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Unformatted text preview: 24. (a) We assume the ﬂux is entirely due to the ﬁeld generated by the long straight wire (which is given
by Eq. 3019). We integrate according to Eq. 313, not worrying about the possibility of an overall
minus sign since we are asked to ﬁnd the absolute value of the ﬂux.
r +b/2 ΦB  = r −b/2 µ0 i
2πr (a dr) = r+
µ0 ia
ln
2π
r− b
2
b
2 . (b) Implementing Faraday’s law involves taking a derivative of the ﬂux in part (a), and recognizing that
dr
dt = v . The magnitude of the induced emf divided by the loop resistance then gives the induced
current:
b
r+ 2
µ0 ia d
E
µ0 iabv
=−
ln
.
iloop =
=
b
R
2πR dt
2πR (r2 − (b/2)2 )
r− 2 ...
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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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