Electromagnetism: Problems and Solutions (Iop Concise Physics) 1681744287, 9781681744285

Table of contents :
PRELIMS.pdf
Preface
Acknowledgements
Author biographies
Carolina C Ilie
Zachariah S Schrecengost
Illustrations
Julia R D’Rozario
CH001.pdf
Chapter 1 Maxwell’s equations—Electrodynamics
Introduction
1.1 Theory
1.1.1 Ohm’s law
1.1.2 Joule heating law
1.1.3 Flux rule for motional electromotive force
1.1.4 Magnetic energy
1.1.5 Maxwell’s equations
1.2 Problems and solutions
Bibliography
CH002.pdf
Chapter 2 Conservation laws
Introduction
2.1 Theory
2.1.1 Electromagnetic energy density
2.1.2 Poynting’s theorem
2.1.3 Maxwell’s stress tensor
2.1.4 Electromagnetic force on a charge in volume V
2.1.5 Electromagnetic momentum density
2.1.6 Continuity equation
2.2 Problems and solutions
Bibliography
CH003.pdf
Chapter 3 Electromagnetic waves
Introduction
3.1 Theory
3.1.1 Intensity
3.1.2 Radiation pressure
3.1.3 Monochromatic plane wave
3.1.4 Electromagnetic waves in matter
3.1.5 Coaxial transmission line
3.2 Problems and solutions
Bibliography
CH004.pdf
Chapter 4 Potentials and fields
Introduction
4.1 Theory
4.1.1 Electric field
4.1.2 Gauge transformations
4.1.3 Retarded potentials
4.1.4 Jefimenko’s equations
4.1.5 Liénard–Wiechert potentials
4.1.6 Moving point charge
4.2 Problems and solutions
Bibliography
CH005.pdf
Chapter 5 Radiation
Introduction
5.1 Theory
5.1.1 Power radiated
5.1.2 Electric dipole radiation
5.1.3 Magnetic dipole radiation
5.1.4 Larmor formula
5.1.5 Radiation fields
5.2 Problems and solutions
Bibliography

Citation preview

Electrodynamics Problems and solutions

Electrodynamics Problems and solutions Carolina C Ilie and Zachariah S Schrecengost State University of New York at Oswego

Morgan & Claypool Publishers

Copyright ª 2018 Morgan & Claypool Publishers All rights reserved. No part of this publication may be reproduced, stored in a retrieval system or transmitted in any form or by any means, electronic, mechanical, photocopying, recording or otherwise, without the prior permission of the publisher, or as expressly permitted by law or under terms agreed with the appropriate rights organization. Multiple copying is permitted in accordance with the terms of licences issued by the Copyright Licensing Agency, the Copyright Clearance Centre and other reproduction rights organisations. Rights & Permissions To obtain permission to re-use copyrighted material from Morgan & Claypool Publishers, please contact [email protected]. ISBN ISBN ISBN

978-1-6817-4931-0 (ebook) 978-1-6817-4928-0 (print) 978-1-6817-4929-7 (mobi)

DOI 10.1088/978-1-6817-4931-0 Version: 20180501 IOP Concise Physics ISSN 2053-2571 (online) ISSN 2054-7307 (print) A Morgan & Claypool publication as part of IOP Concise Physics Published by Morgan & Claypool Publishers, 1210 Fifth Avenue, Suite 250, San Rafael, CA, 94901, USA IOP Publishing, Temple Circus, Temple Way, Bristol BS1 6HG, UK

In the memory of my father. To the people I still learn from every day: my family, mentors, students, and friends—CCI. To my friends, family and mentors—ZSS.

Contents Preface

ix

Acknowledgements

xi

Author biographies

xii

Illustrations

xiii

1

Maxwell’s equations—Electrodynamics

1-1

1.1

Theory 1.1.1 Ohm’s law 1.1.2 Joule heating law 1.1.3 Flux rule for motional electromotive force 1.1.4 Magnetic energy 1.1.5 Maxwell’s equations Problems and solutions Bibliography

1.2

2

Conservation laws

2.1

Theory 2.1.1 Electromagnetic energy density 2.1.2 Poynting’s theorem 2.1.3 Maxwell’s stress tensor 2.1.4 Electromagnetic force on a charge in volume V 2.1.5 Electromagnetic momentum density 2.1.6 Continuity equation Problems and solutions Bibliography

2.2

2-1

3

Electromagnetic waves

3.1

Theory 3.1.1 Intensity 3.1.2 Radiation pressure 3.1.3 Monochromatic plane wave 3.1.4 Electromagnetic waves in matter 3.1.5 Coaxial transmission line Problems and solutions Bibliography

3.2

1-1 1-1 1-1 1-2 1-2 1-2 1-3 1-28

2-1 2-1 2-1 2-2 2-2 2-2 2-2 2-3 2-24 3-1

vii

3-1 3-1 3-1 3-2 3-2 3-2 3-3 3-30

Electrodynamics

4

Potentials and fields

4.1

Theory 4.1.1 Electric field 4.1.2 Gauge transformations 4.1.3 Retarded potentials 4.1.4 Jefimenko’s equations 4.1.5 Liénard–Wiechert potentials 4.1.6 Moving point charge Problems and solutions Bibliography

4.2

5

Radiation

5.1

Theory 5.1.1 Power radiated 5.1.2 Electric dipole radiation 5.1.3 Magnetic dipole radiation 5.1.4 Larmor formula 5.1.5 Radiation fields Problems and solutions Bibliography

5.2

4-1 4-1 4-1 4-1 4-2 4-2 4-2 4-3 4-3 4-25 5-1 5-1 5-1 5-2 5-2 5-2 5-2 5-3 5-24

viii

Preface We wrote this book of problems and solutions as a natural continuation of our first book Electromagnetism: Problems and Solutions, published in November 2016 by Institute of Physics Concise Physics, a Morgan & Claypool publication. As with the first book, we wrote this book for junior or senior undergraduate students, and for graduate students who may have not studied electrodynamics yet and who may want to work on more problems and have an immediate feedback while studying. The authors strongly recommend the textbook of David J Griffiths, Introduction to Electrodynamics as a first source manual, since it is recognized as one of the best books on electrodynamics at the undergraduate level. We consider this book of problems and solutions a companion for the student who would like to work electrodynamics problems more independently in order to deepen their understanding and the problems solving skills and perhaps prepare for graduate school. We add brief theoretical notes and formulae, and for a complete theoretical approach we suggest Griffiths’s book. Every chapter is organized as follows: concise theoretical notes, then the problem text with the solution. Each chapter ends with a brief bibliographical note. We discuss herein the main concepts and techniques related to Maxwell’s equations, conservation laws, electromagnetic waves, potentials and fields, and radiation. For students interested in relativistic electrodynamics we recommend, Introduction to Electrodynamics, by David Griffiths, Classical Electrodynamics, by John David Jackson, Modern Electrodynamics, by Andrew Zangwill, and Relativistic Electrodynamics and Differential Geometry, by Stephen Parrot. We follow here the same notation employed in our first book, Electromagnetism: Problems and solutions and introduced by David Griffiths in Introduction to Electrodynamics. Therefore, we use the r ⃗ for the vector from a source point r ⃗′ to ′ ⃗ the field point r ⃗ . Please note that rˆ = rr = r ⃗ − r ⃗ and, as you see, this notation ∣ r ⃗ − r ′⃗ ∣

simplifies greatly already complex equations, but you need to be careful, especially if you only use cursive or typed letters. Also, we use the same notation s for the distance to z axis in cylindrical coordinates, as is used in Griffiths’ book. The chosen units are SI units—the international system. The reader should be aware that other books may employ either the Gaussian system (CGS) or the Heaviside–Lorentz (HL). Here is the Coulomb force in each of the systems: In the SI system:

F⃗ =

1 q1q2 rˆ 4πεo r 2

In CGS (Gaussian):

F⃗ =

q1q2 rˆ r2

And in Heaviside–Lorentz (HL)

ix

Electrodynamics

F⃗ =

1 q1q2 rˆ 4π r 2

Some of the problems are applications of the Maxwell’s equations for various systems which can be discussed with the available mathematical methods. Several problems which are presented here may appear in a variety of undergraduate textbooks on electrodynamics, as they are classic solvable examples. However, we consider it important to discuss these problems as they are fundamental to the study of electrodynamics. We also present some problems of a more general nature, which may be a bit more challenging. We tried to keep a balance between the two types of problems, and we hope that the readers will enjoy this variation and have as much thrill and excitement as we had while creating and solving these problems.

x

Acknowledgements We want to thank our illustrator, Julia D’Rozario for making all of the figures. We thank Dr Ildar Sabirianov for giving useful suggestions. We appreciate the support of the administration at SUNY Oswego in granting Dr Ilie the sabbatical semester. Dr Ilie is grateful to the Electrodynamics II students Shelby Davis, Vincent DeBiase, Michael Kolacki and Stephen Porter for sharing the enthusiasm for this field. We also thank Victor Sabirianov for contributing to the last minute editing. Dr Ilie is grateful to Dr Peter Dowben, from University of Nebraska at Lincoln for being a wonderful mentor. And many thanks to our editors, Joel Claypool, Publisher at Morgan & Claypool Publishers, Jeanine Burke, Consulting Editor at Morgan & Claypool for the IOP Concise Physics ebook program, and Chris Benson, Production Editor at IOP Publishing. We are also grateful to Brent Beckley, the Direct Marketing Manager of Morgan & Claypool Publishers. Last, but not the least, we thank our families and friends for their love, joy, encouragement, and sense of humor—their constant presence made this journey exciting and rewarding.

xi

Author biographies Carolina C Ilie Carolina C Ilie is an Associate Professor with tenure at the State University of New York at Oswego. She taught Electromagnetic Theory for almost ten years and she designed various problems for her students’ exams, group work and quizzes and was excited to work on the two books on electromagnetism and electrodynamics during the last two years. Dr Ilie received her PhD in Physics and Astronomy at the University of Nebraska at Lincoln, an MSc in Physics at the Ohio State University and another MSc in Physics at the University of Bucharest, Romania. She received the President’s Award for Teaching Excellence in 2016 and the Provost Award for Mentoring in Scholarly and Creative Activity in 2013. She lives in Central New York with her spouse, also a physicist, and their two sons.

Zachariah S Schrecengost Zachariah S Schrecengost is a State University of New York alumnus. He graduated with a BS degree having completed majors in Physics, Software Engineering, and Applied Mathematics. He took the Advanced Electromagnetic Theory course with Dr Ilie and loved to be involved in this project. He brings to the project both the fresh perspective of the student taking electrodynamics, as well as the enthusiasm and talent of an alumnus who is an electrodynamics and upper level mathematics aficionado. Mr Schrecengost worked as a software engineer in Syracuse and is now working on his PhD in Physics.

xii

Illustrations Julia R D’Rozario Julia R D’Rozario is an alumna of the State University of New York at Oswego. She graduated in December 2016 with a BS in Physics and a BA in Cinema and Screen Studies and finished a minor in Astronomy by May 2016. She completed the Advanced Electromagnetic Theory course with Dr Ilie and has much experience in the arts through her cinema studies. Ms D’Rozario contributes her knowledge of electrodynamics and talents in drawing in Inkscape software. She is currently working on her PhD in Microsystems Engineering at the Rochester Institute of Technology. She works for the NanoPower Research Laboratory under the advisement of Dr Seth Hubbard on III–V compound solar cells.

xiii

IOP Concise Physics

Electrodynamics Problems and solutions Carolina C Ilie and Zachariah S Schrecengost

Chapter 1 Maxwell’s equations—Electrodynamics

Introduction In this chapter we will introduce the electromotive force, Ohm’s law, and Faraday’s law along with electromagnetic induction, inductance and energy in a magnetic field. We write Maxwell’s equations in vacuum and matter. At the end of this chapter, we present a problem involving the concept of magnetic charge; we uncover its role in a particular duality transformation and its relationship to the invariance of Maxwell’s equations.

1.1 Theory 1.1.1 Ohm’s law Given conductivity σ , the current density J ⃗ is given by

J ⃗ = σE ⃗ where E ⃗ is the electric field. This can also be expressed in the form

V = IR which relates the potential difference V to current I via the resistance R. 1.1.2 Joule heating law Given a current I flowing through a resistor with resistance R, the power delivered is given by

P=

doi:10.1088/978-1-6817-4931-0ch1

dW = I 2R dt

1-1

ª Morgan & Claypool Publishers 2018

Electrodynamics

1.1.3 Flux rule for motional electromotive force Given a magnetic flux Φ, the electromotive force (emf) generated is

E=−

dΦ dt

where the flux is given by

Φ=

∫

B ⃗ · da ⃗

1.1.4 Magnetic energy The energy stored in a magnetic field B ⃗ is given by

W=

1 2μo

∫

B 2 dτ

all space

Also, the energy stored in an inductor with inductance L , carrying current I, is given by

W=

1 2 LI 2

1.1.5 Maxwell’s equations Gauss’s Law Differential Form

∇ · E⃗ =

ρ ϵo

Integral Form

∮S E ⃗ · da ⃗ =

qenc ϵo

No Name Differential Form

∇ · B⃗ = 0 Integral Form

∮S B ⃗ · da ⃗ = 0 Faraday’s Law Differential Form

∇ × E⃗ = −

1-2

∂B ⃗ ∂t

Electrodynamics

Integral Form

∮P E ⃗ · dl⃗ = − ddΦt Ampère–Maxwell Law Differential Form

∇ × B ⃗ = μoJ ⃗ + μoϵo

∂E ⃗ ∂t

Integral Form

∮P B ⃗ · dl⃗ = μoIenc + μoϵo ∮S ∂∂Et

⃗

· da ⃗

1.2 Problems and solutions Problem 1.1. Two concentric spheres of radii a and b, with b > a , are separated by a material with conductivity σ (r ) = kr , where k is a constant with appropriate units. Find the resistance of the material if the potential between the spheres is V .

Solution 1.1. Starting with the current density, we have

J ⃗ = σE ⃗ =

dI ⃗ da

Note our current will only depend on r so

dI = σE da = (kr )(E )(r 2 sin θ dϕ dθ ) = kr 3E sin θ dϕ dθ Thus, 2π 3

I = kr E

π

∫∫ 0

sin θ dθ dϕ = 4πkr 3E

0

1-3

Electrodynamics

and solving for the field yields

E=

I 4πkr 3

Considering a

V=−

∫

a

E ⃗ · d l⃗ = −

b

∫ b

I ⎛1 1⎞ I ⎜ 2 − 2⎟ dr = 3 ⎝ 4πkr 8πk a b ⎠

we have

V=

b2 − a 2 I 8πka 2b 2

Since Ohm’s law states V = IR , we see the resistance is given by

R=

b2 − a 2 8πka 2b 2

Problem 1.2. Two long cylinders of radii a and b, with b > a , are separated by an unknown material of conductivity σ . They are maintained at an electric potential V with a current I flowing from one to another in a length L . Find the conductivity σ .

Solution 1.2. Let us start by calculating the electric field between the cylinders at a radius s , with s between the radii a and b. We consider λ the charge per unit length of the inner cylinder. Gauss’s law states

∮S E ⃗ · da ⃗ =

qenc ϵo

The left side of the equation is

∮S E ⃗ · da ⃗ = E 2πsL While the enclosed charge is simply

qenc = λL It follows that

2πsLE =

1-4

λL ϵo

Electrodynamics

And the electric field is

E⃗ =

λ sˆ 2πϵos

Remember that J ⃗ = σE ⃗ , then the current is calculated as

I=

∫ J ⃗ · da ⃗ = σ ∫ E ⃗ · da ⃗ = σ 2πϵλ os 2πsL = σλϵoL

The electric potential is a

V=−

∫ b

a

E ⃗ · d l⃗ = −

∫ b

b λ λ λ λ ds = − ln s ab = (ln b − ln a ) = ln 2πϵos 2πϵo 2πϵo 2πϵo a

By substituting λ from the current I = potential

V=

Iϵo σL

2πϵo

ln

σλL ϵo

we obtain λ =

Iϵo . σL

Back to the electric

b I b ln = a 2πσL a

Therefore, the conductivity is

σ=

I b ln 2πLV a

This is how we can determine the type of material by measuring the current and the potential difference and, of course the geometric dimensions a , b and L . Problem 1.3. A conducting bar of mass m slides without friction on two parallel conducting rails placed at distance l connected by a resistor R . The system is placed in a uniform magnetic field B ⃗ , out of page. (a) Find the current in the circuit if the bar moves to the left with speed v. (b) Find the force acting on the bar and the velocity of the bar at time t (v0 at t = 0) (c) Verify if the kinetic energy of the bar is fully or partially transferred to the resistor.

1-5

Electrodynamics

Solution 1.3. (a) Considering the motion of the bar, the current (and the resulting forces) are given by

From Faraday’s law,

E=−

dΦ d(B lx ) dx =− = −B l = −B lv dt dt dt

Here the minus sign indicates the direction of the current flow. In the circuit we can use the absolute value

E = IR Therefore, the electric current is

I=

E B lv = R R

and the direction is clockwise. (b) The force acting on the bar can be easily found from F ⃗ = I (l⃗ × B ⃗ ) since l⃗ is perpendicular to B ⃗

F = I lB =

B 2 l 2v B lv lB = R R

On the other hand,

F = ma = m

dv dt

Therefore,

F=−

B 2 l 2v R

where the force is in the opposite direction of the velocity, which will slow down the bar. From here it is easy to obtain a general formula for the velocity v(t ).

m

dv B 2 l 2v =− dt R

1-6

Electrodynamics

v(̇ t ) = −

B 2 l 2 v (t ) Rm

With the notation

α=

B 2l 2 Rm

we can write

v(̇ t ) = −αv(t ) which has the solution B 2l2

v(t ) = v0e−αt = v0e− Rm t mv 2

(c) The initial kinetic energy is K = 2 . This energy is transmitted to the resistor as heating. The power is P = VI = RI 2 . On the other hand, dW P = dt = RI 2 . Let us calculate the energy: ∞

∞

W = ∫ RI 2 dt = ∫ 0

0

=

∞

∞

B 2l 2 2 B 2l 2 B 2 l 2v 2 dt = v0 ∫ e−2αt dt ∫ v02 e−2αt dt = R R 0 R 0

B 2l 2 2 e−2αt v0 R − 2α

∞

= 0

B 2l 2v02 (0 − 1) mv02 = R −2 B 2l2 2 Rm

Therefore, all kinetic energy is transferred to the resistor. Problem 1.4. Consider the track below, which consists of two parallel conducting rails, separated by a distance l , and two metal bars of mass m; all of which is in a magnetic field B pointing into the page. Initially, the right bar (bar 2), which has resistance R , is held in place, while the left bar (bar 1) is moving towards bar 2 with velocity v0 . At time t = 0, both bars are allowed to move freely. Find the minimum distance d between the bars at time t = 0 such that the bars will not collide.

1-7

Electrodynamics

Solution 1.4. At time t = 0, there is a current, I, that flows in the loop. From

E=−

dΦ dx = −B l = −B l( −v0) = B lv0 dt dt

we can see this current is flowing clockwise. This produces a force on each bar that has magnitude

F = I lB On bar 1, this force is directed towards the left; and on bar 2, this force is directed towards the right. This means that for t > 0, bar 1 will decelerate and bar 2 will accelerate. This results in a time dependent current, I (t ), which is given by

I (t ) = and now

dx dt

E B l dx =− R R dt

will depend on the velocities of both bars,

dx = v2(t ) − v1(t ) dt Therefore,

I (t ) = −

Bl [v2(t ) − v1(t )] R

Now we can analyze the motion of each bar. For bar 1, the force is to the left. Thus,

mv1̇ = −F = −I (t )lB =

B 2l 2 (v2 − v1) R

mv2̇ = F = I (t )lB = −

B 2l 2 (v2 − v1) R

For bar 2, we have

We can rewrite these as

dv1 = α(v2 − v1) dt and

dv 2 = −α(v2 − v1) dt where

α=

B 2l 2 mR

1-8

Electrodynamics

Note

dv1 dv =− 2 dt dt so

v1 = −v2 + C where C is a constant. At t = 0, we have v1(0) = v0 and v2(0) = 0. Thus, C = v0 and v1 = v0 − v2 . Now we can use this to solve for v2 ,

dv 2 = −α(v2 − v0 + v2 ) = −α(2v2 − v0) dt Rearranging yields

dv 2 = − α dt 2v 2 − v 0 and integration yields

1 ln( 2v2 − v0∣) = − αt + C 2 ln(v0 − 2v2 ) = − 2αt + C v0 − 2v2 = Ae−2αt where A and C are constants. Now we can solve for v2 , v v2 = 0 + Ae−2αt 2 Since v2(0) = 0, we have

A=−

v0 2

Therefore,

v2 =

v0 (1 − e−2αt ) 2

Using this, we can solve for v1,

v1 = v0 − v2 =

v0 (1 + e−2αt ) 2

We can now use v1 and v2 to solve for x2 − x1. Solving for x1(t ),

v1 =

dx1 v v v = 0 (1 + e−2αt ) ⇒ x1(t ) = 0 t − 0 e−2αt + E dt 2 2 4α

v2 =

dx2 v v v = 0 (1 − e−2αt ) ⇒ x2(t ) = 0 t + 0 e−2αt + G dt 2 2 4α

and x2(t ),

1-9

Electrodynamics

where E and G are constants. At time t = 0, x2(0) − x1(0) = d . Therefore, v v v x2(0) − x1(0) = 0 + 0 + G − E = 0 + G − E = d 4α 4α 2α So,

G−E=d−

v0 2α

Now to solve x2(t ) − x1(t ),

v0 v v v t + 0 e−2αt − 0 t + 0 e−2αt + G − E 2 4α 2 4α v0 −2αt = − 1) + d (e 2α

x2(t ) − x1(t ) =

Since we do not want the bars to collide, we require

lim [x2(t ) − x1(t )] > 0

t →∞

so

lim[x2(t ) − x1(t )] = −

t →∞

v0 v +d>0⇒d> 0 2α 2α

Therefore, to ensure the bars will not collide, we must have mRv0 d> 2B 2l 2 Problem 1.5. Consider the wire below, which consists of a circular loop of radius a and a resistor with resistance R . The circular portion is placed in a magnetic field directed into the page with B (t ) = B0 e−kt , where k is a constant. Find the total energy dissipated by the resistor for t > 0.

1-10

Electrodynamics

Solution 1.5. The changing magnetic field induces an emf in the loop that is given by

E=−

dΦ dB = −πa 2 = πa 2kB0e−kt dt dt

This means the current in the loop is given by E πa 2kB0 −kt e I= = R R We know the power is dW P= = I 2R dt so total energy is then ∞

W=

∫

I 2R dt

0

Using our current, we have ∞

W=

∫ 0

2 ⎛ πa 2kB0 ⎞2 (πa 2kB0) e−kt ⎟ R dt = ⎜ ⎝ R ⎠ R

∞

∫

e−2kt dt

0

Therefore, the total energy dissipated by the resistor is

W=

π 2a 4kB02 2R

Problem 1.6. (a) A battery of electromotive force E and internal resistance r is connected to either a system of n identical resistors connected in series, or to n resistors connected in parallel, all having the resistance R . Calculate which circuit (series or parallel) has greater power.

1-11

Electrodynamics

(b) For a single resistor circuit of electromotive force E having the internal resistance r , connected to a resistor R , what value of the internal resistance r maximizes the power? Solution 1.6. (a) The resistance for series circuit: n

Rs =

∑Ri = nR i=1

From Kirchhoff’s law:

E = Is(r + Rs ) = Is(r + nR ) For the circuit with resistors in parallel, we have the corresponding equations: n 1 1 n =∑ = Rp i = 1 Ri R

R n ⎛ R⎞ E = Ip(r + Rp) = Ip⎜r + ⎟ ⎝ n⎠ Rp =

The power is:

E2 nR (r + nR )2 R E2 Pp = I p2Rp = R 2 n r+ Ps = Is2Rs =

(

n

)

Let us calculate the limit for an infinite number of resistors E2

P lim s = lim n →∞ Pp n →∞

(r + nR )2 E2

nR R

(

n2 r + = lim n →∞

R 2 n

)

(r + nR )

2

2 n

n →∞

2 r

n

(r + ) R n

( = lim n(

R 2 n

) + R)

n2 r +

2

=

r2 R2

If r < R, Ps < Pp. If r = R, Ps = Pp. (b) We start with Kirchhoff’s law and the formula for power

E = I (R + r ) ⇒ I = P = I 2R =

E R+r

E2 R (R + r )2

We obtain the minimum by finding the value of the resistance R for which the derivative of the power with respect to the resistance becomes zero

1-12

Electrodynamics

⎛ E2 ⎞ d⎜ R⎟ 2 ⎝ ⎠ ( R + r ) dP (R + r )2 − 2R(R + r ) = = E2 dR dR (R + r ) 4 dP = 0 ⇒ (R + r )(R + r − 2R ) = 0 ⇒ (R + r )(r − R ) = 0 ⇒ r = R dR For r = R , the power in the circuit is maximum. Problem 1.7. Consider the loop of wire below with radius a located above a long straight wire carrying a current I (t ) with ddIt = k . Find the induced electromotive force (emf) in the loop if its center is positioned 2a above the wire.

Solution 1.7. The electromotive force is given by

E=−

dΦ dt

where the flux is

Φ=

∫ B ⃗ · da ⃗

We know the magnetic field due to a long wire is

B⃗ =

μoI ˆ ϕ 2πs

If we consider the loop in the zs-plane, we have

1-13

Electrodynamics

We can see the equation of the loop is

z 2 + (s − 2a )2 = a 2 And we also have

da ⃗ = dz ds ϕˆ Thus,

μoI dz ds 2πs To show how z is bounded, we can rewrite our circle equation as B ⃗ · da ⃗ =

z 2 = a 2 − (s − 2a )2 ⇒ z = ± a 2 − (s − 2a )2 Therefore, our flux is 3a

Φ=

∫ a −

a 2−(s−2a )2

∫ a 2−(s−2a )2

μoI μI dz ds = o 2πs 2π

3a

∫

2 a 2 − (s − 2a )2 s

a

ds = μoa(2 −

3 )I

and our electromotive force is

E=−

dΦ = −μoa(2 − dt

3)

dI = −μoa(2 − dt

3 )k

Problem 1.8. A rectangular loop, of sides a and b is placed in the xz-plane and exposed to a non-uniform time dependent magnetic field of the form B ⃗(x , t ) = kx 2t 3yˆ , where k is a constant. Find the electromotive force induced in the loop.

Solution 1.8. Let us start from the flux rule for motional electromotive force

E=−

dΦ dt

1-14

Electrodynamics

where the flux is given by

Φ=

∫

B ⃗ · da ⃗

In our case, the flux can be calculated as following a

Φ(t ) =

∫

B ⃗ · da ⃗ =

∫ B(x, t ) dx dz = kt ∫ 3

b 2

x dx

∫

0

0

dz = kt 3

a 3b 3

Therefore, the electromotive force is

E(t ) = −

dΦ =− dt

(

d kt 3

a 3b 3

dt

) = −kt a b 2 3

Problem 1.9. Consider a circuit with a battery which supplies a constant electromotive force E 0 , a resistor with resistance R and a solenoid of impedance L . Find the current in the circuit and obtain the time after closing the circuit for which the current will reach 1 of its final equilibrium value. n

Solution 1.9. Kirchhoff’s law in the circuit: dI E0 = L + IR dt E 0 L dI = +I R R dt The solution of the first order differential equation is R E I (t ) = 0 + Ae− L t R where A is a constant to be obtained from the initial conditions. Starting with at t = 0, we have I (0) = 0, so

I (0) =

E0 E + Ae0 = 0 ⇒ A = − 0 R R

1-15

Electrodynamics

Therefore,

I (t ) =

R E0 1 − e− L t R

(

)

When

t → ∞ , I (t ) →

E0 R

Let us find the time t necessary for the circuit to reach the current I (t ) = R E0 E = 0 1 − e− L t nR R

(

Note that

L R

E0 nR

)

= τ , where τ is the time constant.

⎛ 1 1 1⎞ n t t t = 1 − e− τ ⇒ e− τ = 1 − ⇒ − = ln ⎜1 − ⎟ ⇒ t = τ ln ⎝ n−1 n n n⎠ τ For example, for n = 2, t = τ ln 2. Problem 1.10. Find the self-inductance L per unit length for an infinitely long coaxial cable of radii a , b with a < b, carrying a current I , as in the figure. Use energy calculations.

Solution 1.10. The self-inductance is related to the energy as

W=

LI 2 2

Let us calculate the energy of the configuration

W=

1 2μo

∫V B 2 dτ

Next, we need to find the magnetic field. From Ampère’s law, the only significant (non-zero magnetic field region) is for radius s between a and b.

∮ B ⃗ · dl⃗ = μoIenc 1-16

Electrodynamics

where Ienc = I is the enclosed current. The left-hand side is given by

∮ B ⃗ · dl⃗ = B 2πs From here,

μoI ˆ ϕ 2πs The energy of a cylindrical shell of length l , radius s and thickness ds is: B⃗ =

dW =

1 2 B 2π ls ds 2μo

dW =

1 ⎛ μoI ⎞2 ⎜ ⎟ 2π ls ds 2μo ⎝ 2πs ⎠

dW =

μoI 2l ds 4π s

We obtain the total energy by integrating over s from a to b b

W=

∫ a

μoI 2l ds μ I 2l μ I 2l b ln s ba = o ln = o 4π s 4π 4π a

Back to impedance per unit length:

W=

μ b LI 2 L 2W ⇒ = 2 = o ln 2 I l 2π a l

Problem 1.11. A long hollow wire (inner radius a , outer radius b) carries a current in one direction that is proportional to its distance from the axis. This current then returns along the surface of the wire. If the total return current is I , find the wire’s inductance per length.

Solution 1.11. Considering that the total energy is given by 1 1 W= B 2 dτ = LI 2 2 2μo V

∫

finding the magnetic field will allow us to solve the above equation for L . Recall that the field is given by

∮P B ⃗ · dl⃗ = μoIenc 1-17

Electrodynamics

Note we have Ienc = 0 for 0 < s < a and s > b. Therefore, we only have a magnetic field for a < s < b. The current density in this region can be written as J ⃗ = kszˆ We then have s

Ienc =

∫

J ⃗ · da ⃗ =

2π

∫∫ a

(ks′)s′ dϕ ds′ =

0

2πk 3 (s − a 3 ) 3

We can also solve for k considering that b

I=

∫

J ⃗ · da ⃗ =

2π

∫∫ a

(ks′)s′ dϕ ds′ =

0

2πk 3 (b − a 3 ) 3

so

k=

3I ⎛ 1 ⎞ ⎜ ⎟ 2π ⎝ b 3 − a 3 ⎠

Our enclosed current is then

s3 − a3 I b3 − a 3 Returning to our magnetic field, we have Ienc =

3

3

∮P B ⃗ · dl⃗ = μoIenc = μoI bs 3 −− aa3 Noting

∮P B ⃗ · dl⃗ = 2πsB we have

B=

μoI ⎛ s 3 − a 3 ⎞ ⎜ ⎟ 2πs ⎝ b3 − a 3 ⎠

Now we can use this to solve for the total energy using 1 W= B 2 dτ 2μo V

∫

where dτ = s ds dϕ dz . Therefore,

1 W= 2μo

b

2π

l

∫∫∫ a

0

0

⎛ μoI ⎞2 1 ⎛ s 3 − a 3 ⎞2 ⎜ ⎟ 2⎜ 3 ⎟ s dz dϕ ds ⎝ 2π ⎠ s ⎝ b − a 3 ⎠

2

μoI l = 2(2π )(b3 − a 3)2 =

6 1 μol 6a ln 2 2π

b

∫ a

(s 3 − a 3 ) 2 ds s

( ) + 3a b a

6

− 4a 3b3 + b6

6(b3 − a 3)2

1-18

I2

Electrodynamics

Now considering

W=

1 2 LI 2

we can see the inductance per length is given by

( ( ) + 1)a

3 2 ln L = l

b a

6

− 4a 3b3 + b6

12π (b3 − a 3)2

Problem 1.12. A slowly varying infinite surface current K⃗ (t ) = K (t )xˆ is flowing over the xy-plane. Find the induced electric field.

Solution 1.12. Using

∮P B ⃗ · dl⃗ = μoIenc we can see this current produce the field

⎧ μo ⎪− K (t )yˆ , B⃗ = ⎨ μ 2 ⎪ o K (t )yˆ , ⎩2

z>0 z 0 and into the page for z < 0. By first considering z > 0, we have

−

dΦ d =− dt dt

⎛ μ

⎞

∫ B ⃗ · da ⃗ = − ddt (B ⃗ · A⃗ ) = − ddt ⎝− 2o K (t )yˆ · (−zlyˆ)⎠ = − ⎜

⎟

μozl dK (t ) 2 dt

Also,

∮P E ⃗ · dl⃗ = −E l Therefore,

E⃗ =

μoz dK (t ) xˆ 2 dt

Similarly, for z < 0, we have

E⃗ = −

μoz dK (t ) xˆ 2 dt

We can check these by considering

∇ × E⃗ = −

∂B ⃗ ∂t

For z > 0,

∇ × E⃗ =

μ dK ( t ) ⎞ ∂B ⃗ ∂⎛ μ ∂ ⎛ μoz dK (t ) ⎞ ⎜ ⎟yˆ = o yˆ = − ⎜ − o K (t )yˆ ⎟ = − ⎝ ⎠ 2 dt 2 ∂t ∂z ⎝ 2 dt ⎠ ∂t

and for z < 0,

∇ × E⃗ =

μ dK ( t ) ⎞ ∂ ⎛μ ∂B ⃗ ∂ ⎛ μoz dK (t ) ⎞ ⎜− ⎟yˆ = − o yˆ = − ⎜ o K (t )yˆ ⎟ = − ⎠ 2 dt ∂t ⎝ 2 ∂t ∂z ⎝ 2 dt ⎠

as expected. Problem 1.13 Consider the circuit below, which consists of a solenoid, with n turns per length, at the center of, and connected to, a square loop, of side l . The axis of the solenoid is perpendicular to the plane of the loop. The loop contains a source E 0(t )

1-20

Electrodynamics

and a resistor R . Find the current that flows if E 0(t ) = E 0 sin(ωt ) and is connected at time t = 0.

Solution 1.13. The field due to the solenoid is given by

B ⃗(t ) = μonI (t )zˆ This varying field induces an emf in the circuit

Ein(t ) = −

dI d dΦ d = − (B ⃗ · A ⃗ ) = − ⎡⎣(nμoI (t )zˆ ) · (l 2zˆ )⎤⎦ = −μonl 2 dt dt dt dt

This opposes the electromotive force of the source, so the current is given by

I=

dI E 0 + Ein ⇒ IR = E 0 sin(ωt ) − μonl 2 dt R

Some rearrangement yields,

μonl 2

dI dI R E0 sin(ωt ) I= + IR = E 0 sin(ωt ) ⇒ + dt dt μonl 2 μonl 2

Now we can solve this differential equation for I to obtain,

I=

⎡ ⎛ − 2Rt ⎞⎤ 2 l μon − cos(ωt )⎟⎥ l R sin( ω t ) + μ n ω e ⎢ ⎜ o 2 ⎝ ⎠⎦ R2 + (μonωl 2) ⎣ E0

Problem 1.14. Consider a conducting spherical shell of radius a , which rotates about the z-axis with angular velocity ω. The shell is placed in a uniform magnetic field B ⃗ = B0 zˆ . (a) Show that the electromotive force E developed between the ‘north pole’ and the ‘south pole’ is zero. (b) What about the electromotive force between the ‘north pole’ and ‘parallel 45’?

1-21

Electrodynamics

Solution 1.14. (a) Let us start by characterizing the motion:

v ⃗ = ωa sin θ ϕˆ B ⃗ = B0zˆ f⃗ =

F⃗ = v ⃗ × B ⃗ = ωaB0 sin θ (ϕˆ × zˆ ) q

The electromotive force E is given by

E=

∫ f ⃗ · dl⃗

where

dl⃗ = a dθ θˆ Therefore, π

E=

∫ 0

π

( sin θ ) ωa B0( sin θ )(ϕˆ × zˆ ) · θˆ dθ = ωa 2B0 ∫ sinθ cos θ dθ = ωa 2B0 2

2

π

2

0

=0 0

Here we used the triple products properties A⃗ · (B ⃗ × C ⃗ ) = B ⃗ · (C ⃗ × A⃗ ) and the commutativity of the scalar product. Also, θˆ × ϕˆ = rˆ and zˆ · rˆ = cos θ . (b) Similarly, π 4

E = ωa 2B0

∫ 0

sinθ cos θ dθ = ωa 2B0

( sin 2

π θ )2 4

0

⎛ ⎜ = ωa 2B0⎜ ⎜ ⎝

( ) −0⎞⎟⎟ = ωa B 2 2

2

2

2

⎟ ⎠

0

4

Problem 1.15. Compare the ratios of conduction current to displacement current for two water solutions of permittivities ϵ1 and ϵ2 , having the resistivities ρ1 and ρ2 , at 1-22

Electrodynamics

frequency ν . Consider that both solutions have permeability μ = μo . Consider a parallel plate capacitor immersed in each type of water solution and with an applied voltage of V = V0 sin(2πνt ). Solution 1.15. The conduction current is given by

Jc = σE = σ

V d

For a parallel plate capacitor, the electric field is given by

E=

V d

The displacement current is

Jd =

∂ ⎛ V sin(2πνt ) ⎞ ϵV0 ∂(ϵE ) ∂D ⎟= (2πν ) cos(2πνt ) =ϵ ⎜ 0 = ⎠ ∂t ⎝ d d ∂t ∂t

Therefore, the ratio of the amplitudes of the two currents is

R=

Jc V d 1 = 0 = Jd 2πνϵρ ρd 2πνϵV0

Now if we compare the two water solutions

R1 = R2

1 2πνϵ1ρ1 1 2πνϵ2ρ2

=

ϵ2ρ2 ϵ1ρ1

Problem 1.16. Prove that Maxwell’s equations with magnetic charge (a)–(d) and the force law equation (e) are invariant under the duality transformation (*). ρ (a) ∇·E ⃗ = ϵe o

(b) ∇·B ⃗ = μo ρm ∂B ⃗ ∂t ∂E ⃗ μo ϵo ∂t

(c) ∇ × E ⃗ = −μo Jm⃗ − (d) ∇ × B ⃗ = μo Je⃗ +

(e) F ⃗ = qe(E ⃗ + v ⃗ × B ⃗ ) + qm(B ⃗ −

1 v ⃗ × E ⃗) c2

Duality transformation (*)

E ⃗′ = E ⃗ cos α − Bc⃗ sin α cB ⃗′ = E ⃗ sin α + Bc⃗ cos α cqe′ = cqe cos α − qm sin α qm′ = cqe sin α + qm cos α

1-23

Electrodynamics

Solution 1.16. (a) We need to prove that

∇ · E ⃗′ =

ρe′ ϵo

∇ · E ⃗′ = ∇ · (E ⃗ cos α − Bc⃗ sin α ) = ∇ · E ⃗ cos α − c(∇ · B ⃗ ) sin α ρ = e cos α − cμoρm sin α ϵo ρ′ ρ ⎞ 1 1 ⎛⎜ ρe cos α − m sin α⎟ = e = (ρe cos α − cμoϵoρm sin α ) = ⎠ ϵo ϵo ϵo ⎝ c We also use the relationship

c2 =

1 ϵoμo

(b) Let us prove that

∇ · B′⃗ = μoρm′ Note that the equation for ρm′ looks similarly to the one for qm′ .

⎛ E⃗ ⎞ 1 ∇ · B′⃗ = ∇ · ⎜ sin α + B ⃗ cos α⎟ = (∇ · E ⃗ ) sin α + (∇ · B ⃗ ) cos α c ⎝c ⎠ ρ = e sin α + μoρm cos α ϵoc ⎛ ρ ⎞ = μo⎜ e sin α + ρm cos α⎟ = μo(cρe sin α + ρm cos α ) = μoρm′ ⎝ ϵoμoc ⎠ (c) The following relationship to prove is

∂B′⃗ ∇ × E ⃗′ = − μoJm′⃗ − ∂t ⃗ ⃗ ∇ × E ′ = ∇ × (E cos α − Bc⃗ sin α ) = (∇ × E ⃗ ) cos α − c(∇ × B ⃗ ) sin α ⎛ ∂E ⃗ ⎞ ∂B ⃗ cos α − c⎜μoJe⃗ + μoϵo ⎟ sin α = − μoJm⃗ cos α − ∂t ⎠ ∂t ⎝ = − μo(Jm⃗ cos α + cJe⃗ sin α ) −

⎞ 1 ∂B′⃗ ∂⎛ ⃗ ⎜B cos α + E ⃗ sin α⎟ = −μoJm′⃗ − ⎠ c ∂t ∂t ⎝

1-24

Electrodynamics

(d) The last Maxwell’s equation we need to prove is

∂E ′⃗ ∇ × B ′⃗ = μoJe′⃗ + μoϵo ∂t ⎛ E⃗ ⎞ 1 ∇ × B ′⃗ = ∇ × ⎜ sin α + B ⃗ cos α⎟ = (∇ × E ⃗ ) sin α + (∇ × B ⃗ ) cos α ⎝c ⎠ c ⎛ 1⎛ ∂E ⃗ ⎞ ∂B ⃗ ⎞ ⎟ sin α + ⎜μoJe⃗ + μoϵo ⎟ cos α = − ⎜μoJm⃗ + c⎝ ∂t ⎠ ∂t ⎠ ⎝ ⎛ 1 ⎞ ∂ = μo⎜ − Jm⃗ sin α + Je⃗ cos α⎟ + μoϵo ( −cB ⃗ sin α + E ⃗ cos α ) ⎝ c ⎠ ∂t ∂E ⃗′ = μoJe′⃗ + μoϵo ∂t (e) Lastly, let us see if the force law equation is invariant to the duality transformation

⎛ ⎞ 1 F ⃗′ = qe′(E ⃗′ + v ⃗ × B ′⃗ ) + qm′ ⎜B ′⃗ − 2 v ⃗ × E ⃗′⎟ ⎝ ⎠ c ⎡ ⎛ E⃗ ⎞⎤ q ⎛ ⎞ = ⎜qe cos α − m sin α⎟⎢(E ⃗ cos α − Bc⃗ sin α ) + v ⃗ × ⎜ sin α + B ⃗ cos α⎟⎥ ⎝ ⎠⎢⎣ c ⎝c ⎠⎥⎦ ⎤ ⎡⎛ E ⃗ ⎞ 1 + (cqe sin α + qm cos α )⎢⎜ sin α + B ⃗ cos α⎟ − 2 v ⃗ × (E ⃗ cos α − Bc⃗ sin α )⎥ ⎥⎦ ⎢⎣⎝ c ⎠ c = qe[E ⃗(cos α )2 − cB ⃗ cos α sin α + E ⃗(sin α )2 + cB ⃗ cos α sin α ] ⎡⎛ E ⃗ ⎞⎤ E⃗ sin α cos α + B ⃗(sin α )2 ⎟⎥ + qev ⃗ × ⎢⎜ sin α cos α + B ⃗(cos α )2 − c ⎢⎣⎝ c ⎠⎥⎦ ⎡ E⃗ ⎤ E⃗ ⃗ α )2 + sin α cos α + B ⃗(cos α )2 ⎥ + qm⎢ − sin α cos α + B (sin c ⎣ c ⎦ ⎡ E⃗ ⎤ B⃗ E⃗ B⃗ sin α cos α − 2 (cos α )2 + sin α cos α⎥ + qmv ⃗ × ⎢ − 2 (sin α )2 − c c c ⎣ c ⎦ ⎛ ⎞ 1 = qe(E ⃗ + v ⃗ × B ⃗ ) + qm⎜B ⃗ − 2 v ⃗ × E ⃗⎟ = F ⃗ ⎝ ⎠ c

Problem 1.17. Consider an infinitely long straight wire carrying a slowly varying current I (t ). It can be shown (Griffiths 1999, example 7.9) that the induced electric field is given by

1-25

Electrodynamics

⎛ μ dI ⎞ E ⃗(s ) = ⎜ o ln(s ) + K ⎟zˆ ⎝ 2π d t ⎠ where K is a constant with respect to s . If I (t ) = I0 sin(ωt ), show this satisfies Maxwell’s equations and find ρ and J ⃗ . Solution 1.17. We can see this current results in a magnetic field

B (⃗ t ) =

μ I0 μoI (t ) ˆ ϕ = o sin(ωt )ϕˆ 2πs 2πs

Starting with ∇ · B ⃗ = 0, we have

⎛ μ I0 ⎞ 1 ∂ ⎛ μoI0 ⎞ sin(ωt )⎟ = 0 ∇ · B ⃗ = ∇ · ⎜ o sin(ωt )ϕˆ ⎟ = ⎜ ⎝ 2πs ⎠ ⎠ s ∂ϕ ⎝ 2πs as expected. Now to verify Faraday’s Law,

∇ × E⃗ = −

∂B ⃗ ∂t

We can see the right-hand side is given by

−

μ I0ω ∂B ⃗ =− o cos(ωt )ϕˆ ∂t 2πs

The left-hand side is given by

⎡⎛ μ dI ⎞ ⎤ ∇ × E ⃗ = ∇ × ⎢⎜ o ln(s ) + K ⎟zˆ⎥ ⎠ ⎦ ⎣⎝ 2π dt with

dI = I0ω cos(ωt ) dt Thus,

∇ × E⃗ =

μoI0ω cos(ωt )[∇ × (ln(s )zˆ )] + ∇ × Kzˆ 2π

Since K does not depend on s (or z or ϕ for that matter),

∇ × Kzˆ = 0 Also,

⎡1 ∂ ⎤ ⎡∂ ⎤ 1 ln(s )⎥sˆ − ⎢ ln(s )⎥ϕˆ = − ϕˆ ∇ × ln(s )zˆ = ⎢ ⎣ ∂s ⎦ s ⎣ s ∂ϕ ⎦

1-26

Electrodynamics

Therefore,

∇ × E⃗ = −

μoI0ω ∂B ⃗ cos(ωt )ϕˆ = − 2πs ∂t

as expected. Now we can use Gauss’s Law to find ρ,

⎡ μ I0ω ⎤ cos(ωt )(∇ · (ln(s )zˆ )) + ∇ · Kzˆ⎥ ρ = ϵo(∇ · E ⃗ ) = ϵo⎢ o ⎣ 2π ⎦ Here we have

∇ · Kzˆ = 0 using the same argument as before. Also,

∇ · (ln(s )zˆ ) =

∂ (ln(s )) = 0 ∂z

Therefore, ρ = 0, which is expected from the context of the example. Finally, we will use Ampère’s Law to find J ⃗ .

∇ × B ⃗ = μoJ ⃗ + μoϵo

∂E ⃗ ∂t

Starting with the left-hand side,

∇ × B⃗ =

⎡ ⎡ ∂ ⎛1 ⎞ ⎛ 1 ⎞⎤ μ I0 μoI0 1 ∂ ⎛ 1⎞ ⎤ ⎜s ⎟zˆ⎥ = 0 sin(ωt )⎢∇ × ⎜ ϕˆ ⎟⎥ = o sin(ωt )⎢ − ⎜ ⎟sˆ + ⎝ s ⎠⎦ ⎣ ⎣ ∂z ⎝ s ⎠ 2π 2π s ∂s ⎝ s ⎠ ⎦

Therefore,

J ⃗ = −ϵo

∂E ⃗ ∂t

where

⎡ μ I0ω 2 ⎡ μ I0ω dK ⎤ dK ⎤ d ∂E ⃗ o ln(s ) sin(ωt ) + ln(s ) (cos(ωt )) + =⎢ o ⎥zˆ ⎥zˆ = ⎢ − ⎣ 2π dt ⎦ dt ⎦ 2π dt ∂t ⎣ The current density is then,

⎡ ϵoμ I0ω 2 dK ⎤ ln(s ) sin(ωt ) − ϵo J⃗ = ⎢ o ⎥zˆ dt ⎦ ⎣ 2π Note that this, as does E ⃗ , depends on a constant (with respect to space) K which depends on the entire history of the current I (t ) = I0 sin(ωt ). Problem 1.18. Use Maxwell’s equations and the Lorentz force law to derive Coulomb’s Law for two charges of charge q .

1-27

Electrodynamics

Solution 1.18. The force law states

F ⃗ = q(E ⃗ + v ⃗ × B ⃗ ) Since there is no magnetic field,

F ⃗ = qE ⃗ We can now consider Gauss’s Law,

∇ · E⃗ =

ρ ϵo

We can take the integral over all space of both sides ρ ∇ · E ⃗ dτ = dτ V V ϵo

∫

∫

The right-hand side is simply

∫V

1 ρ dτ = ϵo ϵo

∫V ρ dτ = ϵqo

Applying the divergence theorem to the left-hand side yields 2π

∫V

∇ · E ⃗ dτ =

∮S

E ⃗ · da ⃗ =

π

∫∫ 0

E r 2 sin(θ ) dθ dϕ = 4πr 2E

0

By combining everything, we have

∇ · E⃗ =

q ρ ⇒ 4πr 2E = ϵo ϵo

or

E⃗ =

q rˆ 4πϵor 2

Substituting this into the force law yields

F ⃗ = qE ⃗ =

q2 rˆ 4πϵor 2

which is exactly what Coulomb’s Law states.

Bibliography Byron F W and Fuller R W 1992 Mathematics of Classical and Quantum Physics revised edn (New York: Dover) Griffiths D J 1999 Introduction to Electrodynamics 3rd edn (Englewood Cliffs, NJ: Prentice-Hall) Griffiths D J 2013 Introduction to Electrodynamics 4th edn (Cambridge, MA: Pearson) Halliday D, Resnick R and Walker J 2013 Fundamentals of Physics extended 10th edn (New York: Wiley)

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Electrodynamics

Halliday D, Resnick R and Walker J 2010 Fundamentals of Physics extended 9th edn (New York: Wiley) Heras J A 1995 Am. J. Phys. 63 242 Purcell E M and Morin D J 2013 Electricity and Magnetism 3rd edn (Cambridge: Cambridge University Press) Rogawski J 2011 Calculus: Early Transcendentals 2nd edn (San Francisco, CA:Freeman)

1-29

IOP Concise Physics

Electrodynamics Problems and solutions Carolina C Ilie and Zachariah S Schrecengost

Chapter 2 Conservation laws

Introduction After discussing the Maxwell’s equations in chapter 1, it is now the time to focus on important conservation laws: conservation of energy, momentum, and charge. From the law of charge conservation we obtain the continuity equation. Poynting theorem is the work–energy theorem in electrodynamics. Unexpectedly, the third Newton’s law of action–reaction does not hold in electrodynamics due to problems with simultaneity. However, the momentum conservation is a valid law in electrodynamics. Problems involving these conservation laws, as well as the uses of Maxwell’s stress tensor, are presented here.

2.1 Theory 2.1.1 Electromagnetic energy density The energy density of electric and magnetic fields E and B is given by

u em =

1⎛ 1 ⎞ ⎜ϵoE 2 + B 2⎟ 2⎝ μo ⎠

2.1.2 Poynting’s theorem Given mechanical and electromagnetic energy densities, u mech and u em , Poynting’s theorem states

∂ (u mech + u em) = −∇ ⋅ S ⃗ ∂t where S ⃗ is the Poynting vector and is given by

S⃗ =

doi:10.1088/978-1-6817-4931-0ch2

1 ⃗ (E × B ⃗ ) μo 2-1

ª Morgan & Claypool Publishers 2018

Electrodynamics

2.1.3 Maxwell’s stress tensor Given electric and magnetic fields with Cartesian components Ex , Ey , Ez , and Bx , By , Bz , the elements of Maxwell’s stress tensor are given by

⎛ ⎞ ⎞ 1 1⎛ 1 Tij = ϵo⎜EiEj − δijE 2⎟ + ⎜BiBj − δijB 2⎟ ⎝ ⎠ ⎝ ⎠ 2 2 μo where i , j ∊{x , y, z} and δij is the Kronecker delta

⎧1, i = j δij = ⎨ ⎩ 0, i ≠ j 2.1.4 Electromagnetic force on a charge in volume V Given volume V bounded by surface S , the total force on the charges contained in V are given by  d F⃗ = T ⋅ da ⃗ − ϵoμo S ⃗ dτ dt V S  where S ⃗ is the Poynting vector and T is the Maxwell stress tensor.

∮

∫

2.1.5 Electromagnetic momentum density The electromagnetic momentum density in the fields is given by  pem⃗ = ϵoμoS with the total momentum given by

pem⃗ =

∫V pem⃗ dτ

We also have

 ∂ (pem ⃗ + pmech ⃗ )=∇⋅T ∂t

 where −T is the momentum flux density.

2.1.6 Continuity equation The relationship between the current density J ⃗ and the volume charge density ρ is expressed via the continuity equation

∂ρ =0 ∇ ⋅ J⃗ + ∂t It is important to note that this equation follows from the law of charge conservation. 2-2

Electrodynamics

2.2 Problems and solutions Problem 2.1. Consider a long solenoid of radius a and n turns per unit length carrying current I (t ). (a) Find the electric and magnetic fields inside the solenoid. (b) Find the energy density and the Poynting vector inside the solenoid. Check using ∂ (u mech + u em) = −∇ ⋅ S ⃗ ∂t (c) Determine the total energy per length in the solenoid and the total power per length flowing into the solenoid. Solution 2.1. (a) Taking the orientation of the solenoid to be

the magnetic field is simply

B ⃗(t ) = μonI (t )zˆ The electric field can be found using Faraday’s Law

∮P E ⃗ · dl⃗ = − ddΦt where

∮P E ⃗ · dl⃗ = 2πsE and

dΦ dB dI = πs 2 = πs 2μon dt dt dt Therefore,

2πsE = −πs 2μon

dI dt

and

E⃗ = −

μons dI ˆ ϕ 2 dt

(b) The energy density is given by 2-3

Electrodynamics

⎤ 1⎛ 1 ⎞ 1 ⎡ ⎛ μ ns dI ⎞2 1 ⎟ + (μonI )2 ⎥ u em = ⎜ϵoE 2 + B 2⎟ = ⎢ϵo⎜ o 2⎝ μo μo ⎠ 2 ⎣ ⎝ 2 dt ⎠ ⎦ ⎤ ⎛ s dI ⎞2 μ n2 ⎡ ⎟ + I 2⎥ = o ⎢ϵoμo⎜ ⎝ 2 dt ⎠ 2 ⎣ ⎦ Recalling that c 2 =

1 , ϵoμo

u em =

⎤ μon 2 ⎡⎛ s dI ⎞2 ⎢⎜ ⎟ + I 2⎥ 2 ⎣⎝ 2c dt ⎠ ⎦

The Poynting vector is given by

S⃗ =

⎤ μ n 2sI dI 1 ⃗ 1 ⎡⎛ μ ns dI ˆ ⎞ (E × B ⃗ ) = ⎢⎜ − o sˆ ϕ⎟ × (μonIzˆ )⎥ = − o ⎦ 2 dt μo μo ⎣⎝ 2 dt ⎠

Now we can check u em and S ⃗ (here u mech = 0). Starting with −∇ ⋅ S ⃗ , we have

⎛ μ n 2I dI ⎞⎛ 1 ∂ ⎞ dI s 2⎟ = μon 2I −∇ ⋅ S ⃗ = −⎜ − o ⎟⎜ 2 dt ⎠⎝ s ∂s ⎠ dt ⎝ Now to compute

∂ u , ∂t em

we have

⎤ μ n 2 ⎡⎛ s ⎞2 ⎛ dI d2I ⎞ μ n 2 ∂ ⎡⎛ s dI ⎞2 dI ⎤ ∂ ⎢⎜ ⎟ + I 2⎥ = o ⎢⎜ ⎟ ⎜2 ⎥ 2 I u em = o + ⎟ dt ⎦ 2 ∂t ⎣⎝ 2c dt ⎠ 2 ⎣⎝ 2c ⎠ ⎝ dt dt 2 ⎠ ∂t ⎦ Since we are assuming a quasistatic situation, it turns out we have (Larsson)

d2I =0 dt 2 Therefore,

dI ∂ u em = μon 2I = −∇ ⋅ S ⃗ dt ∂t as expected. (c) Starting with the total energy, we have

Uem =

∫V

μ n2 u em dτ = o 2 2

=

μon l2π 2

a

∫ 0

a

l

2π

∫∫∫ 0

0

0

⎡⎛ s dI ⎞2 ⎤ ⎢⎜ ⎟ + I 2⎥s dϕ dz ds ⎣⎝ 2c dt ⎠ ⎦

⎡⎛ 1 dI ⎞2 ⎡⎛ 1 dI ⎞2 a 4 ⎤ a2 ⎤ ⎢⎜ ⎟ s 3 + I 2s⎥ds = μon 2lπ ⎢⎜ ⎟ + I2 ⎥ 2⎦ ⎣⎝ 2c dt ⎠ ⎣⎝ 2c dt ⎠ 4 ⎦

2-4

Electrodynamics

so

⎡⎛ a dI ⎞2 I 2 ⎤ Uem ⎟ + ⎥ = μon 2πa 2⎢⎜ 2⎦ l ⎣⎝ 4c dt ⎠ We can compute the power per unit length using the Poynting vector

S⃗ = −

μon 2sI dI sˆ 2 dt

and

dU = − S ⃗ · da ⃗ dt S Here we have s = a and da ⃗ = a dϕ dz sˆ . Therefore,

∮

⎛ μ n 2aI dI ⎞ dU = −⎜− o ⎟ dt 2 dt ⎠ ⎝

2π

l

∫∫ 0

a dz dϕ =

0

μon 2a 2 2πI l dI 2 dt

and

dU /dt dI = μon 2a 2πI dt l We can check this by taking

( ) and remembering

d Uem dt l

d2I dt 2

= 0,

d ⎛ Uem ⎞ 1 dUem 1 d ⎡ 2 2 ⎡⎛ a dI ⎞2 I 2 ⎤⎤ ⎜ ⎟ = ⎢μ n a π ⎢⎜ ⎟ + ⎥⎥ = dt ⎝ l ⎠ l dt 2 ⎦⎥⎦ l dt ⎢⎣ o ⎣⎝ 4c dt ⎠ Therefore,

dUem /dt dI dU /dt = μon 2a 2πI = dt l l as expected. Problem 2.2. Check that the power calculated via Poynting vector should yield the same result as previously studied simple formula P = VI for the system discussed in problem 1.10, a coaxial cable of radii a and b (with a < b) with potential difference V between the two conductors.

2-5

Electrodynamics

Solution 2.2. We use Gauss’s law and Ampère’s law to obtain the electric and the magnetic field, respectively.

qenc λl λ sˆ ⇒ E 2πsl = ⇒ E⃗ = 2πϵos ϵo ϵo μI ∮ B ⃗ ⋅ dl⃗ = μoIenc ⇒ B 2πs = μoI ⇒ B ⃗ = o ϕˆ 2πs

∮ E ⃗ ⋅ da ⃗ =

Poynting vector

S⃗ =

μI ⎞ 1 ⃗ 1⎛ λ λI (E × B ⃗ ) = ⎜ zˆ sˆ × o ϕˆ ⎟ = 2πs ⎠ 4π 2ϵos 2 μo μo ⎝ 2πϵos

Recall that Poynting vector S ⃗ represents the energy per unit time, per unit area. The power–energy per unit time) is related to S ⃗ as follows: b

P=

∫

S ⃗ ⋅ da ⃗ =

∫ a

λI λI 2πs ds = 2 2 4π ϵos 2πϵo

b

∫ a

1 b λI ds = ln s 2πϵo a

The electric potential V is given by a

V=−

∫

b

E ⃗ ⋅ d l⃗ =

b

∫ a

b λ λ ds = ln 2πϵos 2πϵo a

Indeed, we find the simple formula from introductory physics.

P=I

b λ ln = VI 2πϵo a

Problem 2.3. Determine how long it takes for the charge density of initial value ρ(t = 0) = ρ0 to decrease by a factor of e in a medium of conductivity σ and permittivity ϵ . Solution 2.3. Firstly, let us express the form of the charge density from the conservation law. ∂ρ =0 ∇ ⋅ J⃗ + ∂t with

J ⃗ = σE ⃗ The displacement D⃗ = ϵE ⃗ is related to the charge density by ∇ ⋅ D⃗ = ρ. From here it follows that

∇ ⋅ D ⃗ = ∇ ⋅ ( ϵE ⃗ ) = ϵ∇ ⋅ E ⃗

2-6

Electrodynamics

Back to the conservation law, let us express the first term of the equation: ⎛ D⃗ ⎞ σ σ ∇ ⋅ J ⃗ = ∇ ⋅ (σE ⃗ ) = ∇ ⋅ ⎜σ ⎟ = ∇ ⋅ D⃗ = ρ ϵ ⎝ ϵ⎠ ϵ The conservation law becomes:

∂ρ σ =0 ρ+ ∂t ϵ Let us rearrange this equation:

∂ρ σ =− ρ ∂t ϵ dρ σ = − dt ρ ϵ The initial condition is known, ρ(t = 0) = ρ0 , therefore: ρ

∫ ρ0

dρ′ σ =− ρ′ ϵ

t

∫

dt′ ⇒ ln ρ′

ρ ρ0

0

σ σ = − t ⇒ ln ρ = − t + ln ρ0 ϵ ϵ

σ ρ σ σ ln ρ − ln ρ0 = − t ⇒ ln = − t ⇒ ρ(t ) = ρ0 e− ϵ t ϵ ρ0 ϵ We now have the charge density dependence on time. Back to the problem, let us find the time τ necessary for the charge density to decrease by a factor of e. ρ σ ρ(t = τ ) = ρ0 e− ϵ τ = 0 e ϵ −1 − σϵ τ e =e ⇒τ= σ Problem 2.4. Consider the infinite sheet carrying surface current K⃗ (t ) = K (t )xˆ from problem 1.12. Find the energy density and the Poynting vector. Check using the differential form of Poynting’s theorem.

2-7

Electrodynamics

Solution 2.4. From problem 1.12, our fields are given by

⎧ μo ⎪− K (t )yˆ , z > 0 B ⃗( t ) = ⎨ μ 2 ⎪ o K (t )yˆ , z < 0 ⎩ 2 and

⎧ μoz dK (t ) ⎪ ⎪ 2 dt xˆ , z > 0 ⃗ E (t ) = ⎨ ⎪− μoz dK (t ) xˆ , z < 0 ⎪ ⎩ 2 dt Starting with the Poynting vector

S⃗ =

1 ⃗ (E × B ⃗ ) μo

for z > 0,

S⃗ =

μ zK (t ) dK (t ) ⎞ 1 ⎛ μoz dK (t ) ⎞⎛⎜ μo ⎜ ⎟ − K (t )⎟(xˆ × yˆ ) = − o zˆ ⎠ 4 dt μo ⎝ 2 dt ⎠⎝ 2

and for z < 0,

S⃗ =

μ zK (t ) dK (t ) ⎞ 1 ⎛ μoz dK (t ) ⎞⎛⎜ μo ⎜− ⎟ zˆ K (t )⎟(xˆ × yˆ ) = − o ⎝ ⎠ ⎝ ⎠ 4 dt 2 dt 2 μo

Therefore, the Poynting vector everywhere is

S⃗ = −

μozK (t ) dK (t ) zˆ 4 dt

Now our energy density is given by

u em =

1⎛ 1 ⎞ ⎜ϵoE 2 + B 2⎟ 2⎝ μo ⎠

Noting that

(Eabove )2 = (E below )2 and (Babove )2 = (Bbelow )2 we have

⎡ 2 2 ⎤ μ ⎞2 ⎤ 1 ⎢ ϵoμo z ⎛ dK (t ) ⎞2 1 ⎡ ⎛ μoz dK (t ) ⎞2 1 ⎛⎜ μo ⎟ ⎟ + ⎜ ⎟ + o (K (t ))2 ⎥ K (t ) ⎥ = u em = ⎢ϵo⎜ ⎠ ⎦ 2 ⎢⎣ 4 ⎝ dt ⎠ ⎥⎦ 4 2 ⎣ ⎝ 2 dt ⎠ μo ⎝ 2

2-8

Electrodynamics

Now to check using the differential form of Poynting’s theorem ∂ (u em) = −∇ ⋅ S ⃗ ∂t Starting with the right-hand side, ⎛ μ zK (t ) dK (t ) ⎞ μoK (t ) dK (t ) ∂ μ K ( t ) dK ( t ) (z ) = o zˆ⎟ = −∇ ⋅ S ⃗ = −∇ ⋅ ⎜ − o ⎝ 4 dt 4 dt ⎠ 4 dt ∂z Now to calculate the left-hand side ⎤ 1⎡ d ⎛ dK (t ) ⎞2 d ∂ ⎟ + μo (K (t ))2 ⎥ (u em) = ⎢ϵoμo2 z 2 ⎜ 8⎣ dt ⎝ dt ⎠ dt ∂t ⎦ In the first term,

dK (t ) d2K (t ) d ⎛ dK (t ) ⎞2 ⎜ ⎟ =2 =0 dt dt 2 dt ⎝ dt ⎠ since we are considering quasistatic conditions (Larsson). So dK (t ) ⎞ μoK (t ) dK (t ) 1 ⎛ 1 d ∂ ⎟= (u em) = μo (K (t ))2 = μo⎜2K (t ) = −∇ ⋅ S ⃗ ⎝ dt ⎠ 4 dt 8 8 dt ∂t as expected. Problem 2.5. Let us introduce a simple example with the stress tensor. Consider an infinite parallel plate capacitor with the plates at distance d and carrying a surface density σ on the lower plate, and −σ on the upper plate. (a) By considering z-axis perpendicular to the plates, with the origin placed mid-distance between the plates, calculate the stress tensor in matrix form. (b) What is the Poynting vector? (c) What is the force per unit area on the lower plate? (d) Identify the electromagnetic momentum per unit area, per unit time, crossing any plane parallel to the plates of the capacitor and with d d z ∈ ( − 2 , 2 ). Solution 2.5.

2-9

Electrodynamics

(a) For the parallel plate capacitor Ez = ϵσ , while there is no electric field in the o x- and y-directions, Ex = Ey = 0. From the definition of tensor components:

⎛ ⎞ ⎞ 1 1⎛ 1 Tij = ϵo⎜EiEj − δijE 2⎟ + ⎜BiBj − δijB 2⎟ ⎝ ⎠ μo ⎝ ⎠ 2 2 where i , j ∊{x , y, z} and δij is the Kronecker delta,

⎧1, i = j δij = ⎨ ⎩ 0, i ≠ j Also, E 2 = Ex2 + E y2 + Ez2 . More specifically,

1 1 Txx = ϵo(Ex2 − E y2 − E z2 ) + (Bx2 − B y2 − Bz2 ) 2 2μo 1 Txy = ϵoExEy + BxBy μo 1 Txz = ϵoExEz + BxBz μo And so on. In our case, we do not have a magnetic field, therefore Bx = By = Bz = 0.

Txx =

⎛ σ ⎞2 ⎞ σ2 1 ⎛ ϵo⎜⎜0 − 0 − ⎜ ⎟ ⎟⎟ = − ⎝ ϵo ⎠ ⎠ 2 ⎝ 2ϵo

Txy = 0, as well as all components Tij = 0, with i ≠ j .

σ2 1 Tyy = ϵo(E y2 − Ex2 − E z2 ) = − 2 2ϵo 2 σ 1 Tzz = ϵo(E z2 − Ex2 − E y2 ) = 2 2ϵo The stress tensor is:

⎛ ⎞  σ 2 ⎜− 1 0 0 ⎟ T = 0 − 1 0⎟ 2ϵo ⎜⎝ 0 0 1⎠ (b) Since the magnetic field B ⃗ is zero, so is S ⃗ =

1 (E ⃗ μo

(c) The force per unit area on the lower plate is:  d F⃗ = T · da ⃗ − ϵoμo S ⃗ dτ = dt V S

∮

∫

2-10

× B ⃗ ) = 0.



∮S T · da ⃗

Electrodynamics

2

Fz =

∫ Tzz daz = 2σϵo A

With da ⃗ = dx dy zˆ and A is the area. The force per unit area is

f⃗ =

F⃗ σ2 zˆ = A 2ϵo

 (d) The stress  tensor T represents the stress, which is the force per unit area, but also −T is the momentum current density. Therefore, the electromagnetic moment in the z-direction per unit area per unit time, crossing a plane parallel with the plates of the capacitor and for which

⎛ d d⎞ σ2 z ∈ ⎜ − , ⎟ is −Tzz = − . ⎝ 2 2⎠ 2ϵo Problem 2.6. Consider an infinitely long solid cylinder of radius R and uniform charge density ρ. Determine the net force per length on the right side if we have the following geometry:

Solution 2.6. Considering just the right side, we have

2-11

Electrodynamics

We can see from this that the net force is in the y-direction. We will use  F⃗ = T · da ⃗

∮S

to find the force. We can first look at the outer ‘shell’ area. Our field is given by

E⃗ =

1 ρπR2 ρR sˆ = sˆ 2πϵo R 2ϵo

where sˆ = cos ϕ xˆ + sin ϕ yˆ . So

⎛ ρR ⎞2 Tyx = ϵo(EyEx ) = ϵo⎜ ⎟ cos ϕ sin ϕ ⎝ 2ϵo ⎠ and

ϵo 2 ϵ ⎛ ρR ⎞ E y − Ex2 − E z2 ) = o ⎜ ⎟ (sin2 ϕ − cos2 ϕ) ( 2 2 ⎝ 2ϵo ⎠ 2

Tyy =

Since our area is given by da ⃗ = R dϕ dz sˆ = R(cos ϕ xˆ + sin ϕ yˆ ) dϕ dz , we do not ⃗ , is care about Tyz . So our force on the ‘shell’, Fshell

⃗ = Fshell



∫ ( T ⋅ da ⃗ ) y

where ⎛ ρR ⎞2 ⎡  ⎤ 1 (T · da ⃗ ) y = Tyxdax + Tyyday = ϵo⎜ ⎟ R⎢sin ϕ cos2 ϕ + (sin2 ϕ − cos2 ϕ) sin ϕ⎥ dϕ dz ⎣ ⎦ ⎝ 2ϵo ⎠ 2 ⎛ ρR ⎞2 R = ϵo⎜ ⎟ [sin ϕ cos2 ϕ + sin3 ϕ] dϕ dz ⎝ 2ϵo ⎠ 2

Therefore,

Fshell

⎛ ρR ⎞2 R = ϵo⎜ ⎟ l ⎝ 2ϵo ⎠ 2

π

∫

(sin ϕ cos2 ϕ + sin3 ϕ) dϕ =

0

ρ 2 R3l 4ϵo

where l comes from the integral over dz . Now we will consider the force on the inner ‘sheet’. Our field is given by ρs ρs E⃗ = sˆ = (cos ϕ xˆ + sin ϕ yˆ ) 2ϵo 2ϵo and our area is da ⃗ = −dx dy yˆ ; so we only need Tyy

Tyy =

2 ϵo 2 ϵ ⎛ ρs ⎞ E y − Ex2 − E z2 ) = o ⎜ ⎟ (sin2 ϕ − cos2 ϕ) ( 2 2 ⎝ 2ϵo ⎠

2-12

Electrodynamics

Since the inner sheet is at y = 0, we have

cos ϕ =

x 2

x + y2

= ±1

and

sin ϕ =

y 2

x + y2

=0

So

Tyy = −

2 ϵo ⎛ ρs ⎞ ⎜ ⎟ 2 ⎝ 2ϵo ⎠

where s = x because we are only in the xz-plane. Now

⃗ = Fsheet



∫ ( T · da ⃗ ) y

where

⎛ ϵ ⎛ ρx ⎞2 ⎞  (T · da ⃗ ) y = Tyyday = ⎜⎜ − o ⎜ ⎟ ⎟⎟( −dx dz ) ⎝ 2 ⎝ 2ϵo ⎠ ⎠ Therefore

Fsheet

2 ϵ ⎛ ρ ⎞ = o⎜ ⎟l 2 ⎝ 2ϵo ⎠

R

∫

x 2 dx =

−R

ρ 2 R3l 12ϵo

and our total force is

F = Fsheet + Fshell =

ρ 2 R3l ρ 2 R3l ρ 2 R3l + = 12ϵo 4ϵo 3ϵo

and the force per length is

F ρ 2 R3 = 3ϵo l

Problem 2.7. Consider a solid plate that extends infinitely in the x- and y-directions and from −d to d in the z-direction. If the charge density is ρ = kz , where k is a constant, find the force per area on the top half of the plate.

2-13

Electrodynamics

Solution 2.7. Considering just the top half, we have

We can see the net force on the top half is in the positive z-direction. We will use

F⃗ =



∮S T · da ⃗

to find the force. We will have two surfaces: the top-most plate and the plate that is the xy-plane. Looking first at our top surface, the electric field is given by q E ⃗ · da ⃗ = enc ϵo S

∮

with

∮S E ⃗ · da ⃗ = 4Exy and d

qenc = 4

x

y

∫∫∫ 0

0

kz′dy′dx′dz′ = 4

0

Therefore

E⃗ =

kd 2 zˆ 2ϵo

2-14

kd 2xy 2

Electrodynamics

and

Tzx = ϵo(EzEx ) = 0 Tzy = ϵo(EzEy ) = 0 ϵo 2 ϵ ⎛ kd 2 ⎞ 1 ⎛ kd 2 ⎞ E z − E y2 − Ex2 ) = o ⎜ ⎟ = ⎜ ⎟ ( 2 2 ⎝ 2ϵo ⎠ 2ϵo ⎝ 2 ⎠ 2

Tzz =

2

The force on the top is then given by

⃗ = Ftop



∫ ( T · da ⃗ ) z

with 2  1 ⎛ kd 2 ⎞ (T · da ⃗ )z = Tzxdax + Tzyday + Tzz daz = ⎜ ⎟ dx dy 2ϵo ⎝ 2 ⎠

so we have

1 ⎛ kd 2 ⎞ ⎜ ⎟ xy 2ϵo ⎝ 2 ⎠ 2

Ftop =

Now to find the force of the xy-plane surface. The field inside the plate is

E⃗ =

kz 2 zˆ 2ϵo

where z = 0 at the xy-plane. So E ⃗ = 0 and there is no contribution to the force. Therefore, the total force is simply 2 1 ⎛ kd 2 ⎞ ⃗ ⃗ F = Ftop = ⎜ ⎟ xyzˆ 2ϵo ⎝ 2 ⎠

and the force per area is given by 2 F⃗ 1 ⎛ kd 2 ⎞ = ⎟ zˆ ⎜ A 2ϵo ⎝ 2 ⎠

Problem 2.8. Consider a long solenoid, of radius R and n turns per unit length, carrying current I (t ). Find the momentum stored in a segment of length l (far from the ends to ignore edge effects).

2-15

Electrodynamics

Solution 2.8. We will take the orientation of the solenoid to be

Inside the solenoid, the magnetic field is given by

B ⃗(t ) = μonI (t )zˆ and the electric field is given by (see problem 2.1)

E ⃗ (t ) = −

μons dI ˆ ϕ 2 dt

The electromagnetic momentum density is given by

pem⃗ = μoϵoS ⃗ where S ⃗ is the Poynting vector given by

S⃗ =

1 ⃗ (E × B ⃗ ) μo

Therefore,

ϵoμ2 n 2sI dI ⎛ μ ns dI ⎞ ⎟(μonI )( −ϕˆ × zˆ ) = − o sˆ pem⃗ = ϵo(E ⃗ × B ⃗ ) = ϵo⎜ o ⎝ 2 dt ⎠ 2 dt We can now use pem ⃗ to find the momentum

pem⃗ = ∫ pem⃗ dτ = −

ϵoμo2 n 2I dI

V

=−

ϵoμo2 n 2Iπ l 3

2

dt

R

sˆ

l

2π

∫∫∫ 0

0

s 2 dϕ dz ds = −

0

ϵoμo2 n 2I 2π l dI ⎛ R3 ⎞ ⎜ ⎟sˆ 2 dt ⎝ 3 ⎠

dI sˆ dt

Problem 2.9. Consider the long solenoid from the previous problem (radius R , n turns per length, carrying current I (t )). Find the momentum flux density and check the result using  ∂ (pem ⃗ + pmech ⃗ )=∇⋅T ∂t  Solution 2.9.The momentum flux density is given by −T so we need to find the elements of T . Inside the solenoid, the magnetic field is given by

B ⃗(t ) = μonI (t )zˆ

2-16

Electrodynamics

and the electric field is given by

E ⃗ (t ) = −

μons dI ˆ ϕ 2 dt

where ϕˆ = −sin ϕ xˆ + cos ϕ yˆ . So μ ns dI (sin ϕ xˆ − cos ϕ yˆ ) E ⃗ (t ) = o 2 dt  The elements of T are given by ⎛ ⎞ ⎞ 1 1⎛ 1 Tij = ϵo⎜EiEj − δijE 2⎟ + ⎜BiBj − δijB 2⎟ ⎝ ⎠ ⎝ ⎠ 2 2 μo Note we have

Bz = μonI μ ns dI sin ϕ Ex = o 2 dt μ ns dI cos ϕ Ey = − o 2 dt Bx = By = 0; Ez = 0 Now to compute the elements

1 2 1 ⎛ μons dI ⎞2 1 ϵo 2 2 ⎟ (sin2 ϕ − cos2 ϕ) − μonI ) Ex − E y2 ) − Bz = ϵo⎜ ( ( ⎝ ⎠ 2 2μo 2 2 dt 2μo 2 ⎤ 1 ⎡ ⎛ μ ns dI ⎞ ⎟ (sin2 ϕ − cos2 ϕ) − μon 2I 2⎥ = ⎢ϵo⎜ o 2 ⎣ ⎝ 2 dt ⎠ ⎦

Txx =

⎛ μ ns dI ⎞2 ϵ ⎛ μ ns dI ⎞2 ⎟ sin ϕ ( −cos ϕ) = − o ⎜ o ⎟ sin 2ϕ Txy = ϵo(ExEy ) = ϵo⎜ o ⎝ 2 dt ⎠ 2 ⎝ 2 dt ⎠ Txz = ϵo(ExEz ) = 0 Tyx = ϵo(EyEx ) = Txy = −

Tyy =

ϵo ⎛ μons dI ⎞2 ⎜ ⎟ sin 2ϕ 2 ⎝ 2 dt ⎠

⎤ 1 2 1 ⎡ ⎛ μons dI ⎞2 ϵo 2 ⎟ (cos2 ϕ − sin2 ϕ) − μon 2I 2⎥ E y − Ex2 ) − Bz = ⎢ϵo⎜ ( 2 2μo 2 ⎣ ⎝ 2 dt ⎠ ⎦

Tyz = ϵo(EyEz ) = 0 Tzx = Txz = 0 Tzy = Tyz = 0

2-17

Electrodynamics

Tzz =

⎤ 1 2 1 ⎡ ⎛ μons dI ⎞2 ϵo ⎟ (sin2 ϕ + cos2 ϕ) + μon 2I 2⎥ −Ex2 − E y2 ) + Bz = ⎢ −ϵo⎜ ( 2 2μo 2 ⎣ ⎝ 2 dt ⎠ ⎦

⎤ 1 ⎡ ⎛ μ ns dI ⎞2 ⎟ + μon 2I 2⎥ = ⎢ −ϵo⎜ o 2 ⎣ ⎝ 2 dt ⎠ ⎦ Now 2 ⎛ 2 0 ⎞ − sin 2ϕ ⎛− 1 0 0 ⎞  1 ⎛ μons dI ⎞2 ⎜sin ϕ − cos ϕ ⎟ 1 2 2 2 2 ⎟ T = ϵo⎜ cos ϕ − sin ϕ 0 ⎟ + μon I ⎜⎜ 0 − 1 0 ⎟⎟ − sin 2ϕ 2 ⎝ 2 dt ⎠ ⎜⎜ ⎟ 2 ⎝ 0 0 1⎠ ⎝ 0 0 − 1⎠

and the momentum flux density is

⎞ ⎛ ⎛1 0 0 ⎞  1 ⎛ μons dI ⎞2 ⎜ cos 2ϕ sin 2ϕ 0 ⎟ 1 ⎟ ⎜ sin 2ϕ − cos 2ϕ 0 ⎟ + μon 2I 2⎜⎜ 0 1 0 ⎟⎟ −T = ϵo⎜ 2 ⎝ 2 dt ⎠ 2 ⎝ 0 0 − 1⎠ ⎝ 0 0 1⎠ Now to check using

 ∂ (pem ⃗ + pmech ⃗ )=∇⋅T ∂t Since pmech = 0, we want to show ⃗

 We can rewrite T as

 ∂ (pem ⃗ )=∇⋅T ∂t    T = TE + TB

where

⎛− cos 2ϕ − sin 2ϕ 0 ⎞  1 ⎛ μons dI ⎞2 ⎜ ⎟ ⎜ ⎟ TE = ϵo 0 ⎟ − sin 2ϕ cos 2ϕ ⎜ ⎝ ⎠ 2 2 dt ⎝ 0 0 − 1⎠ and

⎛− 1 0 0 ⎞  1 TB = μon 2I 2⎜⎜ 0 − 1 0 ⎟⎟ 2 ⎝ 0 0 1⎠

  Since TB is constant with respect to spatial coordinates, ∇ ⋅ TB = 0. So   ∇ ⋅ T = ∇ ⋅ TE  We will rewrite TE in terms of Cartesian coordinates using

s2 = x2 + y2

2-18

Electrodynamics

and

⎛y⎞ ϕ = tan−1⎜ ⎟ ⎝x ⎠ So

⎞ ⎛  1 ⎛ μon dI ⎞2 2⎜−cos 2ϕ −sin 2ϕ 0 ⎟ ⎟ s −sin 2ϕ cos 2ϕ 0 ⎟ TE = ϵo⎜ 2 ⎝ 2 dt ⎠ ⎜ ⎝ 0 0 −1⎠ ⎛ ⎞ ⎡ ⎡ ⎛ y ⎞⎤ ⎛ y ⎞⎤ ⎜−cos⎢2tan−1⎜ ⎟⎥ −sin ⎢2tan−1⎜ ⎟⎥ 0 ⎟ ⎝ x ⎠⎦ ⎝ x ⎠⎦ ⎣ ⎣ ⎜ ⎟ 1 ⎛ μon dI ⎞2 2 2 ⎟ ⎟ (x + y )⎜ = ϵo⎜ ⎡ ⎡ ⎛ y ⎞⎤ ⎛ y ⎞⎤ 2 ⎝ 2 dt ⎠ ⎜ −sin⎢2tan−1⎜ ⎟⎥ cos⎢2tan−1⎜ ⎟⎥ 0 ⎟ ⎝ x ⎠⎦ ⎝ x ⎠⎦ ⎣ ⎣ ⎜⎜ ⎟⎟ ⎝ 0 0 −1⎠ Now  1 ⎛ μ n dI ⎞2 ⎟ ∇ ⋅ TE = ϵo ⎜ o 2 ⎝ 2 dt ⎠ ⎡ ⎛ ⎞⎤ ⎡ ⎡ ⎛ y ⎞⎤ ⎛ y ⎞⎤ ⎢ ⎜− (x 2 + y 2 )cos⎢2tan−1⎜ ⎟⎥ − (x 2 + y 2 )sin⎢2tan−1⎜ ⎟⎥ ⎟⎥ 0 ⎝ x ⎠⎦ ⎝ x ⎠⎦ ⎣ ⎣ ⎢ ⎜ ⎟⎥ ⎢⎛ ∂ ∂ ∂ ⎞ ⎜ ⎟⎥ ⎡ ⎤ ⎡ ⎤ ⎢⎜⎝ ∂x ∂y ∂z ⎟⎠ ⎜ − (x 2 + y 2 )sin⎢2tan−1⎛⎜ y ⎞⎟⎥ (x 2 + y 2 )cos⎢2tan−1⎛⎜ y ⎞⎟⎥ ⎟⎥ 0 ⎝ ⎠ ⎝ ⎠ ⎣ ⎣ ⎢ x ⎦ x ⎦ ⎜ ⎟⎥ ⎜ ⎢ 2 + y 2 ⎟⎥ 0 0 x − ⎝ ( ) ⎠⎦ ⎣ =

1 ⎛ μo n dI ⎞2 ⎟ ϵo⎜ 2 ⎝ 2 dt ⎠

⎡ ⎛ ⎛ ⎡ ⎡ ⎡ ⎡ ⎛ y ⎞⎤ ⎛ y ⎞⎤⎞ ⎤ ⎛ y ⎞⎤ ⎛ y ⎞⎤⎞ × ⎢− 4⎜x cos⎢2tan−1⎜ ⎟⎥ + ysin⎢2tan−1⎜ ⎟⎥⎟ − 4⎜x sin⎢2tan−1⎜ ⎟⎥ − y cos⎢2tan−1⎜ ⎟⎥⎟ 0⎥ ⎝ ⎠ ⎝ x ⎠⎦⎠ ⎥⎦ ⎝ ⎠ ⎝ ⎠ ⎣ ⎣ ⎢⎣ ⎝ ⎣ ⎣ x ⎦ x ⎦⎠ x ⎦ ⎝ = −

⎛ dI ⎞2 1 ϵoμo2 n 2⎜ ⎟ ⎡⎣ x cos 2ϕ + y sin 2ϕ x sin 2ϕ − y cos 2ϕ 0 ⎤⎦ ⎝ dt ⎠ 2

Let us look separately at the x and y terms. Starting with x, we have

x cos 2ϕ + y sin 2ϕ = s cos ϕ (cos2 ϕ − sin2 ϕ) + s sin ϕ 2 sin ϕ cos ϕ = s cos ϕ and for y we have

x sin 2ϕ − y cos 2ϕ = s cos ϕ 2 sin ϕ cos ϕ − s sin ϕ (cos2 ϕ − sin2 ϕ) = s sin ϕ

2-19

Electrodynamics

Therefore, we have

 ⎛ dI ⎞2 1 ∇ ⋅ TE = − ϵoμo2 n 2⎜ ⎟ ⎡⎣ s cos ϕ s sin ϕ 0 ⎤⎦ ⎝ dt ⎠ 2 ⎛ dI ⎞2 1 = − ϵoμo2 n 2⎜ ⎟ s(cos ϕ xˆ + sin ϕ yˆ ) ⎝ dt ⎠ 2 =− Now to find

∂ ∂t

1 2 2⎛ dI ⎞2 ϵoμ n ⎜ ⎟ ssˆ 2 o ⎝ dt ⎠

(pem ⃗ ). From the previous problem,

pem⃗ = −

ϵoμo2 n 2sI dI 2

dt

sˆ

so

ϵoμo2 n 2s ⎛ dI dI d 2I ⎞ ∂ (pem + I 2 ⎟sˆ ⎜ ⃗ )=− 2 ⎝ dt dt dt ⎠ ∂t Since we are assuming quasistatic equilibrium,

d2I dt 2

= 0. Therefore,

  1 2 ⎛ dI ⎞2 ∂ (pem ⃗ ) = − ϵoμo n 2⎜ ⎟ ssˆ = ∇ ⋅ TE = ∇ ⋅ T ⎝ dt ⎠ 2 ∂t as expected. Problem 2.10. In chapter 1 we introduced the magnetic charge and proved that the generalized Maxwell’s equations and the force law equation are invariant under a certain duality transformation. Now let us consider again the magnetic charge qm and by starting with the work done per unit time dW , identify the energy density u em dt  ⃗ and the Poynting vector S . What do you notice? Prove that the stress tensor T has the same components Tij . Solution 2.10. Let us start with generalized Maxwell’s equations in the presence of the magnetic charge density ρm and current of magnetic charge Jm⃗ , as well as the electric charge density ρe and the current of electric charge Je⃗ . ρ (i) ∇ ⋅ E ⃗ = ϵe o

(ii) ∇ ⋅ B ⃗ = μo ρm ∂B ⃗ ∂t ∂E ⃗ μo ϵo ∂t

(iii) ∇ × E ⃗ = −μo Jm⃗ − (iv) ∇ × B ⃗ = μo Je⃗ +

2-20

Electrodynamics

Both electric and magnetic charges are conserved, so the continuity equation applies for both:

∂ρe ∂t ∂ρm ∇ ⋅ Jm⃗ = − ∂t ∇ ⋅ Je⃗ = −

The generalized force is

F ⃗ = qe(E ⃗ + v ⃗ × B ⃗ ) + qm(B ⃗ − ϵoμov ⃗ × E ⃗ ) In this case the magnetic field can do work only on magnetic charges. In electrostatics, the energy density is 1⎛ 1 ⎞ u em = ⎜ϵoE 2 + B 2⎟ 2⎝ μo ⎠ Now let us calculate the work dW done by the electromagnetic force acting on the charges in the interval dt. From the generalized Lorentz law: F ⃗ ⋅ dl⃗ = qe (E ⃗ + v ⃗ × B ⃗ ) ⋅ v ⃗ dt + qm(B ⃗ − ϵoμo v ⃗ × E ⃗ ) ⋅ v ⃗ dt = qeE ⃗ ⋅ v ⃗dt + qmB ⃗ ⋅ v ⃗ dt

Here we used that, of course,

(v ⃗ × B ⃗ ) ⋅ v ⃗ = 0 and (v ⃗ × E ⃗ ) ⋅ v ⃗ = 0 Now let us recall that qe =

∫V ρe dτ and qm = ∫V ρm dτ and by replacing ρe v ⃗ by Je⃗ and

ρm v ⃗ by Jm⃗ , the work per unit time is

dW = dt

∫V ρe E ⃗ ⋅ v ⃗ dτ + ∫V ρm B ⃗ ⋅ v ⃗ dτ = ∫V E ⃗ ⋅ Je⃗ dτ + ∫V B ⃗ ⋅ Jm⃗ dτ

From Maxwell’s equations:

Je⃗ =

1 ∂E ⃗ ∇ × B ⃗ − ϵo ∂t μo

Jm⃗ = −

1 1 ∂B ⃗ ∇ × E⃗ − μo μo ∂t

Now let us look at the scalar products under the integral:

1 ∂E ⃗ E ⃗ ⋅ Je⃗ = E ⃗ ⋅ (∇ × B ⃗ ) − ϵoE ⃗ ⋅ ∂t μo 1 1 ∂B ⃗ B ⃗ ⋅ Jm⃗ = − B ⃗ ⋅ (∇ × E ⃗ ) − B ⃗ ⋅ ∂t μo μo

2-21

Electrodynamics

Let us recall now the product rule, as we will find it in the sum of the scalar products:

∇ ⋅ (E ⃗ × B ⃗ ) = B ⃗ ⋅ (∇ × E ⃗ ) − E ⃗ ⋅ (∇ × B ⃗ ) Also,

E⃗ ⋅

1 ∂ 2 ∂E ⃗ (E ) = 2 ∂t ∂t

B⃗ ⋅

1 ∂ 2 ∂B ⃗ (B ) = 2 ∂t ∂t

and

Therefore, the sum of the both scalar products will be

1 1 ∂B ⃗ ∂E ⃗ E ⃗ ⋅ Je⃗ + B ⃗ ⋅ Jm⃗ = [E ⃗ ⋅ (∇ × B ⃗ ) − B ⃗ ⋅ (∇ × E ⃗ )] − ϵoE ⃗ ⋅ − B⃗ ⋅ ∂t ∂t μo μo 1 1 ∂⎛ 1 ⎞ = − ∇ ⋅ (E ⃗ × B ⃗ ) − ⎜ϵoE 2 + B 2⎟ 2 ∂t ⎝ μo μo ⎠ And back to the work per unit time, we obtain, upon the use of the divergence equation

dW = dt

∫V ∇ ⋅ v ⃗ dτ = ∮S v ⃗ ⋅ da ⃗, with v ⃗ a vector: ⎡

⎞⎤

⎛

∫V (E ⃗ ⋅ Je⃗ + B ⃗ ⋅ Jm⃗ ) dτ = ∫V ⎢⎢⎣− μ1 ∇ ⋅ (E ⃗ × B ⃗) − 12 ∂∂t ⎜⎝ϵoE 2 + μ1 B 2⎟⎠⎥⎥⎦ dτ

=−

o

1 d ∮S (E ⃗ × B ⃗ ) · da ⃗ − dt μo

∫V

o

1⎛ 1 ⎞ ⎜ϵoE 2 + B 2⎟ dτ 2⎝ μo ⎠

Note that the equation does not change, and more, that we recognize the work– energy theorem of electrodynamics, the Poynting theorem, which states that the work done on the charges by the electromagnetic force is equal to the rate at which the energy is evading (i.e. negative sign) the volume V through the closed surface S and the rate the total energy stored in the fields decreases (also negative sign). We recognize the Poynting vector S ⃗ = μ1 (E ⃗ × B ⃗ ) and the energy density o

u em =

1⎛ 1 ⎞ ⎜ϵoE 2 + B 2⎟ 2⎝ μo ⎠

Therefore, the rate of the work is

dW =− dt

∮S S ⃗ · da ⃗ − ddt ∫V u em dτ

We notice that the formulae are the same as those we obtain without the magnetic charge density and magnetic current density.

2-22

Electrodynamics

Now let us look at the stress tensor. Since it is related to the force, let us start with the generalized force, as we did before:

F ⃗ = qe(E ⃗ + v ⃗ × B ⃗ ) + qm(B ⃗ − ϵoμov ⃗ × E ⃗ ) =

∫V ⎡⎣ρe (E ⃗ + v ⃗ × B ⃗) + ρm (B ⃗ − ϵoμov ⃗ × E ⃗ )⎤⎦ dτ

The force per unit volume is, with all Maxwell’s equations involved and by using the anticommutativity of the vector product B ⃗ × A⃗ = −(A⃗ × B ⃗ )

⎛ ⎞ 1 f ⃗ = ρe E ⃗ + Je⃗ × B ⃗ + ⎜ρm B ⃗ − 2 Jm⃗ × E ⃗⎟ ⎝ ⎠ c ⎛1 1 ∂E ⃗ ⎞ = ϵo(∇ ⋅ E ⃗ )E ⃗ + ⎜ ∇ × B ⃗ − ϵo ⎟ × B ⃗ + (∇ ⋅ B ⃗ )B ⃗ t μ μ ∂ ⎝ o ⎠ o ⎛ 1 1 ∂B ⃗ ⎞ ⎟ × E⃗ − ϵoμo⎜ − ∇ × E ⃗ − μo ∂t ⎠ ⎝ μo 1 ∂ = ϵo[(∇ ⋅ E ⃗ )E ⃗ − E ⃗ × (∇ × E ⃗ )] + [(∇ ⋅ B ⃗ )B ⃗ − B ⃗ × (∇ × B ⃗ )] − ϵo (E ⃗ × B ⃗ ) ∂t μo ⎡ ⎤ ⎤ 1⎡ 1 1 = ϵo⎢(∇ ⋅ E ⃗ )E ⃗ − ∇(E 2 ) + (E ⃗⋅∇)E ⃗ ⎥ + ⎢(∇ ⋅ B ⃗ )B ⃗ − ∇(B 2 ) + (B ⋅⃗ ∇)B ⃗⎥ ⎣ ⎦ ⎣ ⎦ 2 2 μo

(

− ϵo

)

∂ ⃗ (E × B ⃗ ) ∂t

We also used ∇(A⃗ ⋅ B ⃗ ) = A⃗ × (∇ × B ⃗ ) + B ⃗ × (∇ × A⃗ ) + (A⃗ ⋅ ∇)B ⃗ + (B ⃗ ⋅ ∇)A⃗ For A⃗ = B ⃗ we obtain ∇(A⃗ ⋅ A⃗ ) = ∇(A2 ) = 2[A⃗ × (∇ × A⃗ ) + (A⃗ ⋅ ∇)A⃗ ] In our case,

E ⃗ × (∇ × E ⃗ ) =

1 ∇(E 2 ) − (E ⃗⋅∇)E ⃗ 2

The stress tensor is the same:

⎛ ⎞ ⎞ 1 1⎛ 1 Tij = ϵo⎜EiEj − δijE 2⎟ + ⎜BiBj − δijB 2⎟ ⎝ ⎠ μo ⎝ ⎠ 2 2  And (∇ ⋅ T )j = ϵo[(∇ ⋅ E ⃗ )Ej + (E ⃗ ⋅ ∇)Ej − 12 ∇j E 2 ] +

1 [(∇ μo

1 ⋅ B ⃗ )Bj + (B ⃗ ⋅ ∇)Bj − 2 ∇j B2 ]

The force per unit volume can be expressed in a more elegant way as

 ∂S ⃗ f ⃗ = ∇ ⋅ T − ϵoμo ∂t

2-23

Electrodynamics

Bibliography Griffiths D J 1999 Introduction to Electrodynamics 3rd edn (Englewood Cliffs, NJ: Prentice-Hall) Griffiths D J 2013 Introduction to Electrodynamics 4th edn (Cambridge, MA: Pearson) Jackson J D 1998 Classical Electrodynamics 3rd edn (New York: Wiley) Larsson 2007 Electromagnetics from a quasistatic perspective Am. J. Phys. 75 230 Popescu I M, Iordache D, Fara V, Stan M and Lupascu A 1986 Probleme Rezolvate de Fizica: Electromagnetism. Teoria Relativitatii Restranse. Unde Electromagnetice. Teoria Electromagnetica a Luminii (Optica) (Solved Physics Problems: Electromagnetism. Relativistic Theory. Electromagnetic Waves. The Electromagnetic Theory of Light (Optics) vol II (Editura Tehnica) Zangwill A 2013 Modern Electrodynamics (Cambridge: Cambridge University Press)

2-24

IOP Concise Physics

Electrodynamics Problems and solutions Carolina C Ilie and Zachariah S Schrecengost

Chapter 3 Electromagnetic waves

Introduction Electromagnetic waves constitute the most important application of Maxwell’s theory. The rigorous analysis of electromagnetic waves opens the avenue towards optics, where light has such a peculiar dual wave–particle behavior. Starting with wave equations, we discuss electromagnetic waves as a linear combination of sinusoidal waves. The careful discussion of a wave at boundaries helps us develop the frames of reflections and transmission. We discuss the monochromatic plane wave, polarization, energy and momentum of electromagnetic waves. Moreover, the electromagnetic waves in matter, in particular in conductors, absorption and dispersion are also analyzed. We conclude with rectangular wave guides and transmission lines.

3.1 Theory 3.1.1 Intensity Given an electromagnetic wave with electric field amplitude E0 , the intensity is given by 1 I = cϵoE 02 2

3.1.2 Radiation pressure The pressure (force per area) due to light falling on a perfect absorber is given by I P= c where I is the intensity.

doi:10.1088/978-1-6817-4931-0ch3

3-1

ª Morgan & Claypool Publishers 2018

Electrodynamics

3.1.3 Monochromatic plane wave The real electric and magnetic fields in a monochromatic plane wave are given by

E ⃗(r ⃗ , t ) = E 0 cos(k ⃗ · r ⃗ − ωt )nˆ and

B ⃗( r ⃗ , t ) =

1 ⃗ k × E ⃗(r ⃗ , t ) = B0 cos(k ⃗ · r ⃗ − ωt )(k ⃗ × nˆ ) c

where k ⃗ is the propagation vector, nˆ is the polarization, and E0 and B0 are the electric and magnetic field amplitudes. The amplitudes are related by

B0 1 = E0 c in vacuum and

B0 = E0

ϵμ 1 +

⎛ σ ⎞2 ⎜ ⎟ ⎝ ωϵ ⎠

in matter where σ is the conductivity and ω is the frequency. 3.1.4 Electromagnetic waves in matter In matter, there exists a phase difference between the electric and magnetic fields given by

⎛κ ⎞ ϕ = tan−1⎜ ⎟ ⎝k ⎠ where

⎡ ⎤1/2 ⎛ σ ⎞2 ϵμ ⎢ ⎜ ⎟ 1+ − 1⎥ κ=ω ⎝ ωϵ ⎠ 2 ⎢⎣ ⎥⎦ and

⎡ ⎤1/2 ⎛ σ ⎞2 ϵμ ⎢ 1 + ⎜ ⎟ + 1⎥ k=ω ⎝ ωϵ ⎠ 2 ⎢⎣ ⎥⎦

3.1.5 Coaxial transmission line An electromagnetic field propagating in a coaxial transmission line, whose center axis is oriented in the z-direction, has electric and magnetic fields given by

E⃗ =

A cos(kz − ωt ) sˆ s

and 3-2

Electrodynamics

B⃗ =

A cos(kz − ωt ) ˆ ϕ cs

where A is a constant of integration.

3.2 Problems and solutions Problem 3.1. Determine which of the following functions represent a wave, and check if they satisfy the wave equation:

1 ∂ 2f ∂ 2f = v 2 ∂t 2 ∂z 2 (a) f (z, t ) = A cos[b(z − vt )], with A and b constants with proper units. 2 (b) f (z, t ) = Ae−b(z−av t ), with A, a, and b constants with proper units. Solution 3.1. (a) The function f (z, t ) = A cos[b(z − vt )] depends on (z − vt ), therefore it represents a wave traveling in the z-direction. In this case, it should satisfy ∂ 2f

∂ 2f

the wave equation 2 = 12 2 . ∂z v ∂t The partial derivatives are

∂f ∂z ∂ 2f ∂z 2 ∂f ∂t ∂ 2f ∂t 2

= − Ab sin[b(z − vt )] = − Ab 2 cos[b(z − vt )] = Abv sin[b(z − vt )] = − Ab 2v 2 cos[b(z − vt )]

Let us check the identity:

−Ab 2 cos[b(z − vt )] =

1 { −Ab 2v 2 cos[b(z − vt )]} v2

It is easy to see that, indeed,

1 ∂ 2f ∂ 2f = v 2 ∂t 2 ∂z 2 Therefore the function is a wave and it satisfies the wave equation. 2 (b) The function f (z, t ) = Ae−b(z−av t ) does not depend on (z − vt ), therefore it may not represent a wave. To support our statement, we will prove that the wave equation is not satisfied here. We expect that Again, the partial derivatives are

3-3

∂ 2f ∂z 2

≠

1 ∂ 2f . v 2 ∂t 2

Electrodynamics

∂f ∂z ∂ 2f ∂z 2 ∂f ∂t ∂ 2f ∂t 2

2

= − Abe−b(z−av t ) 2

= Ab 2 e−b(z−av t ) 2

= Aabv 2 e−b(z−av t ) 2

= Aa 2b 2v 4e−b(z−av t )

Note that, indeed: 2

Ab 2 e−b(z−av t ) ≠

1 2 {Aa 2b 2v 4e−b(z−av t )} v2

The function is not a wave and it does not satisfy the wave equation, as expected. Problem 3.2. (a) Check if the standing wave f (z, t ) = A cos(kz ) cos(kvt ) satisfies the wave equation

1 ∂ 2f ∂ 2f = v 2 ∂t 2 ∂z 2 (b) If so, express it as a sum of a wave traveling to the left and a wave traveling to the right. Hint: Use the trigonometric identity cos u cos v = 12 [cos(u + v ) + cos(u − v )]. Solution 3.2. (a) Firstly, let us prove the wave equation

∂f ∂z ∂ 2f ∂z 2 ∂f ∂t ∂ 2f ∂t 2

∂ 2f ∂z 2

=

1 ∂ 2f v 2 ∂t 2

= − Ak sin(kz ) cos(kvt ) = − Ak 2 cos(kz ) cos(kvt ) = − Akv cos(kz ) sin(kvt ) = − Ak 2v 2 cos(kz ) cos(kvt )

Let us check the identity:

3-4

.

Electrodynamics

−Ak 2 cos(kz ) cos(kvt ) =

1 1 ∂ 2f ∂ 2f [ −Ak 2v 2 cos(kz ) cos(kvt )] ⇒ 2 = 2 2 2 v ∂t v ∂z

The wave equation is satisfied. (b) Now let us express the standing wave as a sum of a wave traveling to the left and a wave traveling to the right. We use the trigonometric identity:

cos u cos v =

1 [cos(u + v ) + cos(u − v )] 2

In our case,

A {cos[k (z + vt )] + cos[k (z − vt )]} 2 = h(z + vt ) + g(z − vt )

f (z , t ) = A cos(kz ) cos(kvt ) =

Where

A cos[k (z − vt )] 2 A h(z + vt ) = cos[k (z + vt )] 2

g(z − vt ) =

Problem 3.3. Show that the general wave, in the form ∞

f˜ (z , t ) =

∫

A˜ (k )e i(kz−ωt ) dk

−∞

does in fact satisfy the wave equation. Solution 3.3. The wave equation is given by

1 ∂ 2f ∂ 2f = v 2 ∂t 2 ∂z 2 Since k is not a function of z ,

∂ 2f = ∂z 2 Now to find

∂ 2f = ∂t 2

∞

∫ −∞

∞

∫ −∞

⎤ ⎡ ∂2 A˜ (k )⎢ 2 e i(kz−ωt )⎥ dk = − ⎦ ⎣ ∂z

∞

∫

k 2A˜ (k )e i(kz−ωt ) dk

−∞

∂ 2f , ∂t 2

⎤ ⎡ ∂2 A˜ (k )⎢ 2 e i(kz−ωt )⎥ dk = − ⎦ ⎣ ∂t

∞

∫

∞

ω A˜ (k )e i(kz−ωt ) dk = − 2

−∞

∫ −∞

Now we have

3-5

v 2k 2A˜ (k )e i(kz−ωt )dk

Electrodynamics

∞ ⎛ ∞ ⎞ 2 1 ∂ 2f 1⎜ 2 2 ˜ i(kz−ωt ) ⎟ = − k 2A˜ (k )e i(kz−ωt ) dk = ∂ f = − v k A ( k )e d k 2 2 2⎜ ⎟ ∂z 2 v ∂t v ⎝ ⎠ −∞ −∞

∫

∫

as expected. Problem 3.4. Let us send an incident wave to the left, fI (z + v1t ) down a string. It gives rise to a reflected wave fR (z − v1t ) to the right and a transmitted wave to the left fT (z + v2t ), where v1 is the speed in medium 1 and v2 is the speed in medium 2. By using the boundary conditions f (0−, t ) = f (0+, t ) (continuity at z = 0) and the continuity of the derivative at z = 0, ∂∂fz ∣0− = ∂∂fz ∣0+ , find fR and fT . Solution 3.4. At z = 0:

fI (v1t ) + fR ( −v1t ) = fT (v2t )

(3.1)

The second condition requires the partial derivatives, which are

∂fI 1 ∂fI = v1 ∂t ∂z ∂fR 1 ∂fR =− v1 ∂t ∂z ∂fT 1 ∂fT = v2 ∂t ∂z It follows that the second condition at z = 0 is

1 ∂fI (v1t ) 1 ∂fR ( −v1t ) 1 ∂fT (v2t ) = − v1 ∂t v1 v2 ∂t ∂t Next, we multiply by v1 and integrate and obtain

fI (v1t ) − fR ( −v1t ) =

v1 f (v2t ) + C1 v2 T

where C1 is an integration constant. Adding (3.1) and (3.2) we obtain

⎛ v ⎞ 2fI (v1t ) = fT (v2t )⎜1 + 1 ⎟ + C1 ⎝ v2 ⎠ 2v2 f (v1t ) + C2 fT (v2t ) = v1 + v2 I where

C2 = −C1

v2 v1 + v2

3-6

(3.2)

Electrodynamics

Writing the functions in a single variable,

u = z + v1t , for fI (z , t ) u = z − v1t , for fR (z , t ) u = z + v2t , for fT (z , t ) At z = 0

fT (u ) =

⎛v u ⎞ 2v2 fI ⎜ 1 ⎟ + C2 v1 + v2 ⎝ v2 ⎠

In order to find fR , we need to eliminate fT between equations (3.1) and (3.2). We v multiply (3.1) by v1 and subtract equation (3.2). 2

v v v1 fI (v1t ) + 1 fR ( −v1t ) = 1 fT (v2t ) v2 v2 v2 v − fI (v1t ) + fR ( −v1t ) = − 1 fT (v2t ) − C1 v2 ⎛ v1 ⎞ ⎛v ⎞ − 1⎟ fI (v1t ) + ⎜ 1 + 1⎟ fR ( −v1t ) = −C1 ⎜ ⎠ ⎝ v2 ⎠ ⎝ v2 v v2 − v1 fI (v1t ) − C1 2 fR ( −v1t ) = v1 + v2 v1 + v2 Finally,

fR (u ) =

v2 − v1 f ( − u ) + C2 v1 + v2 I

where

C2 = −C1

v2 v1 + v2

Problem 3.5. Consider a perfect absorber of mass m. Find the electric and magnetic waves required to move the absorber at a constant acceleration a . Assume this occurs in vacuum with the wave striking an area A. Solution 3.5. We know that when light falls on a perfect absorber, a radiation pressure is exerted. The pressure is given by

P=

3-7

I c

Electrodynamics

where I is the intensity of the wave. This is also given by

1 ϵoE 02 2

P= Since pressure is force per area,

P=

1 F ma = = ϵoE 02 2 A A

Solving for E0 yields

E0 =

2ma Aϵo

Using E0, we have waves

E⃗ =

2ma cos(k ⃗ · r ⃗ − ωt )nˆ Aϵo

and

B⃗ =

2ma 1 cos(k ⃗ · r ⃗ − ωt )(k ⃗ × nˆ ) = Aϵo c

2maμo cos(k ⃗ · r ⃗ − ωt )(k ⃗ × nˆ ) A

since 1c = ϵoμo . Returning to the intensity

I=

1 1 2ma cma cϵoE 02 = cϵo = 2 2 Aϵo A

in order to accelerate a 2.0 kg mass, with incident surface area of 25 cm2 , at 1.0 m s−2 , we would require light with intensity I = 2.4 × 1011 W m−2 . Taking the intensity of light hitting the Earth to be 1300 W m−2 , the intensity required to provide this acceleration is approximately 180 million times that delivered to the Earth’s surface by the Sun. Problem 3.6. Consider a monochromatic place wave traveling in the z-direction with a polarization nˆ = cos θ xˆ + sin θ yˆ . Find the new electric and magnetic fields after this wave interacts with a polarization grating that only allows wave polarized in the x-direction through. Use the resulting wave to find the intensity. Solution 3.6. Initially, we have

Ei⃗ = E 0 cos(kz − ωt )nˆ = E 0 cos(kz − ωt )( cos θ xˆ + sin θ yˆ ) and

Bi⃗ = B0 cos(kz − ωt )(zˆ × nˆ ) =

E0 cos(kz − ωt )( − sin θ xˆ + cos θ yˆ ) c

3-8

Electrodynamics

Since the polarization grating only allows waves polarization in the x-direction, we have

Ef⃗ = E 0 cos θ cos(kz − ωt ) xˆ and

Bf⃗ = −

E0 cos θ cos(kz − ωt ) yˆ c

We know the initial intensity is given by

Ii = Si =

1 cϵoE 02 2

so, the final intensity is given by

I f = Sf where

Sf⃗ =

⎛ E ⎞ 1 ⃗ 1 Ef × Bf⃗ = (E 0 cos θ )⎜ − 0 cos θ ⎟ cos2(kz − ωt )zˆ ⎝ c ⎠ μo μo

(

)

= −E 02ϵoc cos2 θ cos2(kz − ωt )zˆ Therefore,

I f = Sf =

E 02ϵoc cos2 θ = Ii cos2 θ 2

Problem 3.7. Suppose that you send an electromagnetic wave at normal incidence on a thin dielectric film between two semi-infinite transparent media (dielectric layer of thickness l ). This layer is made of a medium of permeability μ = μo and with permittivity which decreases from ϵ1 on superior side to ϵ2 on inferior side, following a function ϵ = f (x ), where x is the distance to the superior side. What is the time τ

3-9

Electrodynamics

necessary to pass the layer? The function is: f (x ) = Ae−bx , with A and b constants with appropriate units. Solution 3.7. The speed of the wave is: 1 v= = ϵμo

1 f (x )μo

But also

dx dx ⇒ dx = v dt ⇒ dt = dt v With the speed from the previous relationship, v=

dt =

μo f (x ) dx

The time needed for the wave to pass the layer of thickness l is found by integration: l

t=

μo

∫

f ( x ) dx

0

Now let us obtain the constants A and b for function f (x ) from the boundary conditions.

f (0) = ϵ1, f (l ) = ϵ2 , ϵ2 < ϵ1 f (0) = A = ϵ1, f (l ) = Ae−bl = ϵ2 From the second equation it follows that ϵ1e−bl = ϵ2 , 1 ϵ ϵ ϵ e−bl = 2 ⇒ −bl = ln 2 ⇒ b = − ln 2 ϵ1 ϵ1 l ϵ1 The time is, therefore, l

t=

μo

∫

Ae−bx dx =

0

⎛ 2 bx ⎞ μoA ⎜ − e− 2 ⎟ ⎝ b ⎠ 1

l

= 0

bl 2 μoA 1 − e− 2 b

(

)

ϵ

Back to the constants A = ϵ1, b = − l ln ϵ2 and 1

t=

2 1 ϵ − l ln ϵ2 1

(

μoϵ1 1 −

1 ϵ2 e− 2 ln ϵ1

)=

2l μo ϵ ln ϵ2 1

( ϵ2 −

ϵ1 ) =

2l μo ϵ

ln ϵ1

( ϵ1 −

ϵ2 )

2

Problem 3.8. Consider a linearly polarized plane in free space. Calculate the Poynting vector and the energy flow averaged over a complete cycle. Show that, if you use the complex notation, the correct formula for the energy flow is

S⃗ =

1 ⁎ Re[E ⃗ × B ⃗ ] 2μo

3-10

Electrodynamics

Solution 3.8. (a) Firstly, let us use the trigonometric function: ˆ 0 cos(ωt − kz ) E ⃗ = xE

ˆ 0 cos(ωt − kz ) B ⃗ = yB E B0 = 0 c The Poynting vector: 1 S ⃗ = E ⃗ × B ⃗ = zE ˆ 0B0[cos(ωt − kz )]2 μo For z = 0

S (⃗ z = 0, t ) = zˆ

1 E 0B0([cos(ωt )])2 μo

The energy flow averaged over the complete cycle, taking into account that 1 〈[cos(ωt )]2 〉 = 2 , is given by

S ⃗ = zˆ

1 1 E 0B0 [cos(ωt )]2 = zˆ E 02 μo 2μoc

(b) In this case, we use the complex function, at z = 0

ˆ 0e iωt E ⃗ = xE ˆ 0e iωt B ⃗ = yB S⃗ =

1 1 1 ⁎ ˆ 0e iωt ) × (yB ˆ 0e−iωt )] = Re[zE Re[(xE ˆ 0B0 ] Re[E ⃗ × B ⃗ ] = 2μo 2μo 2μo

Note that, if we try to calculate without the use of the complex conjugate, we may obtain a zero time averaged value of the Poynting vector. For example,

S⃗ =

1 ⃗ 1 1 1 ˆ 0e iωt ) × (yB ˆ 0e iωt )] = Re[zE ˆ 0B0e 2iωt ] E × B ⃗ = Re[E ⃗ × B ⃗ ] = Re[(xE μo μo μo μo 1 = E 0B0 cos 2ωt μo The averaged value for cos 2ωt is zero, so 〈S ⃗〉 = 0, which is not true.

Problem 3.9. Consider a stationary electromagnetic wave along the z-direction of the form

ˆ 0 cos kz cos ωt E ⃗(z , t ) = xE

3-11

Electrodynamics

Find: (a) the magnetic field B ⃗(z, t ); (b) the Poynting vector; (c) the intensity of the wave 〈S ⃗〉. Solution 3.9. (a) We need to find B ⃗ from the Maxwell’s equations

∂B ⃗ ∂t ∂E ⃗ ∇ × B ⃗ = μoϵo ∂t ∇ × E⃗ = −

Since E ⃗ is oriented in the x-direction, the wave on the z-direction, it means that B ⃗ is oriented in the y-direction.

∇ × E⃗ =

xˆ ∂ ∂x Ex

yˆ ∂ ∂y 0

zˆ ∂ ∂z 0

=

∂By ∂(E 0 cos kz cos ωt ) yˆ = −kE 0yˆ sin kz cos ωt = − yˆ ∂z ∂t

From here,

∂By ∂t

= kE 0 sin kz cos ωt

On the other hand,

∇ × B⃗ =

xˆ ∂ ∂x 0

yˆ ∂ ∂y By

zˆ ∂ ∂z 0

=−

∂By xˆ ∂z

So

−

∂By(z , t ) ∂By(z , t ) ∂E ⃗ xˆ = −μoϵoωE 0xˆ cos kz sin ωt xˆ = μoϵo ⇒− ∂z ∂t ∂z 1 c2 ∂By

But ω = kc and μo ϵo =

ω = 2 E 0 cos kz sin ωt c ∂z ∂By ω = E 0 sin kz cos ωt c ∂t From here,

By =

E0 sin kz sin ωt c

3-12

Electrodynamics

B ⃗( z , t ) =

E0 yˆ sin kz sin ωt c

(b) The Poynting vector

S⃗ =

⎛E ⎞ 1 ⃗ 1 E × B ⃗ = (E 0 cos kz cos ωt )⎜ 0 sin kz sin ωt⎟xˆ × yˆ ⎝ ⎠ c μo μo =

E 02 E2 cos kz sin kz cos ωt sin ωt zˆ = 0 sin 2kz sin 2ωt zˆ 4μoc μoc

where we used sin α cos α = (c) The intensity of the wave

1 2

sin 2α .

I = S = Sz = 0 because the average of the sine function for an entire cycle is zero. Problem 3.10. In conductors, the magnetic field lags behind the electric field. Find the maximum and minimum phase difference that can exist between the fields. (Hint: consider the phase difference for good (σ ≫ ωϵ ) and poor (σ ≪ ωϵ ) conductors. Solution 3.10. The phase difference between the waves is given by

⎛κ ⎞ ϕ = tan−1⎜ ⎟ ⎝k ⎠ where

⎡ ⎤1/2 ⎛ σ ⎞2 ϵμ ⎢ ⎜ ⎟ 1+ − 1⎥ κ=ω ⎝ ωϵ ⎠ 2 ⎢⎣ ⎥⎦ and

⎡ ⎤1/2 ⎛ σ ⎞2 ϵμ ⎢ 1 + ⎜ ⎟ + 1⎥ k=ω ⎝ ωϵ ⎠ 2 ⎢⎣ ⎥⎦ Therefore,

⎡ ⎢ 1+ κ =⎢ k ⎢ ⎣ 1+

σ 2 ωϵ

( ) ( )

σ 2 ωϵ

3-13

⎤1/2 − 1⎥ ⎥ + 1 ⎥⎦

Electrodynamics

We will first consider a poor conductor where

σ ≪ ωϵ ⇒

σ ≪1 ωϵ

The inner square roots can be approximated by

1+

⎛ σ ⎞2 1 ⎛⎜ σ ⎞⎟2 ⎜ ⎟ ≈ 1 + ⎝ ωϵ ⎠ 2 ⎝ ωϵ ⎠

Thus,

⎡ ⎢ 1+ κ =⎢ k ⎢ ⎣ 1+ Since

σ ωϵ

≪ 1, 2 +

σ 2 ωϵ

( ) ( )

σ 2 ωϵ

1 σ 2 2 ωϵ

( )

⎤1/2 ⎡ − 1⎥ ⎢1 + ⎥ ≈⎢ ⎢⎣ 1 + + 1 ⎥⎦

⎤1/2 ⎡ − 1⎥ ⎢ ⎥ =⎢ 2 + 1 ⎥⎦ ⎣⎢ 2

1 σ 2 2 ωϵ

( ) ( )

1 σ 2 ωϵ

1 ⎛⎜ σ ⎞⎟2 ⎤ ⎥ 2 ⎝ ωϵ ⎠ 1 ⎛⎜ σ ⎞⎟2 ⎥ + ⎦ 2 ⎝ ωϵ ⎠ ⎥

1/2

≈ 2 so we have

⎡ 1 ⎛ σ ⎞2 ⎤1/2 ⎡⎛ σ ⎞2 ⎤1/2 κ σ ⎢ 2 ⎝ ωϵ ⎠ ⎥ ⎟ ⎥ = ⎢ 2 ⎥ = ⎢⎜ = ⎝ ⎠ k ⎢ 2ωϵ ⎣ 2ωϵ ⎦ ⎥⎦ ⎣ ⎜

⎟

Therefore, the phase difference is given by

⎛κ ⎞ ⎛ σ ⎞ ⎟ ϕ = tan−1⎜ ⎟ = tan−1⎜ ⎝k ⎠ ⎝ 2ωϵ ⎠ Note that since

σ ωϵ

≪ 1, −1 tan x ≈ x . Therefore,

σ 2ωϵ

≪ 1/2. Also, since

σ 2ωϵ

π ≪ 2 , we can approximate using

⎛ σ ⎞ σ ⎟ ≈ ϕ = tan−1⎜ ⎝ 2ωϵ ⎠ 2ωϵ σ which is very small (consider 2ωϵ ≪ 1/2). So the phase difference between the magnetic electric fields in a poor conductor is negligible. This is expected considering ϕ = 0 in vacuum. For a good conductor,

σ ≫ ωϵ ⇒

σ ≫1 ωϵ

3-14

Electrodynamics

Again,

⎡ ⎢ 1+ κ =⎢ k ⎢ ⎣ 1+ The

σ 2 ωϵ

( )

σ 2 ωϵ

( ) ( )

σ 2 ωϵ

⎤1/2 − 1⎥ ⎥ + 1 ⎥⎦

terms dominate the inner square root, so

1+

⎛ σ ⎞2 σ ⎜ ⎟ ≈ ⎝ ωϵ ⎠ ωϵ

and

⎡ ⎢ 1+ κ =⎢ k ⎢ ⎣ 1+ We still have that

σ ωϵ

σ 2 ωϵ

( ) ( )

σ 2 ωϵ

⎤1/2 ⎡ σ − 1 ⎤1/2 − 1⎥ ⎥ ⎥ ≈ ⎢ ωϵ σ ⎢⎣ ωϵ + 1 ⎥⎦ ⎥ + 1⎦

dominates so

⎡ σ ⎤1/2 κ ⎥ =1 = ⎢ ωϵ σ k ⎣⎢ ωϵ ⎥⎦ Therefore, our maximum phase difference is given by ⎛κ ⎞ ϕ = tan−1⎜ ⎟ = tan−1(1) = 45° ⎝k ⎠ and the limits are

0° ⩽ ϕ ⩽ 45°

Problem 3.11. Consider an electromagnetic wave traveling through a conductor. Find the ratio of real amplitudes B0 /E0 for a good conductor (σ ≫ ωϵ ) and a poor conductor (σ ≪ ωϵ ). Comment on the frequency limits and the resulting limits imposed on the ratio. Solution 3.11. The ratio of real amplitudes is given by

B0 = E0

ϵμ 1 +

3-15

⎛ σ ⎞2 ⎜ ⎟ ⎝ ωϵ ⎠

Electrodynamics

First, we will consider a good conductor where we have σ ≫1 σ ≫ ωϵ ⇒ ωϵ So, the inner square root can be approximated by

1+

⎛ σ ⎞2 ⎜ ⎟ ≈ ⎝ ωϵ ⎠

⎛ σ ⎞2 σ ⎜ ⎟ = ⎝ ωϵ ⎠ ωϵ

Therefore,

B0 = E0

ϵμ 1 +

⎛ σ ⎞2 ⎜ ⎟ = ⎝ ωϵ ⎠

ϵμ

σ = ωϵ

σμ ω

For a typical metal, σ∼107 , ϵ = ϵo, μ = μo . Our approximation was based on

σ ≫ ωϵ ⇒ ω ≪

σ 107 = ϵo 8.85 × 10−12

or

ω ≪ 1018 Consider that ω∼1015 for the optical range, let us take

ω ⩽ 1015 From this, we have

10−15 ⩽

1 ⇒ ω

μσ10−15 ⩽

σμ B = 0 ω E0

Since

μσ10−15 ≈ 10−7 we have

10−7 ⩽

B0 E0

In vacuum,

B0 1 = ≈ 10−9 E0 c Therefore, the magnetic field in a good conductor is at least 100 times stronger than the respective electric field. Also note this is the lower limit; decreasing the frequency to the microwave range, where ω∼1010 , yields

10−5 ⩽

3-16

B0 E0

Electrodynamics

Now, considering a poor conductor, we have σ ≪1 σ ≪ ωϵ ⇒ ωϵ Again, our ratio is given by

B0 = E0 Since

σ ωϵ

ϵμ 1 +

⎛ σ ⎞2 ⎜ ⎟ ⎝ ωϵ ⎠

≪ 1, we can take

1+

⎛ σ ⎞2 ⎜ ⎟ ≈ 1 ⎝ ωϵ ⎠

Therefore,

B0 = E0

ϵμ 1 +

⎛ σ ⎞2 ⎜ ⎟ = ⎝ ωϵ ⎠

ϵμ

Unlike good conductors, whose ratio depends on conductivity, frequency, and permeability, the ratio for poor conductors depends on permittivity and permeability. Although the ratio itself does not depend on ω and σ , their values do determine where this approximation is valid. We will consider both pure water and glass. Starting with pure water, we have ϵ = 80 ϵo and σ = 4.0 × 10−6 Ω−1m−1. Our approximation depends on

σ ≪ ωϵ so, we require that the frequency satisfies

4.0 × 10−6 ≪ ω(80ϵo) ⇒ 5.6 × 103 ≪ ω Therefore, for this approximation to be valid with pure water, we would require at least ω∼107 . This is boarding on the TV and FM range. For glass, we have ϵ = 2.25 ϵo and σ = 10−12 Ω−1m−1 and we require the frequency to satisfy

10−12 ≪ ω(2.25ϵo) ⇒ 0.05 ≪ ω This approximation is valid for at least ω∼100 (i.e. practically all frequencies of interest). Note that as σ → 0 (i.e. the worse the material is as a conductor), the more accurate the approximation becomes. Problem 3.12. Gold is an extremely good conductor, but very expensive. If we want to design an experiment to operate at a frequency of 1011 Hz (microwave), calculate the skin depth and decide how thick the gold coating should be. Resistivity of gold is ρ = 2.21 × 10−8 Ωm .

3-17

Electrodynamics

Solution 3.12. For a homogeneous linear medium, the continuity equation for free charge and Ohm’s law and Gauss’s law yield the fact that the initial free charge density ρf (0) dissipates in a characteristic time τ = σϵ = ϵρ, following σ

1 ρf (t ) = ρf (0)e− ϵ t . For a good conductor τ ≪ ω . Notice the difference between the resistivity ρ and the free charge density ρf (t ). The skin depth d is the distance it takes to reduce the amplitude of the wave by a factor of 1e and it is a measure of how deep the wave penetrates the conductor. Using

that

ϵ σ

≪

1 , ω

we can approximate the skin depth as follows: 1 d= κ 1

⎡ ⎤2 ⎛ σ ⎞2 ϵμ ϵμ ⎢ 1 + ⎜ ⎟ − 1⎥ ≃ ω κ=ω ⎝ ⎠ 2 ⎢⎣ 2 ϵω ⎥⎦

σ = ϵω

ωσμ 2

Calculating the skin depth with ϵ = ϵo and μ = μo and recalling that ω = 2πν , with ν the frequency.

d=

1 = κ

2 = ωσμ

2 1 × 4π × 10−7 2π × 1011 × 2.21 × 10−8

= 0.236 × 10−6 m = 0.236 μm We could plate up to 0.25 microns. Gold plating can be easily done within the 0.03–0.3 microns. Problem 3.13. Find the momentum flux density of TEM waves in a coaxial transmission line of inner radius a and outer radius b.

Solution 3.13. Inside the transmission line, for a constant A, we have

E⃗ =

A cos(kz − ωt ) sˆ s

B⃗ =

A cos(kz − ωt ) ˆ ϕ cs

and

3-18

Electrodynamics

 To find the momentum flux density, we need to find which requires finding the −T  ⃗ ⃗ components of T . We can express E and B in terms of Cartesian coordinates using

s = x2 + y2 sˆ = cos ϕ xˆ + sin ϕ yˆ and

ϕˆ = − sin ϕ xˆ + cos ϕ yˆ Thus,

E⃗ =

A cos(kz − ωt ) x2 + y2

(cos ϕ xˆ + sin ϕ yˆ )

and

B⃗ =

A cos(kz − ωt ) c x2 + y2

( −sin ϕ xˆ + cos ϕ yˆ )

We can use these, along with

⎛ ⎞ ⎞ 1 1⎛ 1 Tij = ϵo⎜EiEj − δijE 2⎟ + ⎜BiBj − δijB 2⎟ ⎝ ⎠ μo ⎝ ⎠ 2 2  to solve for the components of T . Txx =

1 1 ϵo(Ex2 − E y2 − E z2 ) + (Bx2 − B y2 − Bz2 ) 2 2μo

⎛ ⎞2 1 ⎜ A cos(kz − ωt ) ⎟ 2 2 = ϵo ⎟ (cos ϕ − sin ϕ) 2 2 2 ⎜⎝ x +y ⎠ ⎛ ⎞2 1 ⎜ A cos(kz − ωt ) ⎟ + (sin2 ϕ − cos2 ϕ) 2μo ⎜⎝ c x 2 + y 2 ⎟⎠ =

A2 cos2(kz − ωt ) ϵoA2 cos2(kz − ωt ) (cos2 ϕ − sin2 ϕ) + (sin2 ϕ − cos2 ϕ) 2 2 2(x + y ) 2μoc 2(x 2 + y 2 )

Recalling

1 = μo ϵo , c2

Txx =

ϵoA2 cos2(kz − ωt ) (cos2 ϕ − sin2 ϕ + sin2 ϕ − cos2 ϕ) = 0 2( x 2 + y 2 )

3-19

Electrodynamics

Txy = ϵoExEy +

1 BxBy μo

⎛ ⎞2 ⎛ ⎞2 A cos(kz − ωt ) ⎟ 1 ⎜ A cos(kz − ωt ) ⎟ ⎜ = ϵo⎜ ⎟ cos ϕ sin ϕ − μ ⎜ ⎟ cos ϕ sin ϕ x2 + y2 c x2 + y2 ⎠ o⎝ ⎝ ⎠ ⎛ ⎞2 A kz − ω t cos( ) ⎟ (cos ϕ sin ϕ − cos ϕ sin ϕ) = 0 = ϵo⎜⎜ ⎟ x2 + y2 ⎝ ⎠

Txz = ϵoExEz +

1 BxBz = 0 μo

Tyx = Txy = 0

1 1 ϵo(E y2 − Ex2 − E z2 ) + (B y2 − Bx2 − Bz2 ) 2 2μo 1 = Txx + ϵo(Ex2 − E y2 ) + (Bx2 − B y2 ) μo = ϵo⎡⎣Ex2 − E y2 + c 2(Bx2 − B y2 )⎤⎦ ⎡⎛ A2 cos2(kz − ωt ) ⎞ = ϵo⎢⎜ ⎟(cos2 ϕ − sin2 ϕ) 2 2 x y + ⎝ ⎠ ⎣

Tyy =

⎤ ⎛ A2 cos2(kz − ωt ) ⎞ + c 2⎜ ⎟(sin2 ϕ − cos2 ϕ)⎥ = 0 2 2 2 ⎝ c (x + y ) ⎠ ⎦ Tyz = ϵoEyEz +

1 ByBz = 0 μo

Tzx = Txz = 0 Tzy = Tyz = 0

3-20

Electrodynamics

1 1 ϵo E z2 − E x2 − E y2 + Bz2 − Bx2 − B y2 2 2μo ⎤ 1⎡ 1 = − ⎢ϵo E x2 + E y2 + Bx2 + B y2 ⎥ 2⎣ μo ⎦ ⎤ ⎛ A2 cos2(kz − ωt ) ⎞ 1 ⎡⎛ A2 cos2(kz − ωt ) ⎞ = − ϵo⎢⎜ ⎟(cos2ϕ + sin2 ϕ ) + c 2⎜ ⎟(sin2 ϕ + cos2 ϕ )⎥ 2 2 2 2 2 ⎥⎦ 2 ⎢⎣⎝ x +y ⎠ ⎝ c (x + y ) ⎠

(

Tyy =

)

(

=−

)

(

(

)

)

ϵoA2 cos2(kz − ωt ) x2 + y2

Therefore, the momentum flux density is given by

⎛0 0 ⎞ ⎛0 0 0 ⎜ ⎟ ⎜0 0  0 0 0 ⎟=⎜ −T = −⎜ 2 2 ⎜ 0 0 − ϵoA cos (kz − ωt ) ⎟ ⎜ 0 0 ⎜ ⎟ ⎜ x2 + y2 ⎝ ⎠ ⎝

⎞ 0 ⎟ 0 ⎟ 2 2 ϵoA cos (kz − ωt ) ⎟ ⎟ x2 + y2 ⎠

We can check this by considering

 ∂ (pem ⃗ )=∇·T ∂t where

pem⃗ = μoϵoS ⃗ = ϵo(E ⃗ × B ⃗ ) =

ϵoA2 cos2(kz − ωt ) zˆ cs 2

So

ϵoA2 ∂ (pem ⃗ ) = 2 2 cos(kz − ωt )( −sin(kz − ωt ))( −ω)zˆ ∂t cs 2ϵoA2 ω cos(kz − ωt ) sin(kz − ωt )zˆ = cs 2 2ϵoA2 ω = cos(kz − ωt ) sin(kz − ωt )zˆ c (x 2 + y 2 ) Now

3-21

Electrodynamics

⎛0 0 ⎞ 0 ⎜ ⎟ ⎛ ⎞  0 ∂ ∂ ∂ 0 0 ⎟ ∇·T =⎜ ⎟⎜ 2 2 ⎝ ∂x ∂y ∂z ⎠⎜ 0 0 − ϵoA cos (kz − ωt ) ⎟ ⎜ ⎟ x2 + y2 ⎝ ⎠ ⎛ ⎞ ϵoA2 ( −2cos(kz − ωt ))( −sin(kz − ωt ))(k )⎟ = ⎜0 0 2 2 x +y ⎝ ⎠ ϵoA2 ( −2cos(kz − ωt ))( −sin(kz − ωt ))(k )zˆ x2 + y2 2ϵ A2 k = 2 o 2 cos(kz − ωt ) sin(kz − ωt )zˆ x +y =

Using k = ω /c,

 ∇·T =

2ϵoA2 ω ∂ cos(kz − ωt ) sin(kz − ωt )zˆ = (pem ⃗ ) 2 2 c (x + y ) ∂t

as expected. Problem 3.14. Let us consider a wave guide of rectangular shape of sides a and b (with a > b), inside which we have a perfect dielectric. We choose the x-axis to coincide with the direction of the hollow pipe. A plane harmonic wave propagates in the x-direction such that Ex = 0 (TE waves).

E˜ (x , y , z , t ) = E˜ 0(y , z )e i(ωt−kx ) B˜(x , y , z , t ) = B˜0(y , z )e i(ωt−kx ) (a) By using the boundary conditions, find the vector components of electric and magnetic fields. (b) Find the maximum wavelength which can be propagated down the tube. Solution 3.14. (a) Let us start with Maxwell’s equations

(i) ∇ · E ⃗ = 0 (ii) ∇ · B ⃗ = 0

∂B ⃗ = −iωB ⃗ ∂t 1 ∂E ⃗ iω (iv) ∇ × B ⃗ = 2 = 2 E⃗ c ∂t c

(iii) ∇ × E ⃗ = −

Since Ex = 0

E˜ 0(y , z ) = Exxˆ + Eyyˆ + Ezzˆ = Eyyˆ + Ezzˆ

3-22

(3.3)

Electrodynamics

From (iii)

∇ × E⃗ =

xˆ ∂ ∂x 0

yˆ ∂ ∂y Ey

zˆ ∂ ∂z Ez

= −iω(Bxxˆ + Byyˆ + Bzzˆ )

We obtain

∂Ey ∂Ez − = −iωBx ∂y ∂z ∂Ez = iωBy ∂x ∂Ey = −iωBz ∂x From (iv)

∇ × B⃗ =

xˆ ∂ ∂x Bx

yˆ ∂ ∂y By

zˆ ∂ ∂z Bz

=

iω (Eyyˆ + Ezzˆ) c2

∂By ∂Bz − =0 ∂y ∂z iω ∂Bz ∂Bx = 2 Ey − c ∂x ∂z ∂By iω ∂Bx = 2 Ez − c ∂y ∂x From (i) and (ii)

⎧ ∂Ey ∂Ez =0 + ⎪ ⎪ ∂z ∂y ⎨ ⎪ ∂Bx + ∂By + ∂Bz = 0 ⎪ ∂z ∂y ⎩ ∂x Also, we can write the vector component as

⎧ Ey = E 0y e i(ωt−kx ) ⎨ ⎩ Ez = E 0z e i(ωt−kx )

3-23

(3.4)

Electrodynamics

⎧ Bx = B0x e i(ωt−kx ) ⎪ ⎨ By = B0y e i(ωt−kx ) ⎪ ⎩ Bz = B0z e i(ωt−kx ) From here, we can obtain the partial derivatives

⎧ ∂Ey ⎪ ⎪ ∂x = −ikEy ⎨ ⎪ ∂Ez = −ikE z ⎪ ⎩ ∂x

(3.5)

⎧ ∂Bx = −ikBx ⎪ ⎪ ∂x ⎪ ∂By ⎨ = −ikBy ⎪ ∂x ⎪ ∂Bz = −ikBz ⎪ ⎩ ∂x

(3.6)

Combining (3.5) and (3.6) in (3.4), we obtain

⎧ ∂Ez ∂Ey − = −iωBx   ⎪ ∂z ⎪ ∂y ⎪ ⎪ kEz = −ωBy   ⎨ ⎪ kEy = ωBz   ⎪ ∂E ∂Ez ⎪ y + = 0  ⎪ ∂z ⎩ ∂y ⎧ ∂B ∂By ⎪ z − =0 ∂z ⎪ ∂y ⎪ ∂Bx iω + ikBz = 2 Ey   ⎪ ⎪ ∂z c ⎨ B iω ∂ x ⎪ ikB + = − 2 Ez   y ⎪ c ∂y ⎪ ⎪ −ikBx + ∂By + ∂Bz = 0  ⎪ ∂z ∂y ⎩

(a) (b) (c) (d) (e)

(3.7)

(f) (g) (h)

From this system of equations, we can express all electric and magnetic field components as a function of one component, for example as a function of Bx . Recall that Ex = 0.

3-24

Electrodynamics

From (3.7b) get Ez and substitute in (3.7g) ω Ez = − By k iω 2 ∂Bx ikBy + = 2 By kc ∂y 2 i⎛ 2 ω ⎞ ∂Bx ⎜k − 2 ⎟By = − k⎝ c ⎠ ∂y Therefore,

By = −

ik ∂Bx ⎛ ω2 ⎞ ∂y ⎜ 2 − k 2⎟ ⎝c ⎠

Now let us find Bz . From (3.7c) and (3.7f) we obtain, by eliminating Ey ω Ey = Bz k iω 2 ∂Bx + ikBz = 2 Bz kc ∂z 2 ⎛ ⎞ i ω ∂Bx − k 2⎟Bz = ⎜ ⎠ k ⎝ c2 ∂z ik ∂Bx Bz = − 2 ⎛ω ⎞ ∂z ⎜ 2 − k 2⎟ ⎝c ⎠ By replacing Bz in (3.7c) we obtain ω iω ∂Bx Ey = Bz = − 2 ⎛ω ⎞ ∂z k ⎜ 2 − k 2⎟ ⎝c ⎠

(3.8)

By replacing By in (3.7b) we obtain

iω ω ∂Bx Ez = − By = 2 ⎛ ⎞ k ∂y ω ⎜ 2 − k 2⎟ ⎝c ⎠ Let us determine Bx . We derive (3.7f) with respect to z and (3.7g) with respect to y and add them as follows:

⎧ ∂ 2Bx iω ∂Ey ∂Bz = 2 ⎪ 2 + ik ⎪ ∂z c ∂z ∂z ⎨ 2 ⎪ik ∂By + ∂ Bx = − iω ∂Ez ⎪ c 2 ∂y ∂y 2 ⎩ ∂y

3-25

(3.9)

Electrodynamics

⎛ ∂By ∂Ey ⎞ ∂ 2Bx ∂ 2Bx ∂Bz ⎞ iω ⎛ ∂Ez + + ik ⎜ + − ⎟ = − 2⎜ ⎟ 2 2 ∂y ∂z ∂z ⎠ ∂z ⎠ c ⎝ ∂y ⎝ ∂y By using (3.7a) and (3.7h)

∂ 2Bx ∂ 2Bx ω2 2 k B Bx − = − + x c2 ∂z 2 ∂y 2 ∂ 2Bx ∂ 2Bx + α 2Bx = 0 + ∂z 2 ∂y 2 where α 2 =

ω2 − k2 c2

⎛ ∂2 ⎞ ∂2 ⎜ 2 + 2 + α 2⎟Bx = 0 ∂z ⎝ ∂y ⎠

(3.10)

A similar equation is satisfied by the amplitude B0x as well,

⎛ ∂2 ⎞ ∂2 ⎜ 2 + 2 + α 2⎟B0x = 0 ∂z ⎝ ∂y ⎠ From the boundary conditions By∣ y=0 = 0

(3.11)

Bz z=0 = 0

y=a

z=b

Since the formulae for By and for Bz are similar

ik

∂Bx ⎛ω ⎞ ∂y ⎜ 2 − k 2⎟ ⎝c ⎠ ∂Bx ik Bz = − ⎛ 2 ⎞ ω ⎜⎜ − k2⎟⎟ ∂z ⎝ c2 ⎠

By = −

2

we have similar boundary conditions for the initial field components, as well

∂B0x ∂y

=0 y=0 y=a

∂B0x ∂z

z=0 z=b

=0

The solution for equation (3.11) is of the form:

B 0 x (y , z ) = Y (y )Z (z ) From equation (3.11)

Z (z )

d2Z (z ) d2Y (y ) ( ) Y y + + α 2 Y (y )Z (z ) = 0 dz 2 dy 2

As expected, we divide by Y (y )Z (z ) and obtain, with −k y2 − k z2 + α 2 = 0

3-26

Electrodynamics

(i)

1 d2Y (y ) = −k y2 2 Y ( y ) dy

(ii)

1 d2Z (z ) = −k z2 Z ( z ) dz 2

The solutions of these equations are

Y (y ) = A cos kyy + Cy

Z (z ) = B cos kzz + Cz From the boundary conditions

∂B0x ∂y

∂B0x ∂z

=0 y=0

=0 z=0

We obtain that Cy = Cz = 0. With B0 = AB ,

B0x = B0 cos kyy cos kzz By again using the boundary conditions, we obtain ky = integers. In conclusion, ⎛ mπ ⎞ ⎛ nπ ⎞ y ⎟ cos⎜ z⎟e i(ωt−kx ) Bx = B0 cos⎜ ⎝ a ⎠ ⎝b ⎠

mπ , a

kz =

nπ , b

m and n

Since we have Bx now, we know all the components of the electric and magnetic fields.

⎛ mπ ⎞ ⎛ nπ ⎞ ∂Bx ik ik mπ y ⎟ cos⎜ z⎟e i(ωt−kx ) = 2 B0 sin⎜ ⎝ ⎠ ⎝b ⎠ ω2 a ∂ y a α − k2 c2 ⎛ mπ ⎞ ⎛ nπ ⎞ i(ωt−kx ) ik ik nπ ∂Bx y ⎟ sin⎜ z⎟e Bz = − = 2 B0 cos⎜ 2 ⎝ a ⎠ ⎝b ⎠ ω z b ∂ α 2 −k c2 ⎛ mπ ⎞ ⎛ nπ ⎞ i(ωt −kx ) iω iω nπ ∂Bx y ⎟ sin⎜ z⎟e Ey = − = 2 B0 cos⎜ 2 ⎝ a ⎠ ⎝b ⎠ ω ∂z α b 2 k − c2 ⎛ mπ ⎞ ⎛ nπ ⎞ iω iω mπ ∂Bx y ⎟ cos⎜ z⎟e i(ωt −kx ) B0 sin⎜ Ez = =− 2 ⎝ ⎠ ⎝b ⎠ ω2 a α a − k 2 ∂y c2 By = −

(

(

)

(

)

(

)

)

3-27

Electrodynamics

Let us look at α

ω2 − k2 c2 ⎛ mπ ⎞2 ⎛ nπ ⎞2 ω2 2 ⎜ ⎟ + ⎜ ⎟ > 0 − = k ⎝ a ⎠ ⎝b⎠ c2

α 2 = k y2 + k z2

α2 =

From here,

⎡⎛ m ⎞2 ⎛ n ⎞2 ⎤ ⎛ ω ⎞2 ⎜ ⎟ − π ⎢⎜ ⎟ + ⎜ ⎟ ⎥ ⎝c⎠ ⎝b⎠ ⎦ ⎣⎝ a ⎠

k=

Discussion (1) If k and α are real, we have a traveling wave. m 2 a

n 2 b

( ) +( )

(2) If ω < cπ

= ωmn , then k is imaginary, and we have exponen-

tially attenuated fields. ωmn is called the cutoff frequency for the TE mode. The lowest cutoff frequency occurs for a TE10 mode: ω10 = caπ . For frequencies lower than ω10 , the wave will not propagate. Lastly, we can write differently the wave number as: k = wave velocity as

ω k

c

= 1−

2

( ωωmn )

1 c

2 and the ω 2 − ωmn

, which is greater than c, the speed of light. We

should not forget, however, that the wave travels at group velocity vg =

1 dk dω

=c 1−

ωmn 2 ω

( )

b), placed in air. (a) What is the condition on the wave frequency ω such that the wave will propagate in the rectangular pipe? (b) If a = 10 cm , b = 5 cm , how many propagation modes exist for a wave of frequency 1 GHz? (c) What about 3.5 GHz? Solution 3.15. (a) From problem 3.14, for TEmn mode, the cutoff frequency is

ωmn = cπ

⎛ m ⎞2 ⎛ n ⎞2 ⎜ ⎟ + ⎜ ⎟ ⎝a⎠ ⎝b⎠

The lowest cutoff frequency for a given wave guide with a > b is TE10 for which m = 1, n = 0, therefore

3-28

Electrodynamics

⎛ 1 ⎞2 ⎛ 0 ⎞2 cπ ω10 = cπ ⎜ ⎟ + ⎜ ⎟ = ⎝a⎠ ⎝b⎠ a Smaller frequencies will not propagate at all. In our case, ω10 = ω = 2πν , so

ν10 =

cπ , and a

c 3 × 108 m s−1 = 1.5 × 109 Hz = 2a 2 × 10−1 m

For waves with frequencies ν < ν10, the wave will not propagate. (b) In general, a wave will propagate if

ω ⩾ ωmn = cπ

⎛ m ⎞2 ⎛ n ⎞2 ⎜ ⎟ + ⎜ ⎟ ⎝a⎠ ⎝b⎠

With ω = 2πν,

2πν ⩾ cπ

⎛ m ⎞2 ⎛ n ⎞2 2ν ⎜ ⎟ + ⎜ ⎟ ⇒ ⩾ ⎝a⎠ ⎝b⎠ c

⎛ m ⎞2 ⎛ n ⎞2 4ν 2 ⎜⎛ m ⎟⎞2 ⎜⎛ n ⎟⎞2 ⎜ ⎟ + ⎜ ⎟ ⇒ ⩾ + ⎝a⎠ ⎝b⎠ ⎝a⎠ ⎝b⎠ c2

For ν = 1 GHz = 109 Hz and a = 10 cm , b = 5 cm , note that a = 2b, and both sides are positive. 2 4ν 2 ⎛⎜ m ⎞⎟2 ⎛⎜ n ⎞⎟ 4a 2ν 2 ⩾ + → ⩾ m 2 + 4n 2 ⎜a⎟ 2 ⎝a⎠ c2 c ⎝2⎠

Here m and n are positive integers. Let us calculate the left side of the inequality (recall that 1 Hz = 1 s−1)

4a 2ν 2 4(10−1 m)2 (109 Hz)2 = = 0.44 2 (3 × 108 m s−1)2 c 0.44 ⩾ m 2 + 4n 2 ⇒ m = 0, n = 0 No propagation in this case, no waveguide. (c) Let us see what the possible propagation modes are when ν = 3.5 GHz.

4a 2 ν 2 ⩾ m 2 + 4n 2 c2 4a 2ν 2 4(10−1 m)2 (3.5 × 109 Hz)2 = 5.44 = 2 c (3 × 108 m s−1)2 5.44 ⩾ m 2 + 4n 2

3-29

Electrodynamics

Solutions (1) m (2) m (3) m (4) m (5) m

= = = = =

0, 0, 1, 1, 2,

n n n n n

= = = = =

0 no wave propagation 1 0 1 0

Bibliography Fogiel M 1987 Staff of Research and Education Association The Electromagnetics Problem Solver (New York: Research and Education Association) Griffiths D J 1999 Introduction to Electrodynamics 3rd edn (Englewood Cliffs, NJ: Prentice-Hall) Griffiths D J 2013 Introduction to Electrodynamics 4th edn (Cambridge, MA: Pearson) Hecht E 2017 Optics 5th edn (Cambridge, MA: Pearson) Jackson J D 1998 Classical Electrodynamics 3rd edn (New York: Wiley) Popescu I M, Iordache D, Fara V, Stan M and Lupascu A 1986 Probleme Rezolvate de Fizica: Electromagnetism. Teoria Relativitatii Restranse. Unde Electromagnetice. Teoria Electromagnetica a Luminii (Optica) (Solved Physics Problems: Electromagnetism. Relativistic Theory. Electromagnetic Waves. The Electromagnetic Theory of Light (Optics) vol II (Editura Tehnica)

3-30

IOP Concise Physics

Electrodynamics Problems and solutions Carolina C Ilie and Zachariah S Schrecengost

Chapter 4 Potentials and fields

Introduction By using scalar potential V and magnetic vector potential A⃗ , the information from the four Maxwell’s equations is concentrated in two equations involving the potentials, the charge density ρ and the current density J ⃗ . Since the potentials which yield the electric and magnetic fields are not uniquely defined, we introduce the gauge transformations. These equations allow changes in potentials such that the same fields are obtained. The Coulomb gauge is interesting in the fact that the scalar potential is determined by the charge distribution at the present moment. The advantage of the Coulomb potential is the easy way the scalar potential is calculated, but the cumbersome calculation of the magnetic vector potential constitutes a disadvantage. The Lorentz gauge has the refined simplicity due to the fact that both potentials are defined by the same operator, the D’Alambertian (which is a generalization of the Laplacian). We discuss continuous distributions, retarded and advanced potentials, Jefimenko’s equation. The moving point charges are also discussed in the light of Liénard–Wiechert potentials, and we end with the electric and magnetic fields.

4.1 Theory 4.1.1 Electric field The electric field due to a scalar potential V and vector potential A⃗ is given by

E ⃗ = −∇V −

∂A ⃗ ∂t

4.1.2 Gauge transformations Given a scalar potential V and vector potential A⃗ , we can transform the potentials using a scalar function λ by doi:10.1088/978-1-6817-4931-0ch4

4-1

ª Morgan & Claypool Publishers 2018

Electrodynamics

V′ = V −

∂λ ∂t

and

A ⃗ ′ = A ⃗ + ∇λ

4.1.3 Retarded potentials Given a charge density ρ(r ′⃗ , tr ), the scalar potential is given by

V (r ⃗ , t ) =

1 4πϵo

∫

ρ(r ⃗′ , tr ) dτ ′ r

Given a current density J ⃗(r ′⃗ , tr ), the vector potential is given by

A ⃗ (r ⃗ , t ) =

μo 4π

∫

J (⃗ r ⃗′ , tr ) dτ ′ r

Note the integrand is evaluated at the retarded time given by r tr = t − c

4.1.4 Jefimenko’s equations Given charge density ρ(r ′⃗ , tr ) and current density J ⃗(r ′⃗ , tr ), the electric field is given by

E ⃗ (r ⃗ , t ) =

1 4πϵo

∫

⎡ ⎤ ⃗̇ ⎢ ρ(r ⃗′ , tr ) rˆ + ρ (̇ r ⃗′ , tr ) rˆ − J (r ⃗′ , tr ) ⎥dτ′ ⎢⎣ r 2 cr c 2r ⎥⎦

and the magnetic field is given by

B (⃗ r ⃗ , t ) =

μo 4π

∫

⎡ ⃗ ⎤ ̇⃗ ⎢ J (r ⃗′ , tr ) + J (r ⃗′ , tr ) ⎥ × r ⃗dτ′ ⎢⎣ r 2 cr ⎥⎦

Again, the integrands are evaluated at the retarded times. 4.1.5 Liénard–Wiechert potentials A point charge q moving at velocity v ⃗ has a scalar potential

V (r ⃗ , t ) =

1 qc 4πϵo (rc − r ⃗ ⋅ v ⃗ )

and vector potential 4-2

Electrodynamics

A ⃗ (r ⃗ , t ) =

v⃗ V (r ⃗ , t ) c2

where

r ⃗ = r ⃗ − w⃗(tr ) and w⃗(tr ) is the position of q at the retarded time, given implicitly by

r = r ⃗ − w⃗(tr ) = c(t − tr ) 4.1.6 Moving point charge Given a point charge q moving with velocity v ⃗ and acceleration a ⃗ , the electric field is given by q r [(c 2 − v 2 )u ⃗ + r ⃗ × (u ⃗ × a ⃗ )] E ⃗ (r ⃗ , t ) = 4πϵo (r ⃗ ⋅ u ⃗ )3 where

u ⃗ = crˆ − v ⃗ The magnetic field is given by

B (⃗ r ⃗ , t ) =

1 rˆ × E ⃗(r ⃗ , t ) c

4.2 Problems and solutions Problem 4.1. Find the fields that correspond to the following potentials

V=0 and

⎛ μ nskt 1⎞ A⃗ = ⎜ o + ⎟ϕˆ ⎝ 2 s⎠ and explain what situation would produce them. Solution 4.1. Starting with the electric field we have μ nsk ˆ ∂ ⎡⎛ μ nskt 1⎞ ⎤ ∂A ⃗ + ⎟ϕˆ ⎥ = − o = − ⎢⎜ o E ⃗ = −∇V − ϕ ∂t ⎣⎝ 2 s⎠ ⎦ 2 ∂t The magnetic field is given by

⎡⎛ μ nskt 1 ⎞ ⎤ 1 ∂ ⎡ ⎛ μonskt 1 ⎞⎤ B ⃗ = ∇ × A ⃗ = ∇ × ⎢⎜ o + ⎟⎥zˆ + ⎟ϕˆ ⎥ = ⎢s⎜ s ⎠⎦ s⎠ ⎦ s ∂s ⎣ ⎝ 2 ⎣⎝ 2 =

⎤ 1 ∂ ⎡ μons 2kt + 1⎥zˆ = μonktzˆ ⎢ s ∂s ⎣ 2 ⎦

4-3

Electrodynamics

Careful inspection would reveal these are the electric and magnetic fields that occur with a solenoid with n turns per length carrying a time dependent current I (t )=ktϕˆ . Problem 4.2. By using the following potential equations, obtain the ‘electric field’ and the ‘magnetic field’

V=0 and

⎧μ k ⎪ o ct − A ⃗ = ⎨ 2c ⎪ ⎩ 0,

(

2

)

(

x 2 + 1 zˆ , for x ∈ − (ct )2 − 1 ,

(ct )2 − 1

)

otherwise

where k is a constant, c =

1 μo ϵo

. Check if Maxwell’s equations are satisfied. Are

these possible electric and magnetic potentials? Solution 4.2.

B ⃗ = ∇ × A⃗ E ⃗ = − ∇V − Since V = 0, B ⃗ = ∇ × A⃗ and E ⃗ = − (a) For x ∈ ( − (ct )2 − 1 ,

E⃗ = −

∂A⃗ . ∂t

(ct )2 − 1 )

∂ ⎡μ k ∂A ⃗ = − ⎢ o ct − ∂t ⎣ 2c ∂t

(

(

2 ⎤ μk x 2 + 1 zˆ⎥ = − o 2 ct − ⎦ 2c

)

(

)

x 2 + 1 zˆ

= − μok ct −

B ⃗ = ∇ × A⃗ =

xˆ

yˆ

zˆ

∂ ∂x

∂ ∂y

∂ ∂z

0

0

=−

μok ∂ ⎡ yˆ ct − 2c ∂x ⎣⎢

=−

μok yˆ 2 ct − 2c

=

∂A ⃗ ∂t

(

(

μok ct − c

(

μok ct − 2c 2⎤ x2 + 1 ⎥ ⎦

( )

x2 + 1

⎛ ⎞ 2x ⎟ x 2 + 1 ⎜− ⎝ 2 x2 + 1 ⎠ x yˆ x2 + 1 x2 + 1

)

)

4-4

2

)

)

x 2 + 1 czˆ

Electrodynamics

(b) For x ∈

[(−∞, −

(ct )2 − 1 )

∪ ( (ct )2 − 1 , ∞)] V =0 A⃗ = 0

so

B ⃗ = ∇ × A⃗ = 0 ∂A ⃗ E⃗ = − =0 ∂t Let us check Maxwell’s equations ∂Ey ∂Ez ∂Ex + = 0, ρ = ϵo∇ ⋅ E ⃗ = 0 + ∂y ∂z ∂x (ii) ∇ ⋅ B ⃗ = 0 ∂B ⃗ (iii) ∇ × E ⃗ = − ∂t  ⎤ ∂ ⎡⎢ μok x x ∂B = ct − x 2 + 1 yˆ ⎥ = μok yˆ 2 2 ∂t ⎢⎣ c ∂t x + 1 ⎥⎦ x +1

(i) ∇ ⋅ E ⃗ =

(

 ∇×E =

)

xˆ yˆ ∂ ∂ ∂x ∂y 0

0

zˆ ∂ ∂z

(

−μok ct −

=− x2 + 1

)

∂ ⎡ −μ k ct − ∂x ⎢⎣ o

⎛ ⎞ x 2x ⎟⎟yˆ = − μ k = μok ⎜⎜ − yˆ o 2 2 ⎝ 2 x +1⎠ x +1 Note that, indeed, ∇ × E ⃗ = − (iv) ∇ × B ⃗ = μo J ⃗ + μo ϵo

∂B ⃗ . ∂t

∂E ⃗ ∂t

4-5

(

⎤ x 2 + 1 yˆ ⎥ ⎦

)

Electrodynamics

xˆ ∂ ⃗ ∇ × B = ∂x

zˆ ∂ ∂z

μok ct − c

x2 + 1

∂ ⎡ μok ⎢ ct − ∂x ⎣ c

x2 + 1

0 = zˆ

yˆ ∂ ∂y

(

(

)

)

x x2 + 1

0

⎤ ⎥ x2 + 1 ⎦ x

⎞ μok ∂ ⎛ x − x⎟ zˆ ⎜ct c ∂x ⎝ ⎠ x2 + 1 2x ⎞ ⎛ x2 + 1 − x ⎟ μok ⎜ 2 x2 + 1 zˆ⎜ct − 1⎟ = 2 c ⎜ x +1 ⎟ ⎠ ⎝ ⎞ μ k ⎛ x2 + 1 − x2 ⎟ 1 = o zˆ⎜⎜ct − ⎟ c ⎝ (x 2 + 1) 23 ⎠ ⎞ μk ⎛ 1 ⎟ 1 = o zˆ⎜⎜ct − ⎟ c ⎝ (x 2 + 1) 23 ⎠ =

∂⎡ ∂E ⃗ = ⎢⎣ −μok ct − ∂t ∂t

(

⎤ x 2 + 1 zˆ⎥⎦ = −μokczˆ

)

The fourth Maxwell equation does not work; therefore, these are not possible electric and magnetic field potentials. Problem 4.3. Consider the following electric potential and magnetic vector potential:

V =0 A ⃗ = A0 zˆ cos(kx − ωt ) Find the electric and magnetic fields, check if they satisfy Maxwell’s equations, and find the relationship between ω and k. Solution 4.3. The electric field is

E ⃗ = −∇V −

∂A ⃗ = 0 + A0 ( −ω)zˆ sin(kx − ωt ) = −ωA0 zˆ sin(kx − ωt ) ∂t

4-6

Electrodynamics

and the magnetic field is

B ⃗ = ∇ × A⃗ =

xˆ yˆ zˆ ∂ ∂ ∂ ∂x ∂y ∂z 0 0 A0 cos(kx − ωt )

⎡∂ ⎤ ⎡∂ ⎤ = xˆ⎢ (A0 cos(kx − ωt ))⎥ − yˆ ⎢ (A0 cos(kx − ωt ))⎥ ⎣ ⎦ ∂x ⎣ ∂y ⎦ ˆ 0 sin(kx − ωt ) = ykA Let us check if the fields satisfy Maxwell’s equations. (i) Gauss Law

∂Ey ∂Ez ∂Ex =0 + + ∂z ∂x ∂y ρ ∇ ⋅ E⃗ = ϵ0

∇ ⋅ E⃗ =

Here ρ = 0, so the law is satisfied. (ii) ∇ ⋅ B ⃗ = 0

∇ ⋅ B⃗ =

∂By ∂Bz ∂Bx =0 + + ∂z ∂x ∂y

as expected. (iii) ∇ × E ⃗ = −

∂B ⃗ ∂t

xˆ yˆ zˆ ∂ ∂ ∂ ∇ × E⃗ = ∂x ∂y ∂z 0 0 −ωA0 sin(kx − ωt ) ∂ [ −ωA0 sin(kx − ωt )] ∂x = yˆ ωA0 k cos(kx − ωt ) = − yˆ

−

∂ ∂B ⃗ ˆ 0 sin(kx − ωt )] = − [ykA ∂t ∂t ˆ 0 ωcos(kx − ωt ) = ykA

Indeed, we obtain the same result.

4-7

Electrodynamics

(iv) Lastly,

∇ × B ⃗ = μoϵo

∂E ⃗ ∂t

The left side of the equality

xˆ yˆ zˆ ∂ ∂ ∂ ∇ × B⃗ = ∂x ∂y ∂z 0 kA0 sin(kx − ωt ) 0 ∂ [kA0 sin(kx − ωt )] ∂x = zk ˆ 2A0 cos(kx − ωt ) = zˆ

The right side of the equality

μoϵo

∂ ∂E ⃗ = μoϵo [ −ωA0 zˆsin(kx − ωt )] = μoϵoω 2A0 zˆ cos(kx − ωt ) ∂t ∂t

The condition for equality is that k 2 = μo ϵoω 2 , and with μo ϵo =

1

c2

, we

obtain ω = ck . Problem 4.4. Are the potentials from problem 4.3 in the Coulomb gauge, or in the Lorentz gauge? Solution 4.4. Recall that

V =0 A ⃗ = A0 zˆ cos(kx − ωt ) In Coulomb gauge we have ∇ ⋅ A⃗ = 0 and ∇2 V = −

ρ and the differential equation ϵ0

for A⃗ is

∇2 A ⃗ − μoϵo

⎛ ∂V ⎞ ∂ 2A ⃗ = −μoJ ⃗ + μoϵo∇⎜ ⎟ 2 ⎝ ∂t ⎠ ∂t

⃗ In Lorentz gauge we have ∇ ⋅ A⃗ = −μo ϵo ∂∂Vt and ∇2 A⃗ − μo ϵo ∂∂tA2 = −μo J ⃗ . There is a 2

2

similar differential equation for the potential ∇2 V − μo ϵo ∂∂tV2 = − ϵρ , where the o

operator is called d’Alambertian: □ = ∇ − 2

4-8

∂2 μo ϵo 2 . ∂t

In our case,

Electrodynamics

∇ ⋅ A⃗ =

∂Ay ∂Az ∂Ax =0 + + ∂z ∂x ∂y

∂V =0 ∂t These equations show that the potentials are in Coulomb gauge, but also in Lorentz gauge. Problem 4.5. (a) Find the fields, and the charge and current distributions, correspond to V (z ) = 0 and

A ⃗ (z , t ) = −

σt zˆ 2ϵo

(b) Use the gauge function

λ=

σzt 2ϵo

to transform the above potentials. Solution 4.5. (a) The electric field is given by

E ⃗ = −∇V −

∂ ⎛ σt ⎞ ∂A ⃗ σ = − ⎜− zˆ⎟ = zˆ ∂t ⎝ 2ϵo ⎠ 2ϵo ∂t

and the magnetic field is given by

⎛ σt ⎞ B ⃗ = ∇ × A⃗ = ∇ × ⎜− zˆ⎟ = 0 ⎝ 2ϵo ⎠ These are simply the fields due to an infinite sheet carrying a surface charge σ . (b) We can transform the scalar potential using ∂ ⎛ σzt ⎞ ∂λ σz V′ = V − =− ⎜ ⎟=− ∂t ⎝ 2ϵo ⎠ 2ϵo ∂t and we can transform the vector potential using

A ⃗ ′ = A ⃗ + ∇λ = −

⎛ σzt ⎞ σt σt σt zˆ + ∇⎜ zˆ + zˆ = 0 ⎟=− ⎝ 2ϵo ⎠ 2ϵo 2ϵo 2ϵo

These are the ‘typical’ potentials of an infinite sheet with surface charge density σ (taking V = 0 at the sheet itself).

4-9

Electrodynamics

Problem 4.6. (a) Find the fields that correspond to the potentials

V=0 and

⎡⎛ ⎛ 2R3 ⎞ R3 ⎞ A ⃗ = −E 0t ⎢⎜1 + 3 ⎟ cos θ rˆ − ⎜1 − 3 ⎟ sin θ ⎝ r ⎠ r ⎠ ⎣⎝

⎤ θˆ⎥ ⎦

(b) Use the gauge function

⎛ R3 ⎞ λ = E 0t⎜r − 2 ⎟ cos θ ⎝ r ⎠ to transform the potentials and comment on the results. Solution 4.6. (a) The electric field is given by

⎡⎛ ⎤⎤ ⎛ 2R3 ⎞ R3 ⎞ ∂⎡ ∂A ⃗ = − ⎢ −E 0t⎢⎜1 + 3 ⎟ cos θ rˆ − ⎜1 − 3 ⎟ sin θ θˆ⎥⎥ ⎝ r ⎠ r ⎠ ∂t ⎢⎣ ∂t ⎣⎝ ⎦⎥⎦ ⎡⎛ ⎤ ⎛ 2R3 ⎞ R3 ⎞ = E 0⎢⎜1 + 3 ⎟ cos θ rˆ − ⎜1 − 3 ⎟ sin θ θˆ⎥ ⎝ r ⎠ r ⎠ ⎣⎝ ⎦

E ⃗ = − ∇V −

and the magnetic field is given by

⎡ ⎡⎛ ⎛ 2R3 ⎞ R3 ⎞ B ⃗ = ∇ × A ⃗ = ∇ × ⎢ −E 0t⎢⎜1 + 3 ⎟ cos θ rˆ − ⎜1 − 3 ⎟ sin θ ⎝ ⎢⎣ r ⎠ r ⎠ ⎣⎝ ⎞ ⎞⎤ 2R3 ⎞ R3 ⎞ E t⎡ ∂ ⎛ ⎛ ∂ ⎛⎛ = − 0 ⎢ ⎜ −r⎜1 − 3 ⎟ sin θ ⎟ − ⎜⎜1 + 3 ⎟ cos θ ⎟⎥ϕˆ r ⎠ r ⎠ r ⎢⎣ ∂r ⎝ ⎝ ⎠ ∂θ ⎝⎝ ⎠⎥⎦ =−

⎤⎤ θˆ⎥⎥ ⎦⎥⎦

⎛ 2R3 ⎞ ⎛ E 0t ⎡ 2R3 ⎞⎤ ⎢sin θ ⎜ − 3 − 1⎟ − ( −sin θ )⎜1 + 3 ⎟⎥ϕˆ = 0 ⎝ r ⎠ ⎝ r ⎣ r ⎠⎦

(b) Our new scalar potential is given by

V′ = V −

⎞ ⎛ R3 ⎞ R3 ⎞ ∂⎛ ⎛ ∂λ = − ⎜E 0t⎜r − 2 ⎟ cos θ ⎟ = −E 0⎜r − 2 ⎟ cos θ ⎝ r ⎠ r ⎠ ∂t ⎝ ⎝ ∂t ⎠

and our new vector potential is given by

A ⃗ ′ = A ⃗ + ∇λ Therefore, we need to find ∇λ , 4-10

Electrodynamics

⎡ ⎛ ⎤ ⎡⎛ ⎤ R3 ⎞ R3 ⎞ ∇λ = ∇⎢E 0t⎜r − 2 ⎟ cos θ ⎥ = E 0t∇⎢⎜r − 2 ⎟ cos θ ⎥ r ⎠ r ⎠ ⎣ ⎝ ⎦ ⎣⎝ ⎦ ⎡ = E 0t⎢cos θ ⎣ ⎡ = E 0t⎢cos θ ⎣

⎤ ⎛ R3 ⎞ ∂ ∂ (r − R3r −2 )rˆ + ⎜1 − 3 ⎟ (cos θ )θˆ⎥ ⎝ r ⎠ ∂θ ∂r ⎦ ⎛ ⎛ 2R3 ⎞ R3 ⎞ ⎤ ⎜1 + 3 ⎟rˆ − sin θ ⎜1 − 3 ⎟θˆ⎥ ⎝ ⎝ r ⎠ ⎦ r ⎠

Now the vector potential is

⎡⎛ ⎛ 2R3 ⎞ R3 ⎞ A ⃗ ′ = − E 0t⎢⎜1 + 3 ⎟ cos θ rˆ − ⎜1 − 3 ⎟ sin θ ⎝ r ⎠ r ⎠ ⎣⎝

⎤ θˆ⎥ ⎦ ⎡ ⎛ ⎛ 2R3 ⎞ R3 ⎞ ⎤ + E 0t⎢cos θ ⎜1 + 3 ⎟rˆ − sin θ ⎜1 − 3 ⎟θˆ⎥ = 0 ⎝ ⎝ r ⎠ ⎦ r ⎠ ⎣

We can see that our gauge function transformed the somewhat complicated potential into

⎛ R3 ⎞ V = −E 0⎜r − 2 ⎟ cos θ ⎝ r ⎠ which is the potential of a classic problem; the potential inside an uncharged metal sphere, of radius R , placed in a uniform electric field E ⃗ = E0zˆ . Problem 4.7. Consider an infinite sheet in the xy-plane carrying surface current

⎧ 0, t ⩽ 0 K ⃗ (t ) = ⎨ ⎩ Kxˆ , t > 0 Find the generated electric and magnetic fields.

4-11

Electrodynamics

Solution 4.7. Since ρ = 0, we have V = 0. At a point P in the figure below, we have vector potential μ μ Kxˆ Kxˆ 1 A⃗ = o da = o da r r 4π 4π

∫

∫

Note that for a time t < cz , the news that the current has been turned on has not arrived yet. So, for t ⩾ cz , the area contained in

r 2 ⩽ (ct )2 − z 2 ⇒ r ⩽

(ct )2 − z 2

is the only part of the sheet that contributes to the fields. We can rewrite our potential as

μ Kxˆ A⃗ = o 4π where r =

(ct )2 − z 2 2π

∫ ∫ rr 0

dϕ dr

0

z 2 + r 2 . Therefore,

μ Kxˆ A ⃗ = o 2π 4π

(ct )2 − z 2

∫ 0

r z2 + r2

dr =

μoKxˆ (ct − z ) 2

Now our electric field is given by

E⃗ = −

μ Kxˆ ∂ μ cKxˆ ∂A ⃗ (ct − z ) = − o =− o 2 ∂t 2 ∂t

and our magnetic field is given by

μoK [∇ × [xˆ(ct − z )]] 2 μK ∂ μK (ct − z )yˆ = − o yˆ = o 2 ∂z 2

B ⃗ = ∇ × A⃗ =

which is what we would expect in the static case.

4-12

Electrodynamics

Problem 4.8. Suppose an infinite wire carries charge density

⎧ 0, λ(t ) = ⎨ ⎩ λ,

t⩽0 t>0

Find the fields that result. Solution 4.8. Let us start with the scalar potential. This is given by 1 λ(tr ) V (r ⃗ , t ) = dz 4πϵo r

∫

Consider a point P, we have

Since the message of charge being ‘turned on’ arrives at the retarded time, tr = t − rc , any time t < cs still sees zero charge. It follows that for time t > cs , the segment of wire given by z ⩽ (ct )2 − s 2 is the only portion of the wire for which news of there being charge has arrived at P . We can now rewrite our potential as (ct )2 − s 2

λ V= 4πϵo

∫ − (ct )2 − s

(ct )2 − s 2

1 2λ dz = r 4πϵo 2

∫ 0

1 dz r

Since this is symmetric about z . Also,

r=

z2 + s2

so (ct )2 − s 2

λ V= 2πϵo

∫ 0

1 z2 + s2

dz =

⎛ ct + λ ln⎜⎜ 2πϵo ⎝

(ct )2 − s 2 ⎞ ⎟ ⎟ s2 ⎠

Now to find the vector potential. This is given by

A ⃗ (r ⃗ , t ) =

μo 4π

∫

J (⃗ r ⃗′ , tr ) dτ ′ r

Now at first glance, we may say that J ⃗(r ′⃗ , tr ) = 0 so A⃗ = 0 which implies B is constant in space, namely zero. But we know that a change in electric field should induce a field. Let us see…

4-13

Electrodynamics

Since we have

⎧ 0, λ(t ) = ⎨ ⎩ λ,

t⩽0 t>0

We can express this as

λ(t ) = λθ (t ) where θ (t ) is the unit step function (Heaviside) given by

⎧ 0, θ (t ) = ⎨ ⎩1,

t0

Also, for a length L ,

λ(t ) =

q(t ) ⇒ q(t ) = Lλ(t ) L

our current is given by

I (t ) =

dq(t ) dθ ( t ) = Lλ = qoδ(t ) dt dt

where qo = Lλ. So our vector potential is now

A ⃗ (r ⃗ , t ) =

μo 4π

∫

μq I (tr )zˆ dz = o o r 4π

∫

δ⎛⎜⎝t − r ⎞⎟⎠ c

r

dz

(ct )2 − s 2 contributing so

Again, we have only the segment ∣z∣ ⩽

(ct )2 − s 2

μ q zˆ A ⃗ (r ⃗ , t ) = o o 2π

∫

δ⎛⎜⎝t − r ⎞⎟⎠ c

r

dz

0

We can use

r 2 = z 2 + s 2 ⇒ r dr = z dz ⇒

dr dz = r z

with

z= When z = 0, r = s , and when z =

A ⃗ (r ⃗ , t ) =

r 2 − s2

(ct )2 − s 2 , r = ct . Therefore,

μoqozˆ 2π

ct

∫ s

4-14

δ⎛⎜⎝t − r ⎞⎟⎠ c

r 2 − s2

dr

Electrodynamics

Using

⎛ 1 ⎞ ⎛ r⎞ δ⎜t − ⎟ = δ⎜ − (r − ct )⎟ = cδ(r − ct ) ⎝ ⎠ ⎝ ⎠ c c we have

A ⃗ (r ⃗ , t ) =

μoqoczˆ 2π

ct

∫

δ(r − ct )

s

r −s 2

2

dr =

μoqoczˆ 2π

1 (ct )2 − s 2

Now we can calculate the fields. Starting with the electric field

∂A ⃗ ∂t ⎡ ⎛ ct + λ ln⎜⎜ = − ∇⎢ ⎢ 2πϵo ⎝ ⎣

E ⃗ = − ∇V −

⎤ ⎡ ⎤ (ct )2 − s 2 ⎞⎥ 1 ⎟ − ∂ ⎢ μoqoczˆ ⎥ ⎟⎥ ∂t ⎢ 2π 2 2 ⎥ s2 ( ) ct s − ⎣ ⎦ ⎠⎦ ⎡ ⎤ ⎡ ⎤ −ct − c 2t λ ⎢ ⎥sˆ − μoqoc ⎢ ⎥zˆ =− 2πϵo ⎢⎣ s (ct )2 − s 2 ⎥⎦ 2π ⎣ ((ct )2 − s 2 )3/2 ⎦ ⎛ ⎞ ⎤ μq ⎡ c 3t ct λ ⎜ ⎟sˆ + o o ⎢ ⎥zˆ = 2πϵos ⎜⎝ (ct )2 − s 2 ⎟⎠ 2π ⎣ ((ct )2 − s 2 )3/2 ⎦ Note that as t → ∞, we have

E⃗ =

λ sˆ 2πϵos

which is exactly what we had in the static case. Now to find the magnetic field,

⎡ μ q czˆ B ⃗ = ∇ × A⃗ = ∇ × ⎢ o o ⎢⎣ 2π

⎤ μ q c⎡ ⎤ −s ⎥= o o ⎢ ⎥ϕˆ 3/2 2π ⎣ ((ct )2 − s 2 ) ⎦ (ct )2 − s 2 ⎥⎦ 1

If we take t → ∞, we have

B⃗ = 0 which is what we would expect in the static case for an infinite line of non-moving charge. Problem 4.9. Consider an infinite wire carrying

⎧ 0, I (t ) = ⎨ ⎩ Io,

t⩽0 t>0

Find the electric and magnetic field generated using Jefimenko’s equations.

4-15

Electrodynamics

Solution 4.9. Starting with the electric field, we have

E ⃗ (r ⃗ , t ) =

1 4πϵo

⎡ ⎤ ⃗̇ ⎢ ρ(r ⃗′ , tr ) rˆ + ρ (̇ r ⃗′ , tr ) rˆ − J (r ⃗′ , tr ) ⎥dτ′ ⎢⎣ r 2 cr c 2r ⎥⎦

∫

Since ρ = 0, the first two terms in the integral are zero. Also, our current can be rewritten as

I (t ) = Ioθ (t ) where θ (t ) is the unit step function

⎧ 0, θ (t ) = ⎨ ⎩1,

t0

So

I (̇ t ) = Io

dθ ( t ) = Io δ(t ) dt

The electric field is now given by

E ⃗ (r ⃗ , t ) = −

Iozˆ 4πϵo

∫

δ (tr ) dz c 2r

The retarded time is given by

tr = t −

1 r = − (r − ct ) c c

so

⎛ 1 ⎞ δ(tr ) = δ⎜ − (r − ct )⎟ = cδ(r − ct ) ⎝ c ⎠ Thus,

E ⃗ (r ⃗ , t ) = −

Iozˆ 4πϵo

∫

I zˆ cδ(r − ct ) dz = − o 2 c r 4πϵoc

∫

δ(r − ct ) dz r

Considering the following figure, and that we are looking for the field at point P,

we have

r 2 = z2 + s2

4-16

Electrodynamics

So

r dr = z dz ⇒

dr dz = r z

Using this, along with

r 2 − s2

z= we have

dr

r −s 2

=

2

dz r

Also consider when z = 0, r = s , and when z =

(ct )2 − s 2 , r = ct , so now we have

(ct )2 − s 2

I zˆ E ⃗ (r ⃗ , t ) = − o 2 4πϵoc =−

Iozˆ 2πϵoc

∫ 0

ct

∫

δ(r − ct ) dz r

δ(r − ct )

s

dr

r 2 − s2

⎛ I zˆ 1 = − o ⎜⎜ 2πϵoc ⎝ (ct )2 − s 2

⎞ ⎟ ⎟ ⎠

Now to find the magnetic field, which is given by

μo 4π μ Io = o 4π

B (⃗ r ⃗ , t ) =

⎡

⎤

∫ ⎢⎣ Iorθ(2tr ) + Iocδr(tr ) ⎥⎦zˆ × r ⃗ dz ⎡

⎤

∫ ⎢⎣ θr(t2r ) + δc(rtr ) ⎥⎦zˆ × r ⃗ dz

We have the same bounds for z as the electric field so

μ Io B ⃗( r ⃗ , t ) = o 2 4π

(ct )2 − s 2

∫ 0

⎡ θ (tr ) δ (tr ) ⎤ ⎢⎣ 2 + ⎥zˆ × r ⃗ dz cr ⎦ r

The cross product can be determined using

so

r ⃗ = ssˆ − zzˆ ⇒ rˆ =

4-17

s z sˆ − zˆ r r

Electrodynamics

and

sˆ ϕˆ zˆ zˆ × rˆ = 0 0 1 s z 0 − r r

=

s ˆ ϕ r

Thus (ct )2 − s 2

μ Iosϕˆ B (⃗ r ⃗ , t ) = o 2π

∫ 0

⎡ θ (tr ) δ (tr ) ⎤ 1 ⎢⎣ 2 + ⎥ dz cr ⎦r r

and using the same substitution as we used for the electric field dz dr dr = = z r r 2 − s2 we have ct ⎤ ⎡ μ Iosϕˆ θ (tr ) δ (tr ) ⎥ dr ⎢ B ⃗( r ⃗ , t ) = o + 2 2 ⎦ 2π ⎣ r 2 r 2 − s2 c s − r r s

∫

We will solve each term in the integral separately. Starting with the second term, ct

∫ s

δ (tr ) cr r 2 − s 2

dr

we know

δ(tr ) = cδ(r − ct ) Thus, ct

∫ s

ct

δ (tr ) cr r 2 − s 2

dr =

∫ s

δ(r − ct )

dr =

r r 2 − s2

1 ct (ct )2 − s 2

Evaluating the first term ct

∫ s

θ (tr )

r

2

r 2 − s2

dr

we have

⎛ 1 ⎞ ⎛ r⎞ θ (tr ) = θ ⎜t − ⎟ = θ ⎜ − (r − ct )⎟ ⎝ ⎝ c ⎠ c⎠ ⎛1 ⎞ = 1 − θ ⎜ (r − ct )⎟ = 1 − θ (r − ct ) ⎝c ⎠ which can be expressed as

⎧ 0, 1 − θ (r − ct ) = ⎨ ⎩1,

r > ct r < ct

Since the bounds of our integral are less than ct , θ (tr ) = 1. Now 4-18

Electrodynamics

ct

∫ s

ct

θ (tr )

r 2 r 2 − s2

dr =

1

∫

r 2 r 2 − s2

s

dr =

(ct )2 − s 2 cts 2

Substituting everything back into our expression for B ⃗ , we have ⎡ ⎤ μ Iosϕˆ ⎢ (ct )2 − s 2 1 ⎥ + B (⃗ r ⃗ , t ) = o cts 2 2π ⎢⎣ ct (ct )2 − s 2 ⎥⎦ Factoring out

1 s 2 (ct )2 − s 2

yields

B (⃗ r ⃗ , t ) =

⎡ (ct )2 − s 2 μoIosϕˆ 1 s2 ⎤ + ⎥ ⎢ 2π s 2 (ct )2 − s 2 ⎣ ct ct ⎦

Therefore,

B ⃗( r ⃗ , t ) =

μoIo ct ϕˆ 2π s (ct )2 − s 2

Note that using the potential formulation (Griffiths 1999, example 10.2) gives the same results with much less effort. Problem 4.10. Prove that the advanced potentials ⎧ 1 ρ(r ⃗′ , ta ) dτ ′ ⎪Va(r ⃗ , t ) = ⎪ 4πϵo r ⎨ μ J ⃗(r ⃗′ , ta ) ⎪ ⃗ Aa (r ⃗ , t ) = o dτ ′ ⎪ ⎩ 4π r

∫

∫

where

ta = t +

r c

is the advanced time and r = ∣r ⃗ − r ′⃗ ∣, satisfy (a) the inhomogeneous wave equation The D’Alambertian:

□ 2 = ∇2 − μoϵo

∂2 ∂t 2

□ 2V = −

(b) the Lorentz condition (gauge)

∇ ⋅ A ⃗ = −μoϵo

4-19

∂V ∂t

ρ ϵo

Electrodynamics

Solution 4.10. (a) Let us prove that the advanced potential satisfies the wave equation. Let us start with:

⎡ ⎛ 1 ⎞⎤ 1 ⎢(∇ρ) + ρ  ∇⎜ ⎟⎥ dτ′ ⎝r ⎠⎦ ⎣ r 1 ∇ρ = ρ̇ ∇ta = ρ̇  ∇r c ⎛1⎞ rˆ ∇r = rˆ, ∇⎜ ⎟ = − 2 ⎝r ⎠ r ⎡ 1 ρ̇ rˆ rˆ ⎤ ∇V = − ρ 2 ⎥ dτ ′ ⎢ ⎣c r 4πϵo r ⎦ ∇V =

1 4πϵo

∫

∫

The Laplacian is

∇ ⋅ (∇V ) = ∇2 V =

1 4πϵo

∫

⎧ 1 ⎡ rˆ ⎛ rˆ ⎞⎤ ⎡ rˆ ⎛ rˆ ⎞⎤⎫ ⎨ ⎢ ⋅ (∇ρ̇ ) + ρ̇ ∇ ⋅ ⎜ ⎟⎥ − ⎢ 2 ⋅ (∇ρ) + ρ∇ ⋅ ⎜ 2 ⎟⎥⎬ dτ′ ⎝ r ⎠⎦ ⎣ r ⎝ r ⎠⎦⎭ ⎩ c ⎣r ⎪

⎪

⎪

⎪

∇ρ̇ = ρ ̈ ∇ta =

1 1 ρ ̈ ∇r = ρ r ̈ ˆ c c

Recall the Delta function

⎛ rˆ ⎞ ⎛ rˆ ⎞ 1 ∇ ⋅ ⎜ ⎟ = 2 and ∇ ⋅ ⎜ 2 ⎟ = 4πδ 3(r ⃗ ) ⎝r ⎠ r ⎝r ⎠ Back to the Laplacian, we obtain

∇2 V =

1 4πϵo

∫

=

1 4πϵo

∫

⎡1 ⎤ ρ̇ 1 rˆ rˆ ρ̇ + − 2 ⋅ rˆ − ρ  4πδ 3(r ⃗ )⎥ dτ′ ̈ ˆ⋅ ⎢ 2 ρr 2 ⎣c ⎦ r r cr c 2 ⎡ 1 ρ̈ ⎤ ρ(r ⃗ , t ) 1 ∂V 3 ⎢⎣ 2 − 4πρδ (r ⃗ )⎥⎦dτ′ = 2 2 − ϵo c r c ∂t

The wave equation is satisfied. (b) We want to prove that the Lorentz condition ∇ ⋅ A⃗ = −μo ϵo ∂∂Vt for the advanced potential

Aa⃗ (r ⃗ , t ) =

μo 4π

∫

Let us recall the product rule

4-20

J (⃗ r ⃗′ , ta ) dτ ′ . r

Electrodynamics

⎛J⃗⎞ 1 ⎛ 1⎞ ∇ ⋅ ⎜ ⎟ = (∇ ⋅ J ⃗ ) + J ⃗ ⋅ ⎜∇ ⎟ ⎝ r⎠ ⎝r ⎠ r ⎛J⃗⎞ 1 ⎛ 1⎞ ∇′ ⋅ ⎜ ⎟ = (∇′ ⋅ J ⃗ ) + J ⃗ ⋅ ⎜∇′ ⎟ ⎝ r⎠ ⎝r ⎠ r Also, note that ∇r1 = −∇′r1 since r ⃗ = r ⃗ − r ′⃗ Replacing this in the first product rule equation above, and using the second equation, we obtain

⎛J⃗⎞ 1 ⎛J⃗⎞ ⎛ 1⎞ 1 1 ∇ ⋅ ⎜ ⎟ = (∇ ⋅ J ⃗ ) − J ⃗ ⋅ ⎜∇′ ⎟ = (∇ ⋅ J ⃗ ) + (∇′ ⋅ J ⃗ ) − ∇′ ⋅ ⎜ ⎟ ⎝ r⎠ r r ⎝r ⎠ r ⎝r ⎠ Let us look at ∇ ⋅ J ⃗ , considering that J ⃗ = J ⃗(r ′⃗ , ta ), ta = t + rc and

∂ta ∂x

=

1 ∂r c ∂x

∂Jy ∂ta ∂Jy ∂J ∂t ∂J ∂t ∂J ∂Jx + z a + + z = x a + ∂ta ∂z ∂ta ∂x ∂ta ∂y ∂z ∂x ∂y ∂Jy ∂r 1 ⎛ ∂J ∂r ∂J ∂r ⎞ 1 ∂J ⃗ ⋅ ( ∇r ) + + z ⎟= = ⎜ x c ⎝ ∂ta ∂x ∂ta ∂z ⎠ c ∂ta ∂ta ∂y

∇ ⋅ J⃗ =

∂ρ ∂t

In similar way, and by using the continuity equation

= −∇ ⋅ J ⃗, ρ ̇ =

∂ρ ∂t

1 ∂J ⃗ ⋅ (∇′r ) c ∂ta

∇′ ⋅ J ⃗ = −ρ ̇ +

Back to our equation, and with ∇r = −∇′r

⎤ 1⎡ ⎤ ⎛ J ⃗ ⎞ 1 ⎡ 1 ∂J ⃗ ⎛J⃗⎞ 1 ∂J ⃗ ⋅ (∇′r )⎥ − ∇′ ⋅ ⎜ ⎟ ⋅ ( ∇r ) ⎥ + ⎢ − ρ ̇ + ∇ ⋅ ⎜ ⎟= ⎢ c ∂ta ⎦ r⎣ ⎦ ⎝r ⎠ r ⎣ c ∂ta ⎝r ⎠ ⎛J⃗⎞ 1 = − ρ ̇ − ∇′ ⋅ ⎜ ⎟ r ⎝r ⎠ Finally, let us obtain the divergence of A⃗ :

⎡μ ∇ ⋅ A⃗ = ∇ ⋅ ⎢ o ⎣ 4π μ ⎡ ∂ = o ⎢− 4π ⎢⎣ ∂t = − μoϵo

∫

⎤ μ J (⃗ r ⃗′ , ta ) dτ′⎥ = o r ⎦ 4π

⎛ ⃗

⎤

⎛ ⃗⎞

∫ rρ dτ − ∫ ∇′ ⋅ ⎜⎝rJ ⎟⎠ dτ ⎥⎥

∂⎛ 1 ⎜ ∂t ⎝ 4πϵo

⎦

⎞

μ

∫ rρ dτ⎟⎠ − 4πo ∮ rJ

4-21

⎞

∫ ∇ ⋅ ⎜⎝ J (rr⃗′, ta ) ⎟⎠ dτ′ ⃗

⋅ da ⃗

Electrodynamics

where we used the divergence theorem

∫ ∇ ⋅ v ⃗ dτ = ∮ v ⃗ ⋅ da ⃗, where v ⃗ is a vector.

J⃗

The term ∮ r ⋅ da ⃗ = 0 because the integration is over the surface at infinity, where J ⃗ = 0. The apparent volume τ τ′ = rˆ ⋅ v ⃗ 1− c and also we easily recognize the electric potential

1 4πϵo

∫ rρ dτ = V

In conclusion, ∇ ⋅ A⃗ = −μo ϵo ∂∂Vt , and the Lorentz condition is fulfilled. Problem 4.11. Consider a particle of charge q moving in a parabolic motion

w⃗(t ) = at 2yˆ Find the Liénard–Wiechert potentials for points on the y-axis. Solution 4.11. We will start by finding an expression for the retarded time using

r ⃗ − w⃗(tr ) = c(t − tr ) Since r ⃗ = yyˆ , we have

r ⃗ − w⃗(tr ) = yyˆ − atr2yˆ = y − atr2 so

y − atr2 =  c(t − tr ) ⇒ atr2 − ctr + (ct − y ) = 0 Solving for tr yields

tr =

c±

c 2 − 4a(ct − y ) 2a

=

c±

4ay − 4act + c 2 2a

Consider the following: when t = 0,

tr =

c±

4ay + c 2 2a

Since the retarded time must lag behind the true time, when t = 0, we require tr < 0. To satisfy this, we must choose the minus sign in the expression c ± 4ay + c 2 . Therefore,

4-22

Electrodynamics

tr =

c−

4a(y − ct ) + c 2 2a

Now to find the potentials. Starting with the scalar potential,

V=

1 qc 4πϵo (rc − r ⃗ ⋅ v ⃗ )

We will start by examining the denominator rc − r ⃗ ⋅ v ⃗ . Since w⃗(t ) = at 2yˆ , we have

v (⃗ tr ) = 2atryˆ Also,

r ⃗ = r ⃗ − w⃗(tr ) = yyˆ − atr2yˆ = (y − atr2 )yˆ Noting that

r ⃗ ⋅ v⃗ = r v so

r c − r ⃗ ⋅ v ⃗ = r c − r v = r (c − v ) Looking at c − v , and the quantities found above, we have

⎛c − c − v = c − 2atr = c − 2a⎜⎜ ⎝

4a(y − ct ) + c 2 ⎞ ⎟= ⎟ 2a ⎠

4a(y − ct ) + c 2

and

r (c − v ) = r 4a(y − ct ) + c 2 We must now solve for r , which can be done using

r = r ⃗ − w⃗(tr ) = yyˆ − at 2yˆ ⎛ c − 4a(y − ct ) + c 2 ⎞2 ⎟ = y − a⎜⎜ ⎟ 2a ⎝ ⎠ c =

(

4a(y − ct ) + c 2 + 2at − c

)

2a

Therefore,

c 4a(y − ct ) + c 2 + 2at − c 4a(y − ct ) + c 2 2a c = ⎡⎣4a(y − ct ) + c 2 + (2at − c ) 4a(y − ct ) + c 2 ⎤⎦ 2a

r (c − v ) =

(

)(

so, the scalar potential is given by

4-23

)

Electrodynamics

V=

1 qc q 2a = 2 4πϵo (rc − r ⃗ ⋅ v ⃗ ) 4πϵo 4a(y − ct ) + c + (2at − c ) 4a(y − ct ) + c 2

and the vector potential is given by

A⃗ =

v⃗ V = μoϵov ⃗ V c2

where

(

v ⃗ = 2atryˆ = c −

)

4a(y − ct ) + c 2 yˆ

Therefore,

A⃗ =

(

)

2a c − 4a(y − ct ) + c 2 μoq yˆ 4π 4a(y − ct ) + c 2 + (2at − c ) 4a(y − ct ) + c 2

Problem 4.12. Find the electric and magnetic fields on the y-axis for the point charge moving in a parabolic motion in problem 4.11. Solution 4.12. The motion of this charge is given by

w⃗(t ) = at 2yˆ Starting with the electric field, we have q r [(c 2 − v 2 )u ⃗ + r ⃗ × (u ⃗ × a ⃗ )] E ⃗ (r ⃗ , t ) = ⃗ 4πϵo (r ⋅ u ⃗ )3 with the following quantities

v ⃗ = 2atryˆ a ⃗ = 2ayˆ

r ⃗ = r ⃗ − w⃗(tr ) = yyˆ − atr2yˆ = (y − atr2 )yˆ r = y − atr2 rˆ = yˆ u ⃗ = crˆ − v ⃗ = (c − 2atr )yˆ Now we can see

r ⃗ × (u ⃗ × a ⃗ ) = 0 Because u ⃗ and a ⃗ are in the same direction. Also,

r ⃗ ⋅ u ⃗ = (y − atr2 )(c − 2atr ) Plugging these quantities into our electric field yields

4-24

Electrodynamics

2 q (y − atr ) [(c 2 − 4a 2tr2 )(c − 2atr )yˆ ] 4πϵo (y − atr2 )3(c − 2atr )3 2 q (c 2 − 4a 2tr )yˆ = 4πϵo (y − atr2 )2 (c − 2atr )2

E ⃗ (r ⃗ , t ) =

=

q (c + 2atr )yˆ 4πϵo (y − atr2 )2 (c − 2atr )

Using our tr from problem 4.11,

tr =

4a(y − ct ) + c 2

c−

2a

we have

c + 2atr = 2c −

4a(y − ct ) + c 2

c − 2atr = 4a(y − ct ) + c 2 1 y − atr2 = y − a 2 c 2 − 2c 4a(y − ct ) + c 2 + 4a(y − ct ) + c 2 4a c = ⎡⎣2at − c + 4a(y − ct ) + c 2 ⎤⎦ 2a

(

)

Therefore, our electric field is given by

(

)

(

)

2c − 4a(y − ct ) + c 2 yˆ q E ⃗ (r ⃗ , t ) = ⎤2 4πϵo ⎡ c ⎡ 2⎤ 2 ⎢⎣ ⎣2at − c + 4a(y − ct ) + c ⎦⎥⎦ 4a(y − ct ) + c 2a 4a 2 2c − 4a(y − ct ) + c 2 yˆ q = 4πϵo c 2⎡2at − c + 4a(y − ct ) + c 2 ⎤2 4a(y − ct ) + c 2 ⎣ ⎦ and our magnetic field is given by

B⃗ =

1 r ⃗ × E⃗ = 0 c

since both r ⃗ and E ⃗ are directed along yˆ .

Bibliography Griffiths D J 1999 Introduction to Electrodynamics 3rd edn (Englewood Cliffs, NJ: Prentice-Hall) Griffiths D J 2013 Introduction to Electrodynamics 4th edn (Cambridge, MA: Pearson) Jackson J D 1998 Classical Electrodynamics 3rd edn (New York: Wiley)

4-25

IOP Concise Physics

Electrodynamics Problems and solutions Carolina C Ilie and Zachariah S Schrecengost

Chapter 5 Radiation

Introduction The electromagnetic waves are produced by electric charges in accelerated motion. In this chapter we analyze the radiation, the energy flowing away from the accelerated source to infinity. We discuss the electric dipole radiation in the three approximations, two needed for a perfect dipole (d ≪ r, d ≪ ωc ) and the far field

radiation zone (r ≫ ωc ), all three in one inequality line as (d ≪ ωc ≪ r ). Similarly, we examine the magnetic dipole radiation. We also apply the theory on point charges and analyze the power radiated by a point charge. We finalize with the Larmor formula, and the Liénard formula, and we keep the problems in the non-relativistic region.

5.1 Theory 5.1.1 Power radiated The power radiated from an arbitrary electric source, with electric dipole p ⃗ = pzˆ , is given by

P=

μop2̈ 6πc

and power radiated from an arbitrary current source, with magnetic dipole m⃗ = mzˆ , is given by

P=

doi:10.1088/978-1-6817-4931-0ch5

μom̈ 2 6πc 3

5-1

ª Morgan & Claypool Publishers 2018

Electrodynamics

5.1.2 Electric dipole radiation Given an oscillating electric dipole p ⃗ (t ) = po cos(ωt )zˆ , the resulting electric and magnetic fields are given by

E⃗ = −

⎡ ⎛ μopo ω 2 ⎛ sin θ ⎞ r ⎞⎤ ⎜ ⎟ cos ⎢ω⎜t − ⎟⎥θˆ ⎣ ⎝ 4π ⎝ r ⎠ c ⎠⎦

B⃗ = −

⎡ ⎛ μopo ω 2 ⎛ sin θ ⎞ r ⎞⎤ ⎜ ⎟ cos ⎢ω⎜t − ⎟⎥ϕˆ ⎝ ⎣ 4πc ⎝ r ⎠ c ⎠⎦

and

5.1.3 Magnetic dipole radiation Given an oscillating magnetic dipole m⃗ (t ) = mo cos(ωt ) zˆ , the resulting electric and magnetic fields are given by

E⃗ =

⎡ ⎛ μomoω 2 ⎛ sin θ ⎞ r ⎞⎤ ⎜ ⎟ cos ⎢ω⎜t − ⎟⎥ϕˆ ⎝ ⎣ 4πc ⎝ r ⎠ c ⎠⎦

and

B⃗ = −

⎡ ⎛ μomoω 2 ⎛ sin θ ⎞ r ⎞⎤ ⎜ ⎟ cos ⎢ω⎜t − ⎟⎥θˆ 2 ⎝ ⎣ ⎝ 4πc c ⎠⎦ r ⎠

5.1.4 Larmor formula The total power radiated by a point charge q , moving with acceleration a , is given by

P=

μoq 2a 2 6πc

5.1.5 Radiation fields The electric and magnetic fields associated with radiation from a point charge q are given by r ⃗ = q [r ⃗ × (u ⃗ × a ⃗ )] E rad 4πϵo (r ⃗ · u ⃗ )3 and

⃗ = Brad

1 ⃗ rˆ × E rad c

with u ⃗ = crˆ − v ⃗ 5-2

Electrodynamics

5.2 Problems and solutions Problem 5.1. An insulating circular ring of radius a lies in the xy-plane centered at the origin. The ring has a linear charge density λ(ϕ ) = λ 0 cos ϕ, where λ 0 is a constant and ϕ is the azimuthal angle. The ring is spinning about the z-axis at angular velocity ω. Calculate the total power radiated. Solution 5.1. The total radiated power is ≅

∫ S ⃗ ⋅ da ⃗ =

μ0 p 2̈ 6πc

, where p is the dipole

moment. Let us start with the dipole moment at t = 0. 2π

p(t = 0) = p0 =

∫ λ r ⃗ dl = ∫

(λ 0 cos ϕ)(a cos ϕ xˆ + a sin ϕ yˆ )a dϕ

0

⎛ = λ 0a x ⎜ ⎝ 2⎜ ˆ

2π

∫

2π 2

(cos ϕ) dϕ + yˆ

0

∫ 0

2

⎞ cos ϕ sin ϕ dϕ⎟⎟ ⎠

2

= λ 0a (xˆπ + yˆ 0) = πa λ 0xˆ A rotating electric dipole can be considered as a superposition of two oscillating dipoles, one along the x-axis, and another along the y-axis.

p ⃗ (t ) = p0 [xˆ cos(ωt ) + yˆ sin(ωt )] Now we want to calculate the second derivative of the total electric dipole, in order to find the total radiated power.

p ⃗ ̇ (t ) = p0 [ −ωxˆ sin(ωt ) + ωyˆ cos(ωt )] And the second derivative with respect to time: p ⃗ ̈ (t ) = p [ −ω 2xˆ cos(ωt ) − ω 2yˆ sin(ωt )] 0

= −ω 2p0 [xˆ cos(ωt ) + yˆ sin(ωt )] = −ω 2p ⃗ (t ) Now let us calculate the square of the second derivative of the dipole momentum: 2 [p ⃗ ̈ (t )] = [ −ω 2p ⃗ (t )]2 = ω4[p ⃗ (t )]2 = ω4p02

Note that

[xˆ cos(ωt ) + yˆ sin(ωt )]2 = cos2(ωt ) + sin2(ωt ) = 1 After substituting the value of the dipole momentum at t = 0 2 [p ⃗ ̈ (t )] = ω4π 2λ 02a 4

Therefore, the total radiated power is

P≅

∫ S ⃗ ⋅ da ⃗ =

μ0p2̈ μ ω4πλ 02a 4 = 0 6πc 6c

5-3

Electrodynamics

Problem 5.2. An insulating spherical shell of radius R carries surface charge density σ (θ ) = σo cos θ , where σo is a constant. The shell is sent spinning around the x-axis (ω⃗ = ωxˆ ). Calculate the power radiated.

Solution 5.2. To find the power radiated, we can think of this as a dipole which is given by p ⃗ = po [cos(ωt )zˆ − sin(ωt )yˆ ] and the power radiated is given by

P=

μop2̈ 6πc

Noting

p ⃗ ̈ = − ω 2p ⃗ it follows that p 2̈ =

ω4po2 .

We must now find po from the charge distribution using

po⃗ =

∫ r ⃗′ρ(r ⃗′) dτ′ = ∫ r ⃗′σo cos θ da′

Since we are in spherical coordinates r ⃗ = R [sin θ (cos ϕ xˆ + sin ϕ yˆ ) + cos θ zˆ ] and

da = R2 sin θ dϕ dθ So

2π

π

po⃗ = σo

∫∫ 0

0

⎡ 3⎢ ˆ = σoR x ⎢ ⎣ π

+ zˆ

R3 sin θ cos θ [sin θ (cos ϕ xˆ + sin ϕ yˆ ) + cos θ zˆ ] dϕ dθ π

∫∫ 0 2π

∫∫ 0

2π

0

π 2

sin θ cos θ cos ϕ dϕ dθ + yˆ

0

∫∫ 0

⎤ 2 sin θ cos θ dϕ dθ ⎥ ⎥ ⎦

5-4

2π

0

sin2 θ cos θ sin ϕ dϕ dθ

Electrodynamics

The xˆ and yˆ integrals are both zero and 2π

π

zˆ

∫∫ 0

sin θ cos2 θ dϕ dθ =

0

4π zˆ 3

Therefore,

po⃗ = σo

4π 3 R zˆ 3

and

p2̈ = ω4po2 = ω4σo2

16π 2 6 R 9

The power radiated is then given by

P=

2 μop2̈ μ 16π 2 6 8πμoσo ω4R6 = o ω4σo2 R = 6πc 6πc 9 27c

Problem 5.3. Consider a disk of radius R in the xy-plane, centered at the origin, carrying surface charge σ . The disk is rotation at an angular velocity ωz about the z-axis and ωx about the x-axis. Find the power radiated.

Solution 5.3. To find the power radiated, we can think of this as a dipole which is given by

m⃗ = mo[cos(ωxt )zˆ − sin(ωxt )yˆ ] and the power radiated is given by

P=

μom̈ 2 6πc 3

Noting

m⃗ ̈ = −ωx2m⃗ it follows that m̈ 2 = ωx4m o2 . We must now find mo from the dipole that results from the rotating disk. The magnetic dipole moment is given by

m⃗ o = Ia ⃗ = IπR2zˆ

5-5

Electrodynamics

where I is the current. This is given by

I=

∫ K dl

where K is the surface current density, given by

K⃗ = σωzrϕˆ Using dl = dr , we have R

I = σωz

∫

r dr =

0

1 σωzR2 2

and the dipole is now given by

m⃗ o = IπR2zˆ =

π σωzR 4zˆ 2

It follows that

m̈ 2 = ωx4m o2 =

π2 2 2 4 8 σ ωz ωx R 4

and the power radiated is given by

P=

μom̈ 2 μo π 2 2 2 4 8 μoπσ 2ωz2ωx4R8 = σ ωz ωx R = 6πc 3 6πc 3 4 24c 3

Problem 5.4. Consider an oscillating electric diploe p ⃗ (t ) = po cos(ωt )zˆ . Calculate the momentum flux density of the resulting fields. Do the same for an oscillating magnetic diploe m⃗ (t ) = mo cos(ωt )zˆ . Determine their ratio and comment on the results. Solution 5.4. The electric field due to an oscillating electric dipole is given by

E⃗ = −

μopo ω 2 ⎛ sin θ ⎞ ⎡ ⎛ r ⎞⎤ ⎜ ⎟cos⎢ω⎜t − ⎟⎥θˆ = Ee,dipθˆ 4π ⎝ r ⎠ ⎣ ⎝ c ⎠⎦

where we are using

Ee,dip = −

μopo ω 2 ⎛ sin θ ⎞ ⎡ ⎛ r ⎞⎤ ⎜ ⎟cos⎢ω⎜t − ⎟⎥ 4π ⎝ r ⎠ ⎣ ⎝ c ⎠⎦

to simplify our expressions. The magnetic field due to an oscillation electric dipole is given by

B⃗ = −

Ee,dip μopo ω 2 ⎛ sin θ ⎞ ⎡ ⎛ r ⎞⎤ ⎜ ⎟cos⎢ω⎜t − ⎟⎥ϕˆ = ϕˆ = 4πc ⎝ r ⎠ ⎣ ⎝ c ⎠⎦ c

5-6

μoϵo Ee,dipϕˆ

Electrodynamics

 The elements of the momentum flux density, −T , are given by ⎛ ⎞ ⎞ 1 1⎛ 1 Tij = ϵo⎜EiEj − δijE 2⎟ + ⎜BiBj − δijB 2⎟ ⎝ ⎠ μo ⎝ ⎠ 2 2

We can also convert our electric and magnetic fields to Cartesian coordinates by considering

θˆ = cos θ cos ϕ xˆ + cos θ sin ϕ yˆ − sin θ zˆ and

ϕˆ = −sin ϕ xˆ + cos ϕ yˆ Now to compute the Tij :

1 1 Txx = ϵo(Ex2 − E y2 − E z2 ) + (Bx2 − B y2 − Bz2 ) 2 2μo 1 2 = ϵoE e,dip (cos2 θ cos2 ϕ − cos2 θ sin2 ϕ − sin2 θ ) 2 1 2 + μoϵoE e,dip (sin2 ϕ − cos2 ϕ) 2μo 1 2 [cos2 θ (cos2 ϕ − sin2 ϕ) − sin2 θ − (cos2 ϕ − sin2 ϕ)] = ϵoE e,dip 2 1 = Te[(cos2 θ − 1)(cos2 ϕ − sin2 ϕ) − sin2 θ ] = −Tesin2 θ cos2 ϕ 2 where we have used 2 Te = ϵoE e,dip

Txy = ϵoExEy +

1 BxBy μO

2 = ϵoE e,dip (cos2 θ cos ϕ sin ϕ) +

1 2 ( −cos ϕ sin ϕ) μ ϵoE e,dip μo o

= Te(cos2 θ − 1)(cos ϕ sin ϕ) = −Tesin2 θ cos ϕ sin ϕ 1 2 BxBz = ϵoE e,dip (cos θ cos ϕ)( −sin θ ) μO = − Te cos θ sin θ cos ϕ

Txz = ϵoExEz +

Tyx = Txy = −Tesin2 θ cos ϕ sin ϕ

5-7

Electrodynamics

1 1 Tyy = ϵo(E y2 − Ex2 − E z2 ) + (B y2 − Bx2 − Bz2 ) 2 2μo 1 2 = ϵoE e,dip (cos2 θ sin2 ϕ − cos2 θ cos2 ϕ − sin2 θ ) 2 1 2 + μ ϵoE e,dip (cos2 ϕ − sin2 ϕ) 2μo o 1 = Te[(cos2 θ − 1)(sin2 ϕ − cos2 ϕ) − sin2 θ ] = −Tesin2 θ sin2 ϕ 2 1 ByBz = ϵoE 2e,dip(cos θ sin ϕ)( −sin θ ) μO = − Te cos θ sin θ sin ϕ

Tyz = ϵoEyEz +

Tzx = Txz = −Te cos θ sin θ cos ϕ Tzy = Tyz = −Te cos θ sin θ sin ϕ

1 1 Tzz = ϵo(E z2 − Ex2 − E y2 ) + Bz2 − Bx2 − B y2 ) ( 2 2μo 1 2 [sin2 θ − cos2 θ (cos2 ϕ + sin2 ϕ)] = ϵoE e,dip 2 1 2 − μ ϵoE e,dip (sin2 ϕ + cos2 ϕ) 2μo o 1 = Te(sin2 θ − cos2 θ − 1) = −Te cos2 θ 2 Therefore,

⎛ sin2 θ cos2 ϕ sin2 θ cos ϕ sin ϕ cos θ sin θ cos ϕ ⎞ ⎜ ⎟  − Te = Te⎜sin2 θ cos ϕ sin ϕ sin2 θ sin2 ϕ cos θ sin θ sin ϕ ⎟ ⎜⎜ ⎟⎟ cos2 θ ⎝ cos θ sin θ cos ϕ cos θ sin θ sin ϕ ⎠ Now to compute the momentum flux density for an oscillating magnetic dipole, which generates electric field

μomoω 2 ⎛ sin θ ⎞ ⎡ ⎛ r ⎞⎤ ⎜ ⎟cos⎢ω⎜t − ⎟⎥ϕˆ 4πc ⎝ r ⎠ ⎣ ⎝ c ⎠⎦ ⎛m ⎞ = −Ee,dip⎜ o ⎟ϕˆ = E m,dipϕˆ ⎝ po c ⎠

E⃗ =

5-8

Electrodynamics

and magnetic field

μomoω 2 ⎛ sin θ ⎞ ⎡ ⎛ r ⎟⎞⎤ ˆ ⎜t − ⎜ ⎟ cos ω ⎢ ⎥θ 4πc 2 ⎝ r ⎠ ⎣ ⎝ c ⎠⎦ ⎛m ⎞ E m,dip = − Ee,dip⎜ o2 ⎟θˆ = − θˆ = − μoϵo E m,dipθˆ c ⎝ po c ⎠

B⃗ = −

where

⎛m ⎞ E m,dip = −Ee,dip⎜ o ⎟ ⎝ po c ⎠ Now to compute the Tij :

1 1 Txx = ϵo(Ex2 − E y2 − E z2 ) + (Bx2 − B y2 − Bz2 ) 2 2μo 1 2 = ϵoE m,dip (sin2 ϕ − cos2 ϕ) 2 1 2 + μ ϵoE m,dip (cos2 θ cos2 ϕ − cos2 θ sin2 ϕ − sin2 θ ) 2μo o 1 2 [(sin2 ϕ − cos2 ϕ)(1 − cos2 θ ) − sin2 θ ] = ϵoE m,dip 2 1 = Tm(sin2 ϕ − cos2 ϕ − 1) sin2 θ = −Tmsin2 θ cos2 ϕ 2 where we have used 2 Tm = ϵoE m,dip

Txy = ϵoExEy +

1 BxBy μO

2 ( −sin ϕ)(cos ϕ) + = ϵoE m,dip

1 2 μoϵoE m,dip (cos2 θ cos ϕ sin ϕ) μO

= − Tm cos ϕ sin ϕ (1 − cos2 θ ) = −Tmsin2 θ cos ϕ sinϕ 1 1 2 BxBz = μoϵoE m,dip (cos θ cos ϕ)( −sin θ ) μO μO = − Tm cos θ sin θ cos ϕ

Txz = ϵoExEz +

5-9

Electrodynamics

Tyx = Txy = −Tmsin2 θ cos ϕ sin ϕ

1 1 Tyy = ϵo(E y2 − Ex2 − E z2 ) + (B y2 − Bx2 − Bz2 ) 2 2μo 1 2 = ϵoE m,dip (cos2 ϕ − sin2 ϕ) 2 +

1 2 μ ϵoE m,dip (cos2 θ sin2 ϕ − cos2 θ cos2 ϕ − sin2 θ ) 2μo o

1 = Tm[(cos2 ϕ − sin2 ϕ)(1 − cos2 θ ) − sin2 θ ] 2 1 = Tm(cos2 ϕ − sin2 ϕ − 1)sin2 θ = −Tmsin2 θ sin2 ϕ 2 1 1 2 ByBz = μoϵoE m,dip (cos θ sin ϕ)( −sin θ ) μO μo = −Tm cos θ sin θ sin ϕ

Tyz = ϵoEyEz +

Tzx = Txz = −Tm cos θ sin θ cos ϕ Tzy = Tyz = −Tm cos θ sin θ sin ϕ

1 1 ϵo E z2 − Ex2 − E y2 + Bz2 − Bx2 − B y2 2 2μo 1 1 2 2 = − ϵoE m,dip μ ϵoE m,dip [sin2 θ − cos2 θ(cos2 ϕ + sin2 ϕ)] (sin2 ϕ + cos2 ϕ) + 2 2μo o 1 = − Tm(1 − sin2 θ − cos2 θ ) = − Tmcos2 θ 2

Tzz =

(

)

(

)

Therefore,

⎛ sin2 θ cos2 ϕ sin2 θ cos ϕ sin ϕ cos θ sin θ cos ϕ ⎞ ⎜ ⎟  − Tm = Tm⎜sin2 θ cos ϕ sin ϕ sin2 θ sin2 ϕ cos θ sin θ sin ϕ ⎟ ⎜⎜ ⎟⎟ cos2 θ ⎝ cos θ sin θ cos ϕ cos θ sin θ sin ϕ ⎠

5-10

Electrodynamics

Now if we look at the ratio

⎛ sin2 θ cos2 ϕ ⎜ Tm⎜sin2 θ cos ϕ sin ϕ  ⎜ ⎝ cos θ sin θ cos ϕ Tm −  = ⎛ −Te sin2 θ cos2 ϕ ⎜ 2 Te⎜sin θ cos ϕ sin ϕ ⎜ ⎝ cos θ sin θ cos ϕ

sin2 θ cos ϕ sin ϕ cos θ sin θ cos ϕ ⎞ ⎟ sin2 θ sin2 ϕ cos θ sin θ sin ϕ ⎟ ⎟ cos θ sin θ sin ϕ cos2 θ ⎠ 2 ⎞ sin θ cos ϕ sin ϕ cos θ sin θ cos ϕ ⎟ sin2 θ sin2 ϕ cos θ sin θ sin ϕ ⎟ ⎟ cos θ sin θ sin ϕ cos2 θ ⎠

2 ϵoE m,dip Tm = = 2 Te ϵoE e,dip

=

⎛ mo ⎞2 2 E e,dip ⎜ ⎟ ⎝ po c ⎠ 2 E e,dip

⎛ mo ⎞2 =⎜ ⎟ ⎝ po c ⎠

which is precisely the ratio of the powers radiated 2 Pm ⎛ mo ⎞ =⎜ ⎟ Pe ⎝ po c ⎠

Problem 5.5. In chapter 1, we proved that Maxwell’s equations are invariant to the following duality transformation

E ⃗′ = E ⃗ cos α − Bc⃗ sinα cB′⃗ = E ⃗sin α + Bc⃗ cos α cqe′ = c qe cos α − qm sin α qm′ = c qesin α + qm cos α Use the oscillating electric dipole

E ⃗ = −∇V −

μ p ω 2 ⎛ sin θ ⎞ ⎡ ⎛ r ⎞⎤ ∂A ⃗ ⎟cos⎢ω⎜t − ⎟⎥θˆ =− o 0 ⎜ 4π ⎝ r ⎠ ⎣ ⎝ c ⎠⎦ ∂t

B ⃗ = ∇ × A⃗ = −

μop0 ω 2 ⎛ sin θ ⎞ ⎡ ⎛ r ⎞⎤ ⎜ ⎟cos⎢ω⎜t − ⎟⎥ϕˆ 4πc ⎝ r ⎠ ⎣ ⎝ c ⎠⎦

to determine the fields produced by an oscillating magnetic dipole formed by equal and opposite magnetic charges (not by an alternating current carrying loop).

5-11

Electrodynamics

Solution 5.5. We start with the duality transformation

E ⃗′ = E ⃗ cos α − Bc⃗ sin α cB′⃗ = E ⃗sin α + Bc⃗ cos α cqe′ = cqe cos α − qmsin α qm′ = cqe sin α + qm cos α Let us set α = 90° and obtain from duality transformation:

E ⃗′ = − Bc⃗ E⃗ B′⃗ = c q qe′ = − m c qm′ = cqe The maximum value of the magnetic dipole moment m 0

m 0 = qm′ d = cqed = cp0 where p0 is the maximum value of the dipole moment. Let us calculate the fields E ⃗′ and B′⃗.

⎧ μ p ω 2 ⎛ sin θ ⎞ ⎡ ⎛ r ⎞⎤ ⎫ ⎟cos⎢ω⎜t − ⎟⎥ϕˆ ⎬ E ⃗′ = −Bc⃗ = −c⎨ − o 0 ⎜ 4πc ⎝ r ⎠ ⎣ ⎝ c ⎠⎦ ⎭ ⎩ =

B′⃗ =

μo(cp0 )ω 2 ⎛ sin θ ⎞ ⎡ ⎛ μ m 0ω 2 ⎛ sin θ ⎞ ⎡ ⎛ r ⎞⎤ r ⎞⎤ ⎜ ⎟cos⎢ω⎜t − ⎟⎥ϕˆ = o ⎜ ⎟cos⎢ω⎜t − ⎟⎥ϕˆ 4πc ⎝ r ⎠ ⎣ ⎝ c ⎠⎦ c ⎠⎦ 4πc ⎝ r ⎠ ⎣ ⎝

μ p ω 2 ⎛ sin θ ⎞ ⎡ ⎛ μ m 0ω 2 ⎛ sin θ ⎞ ⎡ ⎛ r ⎞⎤ r ⎞⎤ E⃗ ⎟cos⎢ω⎜t − ⎟⎥θˆ = − o 2 ⎜ ⎟cos⎢ω⎜t − ⎟⎥θˆ =− o 0 ⎜ ⎝ ⎠ ⎝ 4πc ⎝ r ⎠ ⎣ c ⎠⎦ c ⎦ 4πc ⎝ r ⎠ ⎣ c

We notice that E ⃗′ and B′⃗ are identical to the fields determined by Ampère dipole, a wire loop of radius a around which we drive an alternating current I (t ) = I0 cos ωt , which is a model for an oscillating magnetic dipole m⃗ (t ) = πa 2I (t )zˆ. Note that both models determine identical fields in the radiation zone. Moreover, the fields are the same everywhere, except at the position of the dipole. Problem 5.6. Now let us start with the Ampèrian dipole fields:

μom 0ω 2 ⎛ sin θ ⎞ ⎡ ⎛ r ⎞⎤ ⎜ ⎟cos⎢ω⎜t − ⎟⎥ϕˆ ⎝ 4πc ⎝ r ⎠ ⎣ c ⎠⎦ μ m 0ω 2 ⎛ sin θ ⎞ ⎡ ⎛ r ⎞⎤ ⎟cos⎢ω⎜t − ⎟⎥θˆ B⃗ = − o 2 ⎜ ⎝ ⎝ ⎠ ⎣ c ⎠⎦ 4πc r E⃗ =

5-12

Electrodynamics

And by using the duality transformation

E ⃗′ = E ⃗ cos α − Bc⃗ sin α cB′⃗ = E ⃗sin α + Bc⃗ cos α cqe′ = cqe cos α − qmsin α qm′ = cqesin α + qm cos α obtain the field produced by an oscillating electric dipole. What do you expect? Solution 5.6. We would like to see if, by starting with the Ampèrian dipole fields, we will obtain the electric dipole fields after employing the duality transformation. Again, for α = 90° we obtain from duality transformation:

E ⃗′ = −Bc⃗ E⃗ B′⃗ = c q qe′ = − m c qm′ = cqe The maximum value of the dipole moment p0

p0 = qe′d = −

E ⃗′ = −Bc⃗ =

qmd m =− 0 c c

μom 0ω 2 ⎛ sin θ ⎞ ⎡ ⎛ r ⎞⎤ ⎜ ⎟cos⎢ω⎜t − ⎟⎥θˆ 4πc ⎝ r ⎠ ⎣ ⎝ c ⎠⎦

m0 2 ω ⎛ sin θ ⎞ ⎡ ⎜⎛ r ⎞⎤ c ⎜ ⎟cos⎢ω t − ⎟⎥θˆ = 4π ⎝ r ⎠ ⎣ ⎝ c ⎠⎦ 2⎛ μ p ω sin θ ⎞ ⎡ ⎛ r ⎞⎤ ⎟cos⎢ω⎜t − ⎟⎥θˆ =− o 0 ⎜ c ⎠⎦ 4π ⎝ r ⎠ ⎣ ⎝ μo

This is exactly the field due to oscillating electric dipole. Now let us calculate the magnetic field

μ m 0ω 2 ⎛ sin θ ⎞ ⎡ ⎛ r ⎞⎤ E⃗ ⎟cos⎢ω⎜t − ⎟⎥ϕˆ = o 2 ⎜ 4πc ⎝ r ⎠ ⎣ ⎝ c ⎠⎦ c 2⎛ μ p cω sin θ ⎞ ⎡ ⎛ r ⎞⎤ ⎟cos⎢ω⎜t − ⎟⎥ϕˆ =− o 0 2 ⎜ ⎝ 4πc ⎝ r ⎠ ⎣ c ⎠⎦

B′⃗ =

=−

μop0 ω 2 ⎛ sin θ ⎞ ⎡ ⎛ r ⎞⎤ ⎜ ⎟cos⎢ω⎜t − ⎟⎥ϕˆ ⎝ c ⎠⎦ 4πc ⎝ r ⎠ ⎣

5-13

Electrodynamics

As expected, the fields are exactly the ones due to an oscillating electric dipole. Again, in the radiation zone, the fields due to an Ampèrian dipole and the fields due to an electric dipole are the same. Problem 5.7. Consider the same duality transformation used in problem 5.5. Obtain the electric and magnetic fields of a moving magnetic monopole qm and find an equivalent of the Larmor formula for the radiated power.

E ⃗′ = E ⃗ cos α − Bc⃗ sin α cB′⃗ = E ⃗sin α + Bc⃗ cos α cqe′ = cqe cos α − qm sin α qm′ = cqesin α + qm cos α Solution 5.7. As before, with α = 90° we obtain the following:

E ⃗′ = −Bc⃗ B′⃗ =

E⃗ c

qe′ = −

qm c

qm′ = cqe The electric and magnetic fields associated with radiation from a point charge q are given by

⃗ (r ⃗ , t ) = E rad

q r [(c 2 − v 2 )u ⃗ + r ⃗ × (u ⃗ × a ⃗ )] 4πϵo (r ⃗ · u ⃗ )3

and

1 ⃗ (r ⃗ , t ) rˆ × E rad c

⃗ (r ⃗ , t ) = Brad

The total power radiated by a point charge q , moving with acceleration a , is given by Larmor formula:

P=

μoq 2a 2 6πc

5-14

Electrodynamics

Now back to the fields obtained from the duality transformation

⎡1 ⎤ E ⃗′ = −Bc⃗ = −c⎢ rˆ × E ⃗ ⎥ = −rˆ × E ⃗ = −rˆ × (cB′⃗) = −c(rˆ × B′⃗) ⎣c ⎦ ⎫ r 1⎧ q E⃗ [(c 2 − v 2 )u ⃗ + r ⃗ × (u ⃗ × a ⃗ )]⎬ B′⃗ = = ⎨ e 3 c c ⎩ 4πϵo (r ⃗ · u ⃗ ) ⎭ ⎧ ⎛ q′ ⎞ ⎫ ⎪⎜ m ⎟ ⎪ ⎪ 1⎪⎝ c ⎠ r [(c 2 − v 2 )u ⃗ + r ⃗ × (u ⃗ × a ⃗ )]⎬ = ⎨ 3 c ⎪ 4πϵo (r ⃗ · u ⃗ ) ⎪ ⎪ ⎪ ⎩ ⎭ μoqm′ r [(c 2 − v 2 )u ⃗ + r ⃗ × (u ⃗ × a ⃗ )] = 4π (r ⃗ · u ⃗ )3 The equivalent of the Larmor formula for the power radiated is

P=

μoqe2a 2 6πc

=

μoa 2 (q ′ )2 6πc 3 m

Therefore, the equivalent magnetic and electric fields and power radiated are (no primes needed): μq r [(c 2 − v 2 )u ⃗ + r ⃗ × (u ⃗ × a ⃗ )] B⃗ = o m 4π (r ⃗ · u ⃗ )3

E ⃗ = −c(rˆ × B ⃗ ) with u ⃗ = crˆ − v ⃗

P=

μoa 2 2 q 6πc 3 m

Problem 5.8. An electron starts from rest and moves through a potential difference of −Vo with acceleration a . If the electron travels a distance d , how much energy is required? Solution 5.8. This movement of the electron involves energy that can be broken into three pieces: energy associated with moving a charged particle though a potential difference,

Upot = qV kinetic energy,

Ukin =

1 2 mv 2

5-15

Electrodynamics

and energy to account for radiation losses,

Urad = Pradt We will consider v ≪ c so the Larmor formula can be used.

Prad =

μoq 2a 2 6πc

The initial kinetic energy is zero and the final is given by

Ukin =

1 1 me v 2 = me (2ad ) = me ad 2 2

where we use

v 2 = vo2 + 2ad with vo = 0. For the potential energy, q = −e and V = − Vo , so

Upot = ( −e )( −Vo) = eVo For the radiation energy, we have

Prad =

μoq 2a 2 μ e 2a 2 = o 6πc 6πc

and considering,

a=

v v ⇒t= = a t

2ad = a

2d a

we have

Urad = Pradt =

μoe 2a 2 2d μ e 2a 3/2 d = o 6πc a 3πc 2

The total energy is then given by

U = Ukin + Upot + Urad = me ad + eVo +

μoe 2a 3/2 d 3πc 2

⎛ μ e2 a ⎞ = ad ⎜me + o ⎟ + eVo 3πc 2d ⎠ ⎝ We can see, even without specific values for a , d , and Vo, that the second term in parenthesis contributes very little to the energy; since me ≈ 9.11 × 10−31 kg and μo e 2 3πc

≈ 5.7 × 10−54 kg s. Therefore, the energy loss due to radiation is negligible and we have

U = adme + eVo which is simply the kinetic and potential energy terms.

5-16

Electrodynamics

Problem 5.9. Consider a particle of charge q , rotating around the z-axis, in the xy-plane, with angular velocity ω and at radius R . Find the electric and magnetic fields due to radiation at point P.

Solution 5.9. The fields due to radiation are given by

⃗ = E rad

q r [r ⃗ × (u ⃗ × a ⃗ )] 4πϵo (r ⃗ · u ⃗ )3

and

⃗ = Brad

1 ⃗ rˆ × E rad c

Starting with the electric field, we will start by finding expressions for r ⃗ , u ⃗ , and a ⃗ . The position of the charge can be expressed as

r ⃗′ = R[cos(ωt )xˆ + sin(ωt )yˆ ] and the position of point P is given by

r ⃗ = zzˆ so

r ⃗ = r ⃗ − r ⃗′ = zzˆ − R[cos(ωt )xˆ + sin(ωt )yˆ ] We also have

u ⃗ = crˆ − v ⃗ with the velocity given by

v ⃗ = r ′⃗ ̇ = Rω[ −sin(ωt )xˆ + cos(ωt )yˆ ] and

rˆ =

r⃗ r

5-17

Electrodynamics

Therefore,

u ⃗ = crˆ − v ⃗ =

c r ⃗ − v⃗ r

c (zzˆ − R[cos(ωt )xˆ + sin(ωt )yˆ ]) − Rω[ −sin(ωt )xˆ + cos(ωt )yˆ ] r ⎛ ⎞ ⎛ ⎞ cz c c = zˆ + R⎜ω sin(ωt ) − cos(ωt )⎟xˆ − R⎜ω cos(ωt ) + sin(ωt )⎟yˆ ⎝ ⎠ ⎝ ⎠ r r r =

Finally, the acceleration is given

a ⃗ = v ⃗ ̇ = −ω 2R[cos(ωt )xˆ + sin(ωt )yˆ ] = −ω 2r ⃗′ ⃗ . Starting with r ⃗ · u ⃗ , we have Now we can start computing the pieces of Erad

r ⃗ · u ⃗ = r ⃗ · (crˆ − v ⃗ ) = cr − r ⃗ · v ⃗ = cr − [zzˆ − R[cos(ωt )xˆ + sin(ωt )yˆ ]] · [Rω[ − sin(ωt )xˆ + cos(ωt )yˆ ]] = cr + R2ω[ −cos(ωt )sin(ωt ) + cos(ωt )sin(ωt )] = cr Next, we will express the triple cross product as

r ⃗ × ( u ⃗ × a ⃗ ) = u ⃗ ( r ⃗ · a ⃗ ) − a ⃗( r ⃗ · u ⃗ ) where we need to find r ⃗ · a ⃗ ,

r ⃗ · a ⃗ = (r ⃗ − r ⃗′) · ( −ω 2r ⃗′) = ω 2(r′)2 = ω 2R2 Therefore,

⎡ cz ⎛ ⎞ c r ⃗ × (u ⃗ × a ⃗ ) = u ω ⃗ 2R2 − ac⃗ r = ω 2R2⎢ zˆ + R⎜ω sin(ωt ) − cos(ωt )⎟xˆ ⎝ ⎠ ⎣r r ⎛ ⎞ ⎤ c − R⎜ω cos(ωt ) + sin(ωt )⎟yˆ ⎥ − cr [ −ω 2R[cos(ωt )xˆ + sin(ωt )yˆ ]] ⎝ ⎠ ⎦ r ⎡ czR ⎡ ⎛ cR2 ⎞⎤ zˆ + ⎢ωR2sin(ωt ) − cos(ωt )⎜ = ω 2R⎢ − cr ⎟⎥xˆ ⎢⎣ r ⎝ r ⎠⎦ ⎣ ⎡ ⎛ cR2 ⎞⎤ ⎤ − ⎢ωR2 cos(ωt ) + sin(ωt )⎜ − cr ⎟⎥yˆ ⎥ ⎝ r ⎠⎦ ⎥⎦ ⎣ =

ω 2R [czRzˆ + [ωR2r sin(ωt ) − c cos(ωt )(R2 − r 2 )]xˆ r − [ωR2r cos(ωt ) + c sin(ωt )(R2 − r 2 )]yˆ ]

Noting

R 2 − r 2 = R 2 − (z 2 + R 2 ) = − z 2

5-18

Electrodynamics

we have

r ⃗ × (u ⃗ × a ⃗ ) =

ω 2R [czRzˆ + (cz 2 cos(ωt ) + ωR2r sin(ωt ))xˆ r + (cz 2sin(ωt ) − ωR2r cos(ωt ))yˆ ]

Therefore, our electric field is given by

q r [r ⃗ × (u ⃗ × a ⃗ )] 4πϵo (r ⃗ · u ⃗ )3 q r ω 2R [czRzˆ + (cz 2 cos(ωt ) + ωR2r sin(ωt ))xˆ = 4πϵo r 3c 3 r + (cz 2sin(ωt ) − ωR2r cos(ωt ))yˆ ]

⃗ = E rad

=

q ω 2R [czRzˆ + (cz 2 cos(ωt ) + ωR2r sin(ωt ))xˆ 4πϵo r 3c 3 + (cz 2sin(ωt ) − ωR2r cos(ωt ))yˆ ]

Now we can compute the magnetic field

⃗ = 1 rˆ × E rad ⃗ = 1 r ⃗ × E rad ⃗ Brad c cr 1 q ω 2R = cr 4πϵo r 3c 3 =

xˆ yˆ zˆ z −R cos(ωt ) −R sin(ωt ) 2 2 2 2 cz cos(ωt ) + ωR r sin(ωt ) cz sin(ωt ) − ωR r cos(ωt ) czR

q ω 2R [( −czR2 sin(ωt ) − cz 3 sin(ωt ) + ωR2r z cos(ωt ))xˆ 4πϵo r 4c 4 + (cz 3cos(ωt ) + ωR2rz sin(ωt ) + czR2 cos(ωt ))yˆ + [( −R cos(ωt ))(cz 2 sin(ωt ) − ωR2r cos(ωt )) + (R sin(ωt ))(cz 2 cos(ωt ) + ωR2r sin(ωt ))]zˆ ]

=

q ω 2R [(ωR2rz cos(ωt ) − czsin(ωt )(R2 + z 2 ))xˆ 4πϵo r 4c 4 + (ωR2rzsin(ωt ) + cz cos(ωt )(z 2 + R2 ))yˆ + ( −Rcz 2 cos(ωt )sin(ωt ) + ωR3r cos2(ωt ) + Rcz 2 cos(ωt )sin(ωt ) + ωR3r sin2(ωt ))zˆ ]

=

q ω 2R r [z(ωR2 cos(ωt ) − cr sin(ωt ))xˆ + z(ωR2sin(ωt ) 4πϵo r 4c 4 + cr cos(ωt ))yˆ + (ωR3cos2(ωt ) + ωR3sin2(ωt ))zˆ ]

5-19

Electrodynamics

Therefore, the magnetic field is given by

⃗ = Brad

q ω 2R [z(ωR2 cos(ωt ) − cr sin(ωt ))xˆ 4πϵo r 3c 4 + z(ωR2sin(ωt ) + cr cos(ωt ))yˆ + ωR3zˆ ]

Problem 5.10. One of the limitations of Rutherford–Bohr atomic model is that an electron travelling on circular path emits radiation, therefore falling in a spiral onto the nucleus (the original term in German was ‘bremsstrahlung’, which means breaking, or deceleration radiation). Let us consider the speed of the electron as much smaller than the speed of light, such that we could use the non-relativistic approach. By using Larmor’s formula, calculate the lifespan of the electron before falling on the nucleus. Consider the radius r = 10−10 m, the mass of the electron is m = 9.11 × 10−31 kg, the electron’s charge in absolute value is e = 1.6 × 10−19 C. Solution 5.10. The Coulomb force between the electron and the proton is equal to the centripetal force acting on the electron which moves on circular trajectory around the nucleus.

1 q2 4πϵo r 2

F=

Fcp = ma = m

v2 r

From here we obtain the speed as

q2 4πϵomr

v=

Now let us check if the numerical values will ensure a speed which is small enough compared to the speed of light in order to allow a non-relativistic treatment.

v=

q2 = 4πϵomr

(1.6 × 10−19 C)2 ⎛ C2 ⎞ 4π × ⎜8.85 × 10−12 ⎟ × (9.11 × 10−31 kg) × (10−10 m) ⎝ N ⋅ m2 ⎠

= 1.59 × 106 m s−1 When we compare to the speed of light

v 1.59 × 106 m s−1 = 5.3 × 10−3 ≪ 1 = c 3 × 108 m s−1 In the Larmor formula, with

a=

v2 r

5-20

Electrodynamics

⎛ 2 ⎞2 2 v μ q ⎟ o ⎜ ⎝r ⎠ μoq 2a 2 μ q 2 ⎛ q 2 ⎞2 = = o ⎜ P= ⎟ 6πc 6πc 6πc ⎝ 4πϵomr 2 ⎠ We want to find the time necessary for the electron to fall on the nucleus, recall the relationship between the total energy of the electron U (kinetic and potential), and power

dU d ⎛ mv 2 q2 ⎞ =− ⎜ − ⎟ dt dt ⎝ 2 4πϵor ⎠ d ⎡ m ⎛ q2 ⎞ q2 ⎤ d ⎛ q2 ⎞ ⎥= ⎜ =− ⎢ ⎜ ⎟− ⎟ dt ⎣ 2 ⎝ 4πϵomr ⎠ 4πϵor ⎦ dt ⎝ 8πϵor ⎠

P =−

=−

q 2 dr 8πϵor 2 dt

Back to the Larmor formula, and we obtain

μoq 2 ⎛ q 2 ⎞2 q 2 dr = − ⎜ ⎟ 6πc ⎝ 4πϵomr 2 ⎠ 8πϵor 2 dt We use that c 2 =

1 ϵoμo 2 q2 ⎛ q2 ⎞ q 2 dr =− ⎜ ⎟ 3⎝ 2⎠ 6πϵoc 4πϵomr 8πϵor 2 dt

Simplify

⎛ q2 ⎞ q2 1 dr ⎜ ⎟=− 3 2⎝ 6πϵoc m 4πϵor 2 ⎠ 2 dt So 2 dr 1 ⎛ q2 ⎞ 1 =− ⎜ ⎟ 2 dt 3c ⎝ 2πϵomc ⎠ r

We integrate on both sides 0

∫ r0

2 1 ⎛ q2 ⎞ r dr = − ⎜ ⎟ 3c ⎝ 2πϵomc ⎠ 2

τ

∫

dt

0

We obtain the time necessary for the electron to fall on the nucleus,

τ=

⎛ ⎞2 3 2πϵomc cr0 ⎜ ⎟ 2 ⎝

q

5-21

⎠

Electrodynamics

Back to the numerical values,

τ = (3 × 108 m s−1)3 ⎛ ⎞2 C2 −31 9.11 10 kg × × ⎜ 2π × 8.85 × 10−12 ⎟ N ⋅ m2 ⎟ × (10−10 m)3⎜ (1.6 × 10−19 C)2 ⎜⎜ ⎟⎟ ⎝ ⎠ = 1.06 × 10−10 s Note that this is a relatively short time. Regarding the limitation of the Rutherford–Bohr atomic model, Sommerfeld worked on improving it by suggesting elliptical trajectories for the electron (as in the solar systems), and after that the model was replaced by Pauli’s quantum mechanical treatment in 1925. Now we still use the hydrogen atomic model of atomic orbitals, developed by Erwin Schrödinger in 1926. Problem 5.11. A particle of mass m and charge q oscillate in vacuum along the x-axis as x = Asin ωt . Obtain: (a) the average radiated power; (b) the time τ needed for the amplitude to decrease by a factor of e; (c) the number of complete oscillations of the particle during time τ . Apply for the following numerical values: m = 9.11 × 10−31 kg , q = 1.6 × 10−19 C, A = 10−10 m , ω = 6 × 10−14 rad s−1 Solution 5.11. (a) Starting with the Larmor formula:

P=

μoq 2a 2 6πc

we want to find the acceleration. The position is

x(t ) = Asin ωt The speed is

v(t ) = x(̇ t ) = ωAcos ωt The acceleration is

a(t ) = v(̇ t ) = x(̈ t ) = −ω 2Asin ωt Back to the radiated power, with t ⁎ = t −

P=

r c

μoq 2a 2 μ q 2[ −ω 2A(sinωt ⁎)]2 μ q 2ω4A2 (sinωt ⁎)2 = o = o 6πc 6πc 6πc

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Electrodynamics

The average radiated power in one period is, considering that 〈(sin ωt ⁎)2 〉 =

P =

μoq 2ω4A2 (sin ωt ⁎)2 6πc

=

μoq 2ω4A2 12πc

Back to numerical values,

P =

μoq 2ω4A2 12πc

N × (1.6 × 10−19 C)2 × (6 × 1014 rad s −1)4(10−10 m)2 2 A = 12π × 3 × 108 m s−1 = 1.8 × 10−15 W 4π × 10−7

Now we need to relate the power to time, by the formula:

P =−

dU dt

With U = Ukin + Upot

Ukin =

Upot =

mv 2 mω 2A2 (cos ωt )2 = 2 2

kx 2 kA2 (sin ωt )2 mω 2A2 (sin ωt )2 = = 2 2 2

The total energy is

mω 2A2 (cos ωt )2 mω 2A2 (sin ωt )2 + 2 2 mω 2A2 [(cos ωt )2 + (sin ωt )2 ] mω 2A2 = = 2 2

U = Ukin + Upot =

P =−

dU d ⎛ mω 2A2 ⎞ dA =− ⎜ ⎟ = −mω 2A dt dt ⎝ 2 ⎠ dt

Therefore,

μoq 2ω4A2 dA = −mω 2A 12πc dt

dt = −

12πmc dA μoq 2ω 2 A

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1 2

Electrodynamics

We integrate on both sides: τ

∫ 0

τ=−

12πmc dt = − μoq 2ω 2

A e

∫ A

dA′ A′

⎞ 12πmc A 12πmc ⎛ A 12πmc 12πmc ⎜ln − lnA⎟ = ln = lne = 2 2⎝ 2 2 2 2 A ⎠ μoq ω e μoq ω μoq ω μoq 2ω 2 e

Back to the numerical values,

12πmc 12π × (9.11 × 10−31 kg) × (3 × 108 m s−1) = N μoq 2ω 2 4π × 10−7 2 × (1.6 × 10−19 C)2 (6 × 1014 rad s−1)2 A = 8.8 × 10−7 s

τ=

c) The number of rotations is obtained from

n=

τ τω (8.8 × 10−7 s) × (6 × 1014 rad s−1) = = = 8.4 × 107 T 2π 2π

Bibliography Griffiths D J 1999 Introduction to Electrodynamics 3rd edn (Englewood Cliffs, NJ: Prentice-Hall) Griffiths D 2013 Introduction to Electrodynamics 4th edn (Cambridge, MA: Pearson) Heras J A 1995 Am. J. Phys. 63 242 Jackson J D 1998 Classical Electrodynamics 3rd edn (New York: Wiley) Popescu I M, Iordache D, Fara V, Stan M and Lupascu A 1986 Probleme Rezolvate de Fizica: Electromagnetism. Teoria Relativitatii Restranse. Unde Electromagnetice. Teoria Electromagnetica a Luminii (Optica) (Solved Physics Problems: Electromagnetism. Relativistic Theory. Electromagnetic Waves. The Electromagnetic Theory of Light (Optics) vol II (Editura Tehnica)

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