PHY 5346
Homework Set 11 Solutions – Kimel
1. 5.26 The system is described by
Since the wires are nonpermeable, μ = μ0. The system is made of parts with cylindrical
symmetry, so we can determine B using Ampere’s law.
∇⃗ × B⃗ = μ0J⃗, or ∫ B⃗ ⋅ dl⃗ = μ0 ∫ J⃗ ⋅ da⃗
On the outside of each wire,
∫ B⃗ ⋅ dl⃗ = B2πρ = μ0I → Bout = μ0I
2πρ
On the inside of each wire
∫ B⃗ ⋅ dl⃗ = B2πρ = μ0I ρ
2
R2
, B in =
μ0I
2π
ρ
R2
with R = a,b
From the right-hand rule, the B from each wire is in the φ̂ direction. From the above figure, using
the general expression for the vector potential, we see A⃗ is in the ±ẑ direction. Since ∇⃗ × A⃗ = B⃗,
Bz = − ∂∂ρ
Az → Az = −∫ Bzdρ
Thus
Az =
− μ0I
2π
ln ρR + C = −
μ0I
4π
ln ρ
2
R2
+ 1 on the outside
− μ0I
4π
ρ2
R2
,
on the inside
where I’ve determined C = 1/2, from the requirement that Az be continuous at ρ = R. Let l be the
length of the wire. Then we know the total potential energy is given by
W = 1
2 ∫ J⃗ ⋅ A⃗d
3x =
l
2 ∫JaAdaa + JbAdab
Consider the second term
l
2 ∫ JbAdab. The system is pictured as
From the figure
ρ⃗a = d⃗ + ρ⃗b, ρa2 = d2 + ρb
2 − 2dρb cosφ
so, since Jb =
I
πb2
l
2 ∫ JbAdab =
l
2
I
πb2
∫Aoutρa + A inρb ρbdρbdφ
=
l
2
I
πb2
μ0I
4π ∫
ln
ρa2
a2
+ 1 − ρb
2
b2
ρbdρbdφ
≃
l
2
I
πb2
μ0I
4π
2π ∫
0
b
ln d
2
a2
+ 1 − ρb
2
b2
ρbdρb
=
l
2
I
πb2
μ0I
4π
2π 1
4
b2 1 + 2 ln d
2
a2
=
l
2
μ0
4π
1
2
+ 2 ln da I
2
The first term
l
2 ∫ JaAdaa is equal to
l
2 ∫ JaAdaa =
l
2
μ0
4π
1
2
+ 2 ln d
b
I2
Thus
W =
l
2
μ0
4π
1 + 2 ln d
2
ab
I2 =
l
2
L
l
I2
or
L
l
=
μ0
4π
1 + 2 ln d
2
ab