Lecture 11 Notes, 95.657 Electromagnetic Theory I, Fall 2010
Dr. Christopher S. Baird, UMass Lowell
1. Magnetic Fields of a Localized Current Distribution
 Similar to the electrostatic multipole expansion, if we have a localized current distribution and
want to know the magnetic fields far away, we can make a magnetostatic multiplole expansion and
only keep the first few terms.
 Consider the BiotSavart Law for the vector potential:
A
=
0
4
∫
J
x
'
∣
x
−
x
'
∣
d
x
'
where
B
x
=∇×
A
and this is a solution to
∇×
B
x
=
0
J
x
 The primed coordinates refer to the localized currents, so that an expansion in powers of the
primed variables will allow us to drop higher order terms:
1
∣
x
−
x
'
∣
=
1
∣
x
∣
x
⋅
x
'
∣
x
∣
3
...
 Using this expansion:
A
=
0
4
1
∣
x
∣
∫
J
x
'
d
x
'
0
4
x
∣
x
∣
3
⋅
∫
x
'
J
x
'
d
x
'
...
 The first term corresponds to the magnetic field generated
by a magnetic monopole moment. We
already know that there are no monopoles, so that this term should be zero. Let us prove it.
 Consider a functions
f
and
g
and make an expansion using integration by parts:
∫
V
g
J
⋅∇
'
f d
x
'
=
∫
S
f g
J
⋅
x
'
da
−
∫
V
f
∇
'
⋅
g
J
d
x
'
 Because the current is localized, if the bounding surface is larger than the current distribution, the
current density will be everywhere zero on the surface, so that:
∫
V
g
J
⋅∇
'
f d
x
'
=−
∫
V
f
∇
'
⋅
g
J
d
x
'
 Expand the right side using a vector identity
∫
V
g
J
⋅∇
'
f d
x
'
=−
∫
V
f
J
⋅∇
'
g d
x
'
−
∫
V
f g
∇
'
⋅
J
d
x
'
 Collect terms:
∫
V
g
J
⋅∇
'
f
f
J
⋅∇
'
g
f g
∇
'
⋅
J
d
x
'
=
0
 This holds for any functions
f
and
g
and vector
J
. We now apply it to our problem. Make
J
the
current density vector. The definition of magnetostatics is that there is no divergence:
∇
'
⋅
J
=
0
∫
V
g
J
⋅∇
'
f
f
J
⋅∇
'
g
d
x
'
=
0
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 Set
f
= 1 and
g
=
x
i
∫
V
J
i
d
x
'
=
0
 If the integral of every component is zero, than the integral of the whole vector is zero and we
have proven that the first term in the expansion is zero. If only the second term in the expansion is
kept and all higher order terms are dropped, the potential far from a localized current density
becomes:
A
=
0
4
1
∣
x
∣
3
∫
x
⋅
x
'
J
x
'
d
x
'
 Express this explicitly in terms of components:
A
=
0
4
1
∣
x
∣
3
∫
∑
j
x
j
x
j
'
∑
i
x
i
J
i
d
x
'
 Break the terms in half and shuffle things around:
A
=−
1
2
0
4
1
∣
x
∣
3
∫
d
x
'
∑
i
x
i
∑
j
x
j
[−
x
j
'
J
i
−
x
j
'
J
i
]
 If we set
f
=
x
i
' and
g
=
x
j
' in the identity on the bottom of the previous page, we get
∫
V
−
x
j
'
J
i
d
x
'
=
∫
V
x
i
'
J
j
d
x
'
 Use this to evaluate the integral in the potential equation:
A
=−
1
2
0
4
1
∣
x
∣
3
∫
d
x
'
∑
i
x
i
∑
j
x
j
[
x
i
'
J
j
−
x
j
'
J
i
]
A
=−
1
2
0
4
1
∣
x
∣
3
∫
d
x
'
x
[
y
x
'
J
y
−
y
'
J
x
−
z
z
'
J
x
−
x
'
J
z
]
y
[
z
y
'
J
z
−
z
'
J
y
−
x
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 Fall '11
 Staff
 Current, Mass, Magnetic Field

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