PHY4324/Fall 09: EM II
HOMEWORK ASSIGNMENT #3:
due by 11:45 p.m. Wednesday 09/16
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 circuit consists of a coil of inductance L and resistance r connected in parallel with resistor R (see Fig. ??).
At time t = 0, switch S is thrown disconnecting circuit ABCD. Find the current in circuit BFGC for t > 0.
Explain why this current can damage resistor R. [50 points]
Before the switch wwas thrown, the coil and the resistor are connected in parallel. The current through the coil
is I0 = E/r. At later times, the coil and the resistore are connected in series. The back e.mf. is equal to the sum
of voltage drops. This situation is described by a differential equation
−L
dI
= (R + r)I
dt
with an initial condition I(0) = I0 . solving this equation, we obtain
R+r
E
R+r
I (t) = I0 exp −
t = exp −
t .
L
r
L
Before the switch was thrown, the current through the load was IR = E/R. At short times after the switch was
thrown, the current is
R+r
R+r
R
E
1−
1−
t + ... = IR
t + ... .
I=
r
L
r
L
If R/r ≫ 1, the current through the load is much larger than it was before–this can damage the load.
2. G7.19 [50 points]
From Example (5.10), the magnetic field inside the coil at distance s from the center is
B(t)=
µ0 N I(t)
.
2πs
The flux of the magnetic field through the cross-section of the coil is
Z a+w
µ0 N I(t)
ds
µ0 N I(t)
a+w
µ0 N I(t) wh
Φ (t) =
h
=
h ln
≈
,
2π
s
2π
a
2π
a
a
where the last step is valid because w ≪ a. The e.m.f due a change in the flux is
−
dΦ(t)
µ0 N d
ah
µ0 N wh
=−
I(t)
=−
k
.
dt
2π dt
w
2π
a
The circulation of the electric field around any countor enclosing the coil is
Z
dΦ(t)
E·dl = −
.
dt
Now consider a magnetic field B′ produced by a line circular current I ′ . The circulation of this field satisfies
Z
E·dl = µ0 I ′ .
2
R
F
G
L,r
B
A
C
ε D
S
FIG. 1: Problem 1
The fields themselves satisfy the differential equations
∂B
∂t
∇ × B′ = µ0 J′ .
∇×E = −
′
Therefore, E is identical to B′ upon replacement µ0 I ′ → − dΦ(t)
dt . The magnetic field B at distance z above the
center of the ring of radius a is
B′ =
a2
µ0 I ′
ẑ.
2 (a2 + z 2 )3/2
Therefore,
E=−
1 dΦ(t)
a2
awh
µ0 N
k
ẑ = −
ẑ
2 dt (a2 + z 2 )3/2
4π (a2 + z 2 )3/2
3. Bonus: G7.23. [25 points]
The field of one wire is B = µ0 I/2πs, evidently, the fields add up. The total flux
Φ = 2ℓ
Z
d−ε
dsB =
ε
µ0 Iℓ
π
Z
ε
d−ε
ds
µ0 Iℓ d − ε
µ0 Iℓ d
=
ln
≈
ln
s
π
ε
π
ε
µ0 ℓ d
ln .
→ L=
π
ε
Evidently, one cannot set ε to 0 for the logarithm blows up in this limit.