ÁË a y ˜+ a ∂x ∂x ∂x ∂x ∂x ¯ Ë z ∂x ∂x Since the unit vectors in Cartesian coordinate system are considered to be constant, we have ∂a y =0 ∂x

∂a x =0 ∂x

∂a z ∂x

\

∂Fy Ê ∂F ∂F ˆ ∂ + az z ˜¯ ( F a + Fy a y + Fz az ) = ÁË ax x + a y ∂x x x ∂x ∂x ∂x

\

∂Fy ∂F ˆ ∂ ∂ ∂ Ê ∂Fx + ay + az z ˜¯ ( F a + Fy a y + Fz az ) = Áa ∂x ∂x x x ∂x Ë x ∂x ∂x ∂x

{

}

= ax

∂ 2 Fx ∂x

2

+ ay

∂2 Fy ∂x

2

+ az

∂2 Fz ∂x 2

(1.80a)

Similarly,

{

}

{

}

∂ 2 Fy ∂ 2 Fx ∂ 2 Fz ∂ ∂ + a + az ( Fx ax + Fy a y + Fz az ) = ax y 2 2 ∂y ∂y ∂y ∂y ∂y 2

(1.80b)

and, ∂ 2 Fy ∂ 2 Fx ∂ 2 Fz ∂ ∂ ( Fx ax + Fy a y + Fz az ) = ax + a + az y 2 2 ∂z ∂z ∂z ∂z ∂z 2 By summation of Eqs. (1.80a) to (1.80c), we get Ê ∂2 F ∂2 Fy ∂2 Fy ∂2 Fz ˆ Ê ∂2 Fx ∂2 Fz ˆ x a a a a a — 2 F = Á ax + + + + + ˜ Á ˜ y z x y z Ë ∂x 2 ∂x 2 ∂x 2 ¯ Ë ∂y 2 ∂y 2 ∂y 2 ¯ Ê ∂2 F ∂2 Fy ∂2 Fz ˆ x a + Á ax + + az ˜ y 2 2 Ë ∂z ∂z ∂z 2 ¯

(1.80c)

62

Electromagnetic Field Theory

Ê ∂2 Fy ∂2 Fy ∂2 Fy ˆ Ê ∂2 F Ê ∂2 F ∂2 Fx ∂2 Fx ˆ ∂2 Fz ∂2 Fz ˆ = Á 2x + + + + + a a + Á 2z + az Á 2 + 2 2 ˜ x 2 2 ˜ y ∂y ∂z ¯ ∂y ∂z ¯ ∂y 2 ∂z 2 ˜¯ Ë ∂x Ë ∂x Ë ∂x = (— 2 Fx )ax + (— 2 Fy )a y + (— 2 Fz )az \ — 2 F = (— 2 Fx ) ax + (— 2 Fy ) a y + (— 2 Fz ) az

(1.81)

Since the unit vectors in cylindrical and spherical coordinate systems are not constants, evaluation of vector Laplacian in these two coordinate systems become tedious. However, we can write the final expressions of vector Laplacian in these two coordinate systems as given. Cylindrical coordinate system Ê Ê ˆ 2 ∂Ff 1 ˆ 2 ∂Fr 1 —2 F = Á —2 Fr - 2 - 2 Fr ˜ ar + Á —2 Ff + 2 - 2 Ff ˜ af + —2 Fz az Ë ¯ Ë ¯ r ∂f r r ∂f r

(1.82)

Spherical coordinate system È ∂Ff ∂Fq ˆ ˘ 2 Ê — 2 F = Í— 2 Fr - 2 Á Fr + cot q Fq + cosec q + ˙a ∂f ∂q ˜¯ ˚˙ r r Ë ÎÍ È ∂Fr ∂Ff ˆ ˘ 1 Ê + Í— 2 Fq - 2 Á cosec 2 q Fq - 2 + 2 cot q cosec q ˙a ∂q ∂f ˜¯ ˙˚ q r Ë ÍÎ È ∂Fr ∂F ˆ ˘ 1 Ê + Í— 2 Ff - 2 Á cosec 2 q Ff - 2 cosec q - 2 cot q cosec q q ˜ ˙ af ∂f ∂f ¯ ˙˚ r Ë ÎÍ

(1.83)

NOTE The Laplacian of a vector field is zero if and only if the Laplacian of each of its components is independently zero.

Example 1.38 (a) F =

Find the Laplacian of the following scalar fields:

x2 + y 2 + z 2

(b) F = rz sin f + r2 + z2 cos2 f (c) F = e–r sin q cos f

Solution (a) F =

x2 + y 2 + z 2 —2 F =

∂2 ∂2 ∂2 ( x2 + y 2 + z 2 ) + 2 ( x2 + y 2 + z 2 ) + 2 ( x2 + y 2 + z 2 ) 2 ∂x ∂y ∂z

Vector Analysis

Now,

∂ ( x2 + y 2 + z 2 ) = ∂x

x x + y2 + z2 2

x2 + y 2 + z 2 - x ∂2 ∂ Ê ( x2 + y 2 + z 2 ) = ∂x Á ∂x 2 Ë

ˆ= 2 2 2 ˜ x +y +z ¯ 2 y + z2 = 2 ( x + y 2 + z 2 )3/2

\

x

x x + y2 + z2 2

( x2 + y 2 + z 2 )

Similarly, x2 + z 2 ∂2 2 2 2 ( ) + + = x y z ( x 2 + y 2 + z 2 )3/2 ∂y 2 x2 + y 2 ∂2 ( x2 + y 2 + z 2 ) = 2 2 ( x + y 2 + z 2 )3/2 ∂z Hence, the Laplacian is given as —2 F =

y2 + z2 x2 + z 2 x2 + y 2 + + = ( x 2 + y 2 + z 2 )3/2 ( x 2 + y 2 + z 2 )3/2 ( x 2 + y 2 + z 2 )3/2

2 x + y2 + z2 2

(b) F = rz sin f + r2 + z2 cos2 f

( ) {

1 ∂ ∂F 1 ∂2 F ∂2 F r + 2 + 2 r ∂r ∂r r ∂f 2 ∂z ∂ 1 ∂ 1 ∂2 2 = r (rz sin f + r + z 2 cos 2 f ) + 2 (rz sin f + r 2 + z 2 cos 2 f ) 2 r ∂r ∂r r ∂f ∂2 + 2 (rz sin f + r 2 + z 2 cos 2 f ) ∂z ∂ 1 ∂ 1 ∂ 2 = {rz sin f + 2r } + 2 (rz cos f - 2 z 2 cos f sin f ) + (r sin f + 2 z cos 2 f ) r ∂r ∂z r ∂f 1 1 = ( z sin f + 4r ) + 2 (- rz sin f - 4 z 2 cos 2f ) + 2 cos 2 f r r z z z2 = 4 + sin f - sin f - 4 2 cos 2f + 2 cos 2 f r r r z2 2 = 4 + 2 cos f - 4 2 cos 2f r

—2 F =

}

63

64

Electromagnetic Field Theory

(c) F = e–r sin q cos f

(

)

∂ Ê 2 ∂F ˆ 1 ∂ ∂F 1 ∂2 F + 2 sin q + 2 2 r Á ˜ ∂r Ë ∂r ¯ r sin q ∂q ∂q r sin q ∂f 2 ∂ ∂ 1 {r 2 (- 1)e - r sin q cos f} + 2 {sin q e - r cos q cos f} ∂r r sin q ∂q ∂ -r 1 - 2 2 (e sin q sin f ) ∂ r sin q f 1 ∂ -r 2 1 1 =- 2 {e r sin q cos f} + 2 e - r cos 2q cos f - 2 2 (e - r sin q cos f ) r ∂r r sin q r sin q 1 e - r cos f = - 2 sin q cos f (2 r e - r - e - r r 2 ) + 2 (cos 2q - 1) r r sin q e - r cos f 2 (1 - cos 2q ) = sin q cos f ÁÊ e - r - e - r ˜ˆ - 2 r Ë ¯ r sin q e - r cos f 2 = sin q cos f ÁÊ e - r - e - r ˜ˆ - 2 2 sin 2 q r Ë ¯ r sin q 1 r2 1 = 2 r

—2 F =

e- r 2 = sin q cos f ÁÊ e - r - e - r ˜ˆ - 2 2 sin q cos f r Ë ¯ r 2 2 = e - r sin q cos f ÁÊ 1 - - 2 ˆ˜ r r ¯ Ë

Example 1.39

Find the Laplacian of the scalar fields:

(a) S = x y + xyz (b) S = r2z(cos f + sin f) (c) S = cos q sin f ln r + r2f 2

Solution

(a) S = x2 y + xyz —2 S =

∂2 S ∂2 S ∂2 S ∂ ∂ 2 ∂ + 2 + 2 = (2 xy + yz ) + ( x + xz ) + ( xy ) = 2 y 2 ∂ x ∂ y ∂ z ∂x ∂y ∂z

(b) S = r2 z(cos f + sin f)

( )

1 ∂ ∂S 1 ∂2 S ∂2 S r + 2 + r ∂r ∂r r ∂f 2 ∂ z 2 ∂ 1 ∂ 1 ∂ 2 = [2r 2 z (cos f + sin f )] + 2 [r z (cos f - sin f )] + [ r 2 (cos f + sin f )] r ∂r ∂z r ∂f = 4 z (cos f + sin f ) - z (cos f + sin f ) = 3z (cos f + sin f )

—2 S =

(c) S = cos q sin f ln r + r2 f —2 S =

(

)

1 ∂ Ê 2 ∂S ˆ 1 ∂ ∂S 1 ∂2 S r + sin q + Á ˜ ∂q r 2 ∂r Ë ∂r ¯ r 2 sin q ∂q r 2 sin 2 q ∂f 2

Vector Analysis

65

1 ∂ 1 ∂ [ r cos q sin f + 2 r 3f ] - 2 [sin 2 q sin f ln r ] r 2 ∂r r sin q ∂q 1 ∂ [cos q cos f ln r + r 2 ] + 2 2 ∂ f r sin q 1 È 1 1 2 ˘ [2 sin q cos q sin f ln r ] - 2 2 [cos q sin f ln r ] = 2 Îcos q sin f + 6 r f ˚ - 2 r r sin q r sin q È cos q sin f ˘ È 2 cos q sin f ln r ˘ È cos q sin f ln r ˘ =Í + 6f ˙ - Í ˙ - Í r 2 sin 2 q ˙ 2 r r2 ˚ Î ˚ Î ˚ Î cos q sin f = (1 - 2 ln r - cosec 2 q ln r ) + 6f r2

\ —2 S =

()

*Example 1.40 Prove that — 2 1 = 0 , with usual meaning of r. r

Solution —2

Now,

∂ Ê ∂x Á Ë

()

1 ∂2 = 2Ê r ∂x ËÁ

x ˆ = ∂ ʈ ÁË ( x 2 + y 2 + z 2 )3/2 ˜¯ ˜ ∂ x x +y +z ¯ 1

2

2

2

= ∂2 Ê ∂y 2 ÁË

∂2 Ê ∂z 2 ËÁ

2

2

=

and,

ˆ ˜ x +y +z ¯ 1

2

x ˆ =2 2 ˜ ( x + y + z 2 )3/2 x +y +z ¯ 2

=-

Similarly,

ˆ + ∂2 Ê 2 x 2 + y 2 + z 2 ¯˜ ∂z ËÁ 1

1

2

∂2 Ê ∂x 2 ÁË

\

ˆ + ∂2 Ê 2 x 2 + y 2 + z 2 ¯˜ ∂y ÁË 1

3 2 x( x 2 + y 2 + z 2 )1/2 2 ( x 2 + y 2 + z 2 )3

( x 2 + y 2 + z 2 )3/2 - x

3x 2 - ( x 2 + y 2 + z 2 ) ( x 2 + y 2 + z 2 )5/2 2 x2 - y 2 - z 2 ( x 2 + y 2 + z 2 )5/2

2 2 2 ˆ = 2y - z - x 2 2 2 5/2 x 2 + y 2 + z 2 ˜¯ ( x + y + z )

1

2 2 2 ˆ = 2z - x - y 2 2 2 5/2 x 2 + y 2 + z 2 ¯˜ ( x + y + z )

1

By addition, we get —2

()

2x2 - y 2 - z 2 2 y 2 - z 2 - x2 2 z 2 - x2 - y 2 1 = 2 + 2 + 2 =0 2 2 5/2 2 2 5/2 r (x + y + z ) (x + y + z ) ( x + y 2 + z 2 )5/2

2

66

Electromagnetic Field Theory

—2

Example 1.41

()

1 =0 r

Given the vector field, E = ( x 2 + y 2 )ax + ( x + y )a y ; Find — 2 E.

Solution — 2 E = (— 2 E x ) a x + (— 2 E y ) a y + (— 2 E z ) a z Ê ∂2 ∂2 ∂2 ˆ where, — 2 = Á 2 + 2 + 2 ˜ ∂y ∂z ¯ Ë ∂x Now, —2 Ey will not survive since Ey is linear. Also, —2 Ez vanishes as Ez does not exist.

\

1.10

∂E x = 2 x; ∂x

∂2 Ex =2 ∂x 2

∂E x = 2 y; ∂y

∂2 Ex =2 ∂y 2

∂E x = 0; ∂z

∂2 Ex =0 ∂z 2

— 2 E = — 2 Ex ax = (2 + 2)ax = 4ax

GAUSS’ DIVERGENCE THEOREM

Statement This theorem states that the divergence of a vector field over a volume is equal to the surface integral of the normal component of the vector through the closed surface bounding the volume. Mathematically,

Ú — ◊ Fdv = Ú F ◊ dS = Ú F ◊ an dS

V

S

(1.84)

S

where V is the volume enclosed by the closed surface S.

Proof By definition of divergence in Cartesian coordinates, —◊F = \

∂Fx ∂Fy ∂Fz + + ∂x ∂y ∂z

Ê ∂F

∂F

∂F ˆ

y x z Ú — ◊ Fdv = Ú ÁË ∂x + ∂y + ∂z ˜¯ dxdydz V V

where, dv = dxdydz We consider an elemental volume as shown in Fig. 1.33.

Fig. 1.33

Elemental volume in Cartesian coordinates

Vector Analysis

67

Let the rectangular volume have the dimensions dx, dy and dz along the x, y and z directions respectively. Now,

∂F

x2

Here,

∂F

È ˘ x x Ú ∂x dxdydz = Ú ÎÍ Ú ∂x dx ˚˙ dydz V S ∂F

x Ú ∂x dx = ( Fx 2 - Fx1 ) x 1

where, Fx1 and Fx2 are the x component of the vector on the back and front side of the element along x axis. ∂F

\

x Ú ∂x dxdydz = Ú ( Fx 2 - Fx1 )dydz = Ú Fx dS x V S S

where, dSx = dydz is the x component of the surface area dS. Thus, the above equation gives the surface integral of Fx with the x component of dS over the whole surface. ∂Fy ∂F Similarly, considering and z and summing up, we get ∂y ∂z \

Ú — ◊ Fdv = Ú ( Fx dS x + Fy dS y + Fz dS z ) = Ú F ◊ dS

V

S

S

where, dS = dS x ax + dS y a y + dS z az and F = Fx ax + Fy a y + Fz az . Thus,

Ú — ◊ Fdv = Ú F ◊ dS

V

S

This is Gauss’ divergence theorem.

NOTE The divergence theorem is a mathematical statement of the physical fact that, in the absence of the creation or destruction of matter, the density within a region of space can change only by having it flow into or away from the region through its boundary. Intuitively, this theorem implies that the sum of all sources minus the sum of all sinks gives the net flow out of a region. The divergence theorem is an important result for the mathematics of engineering, particularly in electrostatics and fluid mechanics.

1.10.1

Green’s Identities

These identities are corollaries of the divergence theorem and can be derived as follows. We consider,

A = S1 — S2

(1.85)

where S1 and S2 are scalar functions continuous together with their partial derivatives of first and second order. \

— ◊ A = — ◊ ( S1 — S2 ) = S1 — 2 S2 + —S1 ◊ —S2

{by vector identity}

68

Electromagnetic Field Theory

Applying divergence theorem

Ú — ◊ Adv = Ú A ◊ dS

V

S

Substituting the value of — ◊ A and A, we get 2 Ú ( S1 — S2 + —S1 ◊ —S2 )dv = Ú ( S1 — S2 ) ◊ dS

V

(1.86)

S

This is the first form of Green’s identity. Now, interchanging the functions S1 and S2, we get 2 Ú ( S2 — S1 + —S2 ◊ —S1 )dv = Ú ( S2 — S1 ) ◊ dS

V

(1.87)

S

Subtracting Eq. (1.87) from Eq. (1.86), we get 2 2 Ú ( S1 — S2 - S2 — S1 )dv = Ú ( S1 — S2 - S2 — S1 ) ◊ dS

V

(1.88)

S

This is the second form of Green’s identity. Using Green’s identities, it can be proved that the specifications of both divergence and curl of a vector with boundary conditions are sufficient to make the function unique (known as Uniqueness theorem).

*Example 1.42 Given that A = 2 xyax + 3a y + yz 2 az , evaluate Ú A ◊ dS , where S is the surface of S

the cube defined by 0 £ x £ 1, 0 £ y £ 1, 0 £ z £ 1. Also, verify the result by using divergence theorem.

Solution We evaluate the surface integrals for the six surfaces as follows.

Fig. 1.34

Surface integral of Example 1.42

Ú

S

A ◊ dS =

1

1

Ú Ú

x=0 y=0

+

1

Az 1

Ú Ú

z =0 x=0

dxdy + z =1 (- Ay )

y=0

1

1

Ú Ú

x=0 y=0

dxdz +

(- Az ) z = 0 dxdy + 1

1

Ú Ú

y=0 z =0

Ax

1

1

Ú Ú Ay

z =0 x=0

dydz + x =1

1

1

y =1

dxdz

Ú Ú (- Ax ) x = 0 dydz

y=0 z =0

Vector Analysis

Ax = 2xy;

where

Ay = 3;

Ú A ◊ dS =

\

S

1

69

Az = yz2 1

( y - 0) dxdy +

Ú Ú

x=0 y=0

1

1

Ú Ú

(3 - 3) dxdz +

z =0 x=0

1

1

Ú Ú (2 y - 0) dydz

y=0 z =0

1 + 0 +1 2 3 = 2 =

Also, — ◊ A =

\

∂ ∂ ∂ (2 xy ) + (3) + ( yz 2 ) = 2 y + 2 yz = 2 y (1 + z ) ∂x ∂y ∂z

1 1 1 1 1 1 1 3 Ú — ◊ Adv = Ú Ú Ú 2 y (1 + z ) dxdydz = Ú Ú 2 y (1 + z ) dydz = Ú (1 + z ) dz = 1 + 2 = 2 z =0 y=0 x=0

v

z =0 y=0

z =0

Since, Ú — ◊ Adv = Ú A ◊ dS , divergence theorem is verified. v

S

Ê

3ˆ

*Example 1.43 Given that A = ÁË 5r ˜¯ ar C/m2 in cylindrical coordinates, evaluate both sides of 4 divergence theorem for the volume enclosed by r = 1 m, r = 2 m, z = 0 and z = 10 m.

Solution Here, dS = rdf dzar \

È 2p

10

Ê

3

˘

ˆ

5r Ú A ◊ dS = Í Ú Ú ÁË 4 ar ˜¯ ◊ (rd f dzar )˙ S ÍÎf = 0 z = 0 ˙˚

È 2p 10 Ê 5r 3 ˆ ˘ ar ˜¯ ◊ (rd f dzar ) ˙ - Í Ú Ú ÁË 4 ÍÎf = 0 z = 0 ˙˚ r =1 r=2

È 2p 10 5r 4 ˘ È 2p 10 5r 4 ˘ d f dz ˙ d f dz ˙ =Í Ú Ú -Í Ú Ú 4 4 ÎÍf = 0 z = 0 ˚˙ r = 2 ÎÍf = 0 z = 0 ˚˙ r =1 =

2p 10

2p 10

80 5 Ú Ú 4 d f dz - Ú Ú 4 d f dz f =0 z =0 f =0 z =0 2p 10

=

75 Ú Ú 4 d f dz f =0 z =0

=

75 ¥ 2p Ú dz 4 z =0

=

75 ¥ 2p ¥ 10 4

10

= 375p Also, — ◊ A =

1 ∂ Ê 5r 3 ˆ 1 ∂ Ê 5r 4 ˆ 5 1 = ¥ ¥ 4r 3 = 5r 2 Ár ˜= r ∂r Ë 4 ¯ r ∂r ÁË 4 ˜¯ 4 r

Electromagnetic Field Theory

70

\

2 Ú (— ◊ A)dv = Ú 5r rdrd f dz = v

v

2

2p 10

Ú Ú Ú

5r 3 drd f dz = 2p ¥ 10

r =1 f = 0 z = 0

2

È r4 ˘ 3 Ú 5r dr = 100p ÍÎ 4 ˙˚ 1 r =1 2

= 375p Since, Ú — ◊ Adv = Ú A ◊ dS , divergence theorem is verified. v

S

*Example 1.44 For the given vector F = ( x 2 - y 2 )ax + 2 xya y + ( x 2 - xy )az , evaluate the surface integral

Ú F ◊ dS over the surface of the cube with centre at the origin and side length a.

S

Fig. 1.35

Surface integral of a vector

Solution We calculate the surface integrals Ú F ◊ dS for all six surfaces and then add up to get the closed surface integral

S

Ú F ◊ dS .

S

For surface abcd

x=

a , dS = dydzax 2

2 2 2 Ú F ◊ dS = Ú [( x - y )ax + 2 xya y + ( x - xy )az ] ◊ (dydzax ) =

abcd

S

=

2

3

y a a¥a-a 4 3

a /2

= – a /2

4

4

a /2

Ú

y = - a /2

a /2

Ê

ˆ

2

a 2 Ú ÁË 4 - y ˜¯ dydz z = - a /2

4

a a a = 4 12 6

For surface efgh a x = - , dS = - dydzax 2

Ú

efgh

F ◊ dS = Ú [( x 2 - y 2 )ax + 2 xya y + ( x 2 - xy )az ] ◊ (- dydzax ) = S

a4 =6

a /2

Ú

y = - a /2

a /2

Ê

2

ˆ

a 2 Ú ÁË 4 - y ˜¯ dydz z = - a /2

Vector Analysis

For surface cdhg y=

a , dS = dxdza y 2 a /2

2 2 2 Ú F ◊ dS = Ú [( x - y )ax + 2 xya y + ( x - xy )az ] ◊ (dxdza y ) = Ú

cdhg

a /2

Ú 2 xydxdz = 0

x = -a/2 z = -a/2

S

For surface abfe a y = - , dS = - dxdza y 2 a /2

2 2 2 Ú F ◊ dS = Ú [( x - y )ax + 2 xya y + ( x - xy )az ] ◊ (- dxdza y ) = - Ú

abfe

a /2

Ú 2 xydxdz = 0

x = - a /2 z = – a /2

S

For surface adhe z=

a , dS = dxdyaz 2

2 2 2 Ú F ◊ dS = Ú [( x - y )ax + 2 xya y + ( x - xy )az ] ◊ (dxdyaz ) =

adhe

S

=

a /2

Ú

a /2

2 Ú ( x - xy ) dxdy

x = - a /2 y = - a /2

a4 12

For surface bcgf a z = - , dS = - dxdyaz 2 a /2

2 2 2 Ú F ◊ dS = Ú [( x - y )ax + 2 xya y + ( x - xy )az ] ◊ (- dxdyaz ) = - Ú

bcgf

a /2

2 Ú ( x - xy ) dxdy

x = - a /2 y = - a /2

S

a4 =12 Addition of all six surface integrals gives

Ú F ◊ dS = 0

S

We can verify the result by using divergence theorem as follows: ∂Fx ∂Fy ∂Fz ∂ 2 ∂ ∂ + + = (x - y2 ) + (2 xy ) + ( x 2 - xy ) ∂x ∂y ∂z ∂x ∂y ∂z = 2x + 2x + 0 = 4x

—◊F =

\

a /2

Ú — ◊ Fdv = Ú v

a /2

Ú

a /2

2 2 a /2 Ú 4 xdxdydz = 2a [ x ]- a /2 = 0

- a /2 - a /2 - a /2

Ú — ◊ Fdv = Ú F ◊ dS , divergence theorem is verified. v

S

71

72

Electromagnetic Field Theory

x3 a , evaluate both sides of the divergence theorem for the 3 x volume of a cube, 1 m on an edge, centered at the origin and with edges parallel to the axes.

Example 1.45

Given that A =

3 Solution Here, A = x ax

3

Ê x3 ˆ 1 A ◊ dS = ÁË ax ˜¯ ◊ (dydzax ) = x3 dydz 3 3

\ È

\

0.5

1 3 Ú 3 x dydz ˙ ÎÍ y = - 0.5 z = - 0.5 ˚˙

S

= Also, — ◊ A =

0.5 È 0.5 ˘ 1 3 -Í Ú Ú 3 x dydz ˙ ÎÍ y = - 0.5 z = - 0.5 ˚˙ x = - 0.5 x = 0.5

1 1 1 ¥1¥1+ ¥1¥1= 24 24 12

∂ Ê x3 ˆ Á ˜ = x2 ∂x Ë 3 ¯ 0.5

\

0.5

Ú (— ◊ A)dv = Ú

Ú

x = - 0.5 y = - 0.5

v

Hence,

˘

0.5

Ú A ◊ dS = Í Ú

0.5

È x3 ˘ 1 Ú x dxdydz = ÍÎ 3 ˙˚ ¥ 1 ¥ 1 = 12 - 0.5 z = - 0.5 0.5

2

Ú A ◊ dS = Ú (— ◊ A)dv; thus, divergence theorem is verified.

S

v

*Example 1.46 Given that F = 10e – r ar - 2 zaz , evaluate both sides of the divergence theorem for the volume enclosed by r = 2, z = 0 and z = 5.

Solution Here, F = 10e- r ar - 2 zaz dS = rd f dzar + drdzaf + rdrdf az \

F ◊ dS = (10e - r ar - 2 za z ) ◊ (rd f dzar + drdzaf + rdrd f a z ) = 10re - r d f dz - 2rzdrd f

The cylinder has three surfaces as follows.

Ú F ◊ dS = Ú F ◊ dS + Ú F ◊ dS + Ú F ◊ dS s

top

bottom

curved

For the top surface, z = 5

Ú

F ◊ dS =

2

2p

Ú Ú

- 2rzdrd f

r =0f =0

top

= - 10 z =5

2

2p

2

2 Ú Ú rdrd f = - 10 ¥ 2p ¥ 2 = - 40p r =0f =0

For the bottom surface, z = 0 2

2p

Ú F ◊ dS = Ú Ú - 2rzdrd f

bottom

r =0f =0

=0 z =0

For the curved surface, r = 2 2p

5

-r Ú F ◊ dS = Ú Ú 10 re d f dz

curved

f =0 z =0

= 20e - 2 ¥ 2p ¥ 5 = 200p e - 2 r=2

Vector Analysis

73

By addition, total surface area of the closed cylinder is given as -2 Ú F ◊ dS = - 40p + 0 + 200p e = - 40.63 s

Also, — ◊ F =

1 ∂ 1 ∂Ff ∂Fz 1 ∂ 10e - r ∂ (rFr ) + (10re - r ) + (- 2 z ) = + = - 10e - r - 2 r ∂r r ∂f ∂z r ∂r ∂z r 2

\

2p

Ê 10e - r

5

Ú (— ◊ F )dv = Ú Ú Ú ÁË r v r =0f =0 z=0

ˆ - 10e - r - 2˜¯ rdrd f dz

2

-r -r Ú (10e - 10re - 2r ) dr = - 40.63

= 2p ¥ 5 ¥

r =0

\

Ú F ◊ dS = Ú (— ◊ F ) dv and hence, divergence theorem is verified. s

v

*Example 1.47 Given the vector D = divergence theorem for (a) the volume enclosed by r = 1 and r = 2. *(b) the volume enclosed by r = 2 and q =

Solution Here, D =

5r 2 a in spherical coordinates. Verify both sides of the 4 r

p . 4

5r 2 a 4 r

\

dS = r 2 sin q dq d f ar

\

Ê 5r 2 ˆ 5 D ◊ dS = ÁË a ˜ ◊ ( r 2 sin q dq d f ar ) = r 4 sin q dq d f 4 r¯ 4

(a) Here, the surface integral is radially outward for the surface at r = 2 and radially inward at r = 1. È

p

˘

È 5 p 2p ˘ - Í Ú Ú r 4 sin q dq d f ˙ 4 ˚˙ r = 2 ÎÍ q = 0 f = 0 ˚˙ r =1

2p

5 4 Ú D ◊ dS = Í 4 Ú Ú r sin q dq d f ˙

\

S

ÎÍ

q =0f =0

= 20 ¥ 2p ¥ (- cos q )p0 = 80p - 5p = 75p Also, — ◊ D = \

( )

1 ∂ 2 1 ∂ 5 4 r = 5r ( r Dr ) = 2 r 2 ∂r r ∂r 4

Ú (— ◊ D)dv = v

\

5 ¥ 2p ¥ (- cos q )p0 4

p

2p

Ú D ◊ dS = Ú (— ◊ D) dv; and thus divergence theorem is verified.

S

v

2

È r4 ˘ 2 Ú Ú Ú 5rr sin q d r dq d f = 5 ÍÎ 4 ˙˚ ¥ 2 ¥ 2p = 75p 1 r =1 q = 0 f = 0 2

74

Electromagnetic Field Theory

(b) Here, the surface integral is non-vanishing only at r = 2. È

p

˘

2p

5 4 Ú D ◊ dS = Í 4 Ú Ú r sin q dq d f ˙

\

ÍÎ

S

˙˚ r = 2

q =0f =0

= 20 ¥ 2p ¥ (- cos q )p0 = 80p 2

È r4 ˘ Also, Ú (— ◊ D)dv = Ú Ú Ú 5rr sin q d r dq d f = 5 Í ˙ ¥ 2 ¥ 2p = 80p Î 4 ˚0 v r=0q =0f =0 2 p /4 2p

2

Thus, divergence theorem is verified.

*Example 1.48 Evaluate both sides of the divergence theorem. Ú A ◊ dS = Ú — ◊ Adv

S

v

For each of the following cases: (a) A = xy 2 ax + y 3 a y + y 2 zaz and S is the surface of the cuboid defined by 0 < x < 1, 0 < y < 1, 0 < z < 1. (b) A = 2r zar + 3 z sin f af - 4r cos f az and S is the surface of the wedge 0 < r < 2, 0 < f < 45°, 0 < z < 5.

Solution (a) (a) Here, A = xy 2 ax + y 3a y + y 2 zaz We evaluate the surface integrals for the six surfaces as follows. 1

1

Ú A ◊ dS = Ú Ú Az x=0 y=0

S

+

1

dxdy + z =1

1

Ú Ú Ay

z =0 x=0

+

1

1

Ú Ú

y=0 z =0

where Ax = xy ; 2

\

Ú

S

Ax

Ay = y ;

1

Ú Ú (- Az ) z = 0 dxdy

x=0 y=0

dxdz +

1

1

Ú Ú (- Ay ) y = 0 dxdz

z =0 x=0

dydz + x =1

1

1

Ú Ú (- Ax ) x = 0 dydz

y=0 z =0

Fig. 1.36

Az = y z

3

A ◊ dS =

y =1

1

2

1

1

Ú Ú

( y 2 - 0) dxdy +

x=0 y=0

1

1

Ú Ú

(1 - 0)dxdz +

z =0 x=0

1 1 = +1+ 3 3 5 = 3 Also, — ◊ A =

∂ ∂ 3 ∂ ( xy 2 ) + ( y ) + ( y2 z) = y2 + 3 y2 + y2 = 5 y2 ∂x ∂y ∂z

1

1

Surface integral of Example 1.48 (a)

2 Ú Ú ( y - 0)dydz

y=0 z =0

Vector Analysis 1

1

1

1

5 2 2 Ú — ◊ Adv = Ú Ú Ú 5 y dxdydz = Ú 5 y dy = 3

\

v

z =0 y=0 x=0

y=0

Since, Ú — ◊ Adv = Ú A ◊ dS , divergence theorem is verified. v

S

(b) Here, A = 2r zar + 3 z sin f af - 4r cos f az We evaluate the surface integrals for the six surfaces as follows. 5

45

Ú A ◊ dS = Ú Ú Ar z =0f =0

S

rd fdz + r=2 +

5

2

z =0 r =0 45∞

2

Ú Ú Az

f =0 r =0

where Ar = 2rz; Af = 3z sin f;

5

2

Ú Ú Af f = 45∞ drdz + Ú Ú (- Af ) f = 0 drdz z =0 r =0

rdrd f + z =5

45∞

2

Ú Ú (- Az ) z = 0 rdrd f

f =0 r =0

Az = –4r cos f

Surface integral of Example 1.48 (b)

Fig. 1.37

5 45 5 2 45∞ 2 45∞ 2 3 zdrdz + 0 + Ú Ú - 4rrdrd f + Ú Ú 4rrdrd f Ú A ◊ dS = Ú Ú 8 zd f dz + Ú Ú

\

S

z =0f =0

2

z =0 r =0

f =0 r =0

f =0 r =0

75 = 25p + 2 —. A=

Also,

1 ∂ 1 ∂ ∂ 3z (2r 2 z ) + (3 z sin f ) + (- 4r cos f ) = 4 z + cos f r ∂r r ∂f ∂z r

(

)

5 45∞ 2 3z 75 Ú — ◊ Adv = Ú Ú Ú 4 z + r cos f rdrd fdz = 25p +

\

v

z =0f =0 r =0

Since Ú — ◊ Adv = Ú A ◊ dS , divergence theorem is verified. v

S

2

75

Electromagnetic Field Theory

76

Example 1.49 Find the total charge inside a cubical volume of 1 m on a side situated in the positive octant with three edges coincident with the x, y and z axis and one corner at the origin, if D = ( x + 3)ax . Solve the problem using (a) divergence theorem, and (b) by integrating D over surface of the cube. Solution Here, the flux density is given as, D = ( x + 3)ax We can find the total charge by two methods: (a) by divergence theorem, or (b) by evaluating the surface integrals of the flux density over surface of the cube. (a) By divergence theorem —◊D= 1

∂ ( x + 3) = 1 ∂x 1

1

Ú — ◊ Adv = Ú Ú Ú 1dxdydz = 1

\

z =0 y=0 x=0

v

(b) By evaluating the surface integrals of the flux density over the surface of the cube Here, as the flux density is having only x-component, the surface integral reduces to 1

1

Ú D ◊ dS = Ú Ú Dx y=0 z =0

S

dydz + x =1

1

1

1

1

1

1

Ú Ú (- Dx ) x = 0 dydz = Ú Ú (4 - 3) dydz = Ú Ú dydz = 1

y=0 z =0

y=0 z =0

y=0 z =0

Thus, we see that in both the methods, the total charge inside the cube is found to be 1 Coulomb.

1.11

STOKES’ THEOREM

Statement This theorem states that the line integral of a vector around a closed path is equal to the surface integral of the normal component of its curl over the surface bounded by the path. Mathematically, (1.89) Ú F ◊ d l = ÚÚ (— ¥ F ) ◊ dS = ÚÚ (— ¥ F ) ◊ an dS L

S

S

where S is the surface enclosed by the path L. The positive direction of dS is related to the positive sense of defining L according to the right-hand rule.

Proof We consider an arbitrary surface S as shown in Fig. 1.38. We divide the surface into a large number of still smaller elements 1, 2, 3, … etc. Taking the line integrals of all such small elements and summing up,

Ú F ◊ dl = Ú F ◊ dl + Ú F ◊ dl + Ú F ◊ dl

L

1

2

3

Ú F ◊ dl

= Â Ú F ◊ dl = Â k k k

k

D Sk

D Sk

where DSk is the area of the k element. th

Fig. 1.38

Arbitrary surface S in the yz plane bounded by closed path L

Vector Analysis

77

(Due to cancelation of every interior path, the sum of the line integrals around kth elements is the same as the line integral around the bounding path L, i.e., Ú F ◊ d l + Ú F ◊ d l + Ú F ◊ d l = Ú F ◊ dl ) 1

2

3

L

As DSk Æ 0, summation approaches integration, S Æ Ú, so that Ê Ú F ◊ dl Á ÂÁ k Ú F ◊ d l = DLim S k Æ0 k Ë D S k L

ˆ Ê Ú F ◊ dl ˜ D S = Lim Á k ˜¯ k Ú D S Æ0 ÁË D S k k S

ˆ ˜ DS ˜¯ k

Since, by definition of curl of a vector, the line integral divided by the surface area is the component of Ê Ú F ◊ dl ˆ ˜ = — ¥ F , we can write the curl normal to the surface, i.e., Lim ÁÁ k ˜ D S k Æ0 Ë D S k ¯ Ê Ú F ◊ dl Á ÂÁ k Ú F ◊ d l = DLim Sk Æ 0 k Ë D S k L \

ˆ Ê Ú F ◊ dl ˜ Ák ˜¯ D Sk = Ú DLim Á Sk Æ 0 Ë D S k S

ˆ ˜ ˜¯ D Sk = Ú (— ¥ F ) ◊ dS S

Ú F ◊ d l = Ú (— ¥ F ) ◊ dS

L

S

NOTE Stokes’ theorem is a special case of Green’s theorem in plane. To obtain Stokes’ theorem from Green’s theorem, we have to make two changes. First, the line integral in two dimensions (Green’s theorem) is changed to a line integral in three dimensions (Stokes’ theorem). Second, the double integral of curl F ◊ ak over a region R in the plane (Green’s theorem) is changed to a surface integral of curl F ◊ an over a surface S floating in space (Stokes’ theorem).

*Example 1.50 Given that F = ax + y 2 za y . Verify Stokes’ theorem for this vector field and the flat surface in the yz plane bounded by (0, 0, 0), (0, 1, 0), (0, 1, 1) and (0, 0, 1). Choose the contour in the clockwise direction.

Solution Here, F ◊ d l = (ax + y 2 za y ) ◊ (dxa x + dya y + dza z ) = (dx + y 2 zdy )

Ú F ◊ dl = Ú F ◊ dl + Ú F ◊ dl + Ú F ◊ dl + Ú F ◊ dl

L

L1

L2

L3

L4

Along L1: dx = dy = 0 1

2 Ú F ◊ d l = Ú (dx + y zdy ) = 0

\

z =0

L1

Along L2: dx = 0, \

Ú

L2

F ◊ dl =

1

Ú

y=0

1

( y 2 zdy ) z =1 =

y3 1 = 3 3 0

78

Electromagnetic Field Theory

Along L3: dx = dy = 0 \

Ú F ◊ dl = 0

L3

Along L4: dx = 0, z = 0 0

\

2 Ú F ◊ d l = Ú (dx + y zdy ) = 0 y =1

L4

Hence, the line integral is given as

Ú F ◊ dl = Ú F ◊ dl + Ú F ◊ dl + Ú F . dl + Ú F ◊ dl

L

L1

L2

L3

L4

Fig. 1.39

Line integral of Example 1.50

1 1 =0+ +0+0= 3 3 To evaluate surface integral, we have the elemental surface given as dS = dydz (- ax ) = - dydzax

Also, — ¥ F =

\

ax

ay

az

∂ ∂x 1

∂ ∂y y2 z

∂ = - y 2 ax ∂z 0

Ú (— ¥ F ) ◊ dS =

S

1

1

Ú Ú

z =0 y=0

(- y 2 ax ) ◊ (- dydzax ) =

1

1

Ú Ú

z =0 y=0

1

y 2 dydz =

y3 1 = 3 3 0

Since Ú (— ¥ F ) ◊ dS = Ú F ◊ d l , Stokes’ theorem is verified. S

L

Given A = r sin f ar + r 2 af in cylindrical coordinates. Verify the Stokes’ theorem for the contour shown in Fig. 1.40.

Example 1.51

Fig. 1.40

Contour of Example 1.51

Vector Analysis

79

Solution Here, A ◊ d l = (r sin f ar + r 2 af ) ◊ (drar + rd f af + dzaz ) = (r sin f dr + r 3d f ) \

È

˘

È p /2 ˘ È 0 ˘ + Í Ú r 3df ˙ + Í Ú r sin f dr ˙ = 0 + 4p - 2 = (4p - 2) ˙˚f = 0 ÍÎf = 0 ˙˚ r = 2 ÍÎ r = 2 ˙˚f = p /2

2

Ú A ◊ d l = Í Ú r sin f dr ˙ ÍÎ r = 0

L

1 a r r ∂ Also, — ¥ A = ∂r r sin f \

af ∂ ∂f r3

1 a r z ∂ = (3r - r cos f )az ∂z 0

p /2 2

p /2

f =0 r =0

f =0

p /2 Ú (— ¥ A) ◊ dS = Ú Ú (3r - cos f )rdrd f = Ú (8 - 2 cos f )d f = (8f - 2 sin f )0 = (4p - 2)

S

Since Ú (— ¥ F ) ◊ d S = Ú F ◊ d l , Stokes’ theorem is verified. S

L

*Example 1.52 Evaluate the line integral of vector function F = xi + x 2 yj + y 2 xk around the square contour ABCD in the xy-plane as shown in Fig. 1.41. Also integrate — ¥ F over the surface bounded by ABCD and verify that Stokes’ theorem holds good.

Fig. 1.41

Contour of Example 1.52

Solution Here, F = xi + x 2 yj + y 2 xk d l = dxi + dyj + dzk \

F ◊ d l = ( xi + x 2 yj + y 2 xk ) ◊ (dxi + dyj + dzk ) = xdx + x 2 ydy + y 2 xdz

80

Electromagnetic Field Theory

The closed line integral is given as B

C

D

A

A

B

C

D

Ú F ◊ dl = Ú F ◊ dl + Ú F ◊ dl + Ú F ◊ dl + Ú F ◊ dl

L

Along the path AB, dy = dz = 0, y = 0, z = 0; x varies from 0 to 2 B

È

2

2 ˘2

x Ú F ◊ d l = Ú xdx = ÎÍ 2 ˚˙0 = 2 A 0

\

Along the path BC, dx = dz = 0, x = 2, z = 0; y varies from 0 to 2 C

2

B

0

2

2 Ú F ◊ d l = Ú x ydy

\

x=2

È y2 ˘ = 4Í ˙ = 8 Î 2 ˚0

Along the path CD, dy = dz = 0, y = 2, z = 0; x varies from 2 to 0 C

0

B

2

Ú F ◊ d l = Ú xdx = - 2

\

Along the path DA, dx = dz = 0, x = 0, z = 0; y varies from 2 to 0 C

2

B

0

2 Ú F ◊ d l = Ú x ydy

\

=0 x=0

By addition, we get the closed line integral as

Ú F ◊ dl = 2 + 8 - 2 + 0 = 8

L

Now, the curl of the vector is i ∂ —¥F = ∂x x

j ∂ ∂y x2 y

k ∂ = 2 xyi - y 2 j + 2 xyk ∂z y2 x 2

2

È x2 ˘ È y 2 ˘ Ú (— ¥ F ) ◊ dS = Ú Ú (2 xyi - y j + 2 xyk ) ◊ (dxdyk ) = Ú Ú 2 xydxdy = 2 ÎÍ 2 ˚˙ ¥ ÍÎ 2 ˙˚ = 8 0 0 S x=0 y=0 x=0 y=0 2

2

2

2

2

Thus,

Ú F ◊ d l = Ú (— ¥ F ) ◊ dS ; Stokes’ theorem is verified.

L

1.12

S

CLASSIFICATIONS OF VECTOR FIELDS

A vector field is uniquely characterised by its divergence and curl. From these properties, vector fields are classified into four categories: 1. 2. 3. 4.

Solenoidal and Irrotational Vector Fields (— ◊ F = 0, — ¥ F = 0) Non-solenoidal and Irrotational Vector Fields (— ◊ F π 0, — ¥ F = 0) Solenoidal and Rotational Vector Fields (— ◊ F = 0, — ¥ F π 0) and Non-solenoidal and Rotational Vector Fields (— ◊ F π 0, — ¥ F π 0) .

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81

1. Solenoidal and Irrotational Vector Fields (— ◊ F = 0, — ¥ F = 0) For a given vector field F, if its divergence and curl are both zero (— ◊ F = 0, — ¥ F = 0), the field is known as solenoidal and irrotational vector fields. Such a vector field has neither source nor sink of flux. Examples of such vector fields are: linear motion of incompressible fluids, electrostatic fields in charge-free region, magnetic fields within a current-free region, gravitational fields in free space. 2. Non-solenoidal and Irrotational Vector Fields (— ◊ F π 0, — ¥ F = 0) For a given vector field F, if its divergence is not zero but curl is zero (— ◊ F π 0, — ¥ F = 0), then the field is known as nonsolenoidal and irrotational vector fields. Examples of such vector fields are: electrostatic fields in charged medium, gravitational force inside a mass, linear motion of compressible fluids. 3. Solenoidal and Rotational Vector Fields (— ◊ F = 0, — ¥ F π 0) For a given vector field F , if its divergence is zero but curl is non-zero (— ◊ F = 0, — ¥ F π 0), then the field is known as solenoidal and rotational vector fields. Examples of such vector fields are: magnetic fields within a current carrying conductor, rotational motion of incompressible fluids. 4. Non-solenoidal and Rotational Vector Fields (— ◊ F π 0, — ¥ F π 0) For a given vector field F , if both the divergence and curl are non-zero, (— ◊ F π 0, — ¥ F π 0) then the field is known as nonsolenoidal and rotational vector fields. Examples of such vector fields: rotational motion of compressible fluids.

Example 1.53

(a) Determine the constant c such that the vector F = ( x + ay )i + ( y + bz ) j + ( x + cz )k will be solenoidal. (b) Find the value of constant Q to make V = ( x + 3 y )i + ( y - 2 z ) j + ( x + Qz )k , solenoidal. (c) Find the constants a, b, c so that the vector V = ( x + 2 y + az )i + (bx - 3 y - z ) j + (4 x + cy + 2 z )k is irrotational.

Solution (a) F = ( x + ay )i + ( y + bz ) j + ( x + cz )k The vector will be solenoidal if its divergence is zero. \

—◊F =

∂ ∂ ∂ ( x + ay ) + ( y + bz ) + ( x + cz ) = 0 ∂x ∂y ∂z

fi

1+1+c=0

\

c = –2

(b) V = ( x + 3 y )i + ( y - 2 z ) j + ( x + Qz )k The vector will be solenoidal if its divergence is zero. \

— ◊V =

∂ ∂ ∂ ( x + 3 y) + ( y - 2 z ) + ( x + Qz ) = 0 ∂x ∂y ∂z

fi

1+1+Q=0

\

Q = –2

82

Electromagnetic Field Theory

(c) V = ( x + 2 y + az )i + (bx - 3 y - z ) j + (4 x + cy + 2 z )k The vector will be irrotational if its curl is zero.

\

i j k ∂ ∂ ∂ — ¥V = =0 ∂x ∂y ∂z ( x + 2 y + az ) (bx - 3 y - z ) (4 x + cy + 2 z )

fi

(c + 1)i + (a - 4) j + (b - 2)k = 0

This vector will be zero if and only if its components are individually zero. \ (c + 1) = 0 fi c = –1 (a – 4) = 0 fi a=4 and (b – 2) = 0 fi b=2 Hence, a = 4, b = 2, c = –1.

*Example 1.54 If a scalar potential is given by the expression f = xyz, determine the potential gradient and also prove that the vector F = grad f = grad f is irrotational. Solution Here, the potential is, f = xyz The potential gradient is given as —f =

∂ ∂ ∂ ( xyz )ax + ( xyz )a y + ( xyz )a z = ( yza x + xza y + xya z ) ∂x ∂y ∂z

Now, we let, F = —f = ( yzax + xza y + xya z )

\

— ¥ F = — ¥ —f =

ax

ay

az

∂ ∂x yz

∂ ∂y xz

∂ = (1 - 1) ax + (1 - 1) a y + (1 - 1) az = 0 ∂z xy

Hence, the vector F = —f is irrotational.

1.13

HELMHOLTZ THEOREM

Statement A vector field is uniquely described within a region by its divergence and curl. Explanation If the divergence and curl of any vector F are given as — ◊ F = rv and — ¥ F = rs

(1.90)

where rv is the source density of F , and rs is the circulation density of F both vanishing at infinity, then, according to Helmholtz theorem, we can write a vector field F as a sum of a component Fs whose divergence is zero — ◊ Fs = 0 (solenoidal) and a component Fi whose curl is zero — ¥ Fi = 0 (irrotational).

Vector Analysis

\

83

F = Fs + Fi

Also, we know that the divergence of curl of a vector is zero and the curl of gradient of any scalar is zero. Using these two null identities, we can write Fs and Fi as follows. Fs = — ¥ A and Fi = —f

(1.91)

where A and j are a vector and a scalar quantities respectively. Thus, any vector can be represented by the Helmholtz theorem as F = —f + — ¥ A

(1.92)

Proof We assume that there exist two vector fields F1 and F2 which show identical results for divergence and curl at least at one point, P (say) in volume v of interest. Besides, F1 and F2 both satisfy the boundary condition at the surface S bounded by volume; such that on the surface we can write F1 ◊ an = F2 ◊ an

( F1 - F2 ) ◊ an = 0

fi

(1.93)

At point P, our assumption requires — ◊ F1 = — ◊ F2 — ◊ ( F1 - F2 ) = 0

fi

(1.94)

and — ¥ F1 = — ¥ F2 — ¥ ( F1 - F2 ) = 0

fi

(1.95)

From Eq. (1.94) and (1.95), we see that there exists a vector field H = ( F1 - F2 ) , which is solenoidal and irrotational (— ◊ H = 0 and — ¥ H = 0) . Therefore, we can write H = —f

(1.96)

where f is a scalar field. From Eq. (1.94) and (1.95), —◊H =0

fi

— ◊ —f = 0

fi

— 2f = 0

Now, by Green’s first identity, we can write 2 Ú ( S1 — S2 + —S1 ◊ —S2 )dv = Ú ( S1 — S2 ) ◊ dS

V

S

If S1 = S2 = f (say), then the identity becomes 2 Ú (f — f + —f ◊ —f )dv = Ú (f —f ) ◊ dS

V

fi

S

2 Ú (—f ) dv = Ú (f —f ) ◊ dS

V

S

{

— 2f = 0}

(1.97)

84

Electromagnetic Field Theory

By using Eq. (1.96), we have

Ú (f —f ) ◊ dS = Ú f H ◊ dS = Ú f H ◊ an dS = 0

S

S

{ by Eq. (1.95), H ◊ an = ( F1 - F2 ) ◊ an = 0}

S

Hence, from Eq. (1.97), we have 2 Ú (—f ) dv = 0

V

fi

2 Ú | H | dv = 0

V

fi

H =0

fi

( F1 - F2 ) = 0

\

F1 = F2

Hence, our initial assumption that F1 and F2 are two different vector fields showing identical values of divergence and curl is wrong. Therefore, we conclude that no two vector fields can show identical divergence and curl everywhere. Hence, divergence and curl specifies a vector field uniquely.

1.13.1

Physical Interpretation of Helmholtz Theorem in Classical Electromagnetism

1. Helmholtz theorem is a profound mathematical theorem which provides the definite relationship between a vector field and mathematically defined source functions. The usual presentations of electromagnetic theory establish the principal sources of the electromagnetic field vectors, but do not give information whether all sources are included. Helmholtz theorem provides the basis for investigating the existence of other possible sources. 2. According to Helmholtz theorem, a general vector field can be written as the sum of a conservative field and a solenoidal field. Thus, we ought to be able to write electric and magnetic fields in this form. Second, a general vector field which is zero at infinity is completely specified once its divergence and its curl are given. Thus, we can guess that the laws of electromagnetism can be written as four field equations — ◊ E = something — ¥ E = something — ◊ H = something — ¥ H = something and we can solve the field equations without even knowing the right-hand sides and any solutions we find will be unique. In other words, there is only one possible steady electric and magnetic field which can be generated by a given set of stationary charges and steady currents. If the right-hand sides of the above field equations are all zero, then the only physical solution is E = H = 0 . This implies that steady electric and magnetic fields cannot generate themselves. Instead, they have to be generated by stationary charges and steady currents. So, if we come across a steady electric field, we know that if we trace the field-lines back, we shall eventually

Vector Analysis

85

find a charge. Likewise, a steady magnetic field implies that there is a steady current flowing somewhere. All of these results follow from Helmholtz theorem prior to any investigation of electromagnetism. 3. Helmholtz theorem also provides a significant result pertaining to the meaning of the inverse square radial fields (such as, Newton’s gravitational field and Coulomb’s electrostatic fields). It can be established that the inverse square relation is determined by more elementary properties of the field and of the source function to which it relates.

Summary A quantity that has only magnitude is said to be a scalar quantity, such as time, mass, distance, temperature, work, electric potential, etc. A quantity that has both magnitude and direction is called a vector quantity, such as force, velocity, displacement, electric field intensity, etc. If the value of the physical function at each point is a scalar quantity, then the field is known as a scalar field, such as temperature distribution in a building. If the value of the physical function at each point is a vector quantity, then the field is known as a vector field, such as the gravitational force on a body in space. A unit vector a A along A is defined as a vector whose magnitude is unity and its direction is along A. In general, any vector can be represented as A = Aa A = | A| a A where A or | A| represents the magnitude of the vector and a A direction of the vector A. Two vectors can be added together by the triangle rule or parallelogram rule of vector addition. The dot product of two vectors A and B, written as, A ◊ B, is defined as A ◊ B = AB cos q AB where, qAB is the smaller angle between A and B, and A = | A| and B = | B | represent the magnitude of A and B, respectively. The cross product of two vectors A and B, written as ( A ¥ B), is defined as A ¥ B = AB sin q AB an where an is the unit vector normal to the plane containing A and B. The direction of the cross product is obtained from a common rule, called the right-hand rule. Three orthogonal coordinate systems commonly used are Cartesian coordinates (x, y, z), cylindrical coordinates (r, f, z) and spherical coordinates (r, q, f). The differential lengths in three coordinate systems are given respectively as d l = dxax + dya y + dza z d l = drar + rd f af + dzaz d l = d r ar + rdq aq + r sin q d f af

86

Electromagnetic Field Theory

The differential areas in three coordinate systems are given respectively as dS = dydzax + dxdza y + dxdya z dS = rd f dzar + drdzaf + rdrdf az dS = r 2 sin q dq d f ar + r sin q d r d f aq + rd r dq af The differential volumes in three coordinate systems are given respectively as dV = dxdydz dV = rdrd f dz dV = r 2 sin q d r dq d f For vector F and a path l, the line integral is given by b

Ú F ◊ d l = Ú | F | cos q dl l

a

If the path of integration is a closed curve, the line integral is the circulation of the vector around the path. If the line integration of a vector along a closed path is zero, i.e., Ú F ◊ d l = 0 , then the vector is l known as conservative or lamellar. For a vector F , continuous in a region containing a smooth surface S, the surface integral or the flux of F through S is defined as y = Ú F ◊ dS = Ú F ◊ an dS = Ú | F | cos q dS S

S

S

where an is the unit normal vector to the surface S. If the surface is a closed surface, the surface integral is the net outward flux of the vector. If the surface integral of a vector over a closed surface is zero, i.e., Ú F ◊ dS = 0 , then the vector is S known as solenoidal vector. The volume integral of a scalar quantity F over a volume V is written as U = Ú Fdv V

The differential vector operator (—) or Del or Nabla, defined in Cartesian coordinates as —=

∂ ∂ ∂ a + a + a ∂x x ∂y y ∂z z

is merely a vector operator, but not a vector quantity. It is a vector space function operator, used for performing vector differentiations. The gradient of a scalar function is both the magnitude and the direction of the maximum space rate of change of that function.

Vector Analysis

87

The gradient of a scalar quantity in three different coordinate systems is expressed respectively as ∂F ∂F ∂F a + a + a ∂x x ∂y y ∂z z 1 ∂F ∂F ∂F a + a + a = ∂r r r ∂f f ∂z z 1 ∂F 1 ∂F ∂F a + a + a = ∂r r r ∂q q r sin q ∂f f

—F =

(Cartesian coordinates) (Cylindrical coordinates) (Spherical coordinates)

The divergence of a three-dimensional vector field at a point is a measure of how much the vector diverges or converges from that point. If the divergence of a vector is zero, then the vector is known as solenoidal vector. The divergence of a vector quantity in three different coordinate systems is expressed respectively as ∂Fx ∂Fy ∂Fz + + ∂x ∂y ∂z 1 ∂ 1 ∂Ff ∂Fz (rFr ) + = + r ∂r r ∂f ∂z

—◊F =

=

(Cartesian coordinates) (Cylindrical coordinates)

1 ∂ 1 1 ∂Ff ∂ ( r 2 Fr ) + ( Fq sin q ) + 2 ∂r r sin q ∂q r sin q ∂f r

(Spherical coordinates)

The curl of a vector field, denoted curl F or — ¥ F, is defined as the vector field having magnitude equal to the maximum circulation at each point and to be oriented perpendicularly to this plane of circulation for each point. If the curl of a vector is zero, then the vector is known as irrotational vector. The curl of a vector quantity in three different coordinate systems is expressed respectively as

—¥F =

=

ax

ay

az

∂ ∂x Fx

∂ ∂y Fy

∂ ∂z Fz

(Cartesian coordinates)

ar

raf

az

1 ∂ r ∂r Fr

∂ ∂f rFf

∂ ∂z Fz

()

(Cylindrical coordinates)

ar

r aq

r sin q af

1 ˆ ∂ = ÊÁ 2 Ë r sin q ˜¯ ∂r Fr

∂ ∂q r Fq

∂ ∂f r sin q Ff

(Spherical coordinates)

The Laplacian operator (—2) of a scalar field is the divergence of the gradient of the scalar field upon which the operator operates.

88

Electromagnetic Field Theory

The Laplacian operator (—2) of a scalar field in three different coordinate systems is expressed respectively as —2 F =

∂2 F ∂2 F ∂2 F + 2 + 2 ∂x 2 ∂y ∂z

(Cartesian coordinates)

( )

=

1 ∂ ∂F 1 ∂2 F ∂2 F r + 2 + 2 r ∂r ∂r r ∂f 2 ∂z

=

1 ∂ Ê 2 ∂F ˆ 1 ∂ ∂F 1 ∂2 F sin q + 2 2 (Spherical coordinates) Á r ∂r ¯˜ + 2 2 ∂r Ë ∂q r r sin q ∂q r sin q ∂f 2

(Cylindrical coordinates)

(

)

The Laplacian of a vector is defined as the gradient of divergence of the vector minus the curl of curl of the vector; i.e., — 2 F = — (— ◊ F ) - — ¥ — ¥ F In the Cartesian coordinate system and only in the Cartesian coordinate system, vector Laplacian is written as — 2 F = (— 2 Fx ) ax + (— 2 Fy ) a y + (— 2 Fz ) az Gauss’ divergence theorem is used to convert volume integral into surface integral and vice versa. According to this theorem, the divergence of a vector field over a volume is equal to the surface integral of the normal component of the vector through the closed surface bounding the volume.

Ú — ◊ Fdv = Ú F ◊ dS = Ú F ◊ an dS

V

S

S

where V is the volume enclosed by the closed surface S. According to Green’s theorem, if F is a two-dimensional vector field, such that, F = ( Fx ax + Fy a y), then Ê ∂F

∂F ˆ

y x Ú ( Fx dx + Fy dy ) = ÚÚ ÁË ∂x - ∂y ˜¯ dxdy C R

where C is a positively oriented boundary of the region R. Stokes’ theorem is used to covert line integral into surface integral and vice versa. According to this theorem, the line integral of a vector around a closed path is equal to the surface integral of the normal component of its curl over the surface bounded by the path

Ú F ◊ d l = ÚÚ (— ¥ F ) ◊ dS = ÚÚ (— ¥ F ) ◊ an dS

L

S

S

where S is the surface enclosed by the path L. The positive direction of dS is related to the positive sense of defining L according to the right-hand rule. Helmholtz theorem states that a vector field is uniquely described within a region by its divergence and curl.

Vector Analysis

89

Important Formulae Dot product

A ◊ B = AB cos q AB

Cross product

A ¥ B = AB sin q AB an

Differential lengths

d l = dxax + dya y + dza z ,

Differential areas

(Cartesian coordinate)

= drar + rd f af + dzaz ,

(Cylindrical coordinate)

= d r ar + r dq aq + r sin q d f af ,

(Spherical coordinate)

dS = dvdzax + dxdza y + dxdyaz ,

(Cartesian coordinate)

= rd f dzar + drdzaf + rdrd f az ,

(Cylindrical coordinate)

= r sin q dq d f ar + r sin q d r d f aq + r d r dq af (Spherical coordinate) 2

Differential volume

dS = dxdydz , = rdrd f dz ,

(Cartesian coordinate) (Cylindrical coordinate)

= r 2 sin q d r dq d f , (Spherical coordinate)

Del operator in Cartesian coordinate Gradient of a scalar

Divergence of a vector

—=

∂ ∂ ∂ a + a + a ∂x z ∂y y ∂z z ∂F ∂F ∂F a + a + a ∂x x ∂y y ∂z z

(Cartesian coordinates)

=

1 ∂F ∂F ∂F a + a + a ∂r r r ∂f f ∂z z

(Cylindrical coordinates)

=

1 ∂F 1 ∂F ∂F a + a + a ∂r r r ∂q q r sin q ∂f f

(Spherical coordinates)

—F =

∂Fx ∂Fy ∂Fz + + ∂x ∂y ∂z

(Cartesian coordinates)

=

1 ∂ 1 ∂Ff ∂Fq (g Fx ) + + r ∂g g ∂f ∂z

(Cylindrical coordinates)

=

1 ∂ 1 ∂ ( r 2 Fq ) + ( F sin q ) r sin q ∂q z r 2 ∂r

—◊F =

+

1 ∂Ff r sin q ∂f

(Spherical coordinates)

90

Electromagnetic Field Theory

Curl of a vector —¥F =

=

Laplacian of a scalar

Gauss’ divergence theorem Stokes’ theorem

ay

az

∂ ∂x Fx

∂ ∂y Fy

∂ ∂z Fz

(Cartesian coordinates)

ax

ay

ax

1 ∂ r ∂r Fr

∂ ∂f Ff

∂ ∂z Fz

()

(Cylindrical coordinates)

aq

raz

r sin q af

1 ˆ ∂ = ÊÁ 2 Ë r sin q ˜¯ ∂r Fr

∂ ∂q r Fq

∂ ∂f r sin q Ff

(Spherical coordinates)

∂2 F ∂2 F ∂2 F + 2 + 2 ∂x 2 ∂y ∂z

(Cartesian coordinates)

=

1 ∂ Ê ∂F ˆ 1 ∂ 2 F ∂ 2 F r + r ∂r ÁË ∂r ˜¯ r 2 ∂f 2 ∂z 2

(Cylindrical coordinates)

=

1 ∂ Ê 2 ∂F ˆ 1 ∂ Ê ∂F ˆ Á sin q ∂q ˜¯ Á r ∂r ¯˜ + 2 r 2 ∂r Ë r sin q ∂q Ë

—2 F =

+ Laplacian of a vector in Cartesian coordinate

ax

∂2 F 1 r 2 sin 2 q ∂f 2

(Spherical coordinates)

—2 F = (—2 Fx )az + (—2 Fy )a y + (—2 Fz )az

Ú — ◊ Fdv = Ú F ◊ dS = Ú F ◊ ax dS

V

S

S

Ú F ◊ d l = ÚÚ (— ¥ F )dS = ÚÚ (— ¥ F )ax dS

L

S

S

Exercises [Note: * marked problems are important university problems] Easy 1. Given the vectors A = 5ax + 3a y + 6az and B = 2ax + a y + 4az , find A ◊ B, A ¥ B and the angle between A and B. [37, (6ax - 8a y - az ), 15.2∞]

Vector Analysis

91

2. Determine the gradient of the following scalar fields: (a) A = x2 y + xyz (b) S = r2 z cos 2f sin q sin f (c) W = r2 È (a ) { y (2 x + z )ax + x( x + z )a y + xyaz }; ˘ Í ˙ 2 Í (b) (2rz cos 2f ar - 2rz sin 2f af + r cos 2f az ); ˙ Í Ê 2 sin q sin f cos q sin f cos f ˆ ˙ ar + aq + 3 af ˜ ˙ Í (c ) Á 3 3 r r r ¯ ˚˙ ÎÍ Ë 3. Determine the divergence of the following vectors: ax (a) A = 2 x + y2 (b) A = r sin f ar + 2r cos f af + 2 z 2 az 5 (c) A = ÊÁ 2 ˆ˜ sin f ar + r cot q aq + r sin q cos f af Ër ¯

x È( a ) ; (b) 4 z; (c) - (1 + sin q ) ˘˙ Í 23/2 2 (x + y ) Î ˚

4. Determine the curl of the following vectors: (a) F = ( x 2 - z 2 ) ax + 2a y + 2 xzaz (b) F = rz sin f ar + 3rz 2 cos f af sin f cos f (c) F = 2 ar - 2 af r r

È (a ) - 4 za y ;(b) - 6rz cos f ar + r sin f af + (6 - 1) z cos f az ; ˘ Í ˙ Í (c) Ê 1 cot q cos f ˆ ar + 1 ÁÊ cos f + cos f ˜ˆ aq ˙ Á 3 ¯ r 3 Ë sin q ¯˜ ÎÍ Ë r ˚˙

5. Determine the Laplacian of the following scalar fields: (a) F = x2 y + xyz (b) F = r2 z cos 2f (c) F = 10 r sin2 q cos f (d) F = (3x + 4y + 2z)(x – 2y + 4z) 10 cos f È ˘ (1 + 2 cos 2q ); ( d ) 6 ˙ ÍÎ (a ) 2 y; (b) 0; (c) r ˚ 6. Show that the vector F = yzax + xza y + xyaz is both solenoidal and irrotational.

()

Medium 1 r *7. For a position vector r , prove that — = - 3. r r 8. If (r, q, f) are spherical polar coordinates, show that, 1 grad (cos q ) ¥ grad f = grad ÊÁ ˆ˜ Ë r¯

rπ0

1 curl v , where v is the linear 2 velocity. Give the physical meaning of the cross product of w and r where r is position vector.

9. If a rigid body is rotating with an angular velocity w , prove that w =

92

Electromagnetic Field Theory

*10. Given point P (–2, 6, 3) and vector A = yax + ( x + z )a y . Express P and A in cylindrical and spherical coordinates. Evaluate A at P in the Cartesian, cylindrical and spherical systems. È P( -2, 6, 3) = P(6.32, 108.43∞, 3) = P(7, 64.62∞, 108.43∞); ˘ Í (6a + a ), ( - 0.95a - 6a ), ( - 0.86 a - 0.41a - 6a ) ˙ y r f r q f Î x ˚ 10 a + r cos q aq + af , in Cartesian and cylindrical coordinates. Find B r r (–3, 4, 0) and B (5, p/2, –2). [( - 2ax + a y ); (2.407 ar + af + 1.167 az )]

11. Express the vector, B =

12. Express the vector, 3 xax - yza y + x 2 zaz in cylindrical coordinates. 13. Express the vector, 2 yax - 2a y + xaz in spherical polar coordinates. 10 x3 a , evaluate both sides of the divergence theorem for the volume of a cube, 3 x 2 m on an edge, centered at the origin and with edges parallel to the axes.

*14. Given that A =

15. Given that F = 30e - r ar - 4 zaz in cylindrical coordinates, evaluate both sides of the divergence theorem for the volume enclosed by r = 5, z = 0 and z = 10. 10 r 3 a in cylindrical coordinates, evaluate both sides of the divergence theorem 4 r for the volume enclosed by r = 1, r = 2, z = 0 and z = 10.

*16. Given that D =

Hard 17. Prove that —2 rn = n(n + 1)rn – 2 and —2

( 1r ) = 0.

18. Given a vector field D = r sin f ar -

1 sin q cos f aq + r 2 af r

determine: (a) D at P (10, 150°, 330°). (b) The component of D tangential to the spherical surface r = 10 at P. (c) A unit vector at P perpendicular to D and tangential to the cone q = 150°. Í(a ) (- 5ar + 0.043aq + 100af ); (b) (0.043aq + 100 af ); (c ) (- 0.999ar - 0.0499 af )˙ Î ˚ 19. Given the vector field in ‘mixed’ coordinate variables as J=

È 3 y cos q 3 y 2 3x 2 ˘ 3 xz a + a y + Í 2 - 2 - 2 ˙ az 2 x r r r r ˚ Î

Convert the vector completely in spherical coordinates. 20. For the vector F = ( x + y )ax - xa y + xaz , evaluate the line integral from point P1 to point P2 along the path L1 and L2 as shown in Fig. 1.42. È- 1 , 1 ˘ ÍÎ 2 2 ˙˚

Vector Analysis

Fig. 1.42

93

Contour of Example 21

Ê e- r ˆ 21. Given the vector A = Á a in spherical coordinates. Verify both sides of Stokes’ theorem for Ë r ˜¯ q the curve bounded by the area shown in Fig. 1.43.

Fig. 1.43

Arrangement of Problem 22

Review Questions 1. What is the concept of field? Define scalar and vector fields by giving suitable example of each. What is the importance of unit vector? 2. (a) Discuss cross product and dot product in detail between two vectors. (b) Discuss the vector representation of a surface. 3. Define the divergence of a vector. Explain the physical significance of the term ‘divergence of a vector field’. 4. Define the curl of a vector. Explain the physical significance of the term ‘curl of a vector field’. 5. What is a ‘gradient’? Give its physical interpretation. 6. Define generalised coordinate system. Find its gradient, divergence and curl equations. 7. Define and give examples for the following vector fields: (a) Solenoidal and irrotational (b) Non-solenoidal and irrotational (c) Solenoidal and rotational (d) Non-solenoidal and rotational 8. State and prove the Divergence theorem.

94

Electromagnetic Field Theory

9. State and prove Stoke’s theorem. 10. State and prove the Helmholtz theorem. What is the physical significance of this theorem?

Multiple Choice Questions 1. Choose the correct statement: (a) Divergence of curl of a vector is zero and curl of grad is zero. (b) Divergence of curl of a vector is zero and curl of grad is non-zero. (c) Divergence of curl of a vector is non-zero and curl of grad is non-zero. (d) Divergence of curl of a vector is non-zero and curl of grad is zero. 2. If the vectors A and B are conservative, then (b) A ¥ B is conservative (a) A ¥ B is solenoidal (c) A + B is solenoidal (d) A - B is solenoidal 3. The value of Ú d l along a circle of radius 2 units is C

(a) zero (b) 2p (c) 4p 4. Two vectors A and B are such that A + B = nA where n is a positive scalar. The angle between A and B is 3p p (a) (b) (c) p (d) 2p 4 2 5. Which of the following equations is correct? (b) (ax ¥ a y ) + (a y ¥ ax ) = 0 (a) ax ¥ ax = | ax |2 (c) ax ¥ (a y ¥ az ) = ax ¥ (az ¥ a y ) (d) ar ◊ aq + aq ◊ ar = 0 6. Match List-I (Term) with List-II (Type) and select the correct answer using the codes given below the lists: List-I (Term)

Codes: (a) A 2 (b) A 4 (c) A 2 (d) A 4

List-II (Type)

A. curl ( F ) = 0

1. Laplace equation

B. div ( F ) = 0

2. Irrotational

C. div Grad (f) = 0

3. Solenoidal

D. div div (f) = 0

4. Not defined

B 3 B 1 B 1 B 3

C 1 C 3 C 3 C 1

D 4 D 2 D 4 D 2

Vector Analysis

7. Which one of the following is a meaningless combination? (a) grad div (b) div curl (c) curl grad 8. Which one of the following is zero? (a) grad div (b) div grad (c) curl grad

(d) div curl (d) curl curl

Answers 1. (a)

2. (a)

3. (a)

4. (c)

5. (b)

6. (a)

7. (c)

8. (c)

95

2 ELECTROSTATICS

Learning Objectives This chapter deals with the following topics: ■ ■ ■ ■ ■

2.1

Sources of electrostatics Basic laws of electrostatics To acquire knowledge of fundamental quantities of electrostatics Boundary conditions in electrostatics Concepts of capacitance

INTRODUCTION

The sources of electromagnetic fields are the presence of electric charges. An electrostatic field is considered to be a special case of electromagnetic field in which the sources are stationary. However, individual charges (e.g., electrons) are never stationary, having random velocities. The charges are referred to be stationary when any elemental macroscopic volume is considered and the net movement of charge through any face of the volume is zero. In this chapter, we will learn the basics of electrostatics in detail.

2.2

ELECTRIC CHARGE

Electric charge (q) is a fundamental conserved property of some subatomic particles, which determines their electromagnetic interaction. An isolated electric charge creates an electric field around it and exerts force on all other charges within that field. The SI unit of charge is Coulomb (C). The charge of an electron is 1.602 ¥ 10-19C. Thus, one 1 Ê ˆ Coulomb charge is defined as the charge possessed by Á -19 ˜ electrons. Ë 1.602 ¥ 10 ¯ \

1 Coulomb charge = charge of 6.24 ¥ 1018 electrons

The total electric charge of an isolated system remains constant regardless of changes within the system itself. This is known as the law of conservation of charge. The law of conservation of charge states that charge can neither be created nor destroyed. A charge can however, be transferred from one body to another body.

Electrostatics 97

2.3

COULOMB’S LAW

Statement This law states that the force between two point charges: 1. acts along the line joining the two charges. 2. is directly proportional to the product of the two charges. 3. is inversely proportional to the square of the distance between the charges.

Explanation We will consider two point charges Q1 and Q2 with separation distance R, as shown in Fig. 2.1 (a) and Fig. 2.1 (b).

Fig. 2.1

Coulomb interaction between two point charges (a) like charges, and (b) unlike charges

The force exerted by Q1 on Q2 is F12 a

Q1Q2 aR12 R2

fi

F12 = k

Q1Q2 aR12 R2

(2.1)

where k is the proportionality constant, which takes into account the effect of the medium in which the charges are placed and aR12 is a unit vector directed from Q1 to Q2. In SI unit, charges expressed in Coulomb (C), the distance expressed in metre (m) and the force expressed in Newton (N), the proportionality constant is given as k=

1 4pe

where e = permittivity of the medium = e 0e r 1 e 0 = permittivity of free space = = 8.854 ¥ 10 -12 F/m 36p ¥ 109 e r = relative permittivity of the medium Thus, Coulomb’s law in SI unit becomes, from Eq. (2.1) F12 =

Q1Q2 aR12 4pe R 2

(2.2)

Similarly, force exerted by Q2 on Q1 is F21 a

Q1Q2 QQ aR 21 fi F21 = k 1 2 2 aR 21 = - F12 2 R R

If two charges have the position vectors of r1 and r2, respectively, as shown in Fig. 2.2, then

Fig. 2.2

Coulomb vector force between two point charges

98

Electromagnetic Field Theory

Force on charge Q2 due to charge Q1 is F12 = where R12 = (r2 - r1 ),

R = | R12 |

\ aR12 = F12 =

\

Q1Q2 aR12 4pe R 2 R12 R

Q1Q2 Q Q (r - r ) R = 1 2 2 13 3 12 4pe R 4pe | r2 - r1 |

(2.3)

Similarly, force on charge Q1 due to charge Q2 is F21 =

Q1Q2 Q Q (r - r ) R = 1 2 1 23 = - F12 3 21 4pe R 4pe | r1 - r2 |

Thus, we see that the force exerted by the charges on each other is equal in magnitude, but opposite in direction. It is also noticeable that the force between two like charges (charges of equal sign) is repulsive whereas the force between two unlike charges (charges of opposite sign) is attractive.

2.4

PRINCIPLE OF SUPERPOSITION OF CHARGES

If there is a number of charges Q1, Q2, …, Qn placed at points with position vectors r1 , r2 , …, rn , respectively, then the resultant force F on a charge Q located at point r is the vector sum of the forces exerted on Q by each of the charges Q1, Q2, …, Qn. \

F=

QQ1 (r - r1 ) QQ2 (r - r2 ) + + 4pe | r - r1 |3 4pe | r - r2 |3 F=

\

+

QQn (r - rn ) 4pe | r - rn |3

Q n Qi (r - ri ) Â 4pe i =1 | r - ri |3

Example 2.1

(2.4)

Two point charges, Q1 = 50mC and Q2 = 10mC are located at (–1, 1, –3) m and (3, 1, 0) m, respectively. Find the force on Q1.

Solution Here, R21 = ( -1, 1, - 3) - (3, 1, 0) = ( - 4i - 3k ) a21 =

\

( - 4i - 3k ) ( - 4) + ( - 3) 2

2

=

- 4i - 3k 5

Hence, the force on the charge Q1 is Q1Q2 50 ¥ 10 - 6 ¥ 10 ¥ 10 - 6 Ê - 4i - 3k ˆ a = ¥ ÁË ˜¯ 21 2 5 4pe 0 R21 10 - 9 ¥ (5) 2 4p ¥ 36p = 0.18( - 0.8i - 0.6k ) N

F21 =

Electrostatics 99

The force has a magnitude of 0.18N and a direction given by the unit vector ( - 0.8i - 0.6k ). In component form F21 = ( - 0.144i - 0.108k )N Two small identical conducting spheres have charges of 2.0 ¥ 10–9 C and –0.5 ¥ 10 C, respectively. When they are placed 4 cm apart, what is the force between them? If they are brought into contact and then separated by 4 cm, what is the force between them?

Example 2.2 –9

Solution The force between the conducting spheres is given as F=

Q1Q2 (2.0 ¥ 10 - 9 ) ¥ ( - 0.5 ¥ 10 - 9 ) = = - 0.5625 ¥ 19- 5 N = 0.5625 ¥ 10- 5 N (attractive) 4p e r 2 4p ¥ 8.854 ¥ 10 -12 ¥ (4 ¥ 10 - 2 ) 2

When the two spheres are brought into contact and then separated, the charge of each sphere is Q1¢ = Q2¢ = Q ¢ =

2.0 ¥ 10 - 9 - 0.5 ¥ 10 - 9 = 0.75 ¥ 10 - 9 C 2

In this case, the force between the conducting spheres is given as F=

Q1Q2 (0.75 ¥ 10 - 9 ) 2 Q2 = = = 0.3164 ¥ 10- 5 N (repulsive) 2 2 4p e r 4p e r 4p ¥ 8.854 ¥ 10 -12 ¥ (4 ¥ 10 - 2 ) 2

Example 2.3 Two particles each of mass ‘m’ and having a charge ‘q’ are suspended by string of length ‘l’ from a common point as shown in Fig. 2.3. Show that the angle ‘q’ which each string makes with the vertical is obtained q2 tan 3 q = . from 1 + tan 2 q 16pe 0 mgl 2 Solution Force of repulsion, F =

q2 4p e 0 (AB) 2

From DOAC, AB/2 = sin q l \

Fig. 2.3

fi

AB = 2l sin q F=

q2 q2 = 2 4p e 0 (2l sin q ) 16p e 0l 2 sin 2 q

Net force at point A = (T sin q - F )i + (T cos q - mg ) j = 0 \ \

Two particles suspended by string from a common point

T sin q = F T cos q = mg tan q =

F mg

(i)

100

Electromagnetic Field Theory

Using Eq. (i), we get q2 = tan q 16p e 0 l 2 sin 2 q mg or

q2 sin 3 q sin 3 q tan 3 q tan 3 q 2 2 = tan q sin q = = cos q = = cos q cos3 q 16p e 0 mgl 2 sec 2 q 1 + tan 2 q

\

q2 tan 3 q = 1 + tan 2 q 16p e 0 mgl 2

If the angle is very small, q l, the above expression reduces to the expression for a point charge given as Ey ª

1 2l l /2 1 ll = 4p e 0 y y 4p e 0 y 2 1 Q = 4p e 0 y 2

On the other hand, if l >> y, then we have Ey ª

1 2l 4p e 0 y

The characteristic behavior of Ey /E0 (with E0 = Q/4pe0t2) as a function of y / l is shown in Fig. 2.19.

Fig. 2.19

Electric field of a non-conducting rod as a function of y/l

Electrostatics

115

*Example 2.18 A line charge of length of 2m has a linear charge density of l. Show that the electric field at a distance r perpendicular to the line charge from its middle point as shown in Fig. 2.20 is l

Er = 2p e 0 r

( ) +1 r m

2

Solution We will consider an elemental length dz of the line.

Fig. 2.20

Arrangement of Example 2.17

Fig. 2.21

Field due to a charge carrying line

From the symmetry it is observed that all the z components of the field will cancel each other. Also, there will be no f component of the field. Hence, the field will be only in the radial direction given as dE = =

l dz ¢ l dz ¢ cos a ar = 4p e 0 R 2 4p e 0 (r 2 + z ¢ 2 )

r r + z ¢2 2

ar

l r dz ¢ ar 4p e 0 (r 2 + z ¢ 2 )3/2

So, the field due to the entire line is obtained as E = Ú dE = =

2l 4p e 0

l = 2p e 0

l 4p e 0

m

m

r dz ¢ ar 2 r z ¢ 2 )3/2 ( + -m

Ú

Ú

r dz ¢ ar (r 2 + z ¢ 2 )3/2

Ú

r dz ¢ ar (r 2 + z ¢ 2 )3/2

0 m 0

Let, z = r tan a fi

dz = r sec2 a d a Èz¢ 0 m ˘ Ía 0 a ˙ 1˚ Î

\

E=

where a1 = tan -1

l 2p e 0

a1

() m r

rr sec 2 a d a

Ú 3 3 ar 0 r sec a a

1 l cos a d a ar Ú 2p e 0 r 0 l = sin a1ar 2p e 0 r l m = a 2p e 0 r r 2 + m 2 r

=

ÏÔ tan a = m \ sin a = 1 1 Ì r ÓÔ

¸Ô ˝ r + m ˛Ô m

2

2

116 Electromagnetic Field Theory

\

E=

l 2p e 0 r

1

( ) +1 r m

2

ar

*Example 2.19 A circular ring of radius ‘R’ carries a uniform charge distribution of line charge density ‘l’ and is placed on the xy plane (z = 0 plane) with the axis the same as the z-axis as shown in Fig. 2.22. (a) Show that l Rz E (0, 0, z ) = az 2e[ R 2 + z 2 ]3/2 (b) What value of ‘z’ gives the maximum value of E? (c) If the total charge on the ring is Q, find E as a Æ 0.

Solution (a) The field point is P (0, 0, z). The distance of the field point from the elemental length of the charge distribution is r ¢ = - Rar + zaz Hence, the electric field is given as E=

l 4p e 0

2p

Rd f

(- Rar + zaz ) ( R + z 2 )3/2 2p È 2p ˘ R l R a d z = f + Í Ú Ú az d f ˙ r 4p e 0 ( R 2 + z 2 )3/2 Î 0 0 ˚ R l = 4p e 0 ( R 2 + z 2 )3/2 2p È ÏÔ 2p ˘ ¸Ô Í- R Ì Ú ax cos f d f + Ú a y sin f d f ˝ + 2p zaz ˙ 0 ˛Ô ÎÍ ÓÔ 0 ˚˙ R l = [0 + 2p zaz ] 4p e 0 ( R 2 + z 2 )3/2 lRz = az 2e 0 ( R 2 + z 2 )3/2

\

Ú

0

2

E=

Fig. 2.22

Field due to a ring of uniform charge distribution

Qz 1 l ¥ 2p R z 1 a = 4p e 0 ( R 2 + z 2 )3/2 z 4p e 0 ( R 2 + z 2 )3/2 E=

Qz 1 a 4p e 0 ( R 2 + z 2 )3/2 z

where Q = l ¥ 2pR is the total charge in the ring. A plot of the electric field as function of z is shown in Fig. 2.23 (b). Notice that the electric field at the centre of the ring vanishes. This is to be expected from symmetry arguments. laz \ E= az 2e 0 (a 2 + z 2 )3/2

Electrostatics

Fig. 2.23

117

(a) Electric field directions, and (b) Electric field along the axis of symmetry of a non-conducting Q ring of radius R, with E0 = 4p e 0 R 2

NOTE At the centre of the loop, z = 0 and the electric field is also zero.

(b) For maximum value of the field intensity d |E| =0 dz d È l Rz ˘=0 dz ÍÎ 2e 0 ( R 2 + z 2 )3/2 ˙˚

fi

3 2 ( R + z 2 )1/2 ¥ 2 z ¥ z 2 =0 ( R 2 + z 2 )3

( R 2 + z 2 )3/2 ¥ 1 -

fi fi

( R 2 + z 2 )1/2 ( R 2 + z 2 - 3 z 2 ) = 0

fi

R2 – 2z2 = 0 R z=± 2

fi

z=±

\

R 2

(c) Since the charge is uniformly distributed, the line charge density is l=

Q 2p R

Hence, the field is given as Q Rz Qz l Rz 2 R p E= a = az = az 2 2 3/2 z 2 2e ( R + z ) 2e ( R + z 2 )3/2 4p e ( R 2 + z 2 )3/2

118 Electromagnetic Field Theory

As, R Æ 0, the field is given as E=

2.7.2

Q az 4pe z 2

Electric Field due to Surface Charge Distribution

For surface charge distribution with charge density s (C/m2), the total charge over a surface is obtained as Q = ÚÚ s (r )dS , so that the field intensity is given as S

E=

1 4pe

s (r )

ÚÚ r 2 dS S

(2.16)

Example 2.20 Find the electric field intensity due to a uniformly charged infinite plane sheet with surface charge density of s. Solution We will consider an elemental surface dS as shown in Fig. 2.24. The field at a point P(0, 0, h) due to this elemental surface is dQ dE = aR 4pe R 2 Here, dQ = s dS = s rdrd f ; R = (- rar + haz ) \ R = | R| =

r 2 + h2

\ aR =

R - rar + haz = R r 2 + h2 Fig. 2.24

Hence, the field is dE =

s rdrd f ( - rar + haz ) dQ aR = 4p e R 2 4p e (r 2 + h 2 )3/2

Determination of electric field due to surface charge distribution

From the symmetry, it is understood that for an infinite plane, the ar components of the field will cancel each other and the field will have only the z-component, given as dEz =

s h rdrd f az 4p e (r 2 + h 2 )3/2

Thus, the total field is given as E = Ú dEz = =

2p

h rdrd f s sh rdr ¥ 2p Ú 2 a = az 4p e f =Ú 0 r =Ú 0 (r 2 + h 2 )3/2 z 4p e ( + r h 2 )3/2 r =0

sh rdr a 2e r =Ú 0 (r 2 + h 2 )3/2 z

Electrostatics

119

\ rdr = pdp

Let (r2 + h2) = p2

E=

\

s h pdp s h dp sh È 1 ˘ s a = a = a = a 2e hÚ p3 z 2e hÚ p 2 z 2e ÍÎ p ˙˚ h z 2e z

\

E=

s a 2e z

In general, for an infinite sheet of charge, the field is given as E=

s a 2e n

where an is the unit vector normal to the plane.

NOTE The field is independent of the distance of the field point from the plane. In case of a parallel plate capacitor, the electric field existing between the two plates of equal and opposite charges is given as -s s s E= a + ( - an ) = an . 2e n 2e e

Example 2.21 A circular disc of radius ‘R’ carries uniform charge distribution of surface charge density ‘s’. If the disc lies on the z = 0 plane, with its axis along the z-axis, (a) Find the electric field at a point P, along the z-axis that passes through the center of the disk perpendicular to its plane. Discuss the limit where R >> z. (b) From this, derive E field due to an infinite sheet of charge on the z = 0 plane. Solution (a) By treating the disc as a set of concentric uniformly charged rings, the problem could be solved by using the result obtained in earlier. We will consider a ring of radius r dr ¢ as shown in Fig. 2.25. By the symmetry argument, the electric field at P points in the +z-direction. Since the ring has a charge dq = s2pr ¢dr ¢, from Eq. (2.10–14), we see that the ring gives a contribution zdq 1 1 z 2ps r ¢ dr ¢ = 4p e 0 (r ¢ 2 + z 2 )3/2 4p e 0 (r ¢ 2 + z 2 )3/2

dEz =

Integrating from r ¢ = 0 to r ¢ = R, the total electric field at P becomes Ez = Ú dEz = sz = 4e 0 =-

sz 2e 0

R2 + z 2

Ú

z2

R

Fig. 2.25

r ¢ dr ¢

Ú (r ¢ 2 + z 2 )3/2 0

R2 + z 2

du s z È u -1/2 ˘ = 3/2 4e 0 ÍÎ (-1/ 2) ˙˚ z 2 u

sz È 1 2e 0 Í R 2 + z 2 ÎÍ

s È z 1 ˘ = e0 Í | z | 2 2 ˙ z ˚˙ ÍÎ

˘ ˙ R + z ˚˙ z

2

2

A uniformly charged disc of radius R

120

Electromagnetic Field Theory

The above equation may be written as z s È ˘, z > 0 12e 0 Í 2 2 ˙ z +R ˚ Î z s È ˘, z < 0 = -12e 0 Í 2 2 ˙ z +R ˚ Î

Ez =

Ez Ê s ˆ z as a function of E = E0 ÁË 0 2e 0 ˜¯ R is shown in Fig. 2.26. When z >> R, we will have the result similar to a point charge. By Taylor series expansion The electric field

1-

Ê R2 ˆ = 1 - Á1 + 2 ˜ Ë z ¯ z 2 + R2 z

1/2

Ê 1 R2 = 1 - Á1 + Ë 2 z2 \

Ez =

2

ˆ 1 R2 ˜¯ ª 2 2 z

Fig. 2.26

Electric field of a non-conducting plane of uniform charge density

Fig. 2.27

Electric field of an infinitely large non-conducting plane

2

s 1R 1 spR 1 Q = = 2e 0 2 z 2 4pe 0 z 2 4pe 0 z 2

This is the result for a point charge. On the other hand, we may also consider the limit where R >> z. Physically, this means that the plane is very large, or the field point P is extremely close to the surface of the plane. The electric field in this limit becomes, in unit-vector notation s E= k, z > 0 2e 0 s =k, z < 0 2e 0 The plot of the electric field in this limit is shown in Fig. 2.27. When we cross the plane, there is a discontinuity in the electric field, given as D Ez = ( Ez + - Ez - ) =

s s ˆ s - Ê= 2e 0 ÁË 2e 0 ˜¯ e 0

which is obvious as has been explained in electric boundary conditions that if a given surface has a charge density s, then the normal component of the electric field across that surface always exhibits a s discontinuity with D En = . e0 (b) At z = 0, the field is given as \

E (0, 0, 0) =

s a 2e z

This is the field due to an infinite sheet of charge on the z = 0 plane.

Electrostatics 121

Example 2.22 A thin annular disc of inner radius a and outer radius b carries a uniform surface charge density s. Determine the electric field intensity at any point on the z axis when z ≥ 0. Solution The field point is P (0, 0, z) as shown in Fig. 2.28. The distance of the field point from the elemental length of the charge distribution is R = - rar + zaz Hence, the electric field is given as E=

s 4p e 0

b 2p

rdrd f

Ú Ú 2 2 3/2 ( - rar + zaz ) a 0 (r + z )

2p

However,

Ú ar df = 0

Fig. 2.28

0

\

E=

s 4p e 0

b 2p

rdrd f

Ú Ú 2 2 3/2 zaz a 0 (r + z )

Field due to uniform charge distribution on an annular disc

b

=

s rdr ¥ 2p z Ú 2 az 4p e 0 + r z 2 )3/2 ( a

=

sz È 1 2e 0 Í a 2 + z 2 ÍÎ

˘a ˙ z b + z ˙˚

E=

1

2

2

È 1 sz Í 2e 0 Í a 2 + z 2 Î

˘ ˙ az b 2 + z 2 ˙˚ 1

NOTE (i) For an annular disc of very large outer radius, b Æ , the field is given as E=

sz È 1 2e 0 Í a 2 + z 2 ÎÍ

˘ ˙ az ˚˙

(ii) For a solid finite disc of outer radius b, (a = 0) the field is given as E=

sz È1 2e 0 Í z ÎÍ

˘ ˙ az b + z ˚˙ 1

2

2

(iii) For an infinite plane of charge, a Æ 0, b Æ , and the field at any point is given as E=

2.7.3

s a 2e 0 z

Electric Field due to Volume Charge Distribution

For volume charge distribution with charge density r (C/m3), the total charge over a surface is obtained as Q = Ú r (r ) dV , so that the field intensity is given as V

122

Electromagnetic Field Theory

E=

r (r ) 1 dV Ú 4p e V r 2

(2.17)

A sphere of radius a carries charge with a uniform volume density r (C/m3). Determine the electric field intensity E at any distance r from the centre of the sphere.

Example 2.23

Solution To find the field, we will consider an elemental volume dv within the sphere as shown in Fig. 2.29. The field at a point P(0, 0, h) due to this elemental volume is dE =

dQ aR 4p e R 2

Here, aR = cos a az + sin a ar Due to symmetry, field components Ex and Ey are zero. \

E z = E ◊ az = Ú dE cos a =

r dv cos a 4p e Ú R 2

(i)

Here, dQ = r dv = r r2 sin qdrdqdf; R2 = z2 + r2 – 2zr cos q r2 = z2 + R2 – 2zR cos a cos a =

z 2 + R2 - r 2 2 zR

cos q =

z 2 + r 2 - R2 2 zr

Fig. 2.29

\ sin q dq =

Field distribution due to volume charge distribution

R dR z r

Hence, the field is Ez =

r dv cos a r 2p a R = z + r z 2 + R2 - r 2 1 2 RdR = d f r dr 4p e Ú R 2 4p e f =Ú 0 r =Ú 0 R =Úz - r zr 2 zR R2

a R= z+r È r z2 - r2 ˘ r 2 p r 1 = ¥ + Í ˙ dRdr = Ú Ú 2 2 8p e z R 4e z 2 ˚ r =0 R= z-r Î

=

\

r 4e z 2

(

a

z+r

È z2 - r2 ˘ r R dr Í Ú Î R ˙˚ z - r r =0 a

)

r 4 3 Q 1 1 4 3 2 Ú 4r dr = 4e z 2 3 a = 4p e z 2 3 p a r = 4p e z 2 r =0 E=

Q az 4p e z 2

This result is obtained at point P (0, 0, z). From the symmetry of the charge distribution, it is clear that the field at any point P (r, q, f) is given as E=

Q ar 4p e r 2

Electrostatics 123

NOTE This result is identical to the field at the same point due to a point charge Q located at the origin or center of the spherical charge distribution.

2.8 2.8.1

ELECTRIC FLUX (DISPLACEMENT) (Y ) AND FLUX Æ DENSITY (DISPLACEMENT DENSITY) ( D) Electric Flux (y )

Electric flux is the flux of the electric field. It is the total number of electric field lines passing through a given surface perpendicularly. It is expressed in Coulomb (C). It is the integral of the normal component of the vector D over a surface.

2.8.2

Æ

Electric Flux Density ( D )

Electric flux density is the total number of electric field lines per unit area passing through the area perpendicularly. It is expressed in Coulomb per square metre (C/m2). In case of a point charge, q, the electric displacement per unit area of a charge q at the centre of a sphere of radius r is q (2.18) D= (Coulomb/m 2 ) 4p r 2 This is known as electric displacement density or flux density. Now, for the same point charge, the electric field intensity at the centre of a sphere of radius r is q E= (2.19) 4p e r 2 From Eqs. (2.18) and (2.19), we get (2.20) D = eE Here, flux density of a vector quantity ( D) and its direction is that of the normal to the surface element, which makes the displacement through the element of area a maximum. For a linear, isotropic medium, D is in the same direction as of E. In terms of flux density, D, electric flux is defined as y = Ú D ◊ dS

(2.21)

S

In an SI unit, one line of electric flux emits from +1C charge and terminates on -1C charge. Hence, electric flux is expressed in Coulomb (C) and flux density in Coulomb per square metre (C/m2).

Properties of Electric Flux 1. It is independent of the medium. 2. Its magnitude depends only upon the charge from which it is originated. 3. If a point charge in enclosed in an imaginary sphere of radius R, the electric flux must pass perpendicularly and uniformly through the surface of the sphere. 4. The electric flux density, i.e., flux per unit area, is then inversely proportional to R2.

124

Electromagnetic Field Theory

*Example 2.24 A point charge of 30nC is located at the origin while the plane y = 2 carries charge 20nC/m2. Find the electric flux density D at point (0, 3, 4). Solution The flux density component due to the point charge is given as Q ( r - r ¢) 30 ¥ 10 - 9{(0,3, 4) - (0,0,0)} Q a = = 2 R 3 4p R 4p | r - r ¢ | 4p |{(0,3, 4) - (0,0,0)}|3 30 ¥ 10 - 9 (3 j + 4k ) = = (0.0573 j + 0.0764 k ) nC/m 2 4p ¥ 53

D1 = e 0 E1 =

The flux density component due to the surface charge distribution is given as D2 = e 0 E2 =

20 ¥ 10 - 9 s s an = j = j = 10 j nC/m 2 2 2 2

Hence, the total flux density due to the point charge and the surface charge distribution is D = D1 + D2 = (0.0573 j + 0.0764 k ) + 10 j = (10.0573 j + 0.0764k ) nC/m 2

2.9

ELECTRIC FIELD LINES (LINES OF FORCE) AND FLUX LINES

2.9.1

Electric Field Lines

These are the imaginary lines drawn in such a way that at every point, it has the direction of the electric field ( E ). The number of lines per unit area is proportional to the magnitude of electric field strength (E).

2.9.2

Electric Flux Lines

These are the imaginary lines drawn in such a way that at every point, it has the direction of the electric flux density vector ( D) . The number of flux lines per unit area is used to indicate the magnitude of the displacement density ( D). In homogeneous, isotropic media, lines of force and lines of flux always have the same direction. The field lines for a positive and a negative charge are shown in Figs. 2.30 (a) and (b). It is observed that the direction of field lines is radially outward for a positive charge and radially inward for a negative charge. For a pair of charges of equal magnitude but opposite sign (an electric dipole), the field lines are shown in Fig. 2.30 (c).

Fig. 2.30

Field lines for (a) positive charge, (b) negative charge, and (c) an electric dipole

Electrostatics 125

Properties of Electric Field Lines 1. The direction of the electric field vector E at a point is tangential to the field lines. 2. Electric field lines never cross each other; otherwise, the field would be pointing in two different directions at the same point. 3. The field lines must begin on positive charges (or at infinity) and must terminate on negative charges (or at infinity). 4. Electric field lines are most dense around objects with the greatest amount of charge. 5. At locations where electric field lines meet the surface of an object, the lines are perpendicular to the surface. 6. The number of lines that originate from a positive charge or terminate on a negative charge must be proportional to the magnitude of the charge. 7. The number of lines per unit area through a surface perpendicular to the line is devised to be proportional to the magnitude of the electric field in a given region.

Example 2.25

Show that for electric lines of force dx dy dz = = Ex E y Ez

Solution We will consider a test charge at point P(x, y, z) in the field E as shown in Fig. 2.31. The force on test charge is directed along E, to point Q(x + Dx, y + Dy, z + Dz). The vector displacement of the test charge is D l = ( D xi + D yj + D zk ) But, this is proportional to the field E. \

E μ Dl

or

( Ex i + E y j + Ez k ) μ ( D xi + D yj + D zk )

Fig. 2.31

Now, two vectors are proportional if and only if their components are proportional by the same amount. Dx D y Dz = = Ex E y Ez

\ In the limiting case dx dy dz = = Ex E y Ez

in Cartesian coordinate

dr rd f dz = = Er Ef Ez

in cylindrical coordinate

d r rdq r sin q d f = = Er Eq Ef

in spherical coordinate

126

Electromagnetic Field Theory

Example 2.26

Obtain the equation of flux line that passes through the point P (-2, 7, 10) in the

field E where (a) E = 2( y - 1)ax + 2 xa y

(b) E = e y ax + ( x + 1)e y a y

Solution (a) In two-dimensional system, the equation of flux lines is to be obtained from the equation dy dx = E y Ex fi

dy E y x 2x = = = dx E x 2( y - 1) y - 1

fi

( y - 1)dy = xdx

Integrating, y 2 - 2 y = x2 + C The constant C is evaluated from the coordinates of the point (-2, 7, 10). (7) 2 - 2 ¥ 7 = ( - 2) 2 + C

fi C = 31

Hence, the equation of the flux lines is given as y 2 - 2 y = x 2 + 31

fi

( y - 1) 2 - x 2 = 32

(b) In this case, dy dx = E y Ex fi

dy E y ( x + 1)e y = = dx Ex ey

fi

dy = ( x + 1)dx

Integrating, 2 y = x2 + 2 x + C The constant C is evaluated from the coordinates of the point (-2, 7, 10). 2 ¥ 7 = ( - 2) 2 + 2 ¥ ( - 2) + C

fi C = 14

Hence, the equation of the flux lines is given as 2 y = x 2 + 2 x + 14

2.10

fi

2 y - ( x + 1) 2 = 13

GAUSS’ LAW

Statement Gauss’ law, also known as Gauss’ flux theorem, states that the total electric displacement or electric flux through any closed surface surrounding charges is equal to the net positive charge enclosed by that surface.

Electrostatics 127

Proof We consider a point charge Q located in a homogeneous isotropic medium of permittivity, e. The electric field intensity at any point at a distance r from the charge will be E=

Q ar 4p e r 2

(2.22)

and the electric flux density is given as, D = eE =

Q ar 4p r 2

(2.23)

Now, the electric flux through some elementary surface area dS as shown in Fig. 2.32 is dy = DdS cos q

(2.24)

where, q is the angle between D and the normal to dS. From Fig. 2.32, dS cos q is the projection of dS normal to the radius vector. By definition of a solid angle,

Fig. 2.32

Determination of net electric flux through a closed surface

dS cos q = r 2 d W

(2.25)

where, dW is the solid angle subtended at Q by the elementary surface of area dS. Thus, total displacement or flux through the entire surface is y = Ú dy = Ú DdS cos q = Ú Dr 2 d W = S

S

S

Q dW 4p Ú

{using Eqs. (2.18), (2.22), (2.24) and (2.25)}

However, from the concept of calculus, the solid angle subtended by any closed surface is 4p steradian. Hence, total displacement or flux passing through the entire surface is, y = Ú D ◊ d S = Q = Ú rdv S

(2.26)

v

This is the integral form of Gauss’ law. Applying divergence theorem in Eq. (2.26), we get

Ú D ◊ d S = Ú rdv fi Ú (— ◊ D)dv = Ú rdv fi — ◊ D = r

S

v

v

v

—◊ D = r

(2.27)

This is the differential form or point form of Gauss’ law.

NOTE The surface to which Gauss’ law is applied is known as Gaussian surface.

Steps useful when applying Gauss’ law

The following steps may be useful when applying

Gauss’ law: 1. First, the symmetry associated with the charge distribution is identified. 2. Then a Gaussian surface is identified and the direction of the electric field is determined.

128

Electromagnetic Field Theory

3. The field space associated with the charge distribution is divided into different regions. For each region, the net charge enclosed by the Gaussian surface, Qenc is calculated. 4. The electric flux y through the Gaussian surface for each region is calculated. 5. The magnitude of the electric field (E) is deduced by equating y with Qenc/e.

NOTE Gaussian Surface—a surface on which the magnitude of the electric field is constant over portions of the surface is known as Gaussian surface. This is a closed surface.

2.10.1

Derivation of Gauss’ Law from Coulomb’s Law

Gauss’ law can be derived from Coulomb’s law, which states that the electric field due to a stationary point charge is: Q E= ar 4p e r 2 where ar is the radial unit vector, r is the radius, | r |, q is the charge of the particle, which is assumed to be located at the origin. Using the expression from Coulomb’s law, we get the total field at r by using an integral to sum the field at r due to the infinitesimal charge at each other point r ¢ in space, to give E=

1 r (r ¢)(r - r ¢) dv 4pe Úv | r - r |3

where r is the charge density. If we take the divergence of both sides of this equation with respect to r , and use the known theorem r¢ — ◊ ÊÁ 3 ˆ˜ = 4pd (r ¢) Ë |r ¢| ¯ where d(r ¢) is the Dirac delta function, the result is —◊ E =

1 r (r ¢)d (r - r ¢)dv e Úv

Using the shifting property of the Dirac delta function, we get r —◊ E = e which is the differential form of Gauss’ law.

2.10.2

Derivation of Coulomb’s Law from Gauss’ Law

Gauss’ law provides information only about the divergence of the electric field intensity, E and does not give any information about the curl E. For this reason, Coulomb’s law cannot be derived from Gauss’

Electrostatics 129

law alone. However, we assume that the electric field from a stationary point charge has spherical symmetry. With this assumption, which is exactly true if the charge is stationary, and approximately true if the charge is in motion, Coulomb’s law can be proved from Gauss’ law. We consider a spherical surface of radius r, centered at a point charge Q. Applying Gauss’ law in integral form, we have Q

Ú E ◊ dS = e S

By the assumption of spherical symmetry, the integrand is a constant and can be taken out of the integral as Q 4p r 2 ar E = e where ar is a unit vector directed radially away from the charge. Again, by spherical symmetry, E is also in radially outward direction, and so we get E=

Q ar 4pe r 2

If another point charge q is placed on the surface, the force on that charge due to the charge Q is given as Qq F = qE = ar 4pe r 2 which is essentially equivalent to Coulomb’s law. Thus, the inverse-square law dependence of the electric field in Coulomb’s law follows from Gauss’ law.

NOTE Coulomb’s law applies only to stationary charges; whereas, Gauss’ law holds for moving charges as well, and in this respect Gauss’ law is more general than Coulomb’s law.

2.11

APPLICATIONS OF GAUSS’ LAW

Gauss’ law provides an expedient tool for electric field computation. However, this law is applicable only to systems that possess certain symmetry; namely, systems with planar, cylindrical and spherical symmetry. The following table provides some examples of systems in which Gauss’ law is applicable for electric field computation, with the corresponding Gaussian surfaces: System

Symmetry

Gaussian Surface

Point charge

Spherical

Concentric sphere

Infinite rod

Cylindrical

Coaxial cylinder

Infinite plane

Planar

Gaussian “Pillbox”

Sphere, spherical shell

Spherical

Concentric sphere

130

Electromagnetic Field Theory

2.11.1

Electric Field due to a Point Charge

Example 2.27

Determine the electric field at a distance r from a point charge Q. Use Gauss’ law.

Solution We consider a point charge Q located at the origin. In order to determine the field E (or D ) at point P at a distance r from the charge, we follow the steps outlined in section 2.10. 1. This problem possesses spherical symmetry. 2. We imagine a fictitious spherical surface of radius r, so that the point P is on the surface. So, the surface of the sphere with radius r is the Gaussian surface in this case. 3. The amount of charge enclosed by the Gaussian surface is the point charge Q. 4. Since E is perpendicular to the Gaussian surface, i.e., E = Ear or, D = Dar So, total flux through the Gaussian surface is 2p

Fig. 2.33

Gaussian surface of a point charge

p

2 2 Ú D ◊ dS = D Ú Ú r sin q dq df = D 4p r f =0q =0

S

5. Applying Gauss’ law, fi Ú D ◊ dS = Qenc S

fi \ In vector form

D 4p r 2 = Q Q D= 4p r 2 D=

2.11.2

Q ar 4p r 2

or, E =

Q ar 4p e r 2

(2.28)

Electric Field due to Infinite Line Charge

*Example 2.28 Determine the electric field at a distance r from an infinite straight line carrying a uniform line charge distribution with line charge density l. Use Gauss’ law. Solution We will consider an infinitely long wire of negligible radius with a uniform line charge density, l. We calculate the field E at a distance r from the wire. We shall solve the problem by following the steps outlined in Section 2.10. 1. This infinitely long wire possesses cylindrical symmetry. 2. The charge density is uniformly distributed throughout the length of the wire, and therefore, the electric field E must be radially outward from the axis of the wire [Fig. 2.34 (a)]. The magnitude of the electric field is constant on cylindrical surfaces of radius r. Therefore, a coaxial cylinder of any radius r is a Gaussian surface in this problem. 3. The amount of charge enclosed by the Gaussian surface (i.e., a cylinder of radius r and length l) [Fig. 2.34 (b)] is Qenc = ll. 4. The Gaussian surface consists of three parts, as indicated in Fig. 2.34 (b). Those are: (a) Left-end surface S1 (b) Right-end surface S2 (c) Curved surface S3.

Electrostatics 131

Fig. 2.34

(a) Field lines for an infinite uniformly charged wire, and (b) Gaussian surface for a uniformly charged wire

The flux through the Gaussian surface is y =

ÚÚ E ◊ dS = ÚÚ E ◊ dS1 + ÚÚ E ◊ dS2 + ÚÚ E ◊ dS3 = 0 + 0 + E3 S3 = E (2p rl ) S

S1

S2

S3

where E3 = E (say). As can be seen from the figure, no flux passes through the ends since the area vectors dS1 and dS2 are perpendicular to the electric field, which is directed radially outward. 5. Applying Gauss’ law E (2p rl ) =

ll e

fi

E=

l 2pe r

fi

D = eE =

l 2p r

In vector form E=

l a 2pe r r

or

D=

l a 2p r r

(2.29)

This is observed that the result is independent of the length l of the cylinder and only depends on the inverse of the distance r from the symmetry axis. The variation of E as a function of r is plotted in Fig. 2.35.

2.11.3

Fig. 2.35

Electric field due to a uniformly charged rod as a function of r

Electric Field due to an Infinite Plane Sheet of Charge

*Example 2.29 Determine the electric field intensity due to a uniformly charged infinite plane sheet with surface charge density s. Solution We will consider an infinitely large non-conducting plane in the xy-plane with uniform surface charge density s. We want to determine the electric field everywhere in space. We will follow the steps as outlined in Section 2.9. 1. This infinitely large plane possesses a planar symmetry. 2. Since the charge is uniformly distributed on the surface, the electric field E is directed perpendicularly away from the plane E = Eaz . The magnitude of the electric field is constant on planes parallel to the non-conducting plane. Therefore, the Gaussian surface for this problem is a cylinder, which is often referred to as a ‘pillbox’ (Fig. 2.36).

132

Electromagnetic Field Theory

Fig. 2.36

(a) Electric field for an infinite plane of charge, and (b) Gaussian surface for a large plane

3. Since the charge is uniformly distributed over the surface, the charge enclosed by the Gaussian ‘pillbox’ is, Qenc = sS, where S = S1 = S2 is the area of the end-surfaces. 4. The Gaussian pillbox consists of three parts: two end-surfaces S1 and S2, and a curved surface S3. The total flux through the Gaussian pillbox flux is y =

ÚÚ E ◊ dS = ÚÚ E ◊ dS1 + ÚÚ E ◊ dS2 + ÚÚ E ◊ dS3 = E1S1 + E2 S2 + 0 = ( E1 + E2 ) S S

S1

S2

S3

Since the two ends are at the same distance from the plane, by symmetry, the magnitude of the electric field must be the same, E1 = E2 = E. Hence, the total flux can be rewritten as y = 2ES 5. By applying Gauss’ law, we obtain Qenc s S = e e In vector form, the result can be written as 2 ES =

s a , z>0 2e z s =a , z a

With uniform charge density, total charge enclosed is Qenc = Ú rdv = r v

( )

2p

p r 4 3 2 Ú Ú Ú r sin q drdq df = r 3 p r

f =0q =0 r =0

The flux through the Gaussian surface is y = Ú E ◊ dS = E Ú dS = E S

S

2p

f =0q =0

Applying Gauss’ law y = In vector form E=

Qenc e

or, E (4p r 2 ) =

rr a 3e r

p

2 2 Ú Ú r sin q dq df = E (4p r )

r£a

( )

r 4 3 pr e 3

or

E=

rr 3e

(2.31)

Case 2: r ≥ a In this case, the Gaussian surface is a sphere of radius r ≥ a, as shown in Fig. 2.38 (b). Since the radius of the Gaussian surface is greater than the radius of the sphere, all the charge is enclosed in the Gaussian surface, i.e., Qenc = Ú rdv = r v

2p

a

f =0q =0 r =0

( )

4 = r p a3 3

p

2 Ú Ú Ú r sin q drdq df

Fig. 2.39

Electric field due to a uniformly charged sphere as a function of r

134

Electromagnetic Field Theory

Electric flux through the Gaussian surface is given by y = Ú E ◊ dS = E Ú dS = E S

S

2p

p

2 2 Ú Ú r sin q dq df = E (4p r )

f =0q =0

Applying Gauss’ law, we obtain y=

Qenc e

or

E (4p r 2 ) = E=

or

( )

r 4 3 pa e 3

ra3 3e r 2

In vector form E=

ra3 ar 3e r 2

r≥a

(2.32)

The results can be summarized as rr a 3e r 3 ra ar = 3e r 2

E=

0 a: Applying Gauss’ law for region outside the cylinder of Fig. 2.40 Infinite cylinder length l Q ll (2p rl ) ¥ E = enc = e e l \ E= 2pe r l In vector form E = ar 2pe r When r < a: In this case, we have to find out the charge enclosed by the Gaussian surface for a point inside the cylinder. This is obtained as follows. Let r—the volume charge density

Electrostatics 135

\ Charge in the length l of the charged cylinder is l l = p a 2l r fi r = \ Charge enclosed in the Gaussian cylinder is

l p a2

Ê r2 ˆ Qenc = p r 2l r = Á 2 ˜ ll Ëa ¯ Applying Gauss’ law in this region Qenc Ê r 2 ˆ ll =Á 2˜ Ëa ¯ e e lr E= 2pe a 2

(2p rl ) ¥ E = \

E=

In vector form,

lr ar 2pe a 2

The results can be summarised as l a ; r>a 2pe r r lr ar ; r < a = 2pe a 2

E=

*Example 2.32 A thin spherical shell of radius a has a charge +Q evenly distributed over its surface. Find the electric field both inside and outside the shell. Solution 1. In this case, the charge distribution has spherical symmetry, with a surface charge density s=

Q Q = S 4p a 2

2. The electric field E must have radial symmetry and must be directed outward (Fig. 2.41). So, spherical surface is the Gaussian surface. 3. Flux and charge enclosed are determined by considering two regions r £ a and r ≥ a separately. When r £ a: In this case, the Gaussian surface is a sphere of radius r £ a, as shown in Fig. 2.42 (a).

Fig. 2.42

Fig. 2.41

Electric field for uniform spherical shell of charge

Gaussian surface for uniformly charged spherical shell for (a) r £ a, and (b) r ≥ a

136

Electromagnetic Field Theory

Since all the charge is located on the surface of the shell, the charge enclosed by the Gaussian surface is zero, Qenc = 0. Hence, by Gauss’ law, we have E=

Qenc 0 = =0 r a. (c) Show that the maximum value of E is at r = 0.745a.

Solution (a) Total charge Q is obtained by finding the volume integral of charge density function r over the entire volume of sphere of radius r = a. \

Q = ÚÚÚ rdv vol

Electrostatics 137

Considering a thin spherical shell within the charged sphere as radius r and with thickness dr, the charge within the shell. È r2 ˘ dQ = r ¥ 4p r 2 dr = r0 Í1 - 2 ˙ ¥ 4p r 2 dr a ˚ Î

\ \ Total charge is Q=

a

Ú

dQ =

r =0

a

a È È 2 r4 ˘ Ê r3 r2 ˘ r5 ˆ 2 Ú r0 Í1 - a 2 ˙ 4p r dr = 4pr0 Ú Í r - a 2 ˙ dr = 4pr0 ÁË 3 - 5a 2 ˜¯ Î ˚ ˚ 0 r =0 r =0 Î a

Ê a3 a3 ˆ = 4pr0 ÁË - ˜¯ 3 5 8 = pr0 a 3 15 \

Q=

8 pr a 3 15 0

(b) Field intensity at a distance r from the center of the sphere, outside (r > a) the sphere is E=

3 Q 2 r0 a = 4pe r 2 15 e r 2

For field inside the sphere (r £ a), applying Gauss’ law to the spherical surface E ¥ (4p r 2 ) =

r ˘ 1 Ê Ê r3 Q 1È r2 ˆ r5 ˆ = Í 4p r0 Ú Á1 - 2 ˜ r 2 dr ˙ = 4p r0 Á - 2 ˜ Ë Ë 3 5a ¯ e eÎ a ¯ 0 ˚ e

fi

E=

r0 Ê r r3 ˆ - 2˜ Á e Ë 3 5a ¯

r0 Ê r r3 ˆ - 2 ˜ ar ; Á e Ë 3 5a ¯ 3 2 r0 a = a; 15 e r 2 r

E=

(c) For maximum value of E,

r£a r>a

dE = 0. dr d È r0 Ê r r3 ˆ ˘ - 2 ˜˙ = 0 Í Á dr Î e Ë 3 5a ¯ ˚

fi fi fi

1 3r 2 =0 3 5a 2 5 r 2 = a2 9 r = 0.745a

Thus, the maximum value of E is at r = 0.745a.

138

Electromagnetic Field Theory

Example 2.34 A long cylinder (charged) of radius ‘a’ has a volume charge density, r = kr, where ‘k’ is a constant and ‘r’ is the distance from the axis of the cylinder. Show that the electric field is given by k r2 E= a ; r£a 3e r 3 ka = a ; r≥a 3e r r Solution 1. This problem possesses cylindrical symmetry. 2. The cylindrical surface of any radius ‘r’ is the Gaussian surface. 3. Flux and charge enclosed are determined considering two cases: Inside the cylinder (r £ a) We will consider a tubular cylinder of inner radius r and outer radius (r + dr) located coaxially within the charged cylinder of radius a. The charge contained in the tubing per unit length is dQ = r 2p rdr ¥ 1 = kr 2p rdr = 2p k r 2 dr \ Total charge per unit length contained in a cylinder of radius r is r

Q = Ú dQ = Ú 2p k r 2 dr = 0

2p k r 3 3

Applying Gauss’ law to the Gaussian surface of radius r Q 1 2p k r 3 = e e 3 k r2 E= 3e k r2 E= a 3e r

2p r ¥ 1 ¥ E = fi \ Outside the cylinder (r ≥ a) Here, total charge contained in the cylinder is

a

Q = Ú dQ = Ú 2p k r 2 dr = 0

2p k a 3 3

Applying Gauss’ law to the Gussian surface of radius r Q 1 2p k a 3 = e e 3 k a3 E= 3e r k a3 E= a 3e r r

2p r ¥ 1 ¥ E = fi \ Thus, the electric field is given as

k r2 a; 3e r k a3 = a; 3e r r

E=

r£a r≥a

Electrostatics 139

2.12 2.12.1

ELECTRIC POTENTIAL (V) Electric Potential (V )

The potential of a point is the work done to bring a unit positive charge from infinity to that point. The unit of potential is Joule per Coulomb (J/C) or Volt. The dimensions of electric potential in terms of M, L, T and I are [ML2T-3I-1].

2.12.2

Potential Difference

The potential difference between two points is the change in the potential energy per unit charge as the charge tends to zero. This is depicted in Fig. 2.43. B

W VAB = Lim ÊÁ ˆ˜ = - Ú E ◊ d l QÆ0 Ë Q ¯ A

Explanation

We will consider, a point charge Q, be moved from a point A to another point B in an

electric field E. By Coulomb’s law, the force on Q is, F = QE \ Work done for a displacement of d l is, dW = - F ◊ d l = - QE ◊ d l Hence, the total work done in moving the charge Q from A to B, i.e., the potential energy required is B

W = -QÚ E ◊ dl A

ÊW ˆ The potential energy per unit charge Á ˜ , known as Ë Q¯ the potential difference between the two points A and B, denoted by VAB is given as \

VAB =

B

W = - Ú E ◊ dl Q A

(2.33)

Fig. 2.43

Potential difference due to a uniform electric field

Q For example, if E is a field produced by a point charge Q at origin, i.e., E = ar , then the potential 4pe r 2 difference between the points A and B is given as B

Q Q B dr Q È1 1˘ a ◊ dra = Ú r 2 = 4pe Í r - r ˙ = (VB - VA ) r 2 r pe 4 pe r 4 B A˚ Î A A

VAB = - Ú

Thus, the potential difference between the points A and B may be considered to be the potential (or absolute potential) of B with respect to the potential (or absolute potential) of A. In case of a point charge, the reference is taken to be at infinity with zero potential. Hence, potential (or absolute potential) of a point is defined as the work done to bring a unit positive charge from infinity to that point. \

V=

R Q Q W ar ◊ drar = =-Ú Q 4pe 4pe r 2

R

Q

dr Ú r 2 = 4pe R

140

Electromagnetic Field Theory

V=

Q 4pe R

(2.34)

If the point charge Q is not located at the origin but at a point with position vector r , then the potential of the point at a position vector R is V ( R) =

Q 4pe | R - r |

(2.35)

Principle of Superposition of Potentials If there is a number of point charges Q1, Q2, …, Qn, located at position vectors r1 , r2 , º, rn respectively, then the potential at point r is V (r ) =

Q1 Q2 + + 4pe | r - r1 | 4pe | r - r2 | V (r ) =

1 4pe

+ n

Qn 1 n Qi = Â 4pe | r - rn | 4pe i =1 | r - ri | Q

 | r -i r | i i =1

(2.36)

Potential due to Continuous Charge Distribution

If the charge distribution is continuous, the potential at a point P can be found by summing over the contributions from individual differential elements of charge dq. Consider the charge distribution shown in Fig. 2.44. Taking infinity as our reference point with zero potential, the electric potential at P due to dq is dV =

1 dq 4pe r

Summing over contributions from all differential elements, we have 1 dq V= 4pe Ú r

Fig. 2.44

Continuous charge distribution

For three different types of charge distribution, the potential at a point r is given below. 1 l (r ¢)dl ¢ ; for line charge distribution with density l (C/m) 4pe Úl | r - r ¢ |

(2.37)

=

1 s (r ¢)dS ¢ ; for surface charge distribution with density s (C/m 2 ) 4p e SÚ | r - r ¢ |

(2.38)

=

1 r (r ¢)dv ¢ ; for volume charge distribution with density r (C/m3 ) 4p e Úv | r - r ¢ |

(2.39)

V (r ) =

Here, the primed coordinates refer to the source point and the unprimed coordinates refer to the field point (the point where the potential is to be calculated).

Electrostatics 141

NOTE V = - Ú E ◊ d l ; as E is in the radial direction, any contribution from a displacement in q or f direction is l cancelled out by the dot product. Hence, E ◊ d l = Edl cos q = Edr . Thus, the potential is independent

Ú E ◊ dl

of the path. For a closed path,

= 0 . Applying Stoke’s theorem,

l

\

Ú E ◊ dl l

—¥E =0

= Ú (— ¥ E ) ◊ dS = 0. S

Thus, the electrostatic field is conservative or irrotational.

*Example 2.35 Consider a uniformly charged ring of radius R and charge density l. What is the electric potential at a distance z from the central axis? Solution We shall consider a small differential element d l = Rdf ¢ on the ring as shown in Fig. 2.45. The charge carried by this element is dq = l dl = l Rd f ¢ Potential due to this element is dV =

1 dq 1 = 4p e r 4p e

l Rd f ¢ R2 + z 2

Thus, the potential at point P due to the entire ring is V = Ú dV = =

lR

1 4p e

lR 2e R + z 2

2pl R

1

Ú d f ¢ = 4p e R +z 2

2

=

1 4p e

R2 + z 2

2

Q R +z 2

Fig. 2.45 2

V=

A non-conducting ring of radius R with uniform charge density l

lR 2e R 2 + z 2

where Q = 2plR is the total charge on the ring. Since the ring is uniformly charged, the electric field at point P will be along the axis, i.e., in the z direction. Hence, the field is given as E = Ez k = -

Qz ∂V ∂ lR l Rz 1 ˆk = k =- Ê k= 2 2 3/2 2 ∂z ∂z Á 2e R 2 + z 2 ˜ p e 4 2e ( R + z ) ( R + z 2 )3/2 Ë ¯ E=

l Rz k 2e ( R 2 + z 2 )3/2

NOTE Q Q and the field becomes E = k; 4pe z 4pe z 2 which are the potential and field due to a point charge. Thus, if the distance of the field point is very large compared to the radius of the ring, the ring appears as a point charge.

In the limiting case with z >> R, the potential becomes, V =

142

Electromagnetic Field Theory

*Example 2.36 Determine the electric potential at a distance ‘r’ form the centre of the sphere of Example 2.30.

Solution The results for the field intensity for Example 2.30 are as given below. rr a; r a = 3e r 2

E=

The potential for the two cases is obtained as follows. Outside the sphere (r ≥ a) E=

Here, r

Hence, the potential is, V = - Ú Edr = -

r a3 3e

ra3 3e r 2

r

r a3

dr Ú r 2 = 3e r

\

V=

r a3 3e r

E=

rr 3e

Inside the sphere (r £ a) Here, Hence, the potential is r

V = V (r = a ) - Ú Edr = a

ra 2 r r ra 2 r r (3a 2 - r 2 ) rdr = - (r 2 - a 2 ) = Ú 3e 3e a 3e 3e 6e

\

V=

r (3a 2 - r 2 ) 6e

The results are summarised below. r (3a 2 - r 2 ) r ; r < a 6e ra3 = ; r>a 3e r

V=

Fig. 2.46

The variation of the potential with the distance is shown in Fig. 2.46.

Variation of potential with the distance

*Example 2.37 A spherical conductor of radius ‘R’ contains a uniform surface charge density s. Determine the field and potential due to the charge distribution. Solution We will consider two cases: Outside the sphere (r ≥ R) Applying Gauss’ law to the Gaussian surface of radius r 4p r 2 E =

Q 4p R 2s = e e

fi

E=

s R2 er2

Electrostatics 143

E=

In vector form r

Hence, the potential is, V = - Ú Edr = -

s R2 e

s R2 ar er2

r

dr s R Ú r2 = er s R2 er

V=

\

2

Inside the sphere (r £ R) Since there is no charge inside the sphere, E = 0. Therefore, the field intensity everywhere inside the spherical conductor is zero. sR e Since E is zero inside the sphere, it requires no work to move a test charge inside and therefore, the potential is constant, being equal to the value at the surface of the sphere. The results are summarised below. V=

At r = R,

E = 0; r> a, then H =

NI a 2L z

2. When P is at the centre of the solenoid, Here,

cos q 2 =

The magnetic field is given as, H =

L /2 ( L /2) 2 + a 2 NI Ê 2L Á Ë

H=

= – cos q1

ˆ az = ( L /2) + a ¯˜ L /2 2

NI L + 4a 2 2

2

az

NI L + 4a 2 2

az

288

Electromagnetic Field Theory

NOTE If L >> a, then H =

NI a L z

3. When P is at the midway between one end and centre of the solenoid, Here, L /4 L cos q 2 = = 2 2 2 ( L /2) + a L + 16a 2 3L /4 3L cos (p – q1 ) = = 2 2 2 (3L /4) + a 9 L + 16a 2 3L

cos q1 = –

\

The magnetic field is given as, H =

NI Ê 2L Á Ë

9 L + 16a 2 2

L L + 16a 2

2

+

ˆa z 9 L + 16a ˜¯ 3L

2

2

NOTE If L >> a, then cos q1 = – H=

NI Ê 1+ 2L Á Ë

3L 9L + 16a 2 2

ª –1 and the magnetic field is given as,

ˆ az . L + 16a ˜¯ L

2

2

The variation of the field from point to point is shown in Fig. 3.35 (b).

Fig. 3.35 (b) Variation of magnetic field from point to point along the axis of a finite solenoid

3.13.4

Toroid

*Example 3.43 A toroid of length L has N turns and carries a current I. Determine H inside and outside the toroid.

Solution A toroid consists of a circular ring-shaped magnetic core around which wire is coiled. A toroid may be considered as a solenoid wrapped around with its ends connected. Thus, the magnetic field is completely confined inside the toroid and the field points in the azimuthal direction (clockwise due to the way the current flows, as shown in Fig. 3.36).

Magnetostatics

Fig. 3.36

289

A toroid with N turns

Since N wires cut the Amperian path, each carrying a current I, the net current enclosed by the Amperian path is I enc = NI Applying Ampere’s law to the Amperian path, we obtain

Ú H ◊ dl = Ú Hdl = H Ú dl = H (2p r ) = NI l

l

l

H=

\

NI 2p r

(3.71)

where r is the distance measured from the centre of the toroid and is known as the mean radius. If the thickness of the toroid is much less than its mean radius, then r ª R. Hence, H=

NI NI NI = = inside the toroid 2p r 2p R 2 L

where, L is the length of the wire. Outside the toroid, net current enclosed = (NI – NI) = 0, and thus, H = 0. To summarise the results NI ; inside the toroid 2L = 0 ; outside the toroid

H=

Example 3.44 Consider an infinitely long, cylindrical conductor of radius R carrying a current I with a non-uniform current density J = ar where a is a constant. Find the magnetic field everywhere. This is shown in Fig. 3.37.

Solution By Ampere’s law,

Ú B ◊ dS = m0 I enc

Fig. 3.37

Non-uniform current density

290

Electromagnetic Field Theory

Here, I enc = Ú J ◊ dS = Ú (a r ) (2p rdr ) We will consider two cases: (1) For r < R: In this case r

I enc = Ú (a r )(2p rdr ) = Ú 2pa r 2 dr = 0

2 pa r 3 3

Applying Ampere’s law, the magnetic field at any point inside the conductor is given as Bi (2p r ) =

2 m pa r 3 3 0

Bi =

or,

a m0 2 r 3

The direction of the magnetic field Bi is tangential to the Amperian loop that encloses the current. (2) For r > R: In this case I enc = Ú (a r ) (2p rdr ) R

= Ú 2pa r 2 dr 0

=

2 pa R 3 3

Fig. 3.38

Applying Ampere’s law, the magnetic field at any point outside the conductor is given as Bo (2p r ) =

2 m pa R3 3 0

or,

Bo =

The magnetic field as a function of distance away from the conductor

a m0 R3 3r

To summarise, the results are a m0 2 r 3 a m0 R3 = 3r

B=

rR

A plot of B as a function of r is shown in Fig. 3.38.

Example 3.45 A coaxial cable has core of radius ‘a’ and sheath of radius ‘b’. A current ‘I’ flows along the core, uniformly distributed across it, and returns along the sheath, uniformly distributed around it. Find the magnetic field intensity: (i) Within the core (r < a). (ii) Within the core-sheath space (a £ r £ b); and (iii) Outside the sheath (r > b).

Magnetostatics

Solution Within the core (r < a): Applying Ampere’s law

Ú H ◊ dl = I enc = Ú J ◊ dS l

Since, the current is distributed uniformly over the cross-section J=

I az p a2

dS = rdrd f az

and

r 2p

I enc = Ú

\

0

\

Ê

2

2

ˆ

I I r r Ú p a 2 rdrd f = p a 2 ¥ 2p ¥ 2 = I ÁË a 2 ˜¯ 0 2p

Ê

2

ˆ

r Ú H ◊ dl = Hf Ú rd f = I ÁË a 2 ˜¯

fi

Hf =

0

l

Ir 2p a 2

Ir \ H= af 2p a 2 Within the core-sheath space (a £ r £ b): Applying Ampere’s law

Ú H ◊ dl = I enc = Total current = I l

2p

fi

Hf

Ú rd f = I

fi

Hf =

0

\

H=

I 2p r

I a 2p r f

Outside the sheath (r > b): Applying Ampere’s law

Ú H ◊ dl = ( I inner cond. + I outer cond. ) = ( I - I ) = 0 fi

l

Hf = 0

Note If the thickness of the sheath is ‘t’, then for the region, b £ r £ (b + t), I enc = I + Ú J ◊ dS Here, J is the current density of the outer conductor and is along - az . \

\

J =-

I enc = I -

I az p [(b + t ) 2 - t 2 ]

2p r r 2 - b 2 Ô¸ ÔÏ I rdrd f I 1 = Ì ˝ Ú Ú 2 p [(b + t ) 2 - t 2 ] f = 0 r = b ÔÓ t + 2bt Ô˛

291

292

Electromagnetic Field Theory

By Ampere’s law 2p

Hf

Ú

rd f = I enc

fi

Hf =

0

\

H=

r 2 - b2 ¸ I Ï 1- 2 Ì 2p r Ó t + 2bt ˝˛

r 2 - b2 ¸ I Ï 1- 2 a Ì 2p r Ó t + 2bt ˝˛ f

To summarise, the magnetic field is given as Ir af 2p a 2 I = a 2p r f r 2 - b 2 Ô¸ I ÔÏ 1- 2 = a Ì 2p r ÓÔ t + 2bt ˝˛Ô f =0

H=

0£r£a a£r£b b £ r £ (b + t ) r ≥ (b + t )

Example 3.46 An infinitely long coaxial pair of circular conductors are located in free space and carry equal and opposite total static current I. The inner conductor is of radius a while the inner and outer radii of the outer conductor are b and c respectively. Sketch the variation of magnetic flux density over the range (0, c) and show that the field inside the outer conductor (b < r < c) is Fig. 3.39

m I Ê c2 - r 2 ˆ B= 0 Á 2 2p r Ë c - b 2 ˜¯

Variation of magnetic flux density within the sheath of coaxial cable

and zero for r > c.

Solution From Example 3.45, the thickness of the sheath in this problem can be written as t = (c – b). Replacing this value, the magnetic flux density for the region (b < r < c) is obtained as B=

m0 I ÏÔ r 2 - b 2 ¸Ô m0 I È (c - b) 2 + 2b(c - b) - r 2 + b 2 ˘ m0 I Ê c 2 - r 2 ˆ 1- 2 = Ì ˙ = 2p r Á 2 2p r ÓÔ t + 2bt ˝˛Ô 2p r ÍÎ (c - b) 2 + 2b(c - b) Ë c - b 2 ˜¯ ˚

The variation of the flux density is shown in Fig. 3.39.

Example 3.47 Using Ampere’s circuital law in integral form, find H everywhere due to the current density in cylindrical co-ordinates: J = 0, 0