MHT CET · Physics · 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 .
- A \(\mathrm{B} \pi \mathrm{R}^2\)
- B \(\frac{2 B \pi^2 \mathrm{R}^2}{\omega}\)
- C \(\mathrm{B} \pi \mathrm{R}^2 \omega\)
- D \(\frac{\mathrm{BR}^2 \omega}{2}\)
Answer & Solution
Correct Answer
(D) \(\frac{\mathrm{BR}^2 \omega}{2}\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned}
& \text { Induced e.m.f., } \\
& \begin{aligned}
\mathrm{e} & =\frac{-\mathrm{d} \phi}{\mathrm{dt}}=-\frac{\mathrm{d}(\mathrm{BA})}{\mathrm{dt}} \\
& =-\mathrm{B} \frac{\mathrm{dA}}{\mathrm{dt}} \quad \ldots .(\because \mathrm{B}=\text { constant })
\end{aligned}
\end{aligned}\)
Area swept between axis and the rim, \(\mathrm{dA}=\pi \mathrm{R}^2\) Time during which the change in flux taking place, \(\mathrm{dt}=\frac{2 \pi}{\omega}\)
\(\begin{aligned}
\therefore \quad & e=\frac{-B \pi R^2}{2 \pi / \omega}=\frac{-B \omega R^2}{2} \\
& |e|=\frac{B R^2 \omega}{2}
\end{aligned}\)
& \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, \(\mathrm{dt}=\frac{2 \pi}{\omega}\)
\(\begin{aligned}
\therefore \quad & e=\frac{-B \pi R^2}{2 \pi / \omega}=\frac{-B \omega R^2}{2} \\
& |e|=\frac{B R^2 \omega}{2}
\end{aligned}\)
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