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A copper disc of radius $0.1 \mathrm{~m}$ rotates about an axis passing through its centre and perpendicular to its plane with 10 revolutions per second in a uniform transverse magnetic field of $0.1 \mathrm{~T}$. The emf induced across the radius of the disc is
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The correct answer is:
$10 \pi \mathrm{mV}$
Consider a small radial segment of $\mathrm{dx}$ at a distance from the centre of the disc.
Velocity of this segment, $\mathrm{v}=\omega \mathrm{x}$ emf induced across this segment
$$
\begin{aligned}
& \mathrm{d} \in==\mathrm{Blv} \\
& =\mathrm{B} \mathrm{x} \omega \mathrm{dx}
\end{aligned}
$$
Total emf induced across radius of disc $r$.
$$
\epsilon=\int_0^{\mathrm{r}} \mathrm{B} x \omega \mathrm{dx}=\frac{\mathrm{B} \omega \mathrm{r}^2}{2}
$$
Here, $\mathrm{B}=0.1 \mathrm{~T} \mathrm{w}=10$ revolution $/ \mathrm{s}=20 \pi \mathrm{rads}$. $\mathrm{r}=0.1 \mathrm{~m}$
$$
\begin{aligned}
& \epsilon=\frac{0.1 \times 20 \pi \times(0.1)^2}{2} \\
& =\pi \times 0.1 \times 0.1 \\
& \epsilon=10 \pi \mathrm{mv}
\end{aligned}
$$
Velocity of this segment, $\mathrm{v}=\omega \mathrm{x}$ emf induced across this segment
$$
\begin{aligned}
& \mathrm{d} \in==\mathrm{Blv} \\
& =\mathrm{B} \mathrm{x} \omega \mathrm{dx}
\end{aligned}
$$
Total emf induced across radius of disc $r$.
$$
\epsilon=\int_0^{\mathrm{r}} \mathrm{B} x \omega \mathrm{dx}=\frac{\mathrm{B} \omega \mathrm{r}^2}{2}
$$
Here, $\mathrm{B}=0.1 \mathrm{~T} \mathrm{w}=10$ revolution $/ \mathrm{s}=20 \pi \mathrm{rads}$. $\mathrm{r}=0.1 \mathrm{~m}$
$$
\begin{aligned}
& \epsilon=\frac{0.1 \times 20 \pi \times(0.1)^2}{2} \\
& =\pi \times 0.1 \times 0.1 \\
& \epsilon=10 \pi \mathrm{mv}
\end{aligned}
$$
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