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60. (a) Eq. 3022 applies for
r<c
. Our sign choice is such that
i
is positive in the smaller cylinder and
negative in the larger one.
B
=
µ
0
ir
2
πc
2
for
r
≤
c.
(b) Eq. 3019 applies in the region between the conductors.
B
=
µ
0
i
2
πr
for
c
≤
r
≤
b.
(c) Within the larger conductor we have a superposition of the Feld due to the current in the inner
conductor (still obeying Eq. 3019) plus the Feld due to the (negative) current in the that part of
the outer conductor at radius less than
r
(see part (a) of problem 59 for more details). The result
is
B
=
µ
0
i
2
−
µ
0
i
2
µ
r
2
−
b
2
a
2
−
b
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This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.
 Fall '08
 SPRUNGER
 Physics

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