Homework 11 Answers, 95.657 Electromagnetic Theory I
Dr. Christopher S. Baird, UMass Lowell
Jackson 5.18
A circular loop of wire having a radius
a
and carrying a current
I
is located in vacuum with its center a
distance
d
away from a semiinfinite slab of permeability
μ
. Find the force acting on the loop when
(a) the plane of the loop is parallel to the face of the slab,
SOLUTION:
We can place the surface of the semiinfinite slab along the
z
= 0 plane, and center the loop on the z
axis. We can split the problem into to two separate regions and find the solution in each region and then
link them using boundary conditions. In the region where
z
>
0, there is no magnetic material, so we
can use all the regular equations that do not take materials into effect. We use the method of images in
this region to simulate the effects of the boundary. Place a circular loop of wire having a radius
a
and
carrying a current
I'
centered at the location
z
= 
d
.
The vector potential due to a circular loop in vacuum located at
z
= 0 was found previously to be:
A
=
0
I a
4
∫
0
2
d
'
cos
'
a
2
r
2
−
2
a r
sin
cos
'
Switch this into Cartesian coordinates:
A
=
0
I a
4
∫
0
2
d
'
cos
'
a
2
x
2
y
2
z
2
−
2
a
x
2
y
2
cos
'
Now use a translated form of this for each ring and switch back to spherical coordinates:
A
upper
=
0
a
4
∫
0
2
d
'cos
'
[
I
1
a
2
d
2
r
2
−
2
r
cos
d
−
2
ar
sin
cos
'
I
'
1
a
2
d
2
r
2
2
r
cos
d
−
2
a r
sin
cos
'
]
For the region
z
< 0, there is no current, so we only need a image current
I''
at
z
=
d
to simulate the
effects of the boundary, leading to the vector potential:
A
lower
=
a
4
∫
0
2
d
'cos
'
I
''
1
a
2
d
2
r
2
−
2
r
cos
d
−
2
a r
sin
cos
'
Now match up both regions using the boundary conditions
[
B
upper
⋅
n
=
B
lower
⋅
n
]
on
S
[∇×
A
upper
⋅
=∇×
A
lower
⋅
]
=/
2
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∂
∂
r
r A
,
upper
=
∂
∂
r
r A
,
lower
]
=/
2
After much algebra, this leads to:
0
I
I
'
=
I
''
The other boundary condition is:
[
n
×
H
upper
=
n
×
H
lower
]
on
S
(because there is no free current at the boundary surface)
[
1
0
×∇×
A
upper
=
1
×∇×
A
lower
]
=/
2
[
1
0
∂
∂
sin
A
,
upper
=
1
∂
∂
sin
A
,
lower
]
=/
2
After much algebra, this leads to:
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 Spring '11
 Avery
 Current, Magnetism, Mass, Work, Magnetic Field, Cos, cos cos, Aupper

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