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23. (a) Eq. 2910 gives the field at the center of the large loop with
R
= 1.00 m and
current
i
(
t
). This is approximately the field throughout the area (
A
= 2.00
×
10
–4
m
2
)
enclosed by the small loop. Thus, with
B
=
µ
0
i
/2
R
and
i
(
t
) =
i
0
+
kt
, where
i
0
= 200 A and
k
= (–200 A – 200 A)/1.00 s = – 400 A/s,
we find
(a)
()
( )
7
4
00
4
10 H/m
200A
(0
)
1
.
2
6
1
0T
,
2
2 1.00m
i
Bt
R
−
−
π×
==
=
= ×
(b)
( )( )
7
4
10 H/m
200A
400A/s
0.500s
(
0.500s)
0,
21
.00m
−
−
⎡⎤
⎣⎦
=
and
(c)
( )( )
7
4
4
10 H/m
200A
400A/s 1.00s
(
1.00s)
1.26 10 T,
−
−
−
=
−
×
or
4
 (
1.00s) 1.26 10 T.
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 Spring '08
 Any
 Physics, Current

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