A metal rod of length ' 1 ' rotates about one of its ends in a plane perpendicular to a magnetic field of induction ' \(B\) '. If the e.m.f. induced between the ends of the rod is ' \(\mathrm{e}\) ' then the number of revolutions made by the rod per second is

The induced emf e \(=\frac{\mathrm{d} \phi}{\mathrm{dt}}=\frac{\mathrm{d}}{\mathrm{dt}}(\mathrm{AB})=\mathrm{B} \frac{\mathrm{dA}}{\mathrm{dt}}\)
In one rotation the rod traces out a circle of radius 1 or an area \(\mathrm{A}=\pi \mathrm{l}^2\)
So, in \(n\) rotations it will trace out an area of \(n \pi l^2\)
\(\therefore \mathrm{e}=\mathrm{B} \frac{\Delta \mathrm{A}}{\Delta \mathrm{t}}=\mathrm{B} \cdot \frac{\mathrm{n} \pi \mathrm{l}^2}{\Delta \mathrm{t}} \)
\( \text {But } \frac{\mathrm{n}}{\Delta \mathrm{t}}=\frac{\text {No.of rev.performed }}{\text { time }}=\) \(\text { frequency } \mathrm{f} \)
\( \therefore \mathrm{e}=\mathrm{Bf} \pi \mathrm{l}^2=\text { Bf.A } \)
\( \therefore \mathrm{f}=\frac{\mathrm{e}}{\mathrm{B} \pi \mathrm{l}^2}\)