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A metal rod of length $L$ is placed normal to a magnetic field and rotated in a circular path with frequency $f$. The potential difference between its ends will be

1. $\pi L^2 Bf$
2. $BL/f$
3. $\pi L^2 B/f$
4. $fBL$

Correct answer: $\pi L^2 Bf$

Solution

When the rod rotates in a magnetic field, it sweeps out a circular area of radius $L$. The flux change per revolution is $B\pi L^2$, and with frequency $f$, the average emf is $B\pi L^2 f$. Therefore the potential difference is $\pi L^2 Bf$.

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