3H Electromagnetism: Magnetostatics, time-dependent elds ...dma0sfr/EM/Solutions/21-53.pdf · 3H...

3H Electromagnetism: Magnetostatics, time-dependent fields, Maxwell’s equations page 1

21 Using the quoted result, B is given by

B = µ∇ ·(

r

|r|3

)− µ ·∇

(r

|r|3

).

Now ∇ · (r/|r|3) = 0 for r 6= 0 (this is a result we already used a fair bit in electrostatics) whilethe ith component of the second term is

−µj∂

∂xj

(xi

(xkxk)3/2

)= −µj

(δij

(xkxk)3/2− 3

2

2xixj(xkxk)5/2

)= − µi

(xkxk)3/2+

3xi µjxj(xkxk)5/2

from which the result follows almost immediately.


23 Using the component form of the Stokes’s theorem variant given on the vector calculus handout,with ψ = εijk

∂Bk

∂xl|x=0 , we have∮

C

εijk dlj xl∂Bk

∂xl

∣∣x=0

=

∫S

εjpq np∂

∂xq

(εijk xl

∂Bk

∂xl

∣∣x=0

)da

=

∫S

εjpq np εijk δql∂Bk

∂xl

∣∣x=0

da

=

∫S

εpqj εijk np∂Bk

∂xq

∣∣x=0

da

=

∫S

(δpkδqi − δpiδqk)np∂Bk

∂xq

∣∣x=0

da

=

∫S

[nk

∂Bk

∂xi

∣∣x=0− ni

∂Bk

∂xk

∣∣x=0

]da .

Since ∇ ·B = 0, the second term in the square brackets vanishes and the quoted result follows.Now consider a current loop near the origin. Expanding B(x) to leading order, we have B(x) =B|x=0 + (x ·∇)B|x=0 + . . . , and so

F =I

c

∮C

dl ∧ (B|x=0 + (x ·∇)B|x=0) + . . .

The first term vanishes as the line integral of a constant round closed curve. Writing thecontribution from the second term in components we have

Fi =I

c

∮C

εijk dlj xl∂Bk

∂xl

∣∣x=0

=I

c

∫S

nk da ∂Bk

∂xi

∣∣x=0

.

using the result just obtained for the second equality. Recognising that the integral is justc/I times the magnetic moment µ of the loop, and taking the derivative outside everything(remember that µ is a fixed vector!) gives the desired result.


26 The induced emf is given by

E = −1

c

dΦ

dt

where Φ is the flux linking the wire loop,

Φ =

∫S

B · n dS = πR20B0 cosα sinωt.

Hence

E = −πR20

cωB0 cosα cosωt.


28 Suppose for definiteness that the axis of the solenoid lies along the z axis. Then the magneticfield is equal to (4πIN/c) ez inside the solenoid, and zero outside. The cross-sectional areaof the solenoid is πR2, so the total magnetic flux ΦB through a single loop of the coil isΦB = 4π2R2IN/c . To calculate the self-inductance we need the total flux enclosed by thecircuit, which loops a total of Nl times round the solenoid; hence the flux through the circuitis equal to Φtot = NlΦB = 4π2R2N2l I/c . The self-inductance L is defined by Φtot = cLI , andso the formula quoted in the question follows immediately.

29 This case is actually easier than the last one. Suppose the current round the solenoid is I; thenthe field inside it is 4πIN/c while that outside it is 0. The area of the region where the field isnon-zero is πR2, so the total flux through C2 is ΦB = 4π2R2IN/c . The mutual inductance isdefined via ΦB = cMI and so the quoted formula is confirmed.

30 The flux through the ellipse arising from a current I along the wire is

Φ = cMI =

∫S

B · n da.

The magnetic field of an infinitely long wire is azimuthal and has magnitude 2I/ρc, where ρis the radial distance from the wire. If the ellipse is centred at the origin then we can takeρ = d + x, and since the wire and the loop are in the same plane, the field through the planesurface of the ellipse is in the direction n, so that

B =2I

c

1

d+ xn

and thus

M =2

c2

∫S

da

d+ x.

To do the integral over the ellipse we switch to elliptical polar coordinates

x = ar cosφ, y = br cosφ, dx dy = abr dr dφ.

Then

M =2ab

c2

∫ 1

0

∫ π

−π

r

d+ ar cosφdφ dr

=4ab

c2

∫ 1

0

πr√d2 − a2r2

dr

=4πab

c2

[−√d2 − a2r2a2

]10

=4πb

ac2

(d−√d2 − a2

).


31 Ampere’s Law for a circle of radius ρ concentric with the cylinders is∮CB(r) · dl = 2πρB(ρ) =

4π

cIenc(ρ),

where Ienc(ρ) is the current flowing through the disk of radius ρ. This determines the magneticfield as

B =

2ρca2I for ρ < a,

2cρI for a < ρ < b,

2cρ

(1− ρ2−b2

d2−b2

)I for b < ρ < d,

0 for d < ρ.

The energy density is

u =1

8πB2 =

ρ2

2πc2a4I2 for ρ < a,

12πc2ρ2

I2 for a < ρ < b,

12πc2ρ2

(d2−ρ2d2−b2

)2I2 for b < ρ < d,

0 for d < ρ.

The total electromagnetic energy per unit length is the integral of the energy density

dW

dz=

∫ ∞0

∫ 2π

0

u ρ dφ dρ

=I2

c2a4

∫ a

0

ρ3 dρ+I2

c2

∫ b

a

1

ρdρ+

I2

c2

∫ d

b

1

ρ

(d2 − ρ2

d2 − b2

)2

dρ

=I2

c2

(1

4+ ln

b

a+

1

(d2 − b2)2

[d4 ln

d

b− d2(d2 − b2) +

1

4(d4 − b4)

])=

I2

c2

(1

4+ ln

b

a+

[d4

(d2 − b2)2lnd

b− 3d2 − b2

4(d2 − b2)

]).

This gives the required expression for the self-inductance L per unit length.


33 Starting from

∇ ∧ E = −1

c

∂B

∂t− 4π

cjm .

Take the divergence of this and use ∇ · (∇ ∧ E) = 0 to find

0 = −1

c

∂

∂t∇ ·B− 4π

c∇ · jm .

But, ∇ ·B = 4πρm and so, cancelling common factors,

0 =∂ρm∂t

+ ∇ · jm

which is the desired result.


35 Maxwell’s equations in empty space:

∇ ·B =∂

∂y

(−1

c

∂S

∂t

)= 0,

∇ · E =∂

∂x

(∂S

∂z

)+

∂

∂z

(−∂S∂x

)= 0,

∇ ∧B− 1

c

∂E

∂t=

[− ∂

∂z

(−1

c

∂S

∂t

)]ex +

[∂

∂x

(−1

c

∂S

∂t

)]ez

− 1

c

∂

∂t

[∂S

∂zex −

∂S

∂xez

]= 0,

∇ ∧ E +1

c

∂B

∂t=

[∂

∂z

(∂S

∂z

)− ∂

∂x

(−∂S∂x

)]ey +

1

c

∂

∂t

(−1

c

∂S

∂t

)ey

=

(∂2S

∂x2+∂2S

∂z2− 1

c2∂2S

∂t2

)ey = 0,

Try solutionS = e−iωt[K cos(αx) + L sin(αx)][N cos(βz) +M sin(βz)].

This satisfies the wave equation provided(−α2 − β2

)S =

1

c2(−ω2

)S.

The electric field is

E = βe−iωt[K cos(αx) + L sin(αx)][−N sin(βz) +M cos(βz)]ex

− αe−iωt[−K sin(αx) + L cos(αx)][N cos(βz) +M sin(βz)]ez.

The tangential component of the field has to vanish at the planes z = 0, b or x = 0, a. Thisrequires

Ex = 0 at z = 0 : βM = 0,

Ex = 0 at z = b : βN sin βb = 0,

Ez = 0 at x = 0 : αL = 0,

Ez = 0 at x = a : αL sinαa = 0,

So we needM = L = 0, β =

sπ

b, α =

rπ

awith r, s integer.

This covers the special case if α = 0 or β = 0 as well. So the electric field is, up to overallnormalisation,

E = βe−iωt cos(αx) sin(βz)ex − αe−iωt sin(αx) cos(βz)ez,

where (ωc

)2

= α2 + β2, α =rπ

aβ =

sπ

b, r, s ∈ Z.

At least one of r, s has to be non-zero for a non-vanishing electric field. The magnetic field is

B = iω

ce−iωt cos(αx) cos(βz)ey.

Remark: This represents an electromagnetic field propagating in a rectangular wave guide.


36 Maxwell’s equations in empty space are

∇ · E = 0, ∇ ∧ E + 1c∂B∂t

= 0,

∇ ·B = 0, ∇ ∧B− 1c∂E∂t

= 0.

In terms ofF = E + iB = φ(x+ iy)f(z − ct)(ex + iey)

these can be written as

∇ · F = 0, ∇ ∧ F− i

c

∂F

∂t= 0.

Check these equations:

∇ · F =

(∂φ

∂x+ i

∂φ

∂y

)f(z − ct) = 0 since φ is analytic,

∇ ∧ F− i

c

∂F

∂t= i

(∂φ

∂x+ i

∂φ

∂y

)f(z − ct)ez

−iφ(x+ iy)∂f(z − ct)

∂z(ex + iey)

− i

cφ(x+ iy)

∂f(z − ct)∂t

(ex + iey)

= 0 .


38 Substitute into Maxwell’s equations:

∇ ·B′ = a∇ ·B− b∇ · E = 0,

∇ · E′ = a∇ · E+ b∇ ·B = 0,

∇ ∧B′ − 1

c

∂E′

∂t= −b

(∇ ∧ E+

1

c

∂B

∂t

)+ a

(∇ ∧B− 1

c

∂E

∂t

)= 0,

∇ ∧ E′ +1

c

∂B′

∂t= a

(∇ ∧ E+

1

c

∂B

∂t

)+ b

(∇ ∧B− 1

c

∂E

∂t

)= 0.

Transform energy density and Poynting vector:

u′ =1

8π(E′2 +B′2) =

1

8π(a2 + b2)(E2 +B2) = (a2 + b2)u,

S′ =c

4πE′ ∧B′ =

c

4π(aE+ bB) ∧ (−bE+ aB)

=c

4π(a2E ∧B− b2B ∧ E) = (a2 + b2)S.

So energy density and Poynting vector are invariant provided a2 + b2 = 1.


40

∇ ·A′ +1

c

∂φ′

∂t= ∇ · (A + ∇λ) +

1

c

∂

∂t

(φ− 1

c

∂λ

∂t

)= ∇ ·A +

1

c

∂φ

∂t+ ∇2λ− 1

c

∂2λ

∂t2

= ∇ ·A +1

c

∂φ

∂t

using the given property of λ. Thus, if (A, φ) satisfies the Lorentz gauge condition ∇·A+ 1c∂φ∂t

=0, then so does (A′, φ′).

41 First, check that Maxwell’s equations for empty space hold if A satisfies the two conditionsgiven in the question:

∇ ·B = ∇ · (∇ ∧A) = 0,

∇ · E = −1

c

∂

∂t(∇ ·A) = 0,

∇ ∧B− 1

c

∂E

∂t= ∇(∇ ·A)−∇2A +

1

c2∂2A

∂t2= 0,

∇ ∧ E +1

c

∂B

∂t= −1

c∇ ∧ ∂A

∂t+

1

c

∂

∂t(∇ ∧A) = 0.

Next check the given vector potential satisfies the conditions:

∇ ·A =∂Ax∂x

+∂Ay∂y

= 0,

∇2A =∂2A

∂z2= −

(2π

λ

)2

A =1

c2∂2A

∂t2.

The fields are:

E = −1

c

∂A

∂t=

2πa

λ

(− sin

(2π

λ(z − ct)

)ex + cos

(2π

λ(z − ct)

)ey

)B = ∇ ∧A =

2πa

λ

(− cos

(2π

λ(z − ct)

)ex − sin

(2π

λ(z − ct)

)ey

)or, with θ = 2π

λ(z − ct),

E =2πa

λ(− sin θex + cos θey)

B =2πa

λ(− cos θex − sin θey)

Both E and B lie in the x-y plane. The vector E points at an angle θ measured anti-clockwisearound from the positive y-axis, while B points at an angle θ measured anticlockwise aroundfrom the negative y-axis. Notice that the magnitudes of the field vectors are constant, E =B = 2πa/λ, so they trace out a circle as θ revolves through 2π. Hence the field vectors executecircular motion around the z-axis with frequency c/λ. (The rotation is clockwise when viewedfrom the positive z-direction, as θ decreases with time.) Notice also that B lags behind E byπ/2.


42 First, check that Maxwell’s equations for empty space hold if A satisfies the two conditionsgiven in the question:

∇ ·B = ∇ · (∇ ∧A) = 0,

∇ · E = −1

c

∂

∂t(∇ ·A) = 0,

∇ ∧B− 1

c

∂E

∂t= ∇(∇ ·A)−∇2A +

1

c2∂2A

∂t2= 0,

∇ ∧ E +1

c

∂B

∂t= −1

c∇ ∧ ∂A

∂t+

1

c

∂

∂t(∇ ∧A) = 0.

For the second part of the question, the calculation goes as follows. First,

∇2A = ∇ ∧[k∇2

(ψ(r−ct)

r

)]= ∇ ∧

[k

1

r

d2

dr2r

(ψ(r−ct)

r

)]= ∇ ∧

[kψ′′(r−ct)

r

](using the result ∇2f(r) = 1

rd2

dr2(rf(r)) , which can be deduced easily from the vector calculus

handout). Next,

∂2A

∂t2= ∇ ∧

[k∂2

∂t2

(ψ(r−ct)

r

)]= c2∇ ∧

[kψ′′(r−ct)

r

].

Hence,

∇2A− 1

c2∂2A

∂t2= 0

as required.Note: that the given form of the potential satisfies the gauge condition follows simply from thefact that ∇ ·∇ ∧V = 0 for any vector field V.


53 Fαβ , as given on the formula sheet is

Fαβ =

0 −E1 −E2 −E3

E1 0 −B3 B2

E2 B3 0 −B1

E3 −B2 B1 0

;

lowering indices gives

Fαβ =

0 E1 E2 E3

−E1 0 −B3 B2

−E2 B3 0 −B1

−E3 −B2 B1 0

.

Multiplying components together and adding,

FαβFαβ = 2(|B|2 − |E|2

).

This quantity is a Lorentz scalar (there are no free indices left over) and so is the same in allframes. To do the final part, note that light is circularly polarised in a given frame if |B|2 = |E|2at all times. This is the same as FαβFαβ = 0, and so the condition is preserved when we switchto any other frame.

3H Electromagnetism: Magnetostatics, time-dependent elds ...dma0sfr/EM/Solutions/21-53.pdf · 3H...

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Transcript of 3H Electromagnetism: Magnetostatics, time-dependent elds ...dma0sfr/EM/Solutions/21-53.pdf · 3H...