1
PHYS809Class25Notes
Magnetic dipole moment
The dipole part of the scalar potential for a current loop can be written in the form
(
)
0
3
,
0 ,
4
md
r
r
μ
π
⋅
Φ
=
>
m r
(25.1)
where for a circular loop of radius
a
and unit normal
k
,
2
.
a I
π
=
m
k
(25.2)
(Note that the potential in equation (25.1) is such that the magnetic induction
B
is related to the scalar
potential by
.
= ∇Φ
B
Later we will see that it is more natural to relate the magnetic field
H
to the
scalar potential. In vacuum,
0
,
μ
=
B
H
and hence the potentials differ by a constant factor of
0
.
μ
)
The dipole magnetic induction is given by
(
)
(
)
0
3
ˆ ˆ
3
,
0 .
4
md
r
r
μ
π
⋅

=
>
m r r
m
B
(25.3)
For a general loop, the magnetic dipole moment is
,
I
=
m
S
(25.4)
where
S
is the vector area of the loop.
By applying Stokes’ theorem to a vector field
( )
,
=
×
A r
k
r
where
k
is arbitrary but fixed, it can be
shown that
1
.
2
loop
d
=
×
∫
S
r
r
(25.5)
Hence an alternative expression for the magnetic dipole moment is
.
2
loop
I
d
=
×
∫
m
r
r
(25.6)
For a general current density, the equivalent expression is
( )
3
1
.
2
allspace
d x
=
×
∫
m
r
J r
(25.7)
The vector potential of magnetic dipole
The vector potential of a magnetic dipole is
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2
0
3
.
4
md
r
μ
π
×
=
m
r
A
(25.8)
Taking the curl gives the dipole magnetic field
(
)
0
0
3
3
3
.
4
4
md
md
r
r
r
μ
μ
π
π
×
= ∇×
=
∇×
=
∇⋅

⋅∇
m
r
r
r
B
A
m
m
(25.9)
The first term gives a delta function contribution and the second term is the same as for the electric
dipole. The result is
( )
(
)
(
)
( )
0
0
0
3
ˆ ˆ
3
ˆ ˆ
.
4
md
r
μ
μ
δ
μ
δ
π
⋅

=
+

⋅
m r r
m
B
m
r
m r r
r
(25.10)
Hence there is a difference between the singular part of the dipole field from a current loop and the
dipole field from a pair of oppositely charged particles. This difference is relevant to atomic and
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 Fall '11
 MacDonald
 Current, Magnetic Field, magnetic dipole, aφ

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