PHY4324/Fall 09: EM II
HOMEWORK ASSIGNMENT #4: Solutions
due by 11:45 p.m. Wednesday 09/23
Instructor: D. L. Maslov
maslov@phys.ufl.edu 392-0513 Rm. 2114
Please help your instructor by doing your work neatly. Every (algebraic) final result
must be supplemented by a check of units. Without such a check, no more
than 75% of the credit will be given even for an otherwise correct solution.
1. A parallel-plate capacitor carrying charge Q0 is filled with sea water (resistivity ρ, dielectric permittivity ǫ, and
magnetic permeability µ0 ). The capacitor is slowly discharging due to a current flowing through the inter-plate
distance (and out to the external circuit).
a) Find the dependence of the charge on the capacitor on time. [25 points]
Q
dQ
= RI = −R
→
C
dt
Q = Q0 exp (−t/RC)
b) Find the magnetic field in between the plates. [20 points]
The conduction current I = −dQ/dt. The displacement current
R Id = AdD/dt = Adσ/dt = dQ/dt, where
A is the area and σ is the surface charge density. Therefore, B·dl = µ0 (I + Id ) = 0 → B = 0.
2. G7.28 [25 points] Hint: Calculate the energy stored in this configuration.
Inside the inner cable of radius a is
B1 =
µ0 Ir
.
2πR2
Inside the insulating sheath (inner radius a, outer radius a + ε)
B2 =
µ0 I
.
2πr
Total energy
1
W =
2µ0
Z
2
B12 dτ1
+
Z
B22 dτ2
Z a+ε
1
r2
rdr
+
rdr
4
r2
a
0 R
2
(µ0 I)
a+ε
1
.
=
ℓ2π
+ ln
2µ0
4
a
=
(µ0 I)
ℓ2π
2µ0
Z
a
2
Since ε ≪ a, one can neglect ln a+ε
a ≈ ε/a ≪ 1. Equating the resulting expression for the energy with LI /2
yields
L/ℓ =
µ0
.
8π
3. G7.30 [25 points]
a) The magnetic field of the first loop, treated as a magnetic dipole m1 =Ia1 , is
B1 =
µ0
I1 (3(a1 · r̂)r̂ − a1 ) .
4πr3
2
The flux through the 2nd loop is then
Φ2 = B1 · a2 =
M =
µ0
I1 (3(a1 · r̂)(a2 · r̂) − a1 · a2 ) ≡ M I1 →
4πr3
µ0
(3(a1 · r̂)(a2 · r̂) − a1 · a2 )
4πr3
b)
E1 = −
dM2
dt
dI2
dW1
= −E1 I1 = M I1
.
dt
dt
Since I1 = const, the total work is just W1 = M I1 I2 .
4. G7.42 (a) [9 points]
E=0→∇×E=0→
∂B
= 0.
∂t
5. (b) [8 points]
∂Φ
∂
=
∂t
∂t
Z
∂
B·da = −
∂t
Z
E·dl,
where the line integral is along the conductor. But since the conductor is perfect, E = 0 and
∂Φ
∂t
= 0.
and (c) [8 points]
6.
∇B = µ0 J + µ0 ε0
∂E
,
∂t
where B = 0 and E = 0 → J must be confined to the surface.
7. Bonus: G7.45 [25 points]
Centrifugal froce per charge
f = v×B
v = ωaB0 sin θφ̂
f = ωaB0 sin θ φ̂ × ẑ
f must be integrated over a great circle of the sphere. A line element on the great circle dl =adθθ̂.
i
h f ·dl = ωa2 B0 sin θ θ̂ · φ̂ × ẑ = ωa2 B0 sin θẑ · θ̂ × φ̂ = ωa2 B0 sin θẑ · r̂
= ωa2 B0 sin θ cos θ
Z
Z
2
E =
f ·dl =ωa B0
0
π/2
2
sin θ cos θ = ωa B0
Z
0
1
dxx =
1 2
ωa B0
2