A) \[\frac{T}{8}\]
B) \[\frac{T}{4}\]
C) \[\frac{3T}{8}\]
D) \[\frac{T}{2}\]
Correct Answer: D
Solution :
Assume the area of the flat coil of copper wire to be A. Then the flux linking the coil is \[\phi =BA={{B}_{0}}A\sin \left( \frac{2\pi t}{T} \right)\] Thus, the induced emf in the coil is given by \[e=\frac{d\phi }{dt}=-\frac{2\pi {{B}_{0}}A}{T}\cos \left( \frac{2\pi t}{T} \right)\] Maximum of E occurs when \[cos\frac{2\pi t}{T}=-1\] Since\[e=-\frac{d\phi }{dt}=-A=\frac{dB}{dt}\]is the maximum it is obvious from the graph that this occurs at \[t=T/2\]You need to login to perform this action.
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