c
John
C. Young, June 9, 2010
1
1
Magnetostatic
For the magnetostatic case, Maxwell’s equations reduce to
∇×H = J
∇·B = 0 .
(1)
Since the divergence of H is zero, it can be written in terms of the magnetic vector potential A as
H=
1
∇×A ,
µ
(2)
which leads to
∇×∇×A = µJ .
(3)
∇×∇×A = ∇∇·A − ∇2 A = µJ .
(4)
Since ∇×∇×A = ∇∇·A − ∇2 A, we have
The obvious choice for the divergence of A is
∇·A = 0 .
(5)
The electric current J is φ-symmetric, φ̂-driected, and located at (ρ′ , z ′ ) = (r ′ , θ ′ ). Several of the multiple
methods to solve for the field are derived. If the electric current is assumed to be φ-symmetric, then the
magnetic field is seen to be T Mφ and T Er in spherical coordinates and T Mφ and T Ez is cylindrical coordinates. Hence, we may use a magnetic vector potential Aφ in both spherical and cylindrical coordinates,
an electric vector potential Fr in spherical cooridnates, or an electric vector potential Fz in cylindrical
coordinates to represent the fields.
1.1
Spherical Coordinate Solution
In spherical coordinates, the electric current J is
J(r) = Jφ φ̂ =
1
δ(r − r ′ )δ(θ − θ ′ )φ̂ .
r′
(6)
To verify that this is the proper form for J , we integrate over a surface S perpindicular to φ̂
ZZ
1.1.1
J(r) · ds =
S
Z
r ′ +∆ Z θ ′ +∆
r ′ −∆
θ ′ −∆
1
δ(r − r ′ )δ(θ − θ ′ )φ̂ · r φ̂dθdr = 1 .
r′
(7)
Magnetic Vector Potential
Since the field is T Mφ and J is φ̂-directed, the solution can be written in terms of the magnetic vector
potential A = Aφ φ̂. The differential equation satisfied by Aφ is
∇2 (Aφ φ̂) = −µJ
since the divergence of A is identically zero. The φ-component of the vector ∇2 operator is
1
1
∂Ar
cos θ ∂Aθ
φ̂ · ∇2 A = ∇2 Aφ + 2
−
Aφ + 2
+2
.
r sin θ
sin θ
∂φ
sin θ ∂φ
(8)
(9)
c
John
C. Young, June 9, 2010
2
Since the field is φ-symmetric, the differential equation reduces to
∇2rθ Aφ −
1
r 2 sin2 θ
1
Aφ = −µJφ = −µ ′ δ(r − r ′ )δ(θ − θ ′ ) .
r
(10)
At first glance, this does not appear to be the wave equation, but a careful inspection shows that it is. The
solution may be written in terms of a sum of solutions of the homogeneous equation. If Aφ = R(r)Θ(θ),
then the homogeneous equation is
1 ∂
∂R
1
∂
∂Θ
1
r2
+
sin θ
−
=0,
(11)
R ∂r
∂r
Θ sin θ ∂θ
∂θ
sin2 θ
which is a homogeneous wave equation. Solutions are r n Pn1 (cos θ) and r n Q1n (cos θ). Since the field must
be finite away from the source, the field may be represented as
Aφ (r, θ) =
=
∞
X
n=1
∞
X
Bn r n Pn1 (cos θ) , r < r ′
Cn
n=1
1
r n+1
Pn1 (cos θ) , r > r ′
(12)
The unknown coefficients can be determined as follows. The potential should be continuous at r = r ′ ,
which can be enforced if
∞
X
r n
Pn1 (cos θ) , r < r ′
′
r
n=1
′ n+1
∞
X
r
=
An
Pn1 (cos θ) , r > r ′
r
Aφ (r, θ) =
An
(13)
n=1
The differential equation is
∂Aφ
1
∂
1
1
1 ∂
2 ∂Aφ
r
+ 2
sin θ
− 2 2 Aφ = −µ ′ δ(r − r ′ )δ(θ − θ ′ ) .
2
r ∂r
∂r
r sin θ ∂θ
∂θ
r
r sin θ
We note that a solution ψn to the homogeneous wave equation must satisfy
1
∂
∂ψn
1
ψn
sin θ
− 2 2 ψn = −n(n + 1) 2 ,
2
r sin θ ∂θ
∂θ
r
r sin θ
so that the differential equation becomes
Aφ
1 ∂
1
2 ∂Aφ
r
− n(n + 1) 2 = −µ ′ δ(r − r ′ )δ(θ − θ ′ ) .
2
r ∂r
∂r
r
r
Next, the orthogonality relation
Z
Tnm =
π
0
1
sin θPn1 (cos θ)Pm
(cos θ)dθ = δnm
2 (n + 1)!
2n + 1 (n − 1)!
(14)
(15)
(16)
(17)
is used to obtain
∂
Tnn
∂r
r
2 ∂An (r)
∂r
− n(n + 1)Tnn An (r) = −µr 2
sin θ ′
δ(r − r ′ )Pn1 (cos θ ′ ) .
r′
(18)
c
John
C. Young, June 9, 2010
3
Rπ
by integrating the differential equation with 0 (·) Pn1 (cos θ) sin θdθ. If we apply the so-called ”jumpcondition” and integrate with respect to r over (r ′ − ∆, r ′ + ∆) and let ∆ → 0, we obtain
′+
1 sin θ ′ 1
∂An (r) r
=
−µ
P (cos θ ′ ) .
∂r r′−
Tnn r ′ n
This gives
−An
(19)
2n + 1
1 sin θ ′ 1
=
−µ
P (cos θ ′ ) ,
r′
Tnn r ′ n
(20)
µ (n − 1)!
sin θ ′ Pn1 (cos θ ′ ) .
2 (n + 1)!
(21)
or
An =
Therefore, the solution is
∞
X
µ ′
(n − 1)! 1 r n 1
r sin θ ′
Pn (cos θ)Pn1 (cos θ ′ ) , r < r ′
′ r′
2
(n
+
1)!
r
n=1
∞
X
(n − 1)! 1 r ′ n+1 1
µ ′
′
= r sin θ
Pn (cos θ)Pn1 (cos θ ′ ) , r > r ′ ,
2
(n + 1)! r ′ r
Aφ (r, θ) =
(22)
n=1
which is written compactly as
Aφ (r, θ) =
∞
X
µ r′
(n − 1)! r< n 1
sin θ ′
Pn (cos θ)Pn1 (cos θ ′ ) .
2 r>
(n
+
1)!
r
>
n=1
(23)