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12. (a) Here, the enclosed electric flux is found by integrating
3
2
00
1
2
(0.500 V/m s)(2 )
1
23
rr
E
E
rdr
t
rdr
t
r
R
R
ππ
π
⎛⎞
Φ=
=
⋅
−
=
−
⎜⎟
⎝⎠
∫∫
with SI units understood.
Then (after taking the derivative with respect to time) Eq. 323
leads to
3
2
1
(2
)
r
Br
r
R
πε
µ
=−
.
For
r
= 0.0200 m and
R
= 0.0300 m, this gives
B
= 3.09
×
10
−
20
T.
(b) The integral shown above will no longer (since now
r
>
R
) have
r
as the upper limit;
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 Spring '08
 Any
 Physics

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