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A metal disc of radius $30 \mathrm{~cm}$ rotates with a constant angular velocity $\omega=100 \mathrm{rad} / \mathrm{s}$ about its axis. Find the magnitude of potential difference between the centre and the rim of the disc of external uniform magnetic field of induction $\mathrm{B}=4 \mathrm{mT}$ is directed perpendicular to the disc.
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Verified Answer
The correct answer is:
$18 \mathrm{mV}$
Let us consider a small circular strip at distance ' $\mathrm{r}$ ' from centre having width ' $\mathrm{dr}$ '.
So, area of strip $=2 \pi \mathrm{rdr}$ $\mathrm{d} \phi=\mathrm{B} \times 2 \pi \mathrm{r} \mathrm{dr}$
$$
\phi=\mathrm{B} \times 2 \pi \times \int_0^{\mathrm{R}} \mathrm{rdr}=2 \pi \mathrm{B} \cdot \frac{\mathrm{R}^2}{2}=\mathrm{B} \pi \mathrm{R}^2
$$
So, $\mathrm{e}=\frac{\phi}{\mathrm{T}}=\frac{\mathrm{B} \pi \mathrm{R}^2}{\mathrm{~T}}=\frac{\mathrm{B} \times \pi \mathrm{R}^2}{2 \pi} \times \omega=\frac{1}{2} \mathrm{BR}^2 \omega$
Putting the value of $B, R$ and $\omega$, we get $e=18 \mathrm{mV}$
So, area of strip $=2 \pi \mathrm{rdr}$ $\mathrm{d} \phi=\mathrm{B} \times 2 \pi \mathrm{r} \mathrm{dr}$
$$
\phi=\mathrm{B} \times 2 \pi \times \int_0^{\mathrm{R}} \mathrm{rdr}=2 \pi \mathrm{B} \cdot \frac{\mathrm{R}^2}{2}=\mathrm{B} \pi \mathrm{R}^2
$$
So, $\mathrm{e}=\frac{\phi}{\mathrm{T}}=\frac{\mathrm{B} \pi \mathrm{R}^2}{\mathrm{~T}}=\frac{\mathrm{B} \times \pi \mathrm{R}^2}{2 \pi} \times \omega=\frac{1}{2} \mathrm{BR}^2 \omega$
Putting the value of $B, R$ and $\omega$, we get $e=18 \mathrm{mV}$
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