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62. (a) Eq. 3022 gives
0
2
222 T
2
ir
B
R
µ
π
==
.
(b) Eq. 3019 (or Eq. 306) gives
0
167 T
2
i
B
r
.
(c) As in part (b), we obtain a field of
0
22.7 T
2
i
B
r
.
(d) Eq. 3216 (with Eq. 3215) gives
0
2
1.25 T
2
d
ir
B
R
.
(e) As in part (d), we get
0
2
3.75 T
2
d
B
R
.
(f) Eq. 3217 yields
B
= 22.7
µ
T.
(g) Because the displacement current in the gap is spread over a larger crosssectional
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This note was uploaded on 06/03/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.
 Spring '08
 Any
 Physics

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