PHYSICS 505: CLASSICAL ELECTRODYNAMICS HOMEWORK 8
3
It is clear that when we take the cross product of this with J × B, we will obtain
a horrendous expression of trigonometric functions. Luckily, as in the case above,
many of these will automatically vanish under integration over ϕ. Despite our nobel
intentions, we decided to evaluate this using a computer algebra package. After a
surprisingly long amount of computation, it determined that


Z 2π
− cos θ0
.
cos θ0
N=
x(ϕ) × (J(ϕ) × B(ϕ))dϕ = aIB0 π 
0
sin θ0 (cos φ0 − sin φ0 )
‘
’
óπ²ρ
²́δ²ι
δ²ιξαι
Now, we remark that based on this calculation, it is only important that the magnetic
dipole moment m remain unchanged. This will be the case for any planar loop of
current lying in the same plane as our circle with equal area. Therefore, the shape is
irrelevant as long as the area remains constant and the figure lies in the same plane.
Problem 5.13
Let us consider a sphere of radius a which carries a uniform surface charge distribution σ(θ). If the
sphere is rotated at constant angular velocity ω, then the associated current will induce a magnetic
potential. We are to determine the vector potential inside and outside the sphere.
Confined to the surface of the sphere (although we are leaving out the δ-functions for brevity), we see
that the current will be
J(θ) = ρv = σω sin θφ̂.
Now, following Jackson’s discussion near equation (5.32), we know that the vector potential can be
expressed
Z
Z
µ0
J(r0 ) 3 0
µ0
sin θ0 φ̂0
3
A(r) =
d
x
=
σωa
dΩ.
0
4π |r − r |
4π
|r − r0 |
We can expand 1/|r − r0 | of the integrand in terms of the spherical harmonics. Doing this, we see that
ZX
∞ X̀
`
µ0
4π r−
∗
3
(θ0 , φ0 ) [sin θ0 (− sin φ0 x̂ + cos φ0 ŷ)] dΩ.
A(r) =
Y`m (θ, φ)Y`m
σωa
`+1
4π
2` + 1 r+
`=0 m=−`
After a lot of initial frustration, it becomes clear that the the expression in square-brackets can be
expressed in terms of the spherical harmonics too. Therefore, by the orthogonality conditions, only one
term will remain in the sum. Notice that
r
3
Y11 (θ, φ) = −
sin θe−φ ,
8π
and therefore,
r
r
8π
8π
0
0
0
0
0
0
sin θ sin φ = −
Im [Y11 (θ , φ )]
and
sin θ cos φ = −
Re [Y11 (θ0 , φ0 )] .
3
3
Hence, we can compute directly,
ZX
∞ X̀
`
r−
1
∗
A(r) = µ0 σωa3
Y (θ, φ)Y`m
(θ0 , φ0 ) [sin θ0 (− sin φ0 x̂ + cos φ0 ŷ)] dΩ,
`+1 `m
2` + 1 r+
`=0 m=−`
r
Z ∞
n
o
`
r−
8π X X̀
1
∗
0
0
0
0
0
0
Y
(θ,
φ)Y
(θ
,
φ
)
−
Im
[Y
(θ
,
φ
)]
x̂
+
Re
[Y
(θ
,
φ
)]
ŷ
dΩ,
= −µ0 σωa3
11
11
`m
`+1 `m
3
2` + 2 r+
`=0 m=−`
r
Z
n
o
1
8π
r−
3
∗
0
0
0
0
0
0
= − µ0 σωa
Y
(θ,
φ)Y
(θ
,
φ
)
−
Im
[Y
(θ
,
φ
)]
x̂
+
Re
[Y
(θ
,
φ
)]
ŷ
dΩ,
11
11
11
11
2
3
3
r+
1
r−
= µ0 σωa3 2 sin θ (− sin φx̂ + cos φŷ) ,
3
r+
∴ A(r) =
r−
1
µ0 σωa3 2 sin θφ̂.
3
r+
4
JACOB LEWIS BOURJAILY
Inside
Inside the sphere, we have that r+ = a and r− = |r|. Inserting this directly into the expression above,
we wee that
1
A(r) = µ0 σωa sin θrφ̂.
3
The magnetic flux density is simply the curl of the vector potential, which is trivially computed to be
B(r) = ∇ × A(r),
1
= µ0 σωa∇ × sin θrφ̂,
3
·µ
µ
¶¶
µ
¶ ¸
¢
¢
1
1
∂ ¡
1 ∂ ¡ 2
= µ0 σωa
r sin2 θ
r̂ −
r sin θ θ̂ ,
3
r sin θ ∂θ
r ∂r
·µ
¶
¸
1
1
= µ0 σωa
2r sin θ cos θ r̂ − 2 sin θθ̂ ,
3
r sin θ
³
´
2
∴ Bint (r) = µ0 σωa cos θr̂ − sin θθ̂ .
3
‘
’
óπ²ρ
²́δ²ι
δ²ιξαι
Outside
Outside the sphere, we have that r+ = |r| and r− = a. Inserting this directly into the expression
above, we wee that
1
1
A(r) = µ0 σωa4 2 sin θφ̂.
3
r
The magnetic flux density is simply the curl of the vector potential, which is trivially computed to be
B(r) = ∇ × A(r),
sin θ
1
µ0 σωa4 ∇ × 2 φ̂,
3
r
·µ
µ µ 2 ¶¶¶
µ
µ
¶¶ ¸
1
1
∂ sin θ
1 ∂ sin θ
= µ0 σωa4
r̂
−
θ̂ ,
3
r sin θ ∂θ
r2
r ∂r
r
·µ
¶
¸
1
1 2 sin θ cos θ
sin θ
= µ0 σωa4
r̂
+
θ̂
,
3
r sin θ
r2
r3
´
1
a4 ³
∴ Bint (r) = µ0 σω 3 2 cos θr̂ + sin θθ̂ .
3
r
=
‘
’
óπ²ρ
²́δ²ι
δ²ιξαι
Problem 5.15
Let us consider two long, straight, parallel wires separated by a distance d, carrying current I in
opposite directions. We are to describe the magnetic field H in terms of a magnetic scalar potential ϕM
where H = −∇ϕM .
a) In the limit of d → 0, we are to show that the two-dimensional dipole approaches
¡
¢
Id sin φ
+ O d2 /ρ2 ,
ϕM ≈ −
2πρ
where ρ and φ are the standard polar coordinates.
From our work in class, we know that the magnetic scalar potential of a single infinitely
Iφ
straight wire along the z-axis is simply given by ϕM = − 2π
. This is also obvious from
Ampère’s law. Therefore, because the scalar potentials will linearly superimpose, the
the potential from the two wires is simply given by
I
(φ2 − φ1 ) ,
2π
where φ1 (φ2 ) is the polar angle from wire 1 (2). However, this is not quite adequate
for our discussion because we are working in a frame where neither of the two wires
are collinear with the z-axis. Therefore, we should express this result in terms of the
total polar angle φ.
ϕM =