A) \[2\pi fNB\]
B) \[2fNBA\]
C) \[2\pi fBA\]
D) \[2\pi fNBA\]
Correct Answer: D
Solution :
The maximum emf induced in the coil is given by \[e=NBA(2\pi f)\sin 90{}^\circ =nBA(2\pi f)\times 1\]\[=NBA(2\pi f)\]
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