Page 172 - Electromagnetics

ˆ
Because φ =−ˆ x cos φ + ˆ y sin φ we find that

µIa 2π cos φ
ˆ

A(r) = φ 1/2 dφ .
4π 0 ρ + a + z − 2aρ cos φ
2
2
2
We put the integral into standard form by setting φ = π − 2x:

2
µIa π/2 1 − 2 sin x
ˆ
A(r) =− φ 2 dx.
4π −π/2 ρ + a + z + 2aρ(1 − 2 sin x) 1/2
2

2
2
2
Letting
4aρ
2 2 2 2
k = , F = (a + ρ) + z ,
2
(a + ρ) + z 2
we have
2
µIa 4 π/2 1 − 2 sin x
ˆ
A(r) =− φ 2 dx.
2
4π F 0 [1 − k sin x] 1/2
Then, since
2
2
1 − 2 sin x k − 2 2 2 −1/2 2 2 2 1/2
= [1 − k sin x] + [1 − k sin x] ,
2
2
[1 − k sin x] 1/2 k 2 k 2
we have
µI a 1 2 2 2
ˆ
A(r) = φ 1 − k K(k ) − E(k ) . (3.138)
πk ρ 2
Here
π/2 π/2

2 du 2 2 2 1/2
K(k ) = 2 2 1/2 , E(k ) = [1 − k sin u] du,
0 [1 − k sin u] 0
are complete elliptic integrals of the first and second kinds, respectively.
2
2
2
2
2
We have k 1 when the observation point is far from the loop (r = ρ + z a ).
Using the expansions [47]
π 1 9 π 1 3
2 2 4 2 2 4
K(k ) = 1 + k + k +· · · , E(k ) = 1 − k − k − ··· ,
2 4 64 2 4 64
in (3.138)and keeping the first nonzero term, we find that
µI 2
ˆ
A(r) ≈ φ (πa ) sin θ. (3.139)
4πr 2
Defining the magnetic dipole moment of the loop as
2
m = ˆ zIπa ,
we can write (3.139)as
µ m × ˆ r
A(r) = . (3.140)
4π r 2
Generalization to an arbitrarily-oriented circular loop with center located at r 0 is accom-
plished by writing m = ˆ nIA where A is the loop area and ˆ n is normal to the loop in the
right-hand sense. Then
µ r − r 0
A(r) = m × .
4π |r − r 0 | 3
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