MHT CET · Physics · Electromagnetic Induction
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
- A \(\pi \mathrm{BR}^2\)
- B \(\frac{2 \pi^2 \mathrm{BR}^2}{\omega}\)
- C \(\pi \mathrm{BR}^2 \omega\)
- D \(\frac{\mathrm{BR}^2 \omega}{2}\)
Answer & Solution
Correct Answer
(D) \(\frac{\mathrm{BR}^2 \omega}{2}\)
Step-by-step Solution
Detailed explanation
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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