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Homework 11 Answers, 95.657 Fall 2011, Electromagnetic Theory I
Dr. Christopher S. Baird, UMass Lowell
Problem 1
Jackson 5.17
A current distribution
J
(
x
) exists in a medium of unit relative permeability adjacent to a semiinfinite
slab of material having relative permeability
μ
r
and filling the halfspace,
z
< 0.
(a) Show that for
z
> 0 the magnetic induction can be calculated by replacing the medium of
permeability
μ
r
by an image current distribution
J
*, with components,
r
−
1
r
1
J
x
x , y ,
−
z
,
r
−
1
r
1
J
y
x , y ,
−
z
,
−
r
−
1
r
1
J
z
x , y ,
−
z
(b) Show that for z < 0 the magnetic induction appears to be due to a current distribution [2
μ
r
/(
μ
r
+ 1)]
J
in a medium of unit relative permeability.
SOLUTION:
(a) Using the method of images, we replace the effects of the actual currents on the material interface
with an image current deep within the magnetic material. We will label the original current distribution
J
(
x
) and the image current distribution as
J
*(
x
), where
J
is known but its image at this point needs to
be determined. Because the mirror surface is a flat plane, we can safely assume that a piece of current
in
J
at (
x
,
y
,
z
) will be mirrored by a piece of current in
J
* at (
x
,
y
, 
z
). Therefore:
J
i
*
x , y , z
=
A
i
J
i
x , y ,
−
z
for each component, so that
i
=
x
,
y
,
z
The magnetic
B
field for
z
> 0 created by these currents are:
B
z
0
x
=
0
4
∫
J
x
'
J
*
x
'
×
x
−
x
'
∣
x
−
x
'
∣
3
d
3
x
'
For the
z
< 0 region, place an additional image current distribution
J
** =
a
J
in its positive
z
location,
which creates the field:
B
z
0
x
=
0
r
a
4
∫
J
x
'
×
x
−
x
'
∣
x
−
x
'
∣
3
d
3
x
'
Now apply boundary conditions:
[
B
2
−
B
1
⋅
n
=
0
]
on S
[
B
z
0
⋅
z
=
B
z
0
⋅
z
]
z
=
0
[
0
4
∫
J
x
'
J
*
x
'
×
x
−
x
'
∣
x
−
x
'
∣
3
d
3
x
'
]
z
=
0
⋅
z
=
[
0
r
a
4
∫
J
x
'
×
x
−
x
'
∣
x
−
x
'
∣
3
d
3
x
'
]
z
=
0
⋅
z
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View Full DocumentBecause each piece mirrors each other piece, and because we are on the z = 0 plane, which is
equidistant from each current piece and its image, the integrands must be equal.
[
J
x
'
J
*
x
'
×
x
−
x
'
∣
x
−
x
'
∣
3
]
z
=
0
⋅
z
=
[
r
a
J
x
'
×
x
−
x
'
∣
x
−
x
'
∣
3
]
z
=
0
⋅
z
[
J
x
'
J
*
x
'
×
x
−
x
'
]
z
=
0
⋅
z
=
[
r
a
J
x
'
×
x
−
x
'
]
z
=
0
⋅
z
If we break each current into a
z
component and transverse component,
J
=
J
z
z
J
t
, we have:
[
J
z
z
J
t
J
z
*
z
J
t
*
×
x
−
x
'
]
z
=
0
⋅
z
=
[
r
a
J
z
z
J
t
×
x
−
x
'
]
z
=
0
⋅
z
Use the identity
a
⋅
b
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 Fall '11
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 Current, Mass, Work

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