PHY 5346
Homework Set 9 Solutions – Kimel
1. 5.10
a) From Eq. (5.35)
Aφr,θ =
μ0
4π
I
a ∫
r ′2dr ′dΩ ′ sinθ ′ cosϕ ′δcosθ ′δr ′ − a
|x⃗ − x⃗ ′ |
Using the expansion of 1/|x⃗ − x⃗ ′ | given by Eq. (3.149),
1
|x⃗ − x⃗ ′ |
= 4π ∫
0
∞
dkcoskz − z ′ 1
2
I0kρ<K0kρ> +∑
m=1
∞
cosmϕ − ϕ ′ Imkρ<Kmkρ>
We orient the coordinate system so ϕ = 0, and because of the cosϕ ′ factor, m = 1. Thus,
Aφr,θ =
μ0
4π
I
a
4π
π ∫
0
∞
dk ∫ r ′2dr ′dcosθ ′ sinθ ′δcosθ ′δr ′ − acoskzI1kρ<K1kρ>
Aφr,θ =
μ0
π aI ∫
0
∞
dkcoskzI1kρ<K1kρ>
where ρ<ρ> is the smaller (larger) of a and ρ.
b) From problem 3.16 b),
1
|x⃗ − x⃗ ′ |
= ∑
m=−∞
∞
∫
0
∞
dke imφ−φ
′JmkρJmkρ ′e−k|z|
Note z ′ = 0, and φ = 0, so
Aφ =
μ0Ia
2 ∫0
∞
dke−k|z|J1kρJ1ka