A metal rod of length ' l ' rotates about one of its ends in a...

The induced emf $e$ across the ends of a metal rod of length $l$ rotating with angular velocity $\omega$ in a perpendicular magnetic field $B$ is given by:

$e = \dfrac{1}{2} B \omega l^2$

The angular velocity $\omega$ is related to the frequency $f$ (number of revolutions per second) by:

$\omega = 2 \pi f$

Substituting the value of $\omega$ in the expression for emf:

$e = \dfrac{1}{2} B (2 \pi f) l^2$

$e = B \pi f l^2$

Rearranging for $f$, we get:

$f = \dfrac{e}{B \pi l^2}$

Answer: $\dfrac{e}{B\pi l^2}$