A metal rod of length ' $l$ ' rotates about one of its ends in a plane perpendicular to a magnetic field of…

A metal rod of length ' $l$ ' 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 ' $e$ ', then the number of revolutions made by the rod per second is
  1. $\frac{\mathrm{e}}{\mathrm{B} \pi^2 l}$
  2. $\frac{\mathrm{e}}{\mathrm{B} \pi l^2}$
  3. $\frac{\mathrm{B}^2}{\mathrm{e} \pi l}$
  4. $\frac{\pi l^2}{\mathrm{eB}}$

Solution

A conducting rod of length ' $l$ ' whose one end is fixed, is rotated about the axis passing through its fixed end and perpendicular to its length with constant angular velocity $\omega$.
$\begin{aligned} & \mathrm{e}=\mathrm{B} \pi l^2 \mathrm{n} \\ \therefore \quad \mathrm{n} & =\frac{\mathrm{e}}{\pi \mathrm{B} l^2}\end{aligned}$

Asked in: MHT CET 2024 (09 May Shift 1)

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