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42 82 1
ππ
2
aR
R
=>
,
as one can check numerically (that
82
1
π
2
>
).
75. Let the length of each side of the square be
a
. The center of a square is a distance
a
/2
from the nearest side. There are four sides contributing to the field at the center. The
result is
()
0
center
2
2
22
4.
42
ii
a
B
aa
µ
π
⎛⎞
==
⎜⎟
⎝⎠
+
0
π
On the other hand, the magnetic field at the center of a circular wire of radius
R
is
0
/2
iR
(e.g., Eq. 2910). Thus, the problem is equivalent to showing that
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 Spring '08
 Any
 Physics

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