Physics 416G : Solutions for Problem set 12
Due : December 9, 2015
Please read Griffiths sections 7.3.1-7.3.3, 8.1, 9.1, 9.2.
We covered the material from slightly different point of view in the class.
Please show all the details of your computations including intermediate steps.
1
Problem 1
An alternating current I = I0 cos(ωt) flows down a long straight wire, and returns along a coaxial
conducting tube of radius R.
a) In what direction does the induced electric field point (radial, circumferential, or longitudinal) in
the quasistatic regime?
~ t).
b) Assuming that the field goes to zero as ρ → ∞, find E(ρ,
~
c) Find the displacement current density Jd .
d) Integrate it to get the total displacement current.
e) Compare Id and I. (What’s their ratio?)
f) If the outer cylinder were, say, 2mm in diameter, how high would the frequency have to be, for
Id to be 1% of I? [This problem is designed to indicate why Faraday never discovered displacement
currents, and why it is ordinarily safe to ignore them unless the frequency is extremely high.]
Sol: a) The magnetic field is “circumferential” in the quasistatic approximation. From the equation
~ ×E
~ = − ∂ B~ , the relation between the electric and magnetic fields is analogous to the magnetic field
∇
∂t
~ ×B
~ = µ0 J.
~ Thus the direction of the electric field is longitudinal.
and current in the equation ∇
R
H
R
~ ∇×
~ E)
~ = d~l·E
~ = − d dS·
~ B
~ = − d Φ.
b) This problem can be solved using the Stokes’ theorem dS·(
dt
dt
Using the ‘ampere loop’ shown in the figure (with replacing s → ρ, a → R), we compute the electric
field inside the cylinder as
Z
d R µ0 I
0
~ = − µ0 ln R dI ẑ = µ0 I0 ω sin(ωt) ln R ẑ .
ldρ
,
→
E
El = −
dt ρ 2πρ0
2π
ρ dt
2π
ρ
The electric field outside vanishes.
c) The displacement current is
~
µ0 0 I0 ω 2
R
µ0 0 Iω 2
R
∂E
=
cos(ωt) ln
ẑ =
ln
ẑ .
J~d = 0
∂t
2π
ρ
2π
ρ
d) The displacement current Id is given by
Z
Z
µ0 0 Iω 2 R
µ0 0 Iω 2 R2
R
~
~
Id = dS · Jd =
ẑ =
.
2πρdρ ln
2π
ρ
4
0
e) The ratio goes as
Id
µ0 0 ω 2 R2
=
.
I
4
2
3×108 m/sec
1
1
2c
f) The ratio goes as IId = ωR
= 100
. This gives ωR
2c
2c = 10 , and thus ω = 10R = 5×10−3 m =
0.6 × 1011 /sec. And f = ω/2π ≈ 1010 Hz. This turns out to be in microwave region, far above the
radio frequency!
2
Problem 2
Write down the real electric and magnetic fields for a monochromatic plane wave of amplitude E0 ,
frequency ω, and phase angle zero that is
a) traveling in the ~x = (−1, 0, 0) direction and polarized in the ~x = (0, 0, 1) direction;
b) traveling in the direction from the origin to the point ( √13 , √13 , √13 ), with polarization parallel to the
xz plane.
c) In each case, sketch the wave, and give the explicit Cartesian components of the propagating vector
~k and polarization vector n̂.
Sol: a) The propagating vector ~k and polarization vector n̂ are ~k = −ω/cx̂, n̂ = ẑ. And ~k · ~r = −ωx/c.
Thus
~ x, t) = E0 ŷ cos ω x + ωt ,
~ x, t) = E0 ẑ cos ω x + ωt ,
B(~
E(~
c
c
c
~
where we used k × n̂ = ŷ and cos(−a) = cos a. ω x̂+ŷ+ẑ
~
~
√
b) The propagating vector k is k = c
. We need to normalize the vector such that the
3
magnitude is ω/c. And the normalized polarization vector n̂ = αx̂ + β ẑ can be obtained by requiring
√ . And ~
that it is perpendicular to ~k. Thus, n̂ = x̂−ẑ
k · ~r = √ωx
(x + y + z). Thus
2
3c
ω
√ (x + y + z) − ωt ,
3c
E0 −x̂ + 2ŷ − ẑ
ω
~
√
√
B(~x, t) =
cos
(x + y + z) − ωt ,
c
6
3c
− ẑ
~ x, t) = E0 x̂√
cos
E(~
2
√
where we use ~k × n̂ = −x̂+2ŷ−ẑ
.
6
c) The waves have the following pattern
3
Problem 3
During the class we studied the complex notation. We noted that we need to be careful when we
~ and B.
~ Yet, there is a clever device for finding the time
compute something which is not linear in E
~
average of a product. Suppose f (~r, t) = A cos(k · ~r − ωt + δa ) and g(~r, t) = B cos(~k · ~r − ωt + δb ).
a) Show that hf gi = (1/2)Re(f˜g̃∗), where the star denotes complex conjugation and à represents a
complex function of A.
Hint : Note that this only works if the two waves have the same ~k and ω, but they need not have the
same amplitude or phase.
b) Write the electric and magnetic fields in part b) in problem 1 or part b) in problem 2 in terms of
complex functions.
c) Compute the energy density and Poynting vector
1
~˜ · E
~˜ ∗ + 1 B
~˜ · B
~˜ ∗ ,
~ = 1 Re E
~˜ × B
~˜ ∗ ,
hui = Re 0 E
hSi
4
µ0
2µ0
for the wave given in part b) in problem 1 or part b) in problem 2.
Sol: a) The time average of real functions are
Z
1 T
hf gi =
A cos(~k · ~r − ωt + δa )B cos(~k · ~r − ωt + δb )dt
T 0
Z
i
AB T h
AB
=
cos(2~k · ~r − 2ωt + δa + δb ) + cos(δa − δb ) dt =
cos(δa − δb ) .
2T 0
2
~
~
In the complex notation, we have f (~r, t) = Ãei(k·~r−ωt) and g̃(~r, t) = B̃ei(k·~r−ωt) with à = Aeiδa , B̃ =
Beiδb . Thus
1
1
1
1
~
~
Re(f˜g̃∗) = Re Ãei(k·~r−ωt) B̃ ∗ e−i(k·~r−ωt) = Re ÃB̃ ∗ = AB cos(δa − δb ) .
2
2
2
2
Thus they are the same.
b) We are given electric field and the corresponding magnetic field can be obtained using the Maxwell
R
~ = − dt∇
~ × E.
~ The electric and magnetic fields for the problem 2 b) in complex notation
equation B
are
− ẑ √ (x+y+z)−ωt
~ x, t) = E0 x̂√
e 3c
,
E(~
2
ω
ω
~ x, t) = E0 −x̂ +√2ŷ − ẑ e √3c (x+y+z)−ωt .
B(~
c
6
c) Using the complex notation we show for the Problem 2, part b), we get
1 ~˜ ~˜ ∗
1
1 E02
1
1
˜ ~˜ ∗
2
~
B·B
= Re 0 E0 +
= 0 E02 ,
hui = Re 0 E · E +
2
4
µ0
4
µ0 c
2
2
2
x̂
−
ẑ
x̂
+
ŷ
+
ẑ
1
E
−x̂
+
2ŷ
−
ẑ
1
E
0
0
~ =
√ ×
√
√
hSi
Re
=
,
2µ0
c
2µ0
c
2
6
3
where we use that the electric and magnetic fields are normalized and c2 = 1/(0 µ0 ).
For the rest of the problem, let us consider the electric field given in problem 1 b).
b) We are given electric field and the corresponding magnetic field can be obtained using the Maxwell
R
~ = − dt∇
~ ×E
~ in the cylindrical coordinate. The electric and magnetic fields for the
equation B
problem 1 b) in complex notation are
µ0 I0 ω i(ωt−π/2)
R
˜
~
~˜ x, t) = i µ0 I0 ei(ωt−π/2) ϕ̂ .
E(~x, t) =
e
ln
ẑ ,
B(~
2π
ρ
2πρ
first of all, we see that the electric and magnetic fields are not in the form of wave. Thus we suspect
that the poynting vector would vanish.
c) Using the complex notation we show for the problem 1) part b) as
2
2
2 !
1
1
1
µ
I
R
1
1
˜
˜
˜
˜
0
0
~ ·E
~∗ +
~ ·B
~∗ =
hui = Re 0 E
B
0 ω ln
+
,
4
µ0
4
2π
ρ
µ0 ρ
~ = 1 Re µ0 I0 ω ln R (−i) µ0 I0 ρ̂ = 0 ,
hSi
2µ0
2π
ρ
2πρ
where the Poynting vector is pure imaginary and its real part vanishes.
4
Problem 4
~ = ŷE0 ei(hz−ωt)−κx .
Consider an electric field of a wave propagating in vacuum, E
a) How are the real parameters h, κ, and ω related to one another?
~
b) Find the associated magnetic field B.
c) Under what conditions is the polarization of the magnetic field close to circular?
d) Compute the time-averaged Poynting vector.
Sol: a) The electric field satisfies the wave equation
~
1 ∂2E
~ 2E
~ =0,
−∇
c2 ∂t2
2
and the corresponding dispersion relation is − ωc2 + h2 − κ2 = 0.
~ ×E
~ = − ∂ B~ gives
b) Faraday’s law ∇
∂t
~ ×E
~ = iE0 (ihx̂ + κẑ)E0 ei(hz−ωt)−κx .
~ =−i∇
B
ω
ω
c) The magnetic field will be nearly circularly polarized if the polarization vector has a phase eiπ/4 .
This can be achieved by imposing h ≈ κ. From a) we can also have the following condition h, κ ω/c.
d) As we done in the previous problem, we can get
~ =
hSi
2
2
1
~˜ × B
~˜ ∗ = E0 Re (hẑ − iκx̂) e−2κx = E0 hẑe−2κx ,
Re E
2µ0
2µ0 ω
2µ0 ω
where we drop the term proportional to x̂ because it is purely real.
5
Problem 5
~ ·E
~ = 0, ∇
~ ×E
~ = − ∂ B~ , write down the electric field as an integral in terms
a) From the equations ∇
∂t
~
of time derivative of B.
~ ·B
~ = 0, ∇
~ ×B
~ = µ0 J.
~ And then compare the
Hint: First write down the Biot-Savart solution for ∇
mathematical expressions to write down the solution.
~ ·A
~ = 0, ∇
~ ×A
~ = B,
~ write down the vector potential A
~ as an integral
b) Similarly, from the equations ∇
~
in terms of B.
~ in terms of A.
~ Check the result by taking a curl for
c) By comparing the results of a) and b), Write E
both sides.
Now we would like to compute the electric field for a spherical shell of radius R carries a uniform
surface charge σ and also spinning around a fixed axis with an angular velocity ω(t) that changes slowly
with time. There are two contributions here: the Coulomb field due to the charge, and the Faraday
~
field due to the changing B.
d) Find the electric field inside and outside the sphere due to the charge.
e) Find the electric field inside and outside the sphere due to the change of magnetic field. Combine
the results with d). Hint: One can first evaluate the vector potential of the spinning spherical shell,
followed by taking a time derivative. The answer is given in the Problem 5 below. You can take the
results from there.
R ~ x0 )×(~x−~x0 ) 3 0
~ is given by B
~ = µ0 J(~
Sol: a) The Biot-Savart solution for B
d x . Thus the electric field
4π
|~
x−~
x0 |3
has the form
Z ~ 0
B(~x , t) × (~x − ~x0 ) 3 0
~ =− 1 ∂
d x .
E
4π ∂t
|~x − ~x0 |3
b) Similarly, we have
~= 1
A
4π
Z ~ 0
B(~x , t) × (~x − ~x0 ) 3 0
d x .
|~x − ~x0 |3
~
~
~
~ = − ∂ A . By taking a curl for both sides, we get ∇×
~ E
~ = − ∂ ∇×A =
c) We find a simple connection as E
∂t
∂t
~
− ∂∂tB . This is nothing but the Faraday’s law.
d) Gauss’ law with spherical Gaussian surfaces readily give the electirc field
~ =
E
0,
σR2 R2
0 r 2 r̂
,
r<R
r>R
.
e) The electric field has the Coulomb part and the time derivative of the vector potentials. And the
result is
(
− µ0 R3 ω̇σ r sin θφ̂ ,
r<R
~ =
E
.
2 2
µ0 R4 ω̇σ sin θ
σR R
r>R
0 r 2 r̂ −
3
r 2 φ̂ ,
6
Problem 6 : Vector potential (Bonus problem, optional)
A spherical shell, of radius R, carrying a uniform surface charge σ, is set spinning at angular velocity
ω(t). Find the vector potential it produces at point ~x as
(
µ0 Rωσ
r sin θφ̂ ,
r<R
3
~=
A
.
µ0 R4 ωσ sin θ
r>R
3
r 2 φ̂ ,
Sol: It might seem natural to align the polar axis along ω, but the integration is easier if we let ~r lie
on the z axis, so that ω is tilted at an angle ψ. We may as well orient the x axis so that ω lies in the
xz plane. Then
~ r) = µ0
A(~
4π
Z
dS 0
~ r0 )
K(~
,
|~r − ~r0 |
where
~ = σ~v = σω × ~r0 = σ(ω sin ψx̂ + ω cos ψẑ) × (R sin θ0 cos φ0 x̂ + R sin θ0 sin φ0 ŷ + R cos θ0 ẑ)
K
= σRω[− cos ψ sin θ0 sin φ0 x̂ + (cos ψ sin θ0 cos φ0 − sin ψ cos θ0 )ŷ + sin ψ sin θ0 sin φ0 ẑ] ,
√
and |~r − ~r0 | = R2 + r2 − 2Rr cos θ0 and dS 0 = R2 sin θ0 dθ0 dφ0 . Using the fact that sin φ0 , cos φ0
integrals vanish if done over the period, we get
Z
cos θ0 sin θ0
µ0 R3 σω sin ψ π 0
~
A(~r) = −
dθ √
.
2
R2 + r2 − 2Rr cos θ0
0
This integral can be done using u = cos θ0 , which gives
Z
0
π
1
u
R2 + r2 + Rru p 2
2
du √
=−
R + r − 2Rru
2
2
2
2
3R r
R + r − 2Rru
−1
−1
1 2
= − 2 2 (R + r2 + Rr)|R − r| − (R2 + r2 − Rr)|R + r|
3R r
2r
r<R
3R2 ,
=
.
2R
r>R
3r 2 ,
cos θ0 sin θ0
dθ √
=
R2 + r2 − 2Rr cos θ0
0
Z
1
Using ω
~ × ~r = −ωr sin ψ ŷ, we get
(
~=
A
µ0 Rσ
ω × ~r) ,
3 (~
µ0 R4 σ
(~
ω × ~r) ,
3r 3
r<R
r>R
.
Now let us return to the original coordinate, where ω
~ coincide with z coordinate and the ~r is at (r, θ, φ).
Then we get
(
µ0 Rωσ
r sin θφ̂ ,
r<R
3
~=
A
.
µ0 R4 ωσ sin θ
φ̂
,
r
>R
3
r2