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Chapter 1

Maxwell’s equations—Electrodynamics

IntroductionIn this chapter we will introduce the electromotive force, Ohm’s law, and Faraday’slaw 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, wepresent a problem involving the concept of magnetic charge; we uncover its role in aparticular duality transformation and its relationship to the invariance of Maxwell’sequations.

1.1 Theory1.1.1 Ohm’s law

Given conductivity σ , the current density J is given by

σ = J E

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 isgiven by

= =PWt

I Rdd

2

doi:10.1088/978-1-6817-4931-0ch1 1-1 ª Morgan & Claypool Publishers 2018

1.1.3 Flux rule for motional electromotive force

Given a magnetic flux Φ, the electromotive force (emf) generated is

E = − Φt

dd

where the flux is given by

∫Φ = · B ad

1.1.4 Magnetic energy

The energy stored in a magnetic field B is given by

∫μτ=W B

12

do all space

2

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

=W LI12

2

1.1.5 Maxwell’s equations

Gauss’s LawDifferential Form

ρϵ

∇ · =Eo

Integral Form

S∮ ϵ

· =E aq

do

enc

No NameDifferential Form

∇ · =B 0

Integral Form

S∮ · =B ad 0

Faraday’s LawDifferential Form

∇ × = − ∂ ∂

EBt

Electrodynamics

1-2

Integral Form

lP

∮ · = − ΦE

td

dd

Ampère–Maxwell LawDifferential Form

μ μ ϵ∇ × = + ∂ ∂

B JEto o o

Integral Form

lP S

∮ ∮μ μ ϵ · = + ∂ ∂

· B IEt

ad do o oenc

1.2 Problems and solutionsProblem 1.1. Two concentric spheres of radii a and b, with >b a, are separated by amaterial 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 isV .

Solution 1.1. Starting with the current density, we have

σ = =

J EIa

dd

Note our current will only depend on r so

σ θ ϕ θ θ ϕ θ= = =I E a kr E r kr Ed d ( )( )( sin d d ) sin d d2 3

Thus,

∫ ∫ θ θ ϕ π= =π π

I kr E kr Esin d d 43

0

2

0

3

Electrodynamics

1-3

and solving for the field yields

π=E

Ikr4 3

Considering

⎛⎝⎜

⎞⎠⎟l∫ ∫ π π

= − · = − = −V EIkr

rIk a b

d4

d8

1 1

b

a

b

a

3 2 2

we have

π= −

Vb a

ka bI

8

2 2

2 2

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

π= −

Rb a

ka b8

2 2

2 2

Problem 1.2. Two long cylinders of radii a and b, with >b a, are separated by anunknown material of conductivity σ . They are maintained at an electric potentialVwith 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 aradius s, with s between the radii a and b. We consider λ the charge per unit length ofthe inner cylinder. Gauss’s law states

S∮ ϵ

· =E aq

do

enc

The left side of the equation is

S∮ π · =E a E sLd 2

While the enclosed charge is simply

λ=q Lenc

It follows that

π λϵ

=sLEL

2o

Electrodynamics

1-4

And the electric field isλ

πϵ = ˆE

ss

2 o

Remember that σ = J E , then the current is calculated as

∫ ∫σ σ λπϵ

π σλϵ

= · = · = =I J a E as

sLL

d d2

2o o

The electric potential is

l∫ ∫ λπϵ

λπϵ

λπϵ

λπϵ

= − · = − = − = − =V Es

s s b aba

d2

d2

ln2

(ln ln )2

lnb

a

b

a

o oba

o o

By substituting λ from the current = σλϵ

I L

owe obtain λ = ϵ

σI

Lo . Back to the electric

potential

πϵ πσ= =

ϵσV

ba

IL

ba2

ln2

lnI

Lo

o

Therefore, the conductivity is

σπ

= ILV

ba2

ln

This is how we can determine the type of material by measuring the current and thepotential 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 parallelconducting rails placed at distance l connected by a resistor R. The system is placedin 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.

Electrodynamics

1-5

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

given by

From Faraday’s law,

E ll l= − Φ = − = − = −

tB x

tB

xt

B vdd

d( )d

dd

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

E = IR

Therefore, the electric current is

E l= =IR

B vR

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

perpendicular to B

ll

ll= = =F I B

B vR

BB v

R

2 2

On the other hand,

= =F ma mvt

dd

Therefore,

l= −FB v

R

2 2

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

l= −mvt

B vR

dd

2 2

Electrodynamics

1-6

l = −v tB v t

Rm( )

( )2 2

With the notation

lα = BRm

2 2

we can write

α = −v t v t( ) ( )

which has the solution

l= =α− −v t v v( ) e et B

Rmt

0 0

2 2

(c) The initial kinetic energy is =K mv2

2. This energy is transmitted to the

resistor as heating. The power is = =P VI RI 2. On the other hand,= =P RIW

tdd

2. Let us calculate the energy:

l l l

l l

l

∫ ∫ ∫ ∫

α

= = = =

=−

= −

−=

α α

α

∞ ∞ ∞−

∞−

− ∞

W RI tB v

Rt

BR

v tB

Rv t

BR

vB v

Rmv

d d e d e d

e2

(0 1)

2 2BRm

t t

t

0

2

0

2 2 2 2 2

002 2

2 2

02

0

2

2 2

02

2

0

2 202

2 202

Therefore, all kinetic energy is transferred to the resistor.

Problem 1.4. Consider the track below, which consists of two parallel conductingrails, separated by a distance l, and two metal bars of mass m; all of which is in amagnetic field B pointing into the page. Initially, the right bar (bar 2), which hasresistance R, is held in place, while the left bar (bar 1) is moving towards bar 2 withvelocity v0. At time =t 0, both bars are allowed to move freely. Find the minimumdistance d between the bars at time =t 0 such that the bars will not collide.

Electrodynamics

1-7

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

E l l l= − Φ = − = − − =t

Bxt

B v B vdd

dd

( )0 0

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

l=F I B

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

E l= = −I tR

BR

xt

( )dd

and now xt

dd

will depend on the velocities of both bars,

= −xt

v t v tdd

( ) ( )2 1

Therefore,

l= − −I tBR

v t v t( ) [ ( ) ( )]2 1

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

ll = − = − = −mv F I t B

BR

v v( ) ( )1

2 2

2 1

For bar 2, we have

ll = = = − −mv F I t B

BR

v v( ) ( )2

2 2

2 1

We can rewrite these as

α= −vt

v vdd

( )12 1

and

α= − −vt

v vdd

( )22 1

where

lα = BmR

2 2

Electrodynamics

1-8

Note

= −vt

vt

dd

dd

1 2

so

= − +v v C1 2

where C is a constant. At =t 0, we have =v v(0)1 0 and =v (0) 02 . Thus, =C v0 and= −v v v1 0 2. Now we can use this to solve for v2,

α α= − − + = − −vt

v v v v vdd

( ) (2 )22 0 2 2 0

Rearranging yields

α−

= −vv v

td

2d2

2 0

and integration yields

α

α

− ∣ = − +

− = − +− = α−

v v t C

v v t C

v v A

12

ln( 2 )

ln( 2 ) 2

2 e t

2 0

0 2

0 22

where A and C are constants. Now we can solve for v2,

= + α−vv

Ae2

t2

0 2

Since =v (0) 02 , we have

= −Av20

Therefore,

= − α−vv2

(1 e )t20 2

Using this, we can solve for v1,

= − = + α−v v vv2

(1 e )t1 0 20 2

We can now use v1 and v2 to solve for −x x2 1. Solving for x t( )1 ,

α= = + ⇒ = − +α α− −v

xt

vx t

vt

vE

dd 2

(1 e ) ( )2 4

et t1

1 0 21

0 0 2

and x t( )2 ,

α= = − ⇒ = + +α α− −v

xt

vx t

vt

vG

dd 2

(1 e ) ( )2 4

et t2

2 0 22

0 0 2

Electrodynamics

1-9

where E and G are constants. At time =t 0, − =x x d(0) (0)2 1 . Therefore,

α α α− = + + − = + − =x x

v vG E

vG E d(0) (0)

4 4 22 1

0 0 0

So,

α− = −G E d

v2

0

Now to solve −x t x t( ) ( )2 1 ,

α α

α

− = + − + + −

= − +

α α

α

− −

−

x t x tv

tv v

tv

G E

vd

( ) ( )2 4

e2 4

e

2(e 1)

t t

t

2 10 0 2 0 0 2

0 2

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

− >→∞

x t x tlim [ ( ) ( )] 0t

2 1

so

α α− = − + > ⇒ >

→∞x t x t

vd d

vlim[ ( ) ( )]

20

2t2 1

0 0

Therefore, to ensure the bars will not collide, we must have

l>d

mRvB2

02 2

Problem 1.5. Consider the wire below, which consists of a circular loop of radius aand a resistor with resistance R. The circular portion is placed in a magnetic fielddirected into the page with = −B t B( ) e kt

0 , where k is a constant. Find the totalenergy dissipated by the resistor for >t 0.

Electrodynamics

1-10

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

E π π= − Φ = − = −

ta

Bt

a kBdd

dd

e kt2 20

This means the current in the loop is given by

E π= = −IR

a kBR

e kt2

0

We know the power is

= =PWt

I Rdd

2

so total energy is then

∫=∞

W I R td0

2

Using our current, we have

⎛⎝⎜

⎞⎠⎟∫ ∫π π= =

∞

−

∞

−Wa kB

RR t

a kBR

te d ( ) e dkt kt

0

20

2 20

2

0

2

Therefore, the total energy dissipated by the resistor is

π=Wa kB

R2

2 402

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 resistorsconnected in parallel, all having the resistance R. Calculate which circuit(series or parallel) has greater power.

Electrodynamics

1-11

(b) For a single resistor circuit of electromotive force E having the internalresistance r, connected to a resistor R, what value of the internal resistance rmaximizes the power?

Solution 1.6.(a) The resistance for series circuit:

∑= ==

R R nRi

n

1

s i

From Kirchhoff’s law:

E = + = +I r R I r nR( ) ( )s s s

For the circuit with resistors in parallel, we have the correspondingequations:

⎜ ⎟⎛⎝

⎞⎠E

∑= =

=

= + = +

=R RnR

RRn

I r R I rRn

1 1

( )

i

n

1p i

p

p p p

The power is:

E

E

= =+

= =+( )

P I Rr nR

nR

P I Rr

Rn

( )

Rn

s s s

p p p

22

2

22

2

Let us calculate the limit for an infinite number of resistors

E

E= =

+

+=

+

+=

→∞ →∞

+

+

→∞ →∞

( ) ( )( )

( )

PP

nR n r

r nR

n r

n R

rR

lim lim lim( )

limn n

r nR

r

Rn

n

Rn

n

Rn

rn

( )s

p

2

2

2

2

22

2

22

2 2

2

2

Rn

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

E E

E

= + ⇒ =+

= =+

I R r IR r

P I RR r

R

( )

( )2

2

2

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

Electrodynamics

1-12

⎜ ⎟⎛⎝

⎞⎠

E

E

= = + − ++

= ⇒ + + − = ⇒ + − = ⇒ =

+PR

R

RR r R R r

R rPR

R r R r R R r r R r R

dd

d

d( ) 2 ( )

( )dd

0 ( )( 2 ) 0 ( )( ) 0

R r( )

2

22

2

4

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 longstraight wire carrying a current I t( ) with = kI

tdd

. Find the induced electromotiveforce (emf) in the loop if its center is positioned a2 above the wire.

Solution 1.7. The electromotive force is given by

E = − Φt

dd

where the flux is

∫Φ = · B ad

We know the magnetic field due to a long wire is

μπ

ϕ = ˆBI

s2o

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

Electrodynamics

1-13

We can see the equation of the loop is

+ − =z s a a( 2 )2 2 2

And we also have

ϕ = ˆa z sd d d

Thus,μπ

· =B aI

sz sd

2d do

To show how z is bounded, we can rewrite our circle equation as

= − − ⇒ = ± − −z a s a z a s a( 2 ) ( 2 )2 2 2 2 2

Therefore, our flux is

∫ ∫ ∫μπ

μπ

μΦ = =− −

= −− − −

− −I

sz s

I a s a

ss a I

2d d

2

2 ( 2 )d (2 3 )

a

a

a s a

a s a

o o

a

a

o

3

( 2 )

( 2 ) 3 2 2

2 2

2 2

and our electromotive force is

E μ μ= − Φ = − − = − −t

aIt

a kdd

(2 3 )dd

(2 3 )o o

Problem 1.8. A rectangular loop, of sides a and b is placed in the xz-plane andexposed to a non-uniform time dependent magnetic field of the form

= ˆB x t kx t y( , ) 2 3 , where k is a constant. Find the electromotive force induced inthe loop.

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

E = − Φt

dd

Electrodynamics

1-14

where the flux is given by

∫Φ = · B ad

In our case, the flux can be calculated as following

∫ ∫ ∫ ∫Φ = · = = =t B a B x t x z kt x x z kta b

( ) d ( , ) d d d d3

a b

3

0

2

0

33

Therefore, the electromotive force is

E = − Φ = − = −( )

tt

kt

tkt a b( )

dd

d

d

a b3

33

2 3

Problem 1.9. Consider a circuit with a battery which supplies a constant electro-motive force E0, a resistor with resistance R and a solenoid of impedance L. Find thecurrent in the circuit and obtain the time after closing the circuit for which thecurrent will reach

n1 of its final equilibrium value.

Solution 1.9. Kirchhoff’s law in the circuit:

E

E

= +

= +

LIt

IR

RLR

It

I

dddd

0

0

The solution of the first order differential equation is

E= + −I tR

A( ) eRL

t0

where A is a constant to be obtained from the initial conditions. Starting with at=t 0, we have =I (0) 0, so

E E= + = ⇒ = −IR

A AR

(0) e 00 0 0

Electrodynamics

1-15

Therefore,

E= − −( )I tR

( ) 1 eRL

t0

When

E→ ∞ →t I tR

, ( ) 0

Let us find the time t necessary for the circuit to reach the current E=I t( )nR

0

E E= − −( )nR R1 e

RL

t0 0

Note that τ=LR

, where τ is the time constant.

⎛⎝⎜

⎞⎠⎟τ

τ= − ⇒ = − ⇒ − = − ⇒ =−

τ τ− −

n nt

nt

nn

11 e e 1

1ln 1

1ln

1t t

For example, for τ= =n t2, ln 2.

Problem 1.10. Find the self-inductance L per unit length for an infinitely longcoaxial cable of radii a, b with <a b, carrying a current I , as in the figure. Useenergy calculations.

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

=WLI

2

2

Let us calculate the energy of the configuration

V∫μ

τ=W B1

2d

o

2

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.

l∮ μ · =B Id o enc

Electrodynamics

1-16

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

l∮ π · =B B sd 2

From here,μπ

ϕ = ˆBI

s2o

The energy of a cylindrical shell of length l, radius s and thickness ds is:

⎛⎝⎜

⎞⎠⎟

l

l

l

μπ

μμπ

π

μπ

=

=

=

W B s s

WI

ss s

WI s

s

d1

22 d

d1

2 22 d

d4

d

o

o

o

o

2

2

2

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

l l l∫ μπ

μπ

μπ

= = =WI s

s

Is

I ba4

d4

ln4

lna

b

o oab o

2 2 2

Back to impedance per unit length:

l l

μπ

= ⇒ = =WLI L W

Iba2

22

lno2

2

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

Solution 1.11. Considering that the total energy is given by

V∫μ

τ= =W B LI1

2d

12o

2 2

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

lP

∮ μ · =B Id o enc

Electrodynamics

1-17

Note we have =I 0enc for < <s a0 and >s b. Therefore, we only have a magneticfield for < <a s b. The current density in this region can be written as

= ˆJ kszWe then have

∫ ∫ ∫ ϕ π= · = ′ ′ ′ = −π

I J a ks s sk

s ad ( ) d d2

3( )

a

s

enc

0

2

3 3

We can also solve for k considering that

∫ ∫ ∫ ϕ π= · = ′ ′ ′ = −π

I J a ks s sk

b ad ( ) d d2

3( )

a

b

0

2

3 3

so⎛⎝⎜

⎞⎠⎟π

=−

kI

b a32

13 3

Our enclosed current is then

= −−

Is ab a

Ienc

3 3

3 3

Returning to our magnetic field, we have

lP

∮ μ μ · = = −−

B I Is ab a

d o oenc

3 3

3 3

Noting

lP

∮ π · =B sBd 2

we have⎛⎝⎜

⎞⎠⎟

μπ

= −−

BI

ss ab a2

o3 3

3 3

Now we can use this to solve for the total energy using

V∫μ

τ=W B1

2d

o

2

where τ ϕ= s s zd d d d . Therefore,

⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

l

l

l

∫ ∫ ∫

∫

μμ

πϕ

μπ

μπ

= −−

=−

−

=+ − +

−

π

( )

WI

ss ab a

s z s

I

b as a

ss

a a a b b

b aI

12 2

1d d d

2(2 )( )( )

d

12 2

6 ln 3 4

6( )

ba

o a

b

o

o

a

b

o

0

2

0

2

2

3 3

3 3

2

2

3 3 2

3 3 2

6 6 3 3 6

3 3 22

Electrodynamics

1-18

Now considering

=W LI12

2

we can see the inductance per length is given by

l π=

+ − +

−( )( )L a a b b

b a

3 2 ln 1 4

12 ( )

ba

6 3 3 6

3 3 2

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

Solution 1.12. Using

lP

∮ μ · =B Id o enc

we can see this current produce the field

⎧⎨⎪

⎩⎪

μ

μ =− ˆ >

ˆ <B

K t y z

K t y z

2( ) , 0

2( ) , 0

o

o

Now we can use

lP

∮ · = − ΦE

td

dd

Electrodynamics

1-19

to find the field. Looking at the xz-plane, we have

where B is directed out of the page for >z 0 and into the page for <z 0. By firstconsidering >z 0, we have

⎜ ⎟⎛⎝

⎞⎠l

l∫ μ μ− Φ = − · = − · = − − ˆ · − ˆ = −

t tB a

tB A

tK t y z y

z K tt

dd

dd

ddd

( )dd 2

( ) ( )2

d ( )d

o o

Also,

l lP

∮ · = −E Ed

Therefore,

μ = ˆEz K t

tx

2d ( )

do

Similarly, for <z 0, we have

μ = − ˆEz K t

tx

2d ( )

do

We can check these by considering

∇ × = − ∂ ∂

EBt

For >z 0,

⎜ ⎟⎛⎝⎜

⎞⎠⎟

⎛⎝

⎞⎠

μ μ μ∇ × = ∂

∂ˆ = ˆ = − ∂

∂− ˆ = − ∂

∂E

z

z K tt

yK t

ty

tK t y

Bt2

d ( )d 2

d ( )d 2

( )o o o

and for <z 0,

⎜ ⎟⎛⎝⎜

⎞⎠⎟

⎛⎝

⎞⎠

μ μ μ∇ × = ∂

∂− ˆ = − ˆ = − ∂

∂ˆ = − ∂

∂E

z

z K tt

yK t

ty

tK t y

Bt2

d ( )d 2

d ( )d 2

( )o o o

as expected.

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

Electrodynamics

1-20

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

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

μ = ˆB t nI t z( ) ( )o

This varying field induces an emf in the circuit

⎡⎣ ⎤⎦E l lμ μ= − Φ = − · = − ˆ · ˆ = −( )tt t

B At

n I t z z nIt

( )dd

dd

( )dd

( ) ( )dd

in o o2 2

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

E E E lω μ= + ⇒ = −IR

IR t nIt

sin( )dd

ino

00

2

Some rearrangement yields,

E El

l lμ ω

μ μω+ = ⇒ + =n

It

IR tIt

Rn

In

tdd

sin( )dd

sin( )oo o

20 2

02

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

⎡⎣⎢

⎛⎝⎜

⎞⎠⎟⎤⎦⎥

El

l l

μ ωω μ ω ω=

++ −μ

−

( )I

R nR t n tsin( ) e cos( )

oo

Rtn0

2 2 22

o2

Problem 1.14. Consider a conducting spherical shell of radius a, which rotates aboutthe z-axis with angular velocity ω. The shell is placed in a uniform magnetic field

= ˆB B z0 .(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’?

Electrodynamics

1-21

Solution 1.14.

(a) Let us start by characterizing the motion:

ω θ ϕ

ω θ ϕ

= ˆ

= ˆ

== × = ˆ × ˆ

v a

B B z

fFq

v B aB z

sin

sin ( )

0

0

The electromotive force E is given by

E l∫= · f d

where

l θ θ = ˆad d

Therefore,

E ∫ ∫ω θ ϕ θ θ ω θ θ θ ωθ= ˆ × ˆ · ˆ = = =

π π π

a B z a B a B( sin )( ) d sin cos d( sin )

20

0

20

20

0

20

2

0

Here we used the triple products properties · × = · × A B C B C A( ) ( ) andthe commutativity of the scalar product. Also, θ ϕˆ × ˆ = r and θˆ · ˆ =z r cos .

(b) Similarly,

⎛

⎝

⎜⎜⎜

⎞

⎠

⎟⎟⎟E ∫ω θ θ θ ω θ ω ω= = = − =

π π ( )a B a B a B

a Bsin cos d

( sin )2 2

04

222

0

0

42

0

2

0

42

0

2

20

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

Electrodynamics

1-22

frequency ν. Consider that both solutions have permeability μ μ= o. Consider aparallel plate capacitor immersed in each type of water solution and with an appliedvoltage of πν=V V tsin(2 )0 .

Solution 1.15. The conduction current is given by

σ σ= =J EVd

c

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

=EVd

The displacement current is

⎛⎝⎜

⎞⎠⎟

ϵ ϵ πν ϵ πν πν= ∂∂

= ∂∂

= ∂∂

=JDt

Et t

V td

Vd

t( ) sin(2 )

(2 ) cos(2 )d0 0

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

ρ πνϵ πνϵρ= = =R

JJ

Vd

dV2

12

c

d

0

0

Now if we compare the two water solutions

ϵ ρϵ ρ

= =πνϵ ρ

πνϵ ρ

RR

12

12

1

2

2 2

1 1

1 1

2 2

Problem 1.16. Prove that Maxwell’s equations with magnetic charge (a)–(d) and theforce law equation (e) are invariant under the duality transformation (*).

(a) ∇· = ρϵ

E e

o

(b) μ ρ∇· =B o m

(c) μ∇ × = − − ∂ ∂

E J Bto m

(d) μ μ ϵ∇ × = + ∂ ∂

B J Eto e o o

(e) = + × + − × F q E v B q Bc

v E( ) (1

)e m 2

Duality transformation (*)

α αα α

α αα α

′ = − = + ′ = −′ = +

′E E Bc

cB E Bccq cq q

q cq q

cos sinsin coscos sin

sin cose e m

m e m

Electrodynamics

1-23

Solution 1.16.(a) We need to prove that

⎜ ⎟⎛⎝

⎞⎠

ρϵ

α α α αρϵ

α μ ρ α

ϵρ α μ ϵ ρ α

ϵρ α

ρα

ρϵ

∇ · ′ = ′

∇ · ′ = ∇ · − = ∇ · − ∇ ·

= −

= − = − =′

E

E E Bc E c B

c

cc

( cos sin ) cos ( ) sin

cos sin

1( cos sin )

1cos sin

e

o

e

oo m

oe o o m

oe

m e

o

We also use the relationship

ϵ μ=c

1

o o

2

(b) Let us prove that

μ ρ∇ · ′ = ′B o m

Note that the equation for ′ρm looks similarly to the one for ′q .m

⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

α α α α

ρϵ

α μ ρ α

μρ

ϵ μα ρ α μ ρ α ρ α μ ρ

∇ · ′ = ∇ ·

+ = ∇ · + ∇ ·

= +

= + = + = ′

BEc

B Ec

B

c

cc

sin cos ( )1

sin ( ) cos

sin cos

sin cos ( sin cos )

e

oo m

oe

o om o e m o m

(c) The following relationship to prove is

⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

μ

α α α α

μ α α μ μ ϵ α

μ α α α α μ

∇ × ′ = − ′ − ∂ ′∂

∇ × ′ = ∇ × − = ∇ × − ∇ ×

= − − ∂ ∂

− + ∂ ∂

= − + − ∂∂

+ = − ′ − ∂ ′∂

E JBt

E E Bc E c B

JBt

c JEt

J cJt

Bc

E JBt

( cos sin ) ( ) cos ( ) sin

cos cos sin

( cos sin ) cos1

sin

o m

o m o e o o

o m e o m

Electrodynamics

1-24

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

⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

μ μ ϵ

α α α α

μ α μ μ ϵ α

μ α α μ ϵ α α

μ μ ϵ

∇ × ′ = ′ + ∂ ′∂

∇ × ′ = ∇ ×

+ = ∇ × + ∇ ×

= − + ∂ ∂

+ + ∂ ∂

= − + + ∂∂

− +

= ′ + ∂ ′∂

B JEt

BEc

Bc

E B

cJ

Bt

JEt

cJ J

tcB E

JEt

sin cos1

( ) sin ( ) cos

1sin cos

1sin cos ( sin cos )

o e o o

o m o e o o

o m e o o

o e o o

(e) Lastly, let us see if the force law equation is invariant to the dualitytransformation

⎜ ⎟

⎛⎝⎜

⎞⎠⎟

⎛⎝

⎞⎠⎡⎣⎢⎢

⎛⎝⎜

⎞⎠⎟⎤⎦⎥⎥

⎡⎣⎢⎢⎛⎝⎜

⎞⎠⎟

⎤⎦⎥⎥

⎡⎣⎢⎢⎛⎝⎜

⎞⎠⎟⎤⎦⎥⎥

⎡⎣⎢

⎤⎦⎥

⎡⎣⎢

⎤⎦⎥

⎛⎝⎜

⎞⎠⎟

α α α α α α

α α α α α α

α α α α α α

α α α α α α

α α α α α α

α α α α α α

′ = ′ ′ + × ′ + ′ ′ − × ′

= − − + ×

+

+ +

+ − × −

= − + +

+ ×

+ −

+

+ −

+ +

+

+ × −

−

−

+

= + × + − × =

F q E v B q Bc

v E

qq

cE Bc v

Ec

B

cq qEc

Bc

v E Bc

q E cB E cB

q vEc

BEc

B

qEc

BEc

B

q vEc

Bc

Ec

Bc

q E v B q Bc

v E F

( )1

cos sin ( cos sin ) sin cos

( sin cos ) sin cos1

( cos sin )

[ (cos ) cos sin (sin ) cos sin ]

sin cos (cos ) sin cos (sin )

sin cos (sin ) sin cos (cos )

(sin ) sin cos (cos ) sin cos

( )1

e m

em

e m

e

e

m

m

e m

2

2

2 2

2 2

2 2

22

22

2

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

Electrodynamics

1-25

⎛⎝⎜

⎞⎠⎟

μπ

= + ˆE sIt

s K z( )2

dd

ln( )o

where K is a constant with respect to s. If ω=I t I t( ) sin( )0 , show this satisfiesMaxwell’s equations and find ρ and J .

Solution 1.17. We can see this current results in a magnetic field

μπ

ϕμ

πω ϕ = ˆ = ˆB t

I t

s

I

st( )

( )

2 2sin( )o o 0

Starting with ∇ · =B 0, we have

⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

μπ

ω ϕϕ

μπ

ω∇ · = ∇ · ˆ = ∂∂

=BI

st

s

I

st

2sin( )

12

sin( ) 0o o0 0

as expected. Now to verify Faraday’s Law,

∇ × = − ∂ ∂

EBt

We can see the right-hand side is given by

μ ωπ

ω ϕ− ∂ ∂

= − ˆBt

I

st

2cos( )o 0

The left-hand side is given by

⎡⎣⎢⎛⎝⎜

⎞⎠⎟

⎤⎦⎥

μπ

∇ × = ∇ × + ˆEIt

s K z2

dd

ln( )o

with

ω ω=It

I tdd

cos( )0

Thus,

μ ωπ

ω∇ × = ∇ × ˆ + ∇ × ˆEI

t s z Kz2

cos( )[ (ln( ) )]o 0

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

∇ × ˆ =Kz 0

Also,

⎡⎣⎢

⎤⎦⎥

⎡⎣⎢

⎤⎦⎥ϕϕ ϕ∇ × ˆ = ∂

∂ˆ − ∂

∂ˆ = − ˆs z

ss s

ss

sln( )

1ln( ) ln( )

1

Electrodynamics

1-26

Therefore,

μ ωπ

ω ϕ∇ × = − ˆ = − ∂ ∂

EI

st

Bt2

cos( )o 0

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

⎤⎦⎥ρ ϵ ϵ

μ ωπ

ω= ∇ · = ∇ · ˆ + ∇ · ˆEI

t s z Kz( )2

cos( )( (ln( ) ))o oo 0

Here we have

∇ · ˆ =Kz 0

using the same argument as before. Also,

∇ · ˆ = ∂∂

=s zz

s(ln( ) ) (ln( )) 0

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

μ μ ϵ∇ × = + ∂ ∂

B JEto o o

Starting with the left-hand side,

⎡⎣⎢

⎛⎝⎜

⎞⎠⎟⎤⎦⎥

⎡⎣⎢

⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

⎤⎦⎥

μπ

ω ϕμ

πω∇ × = ∇ × ˆ = − ∂

∂ˆ + ∂

∂ˆ =B

It

s

It

z ss

s ss

sz

2sin( )

12

sin( )1 1 1

0o o0 0

Therefore,

ϵ = − ∂ ∂

JEt

o

where

⎡⎣⎢

⎤⎦⎥

⎡⎣⎢

⎤⎦⎥

μ ωπ

ωμ ω

πω∂

∂= + ˆ = − + ˆE

t

Is

tt

Kz

Is t

Kt

z2

ln( )dd

(cos( ))ddt 2

ln( ) sin( )dd

o o0 02

The current density is then,

⎡⎣⎢

⎤⎦⎥

ϵ μ ωπ

ω ϵ = − ˆJI

s tKt

z2

ln( ) sin( )dd

o oo

02

Note that this, as does E , depends on a constant (with respect to space) K whichdepends on the entire history of the current ω=I t I t( ) sin( )0 .

Problem 1.18. Use Maxwell’s equations and the Lorentz force law to deriveCoulomb’s Law for two charges of charge q.

Electrodynamics

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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,ρϵ

∇ · =Eo

We can take the integral over all space of both sides

V V∫ ∫τ ρ

ϵτ∇ · =E d d

o

The right-hand side is simply

V V∫ ∫ρ

ϵτ

ϵρ τ

ϵ= = q

d1

do o o

Applying the divergence theorem to the left-hand side yields

V S∫ ∮ ∫ ∫τ θ θ ϕ π∇ · = · = =

π π

E E a E r r Ed d sin( ) d d 40

2

0

2 2

By combining everything, we haveρϵ

πϵ

∇ · = ⇒ =E r Eq

4o o

2

or

πϵ = ˆE

qr

r4 o

2

Substituting this into the force law yields

πϵ = = ˆF qE

qr

r4 o

2

2

which is exactly what Coulomb’s Law states.

BibliographyByron 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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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 242Purcell 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)

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