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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. 30-19). We integrate according to Eq. 31-3, 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+
2 . (b) Implementing Faraday’s law involves taking a derivative of the ﬂux in part (a), and recognizing that
dt = v . The magnitude of the induced emf divided by the loop resistance then gives the induced
µ0 ia d
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