PHY4324/Fall 09
Test I: SOLUTIONS
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. However, just a
correct dimensional estimate for the result can give you up to 25 % of the full credit.
1. A long solenoid is made of thin wire (diameter d) wrapped tightly around a cylinder of radius a. The conductivity
of the wire is σ. At t = 0, the solenoid is connected to a battery. Find a characteristic ”delay time” after which
the current in the solenoid approaches its steady-state value.
If X is the length of the solenoid, the number of turns the wire makes around the solenoid is X/d. The total
length of the wire
ℓ = 2πaX/d.
The resistance of the wire is
R=
ℓ
aX
= 8 3.
2
σπd /4
σd
The magnetic field inside the solenoid B = µ0 N I, where N = 1/d. The energy stored in the solenoid
Z
µ0 2
1
1
B 2 dτ =
N πa2 X I 2 = LI 2
W =
2µ0
2
2
which implies that the self-inductance
L = µ0 N 2 (πa2 X) = µ0
πa2 X
d2
The delay time is
τ = L/R =
π
µ0 σad.
8
Unit check:
[τ ] = [µ0 ][σ]m2 =
m2
J
QV
N
=
=
= s.
2
A Ohm · m
AV
(Q/s)V
2. A time dependent voltage V (t) = V0 cos(ωt) is applied to a capacitor, which consists of two concentric spheres
of radii a and b (a < b). The space in between the spheres is filled with two spherical shells made of different
insulators, so that
ε1 for a < r < c
ε=
ε2 for c < r < b
a) Find the displacement current. If the charge on the inner sphere is Q, the induction D at distance r from
the center is
D=
Q
4πr2
. Displacement current
Id =
dQ
dD
4πr2 =
dt
dt
2
The electric field is
E=
(
D
ε1
D
ε2
=
=
Q
4πε1 r 2
Q
4πε2 r 2
for a < r < c
for c < r < b
Voltage drop
V = V (a) − V (b) = −
Z
a
Edr =
b
Z
a
b
Q
Edr =
4πε1
1 1
−
a c
Q
+
4πε2
1 1
−
c
b
which means that
Id =
dV
dt
1
4πε1
1
a
−
1
c
1
+
1
4πε2
1
c
−
1
b
=−
1
4πε1
1
a
V0 ω sin(ωt)
1
− 1c + 4πε
2
1
c
b) Find the magnetic field produced by the displacement current. Zero by symmetry.
−
1
b
3. A small square loop of side b lies above the center of a circular loop (radius a) so that the angle between the
planes of the loops is α. Assuming that b ≪ a, find the mutual inductance.
The magnetic field at distance z above the center of a circular current is
B=
a2 I
µ0
ẑ
2 (a2 + z 2 )3/2
The flux through a small square loop
Φ=
Z
B · da =
µ0 a2 b2 cos(α)I
2 (a2 + z 2 )3/2
so
M=
µ0 a2 b2 cos(α)
2 (a2 + z 2 )3/2