Post on 23-Sep-2020
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.
3H Electromagnetism: Magnetostatics, time-dependent fields, Maxwell’s equations page 1
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.
3H Electromagnetism: Magnetostatics, time-dependent fields, Maxwell’s equations page 1
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.
3H Electromagnetism: Magnetostatics, time-dependent fields, Maxwell’s equations page 1
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
).
3H Electromagnetism: Magnetostatics, time-dependent fields, Maxwell’s equations page 2
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.
3H Electromagnetism: Magnetostatics, time-dependent fields, Maxwell’s equations page 1
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.
3H Electromagnetism: Magnetostatics, time-dependent fields, Maxwell’s equations page 1
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.
3H Electromagnetism: Magnetostatics, time-dependent fields, Maxwell’s equations page 2
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 .
3H Electromagnetism: Magnetostatics, time-dependent fields, Maxwell’s equations page 1
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.
3H Electromagnetism: Magnetostatics, time-dependent fields, Maxwell’s equations page 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.
3H Electromagnetism: Magnetostatics, time-dependent fields, Maxwell’s equations page 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.
3H Electromagnetism: Magnetostatics, time-dependent fields, Maxwell’s equations page 1
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.