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A metal disc of radius : 'R' rorates with an angular velocity ' $\omega$ ' 'about an axis perpendicular to its plane passing through its centre in a magnetic' field of induction ' $\mathrm{B}$ ' acting perpendicular to the plane of the disc. The magnitude of induced. e.m.f. between the rim and axis of the disc is
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The correct answer is:
$\frac{\mathrm{BR}^2 \omega}{2}$
Induced e.m.f.,
$\mathrm{e}=\frac{-\mathrm{d} \phi}{\mathrm{dt}}=-\frac{\mathrm{d}(\mathrm{BA})}{\mathrm{dt}}$
$=-\mathrm{B} \frac{\mathrm{dA}}{\mathrm{dt}}$
$\ldots .(\because \mathrm{B}=$ constant $)$
Area swept between axis and the rim, $\mathrm{dA}=\pi \mathrm{R}^2$ Time during which the change in flux taking place, $\mathrm{dt}=\frac{2 \pi}{\omega}$
$\therefore \quad \mathrm{e}=\frac{-\mathrm{B} \pi \mathrm{R}^2}{2 \pi / \omega}=\frac{-\mathrm{B} \omega \mathrm{R}^2}{2}$
$\therefore \quad|\mathrm{e}|=\frac{\mathrm{B} \omega \mathrm{R}^2}{2}$
$\mathrm{e}=\frac{-\mathrm{d} \phi}{\mathrm{dt}}=-\frac{\mathrm{d}(\mathrm{BA})}{\mathrm{dt}}$
$=-\mathrm{B} \frac{\mathrm{dA}}{\mathrm{dt}}$
$\ldots .(\because \mathrm{B}=$ constant $)$
Area swept between axis and the rim, $\mathrm{dA}=\pi \mathrm{R}^2$ Time during which the change in flux taking place, $\mathrm{dt}=\frac{2 \pi}{\omega}$
$\therefore \quad \mathrm{e}=\frac{-\mathrm{B} \pi \mathrm{R}^2}{2 \pi / \omega}=\frac{-\mathrm{B} \omega \mathrm{R}^2}{2}$
$\therefore \quad|\mathrm{e}|=\frac{\mathrm{B} \omega \mathrm{R}^2}{2}$
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