A metal disc of radius $R$ rotates with an angular velocity $\omega$ about an axis perpendicular to its…

A metal disc of radius $R$ rotates with an angular velocity $\omega$ about an axis perpendicular to its plane passing through its centre in a magnetic field of induction B acting perpendicular to the plane of the disc. The induced e.m.f. between the rim and axis of the disc is .
  1. $\mathrm{B} \pi \mathrm{R}^2$
  2. $\frac{2 B \pi^2 \mathrm{R}^2}{\omega}$
  3. $\mathrm{B} \pi \mathrm{R}^2 \omega$
  4. $\frac{\mathrm{BR}^2 \omega}{2}$

Solution

$\begin{aligned} & \text { Induced e.m.f., } \\ & \begin{aligned} \mathrm{e} & =\frac{-\mathrm{d} \phi}{\mathrm{dt}}=-\frac{\mathrm{d}(\mathrm{BA})}{\mathrm{dt}} \\ & =-\mathrm{B} \frac{\mathrm{dA}}{\mathrm{dt}} \quad \ldots .(\because \mathrm{B}=\text { constant }) \end{aligned} \end{aligned}$
Area swept between axis and the rim, $\mathrm{dA}=\pi \mathrm{R}^2$ Time during which the change in flux taking place, $\mathrm{dt}=\frac{2 \pi}{\omega}$ $\begin{aligned} \therefore \quad & e=\frac{-B \pi R^2}{2 \pi / \omega}=\frac{-B \omega R^2}{2} \\ & |e|=\frac{B R^2 \omega}{2} \end{aligned}$

Asked in: MHT CET 2024 (15 May Shift 2)

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