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Unformatted text preview: 73. (a) The ﬁeld in this region is entirely due to the long wire (with, presumably, negligible thickness).
Using Eq. 3019,
µ0 iw
B  =
= 4.8 × 10−3 T
2πr
where iw = 24 A and r = 0.0010 m.
(b) Now the ﬁeld consists of two contributions (which are antiparallel) – from the wire (Eq. 3019) and
from a portion of the conductor (Eq. 3022 modiﬁed for annular area):
B  =
= µ0 ienc
µ0 iw
−
2πr
2πr
2
µ0 ic π r2 − πRi
µ0 iw
−
2 − πR2
2πr
2πr πRo
i where r = 0.0030 m, Ri = 0.0020 m, Ro = 0.0040 m and ic = 24 A. Thus, we ﬁnd B  = 9.3 × 10−4 T.
(c) Now, in the external region, the individual ﬁelds from the two conductors cancel completely (since
ic = iw ): B = 0. ...
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 Fall '08
 SPRUNGER
 Physics

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