Example using spherical harmonics— Sp 2017
Magnetic field due to a current loop
.
A circular loop of radius a carries current I. We place the origin at the
center of the loop, with polar axis perpendicular to the plane of the loop. Then
the current density is
δ (r − a)
δ (µ) φ̂
j=I
a
(You can get this most easily by starting with the expression in cylindrical
coordinates
j = Iδ (z) δ (r − a) φ̂
and using z = r cos θ. See also Lea pg 315 Example 6.7.) Then the magnetic
vector potential is (Notes 1 eqn 16)
]
j (x ) 3
µ0
d x
(1)
A (x) =
4π
|x − x |
]
δ (r − a) δ (µ ) φ̂ φ 3
µ0
I
d x
=
4πa
|x − x |
We must take care here, because the unit vector φ̂ is not a constant. We must
re-express it in terms of the constant Cartesian unit vectors,
and thus:
A (x) =
µ0
I
4πa
]
φ̂ φ = − sin φ x̂ + cos φ ŷ
δ (r − a) δ (µ ) − sin φ x̂ + cos φ ŷ d3 x
|x − x |
Now we use our most useful result (J eqn 3.70) to expand the 1/R in the
integrand. With r< = min (r, r ) and similarly for r>,
1
A (x) =
µ0 I
4πa
]
δ (r − a) δ (µ )
∞
l
l
[
[
r<
4π
∗
Ylm (θ, φ) Ylm
×
θ , φ − sin φ x̂ + cos φ ŷ d3 x
l+1
2l
+
1
r
l=0 > m=−l
∞
l
µ0 I [ [ 4π
Ylm (θ, φ) ×
=
4πa
2l + 1
l=0 m=−l
] 2π ] +1 ] ∞
l
r<
2
∗
θ , φ − sin φ x̂ + cos φ ŷ (r ) dr dµ dφ
δ (r − a) δ (µ ) l+1
Ylm
r>
0
−1
0
∞ [
l
π l ] 2π
[
Ylm (θ, φ) r<
∗
, φ − sin φ x̂ + cos φ ŷ dφ
= µ0 aI
Y
lm
l+1
2l + 1 r>
2
0
l=0 m=−l
where we used the sifting property to evaluate the integrals over r and µ . We
now interpret r< as the lesser of r and a, and similarly for r> .
To do the integral over φ , we rewrite the sines and cosines in terms of
exponentials:
] 2π
π ∗
, φ − sin φ x̂ + cos φ ŷ dφ
Ylm
Ilm =
2
0
] 2π v
3
3
3
2l + 1 (l − m)! m
ŷ + ix̂
ŷ − ix̂
Pl (0) e−imφ eiφ
=
+ e−iφ
dφ
4π (l + m)!
2
2
0
The integral is zero unless m = ±1. (Be alert here— if you use the mantra
"axisymmetry so m = 0" you will get into big trouble!) With m = +1 we get:
v
ŷ + ix̂
2l + 1 (l − 1)! 1
P (0)
Il1 =
2π
4π (l + 1)! l
2
and with m = −1
v
ŷ − ix̂
2l + 1 (l + 1)! −1
P (0)
Il,−1 =
2π
4π (l − 1)! l
2
v
(l − 1)! 1
2l + 1 (l + 1)!
(−1)
P (0) (ŷ − ix̂)
= π
4π (l − 1)!
(l + 1)! l
v
2l + 1 (l − 1)! 1
P (0) (ŷ − ix̂)
= −π
4π (l + 1)! l
So
A (x) = µ0 aI
∞
[
l=1
l
r<
π
l+1
2l + 1 r>
v
(Lea 8.54)
2l + 1 (l − 1)! 1
P (0) [(ŷ + ix̂) Yl1 (θ, φ) − (ŷ − ix̂) Yl,−1 ]
4π (l + 1)! l
2
The sum over l starts at 1 now because with l = 0 there is no m = ±1 term.
m ∗
, we get
Then, using Yl,−m = (−1) Ylm
A (x) = µ0 aI
µ0 aI
4
=
∞
[
l=1
∞
[
l=1
l
r<
π
2l + 1 (l − 1)! 1
Pl (0) Pl1 (µ) (ŷ + ix̂) eiφ + (ŷ − ix̂) e−iφ
l+1
2l + 1 r>
4π (l + 1)!
l
r<
1
P 1 (0) Pl1 (µ) (2ŷ cos φ − 2x̂ sin φ)
l+1 l (l + 1) l
r>
∞
l
µ0 aI [ r<
Pl1 (0) Pl1 (µ)
φ̂
2
rl+1 l (l + 1)
l=1 >
=
(2)
We might have expected to find that A is in the φ direction, parallel to j. Check
dimensions: A is current times µ0 , which is consistent with (1).
We can simplify a bit by inserting the value of Pl1 (0) . First note that Pl
is even if l is even, and odd if l is odd. Since Plm is proportional to the mth
derivative of Pl , Plm will be odd if l + m is odd and even if l + m is even. (Lea
pg 385) So Plm (0) = 0 unless l + m is even, or, in this case, l is odd.
Now we can use the recursion relation (J3.29 or Lea 8.37)
lPl (µ) = µPl (µ) − Pl−1 (µ)
1
(0)
lPl (0) = −Pl−1 (0) = Pl−1
(Lea 8.53)
Thus, using Lea 8.47, with l = 2n + 1,
Pl1 (0) = (l + 1) Pl+1 (0) = (l + 1) (−1)(l+1)/2
1
P2n+1
(0) = (−1)n+1
= (−1)n+1
l!!
(l + 1)!!
(2n + 1)!!
(2n)!!
(2n + 1)!!
2n n!
Thus (2) becomes
A (x) =
=
∞
2n+1
1
µ0 aI [ r<
n+1 (2n + 1)!! 1
P2n+1 (µ) φ̂
2n+2 (2n + 1) (2n + 2) (−1)
2 n=0 r>
2n n!
2n+1
∞ µ0 aI [ r<
(2n + 1)!!
(−1)n+1 1
P2n+1 (µ) φ̂
2r> n=0 r>
2 (n + 1) (2n + 1) 2n n!
Separating out the n = 0 term, we have
&
%
2n+1
∞
[
µ0 aI r< 1
(2n − 1)!! 1
n r<
P
P (µ) +
(−1)
(µ)
Aφ = −
4r> r> 1
r>
2n (n + 1)! 2n+1
n=1
(3)
Outside the loop, r > a, and
&
%
∞
2n+1 (2n − 1)!!
[
µ0 aI a 1
n a
1
P (µ) +
P
(−1)
(µ) φ̂ (4)
A (x) = −
4r
r 1
r
2n (n + 1)! 2n+1
n=1
3
For r
a, n = 0 ( = 1) is the dominant term:
A (x)
−
µ0 I a2 1
P (µ) φ̂
4 r2 1
and (Lea 8.53)
P11 = − sin θ
d
(µ) = − sin θ
dµ
so
µ0 a2 I
µ m
µ m×r
sin θ φ̂ = 0 2 sin θ φ̂ = 0
(5)
2
4r
4π r
4π r3
where m = πa2 I ẑ is the magnetic moment of the loop. Compare with Jackson
equation 5.55. Then, in this limit
A (x)
B
θ̂ ∂ µ m
r̂
∂ µ0 m
0
2
sin θ
sin
θ
−
r sin θ ∂θ 4π r2
r ∂r 4π r
r̂ µ0 m
θ̂ µ0 m
2 sin θ cos θ +
sin θ
r sin θ 4π r2
r 4π r2
µ0 m
r̂ 2 cos θ + θ̂ sin θ
4π r3
= ∇×A =
=
=
This is a dipole field, as expected.
Inside the loop, r < a and we have:
&
%
∞
2n+1 (2n − 1)!!
[
µ0 aI r 1
n r
P (µ) +
P1
(−1)
(µ) φ̂
A (x) = −
n (n + 1)! 2n+1
4a
a 1
a
2
n=1
Near the center, r
(6)
a, the n = 0 term dominates again, and we have:
A
µ0 I r 1
P (µ) φ̂
4 a 1
µ0 I r
sin θ φ̂
4 a
−
=
and
B (x)
=
=
%
&
θ̂ ∂ 2
µ0
r̂
∂ 2
I
r sin θ −
r sin θ
4a r sin θ ∂θ
r ∂r
µ0 I r̂ cos θ − θ̂ sin θ
2a
µ0
I ẑ
2a
a uniform field, as expected. Compare with Lea and Burke equation 28.7 with
z = 0.
4
Field on axis:
From LB 28.7, the field on the polar (z−) axis is
B (z) =
µ0 Ia2
2 (z 2 + a2 )3/2
ẑ
So for z > a we can do a binomial expansion to get
−3
µ0 Ia2
3 a2
−5 1 a4
+
+ · · · ẑ
B (z) =
1−
2z 3
2 z2
2
2
2 z4
µ0 Ia2
3 a2 15 a4
=
+
+ · · · ẑ
1
−
2z 3
2 z2
8 z4
(7)
We can also find B from the solution (4) for r > a:
B
= ∇×A
&$
+
#
%
∞
2n+1
[
µ0 aI
(2n − 1)!! 1
r̂
∂
a 1
n+1 a
P
=
(−1)
(µ)
sin θ − 2 P1 (µ) +
4
r sin θ ∂θ
r
r2n+2 2n (n + 1)! 2n+1
n=1
$&,
% #
∞
2n+1 (2n − 1)!!
[
θ̂ ∂
a 1
n+1 a
1
−
P
(−1)
(µ)
− P1 (µ) +
r ∂r
r
r
2n (n + 1)! 2n+1
n=1
0 aI
For µ = 1 (θ = 0) , the theta component is µ4r
times
$
#
∞
2n+1 (2n − 1)!!
[
a 1
n+1 (2n + 1) a
1
P
(−1)
(1)
− 2 P1 (1) +
r
r
r
2n (n + 1)! 2n+1
n=1
But
1
P2n+1
(1) = −
s
d
1 − µ2 P2n+1 (µ)
=0
dµ
µ=1
So the theta component on axis is zero, as we would expect from the symmetry
(azimuthal symmetry about the axis and reflection anti-symmetry about the
plane of the loop).
The r−component is
&$
#
%
∞
2n+1
[
µ0 aI r̂
(2n − 1)!! 1
∂
a 1
n a
P
(−1) 2n+2 n
(µ)
Br = −
sin θ 2 P1 (µ) +
4 r sin θ ∂θ
r
r
2 (n + 1)! 2n+1
n=1
&$
%
#
∞
2n
[
µ0 a2 I ∂ s
a
(2n
−
1)!!
n
1
=
P2n+1
r̂
1 − µ2 P11 (µ) +
(−1) 2n n
(µ)
4r3 ∂µ
r
2
(n
+
1)!
n=1
From the definition of Pl1 (Lea 8.53) and the differential equation for P2n+1 (Lea
8.19), we find
l
d
d d ks
1
1 − µ2
P2n+1 (µ)
1 − µ2 P2n+1
(µ) = −
dµ
dµ
dµ
= (2n + 1) (2n + 2) P2n+1 (µ)
5
Then we evaluate at µ = 1 where P2n+1 (1) = 1
&$
#%
∞
2n
[
µ0 a2 I r̂
(2n − 1)!!
n a
(2n + 1) (2n + 2)
Br (r, 0) =
(−1) 2n n
2+
4 r3
r 2 (n + 1)!
n=1
Changing to the z−coordinate, we get
#
$
∞
2n
µ0 a2 I
1[
(2n + 1)!!
n a
Bz (z) =
ẑ 1 +
(−1) 2n n−1
2z 3
2 n=1
z
2
n!
2
2
4
3a
a 5×3
µ0 a I
+···
ẑ 1 −
+ 4
=
3
2
2z
2z
2z 2 × 2
3 a 2 15 a 4
µ0 a2 I
ẑ
1
−
+
+
·
·
·
=
2z 3
2 z
8 z
which agrees with (7).
6