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A copper disc of radius $0.1\,\text{m}$ is rotated about its centre at 10 revolutions per second in a uniform magnetic field of $0.1\,\text{T}$, with its plane perpendicular to the field. The emf induced across the radius of the disc is:

1. $10\pi\,\text{V}$
2. $10\sqrt{2}\pi\,\text{V}$
3. $2\times 10^{-1}\pi\,\text{V}$
4. $2\times 10^{-2}\pi\,\text{V}$

Correct answer: $2\times 10^{-2}\pi\,\text{V}$

Solution

For a rotating disc, the emf across the radius is $\mathcal{E}=\frac{1}{2}B\omega r^2$. Here $\omega=2\pi\times 10=20\pi\,\text{rad/s}$, so $\mathcal{E}=\frac{1}{2}(0.1)(20\pi)(0.1)^2=0.01\pi\,\text{V}=2\times 10^{-2}\pi\,\text{V}$.

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