Question 238: Electromagnetic Induction

Question

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

Accepted Answer

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

Answer

$\mathrm{B} \pi \mathrm{R}^2$

Answer

$\frac{2 \mathrm{~B} \pi^2 \mathrm{R}^2}{\omega}$

Answer

$\mathrm{B} \pi \mathrm{R}^2 \omega$

Answer

$\frac{\mathrm{BR}^2 \omega}{2}$

Electromagnetic Induction

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

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