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A conducting circular loop is placed in a uniform magnetic field of $0.04 \mathrm{~T}$ with its plane perpendicular to the magnetic field. The radius of the loop starts shrinking at $2 \mathrm{~mm} / \mathrm{s}$. The induced emf in the loop when the radius is $2 \mathrm{~cm}$ is
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The correct answer is:
$3.2 \pi \mu \mathrm{V}$
Induced emf in the loop is given by
$\mathrm{e}=-\mathrm{B} \cdot \frac{\mathrm{d} \mathrm{A}}{\mathrm{dt}}$ where $\mathrm{A}$ is the area of the loop.
$$
\begin{array}{l}
\mathrm{e}=-\mathrm{B} \cdot \frac{\mathrm{d}}{\mathrm{dt}}\left(\pi \mathrm{r}^{2}\right)=-\mathrm{B} \pi 2 \mathrm{r} \frac{\mathrm{dr}}{\mathrm{dt}} \\
\mathrm{r}=2 \mathrm{~cm}=2 \times 10^{-2} \mathrm{~m} \\
\mathrm{dr}=2 \mathrm{~mm}=2 \times 10^{-3} \mathrm{~m} \\
\mathrm{dt}=1 \mathrm{~s} \\
\mathrm{e}=-0.04 \times 3.14 \times 2 \times 2 \times 10^{-2} \times \frac{2 \times 10^{-3}}{1} \mathrm{~V} \\
=0.32 \pi \times 10^{-5} \mathrm{~V} \\
=3.2 \pi \times 10^{-6} \mathrm{~V} \\
=3.2 \pi \mu \mathrm{V}
\end{array}
$$
$\mathrm{e}=-\mathrm{B} \cdot \frac{\mathrm{d} \mathrm{A}}{\mathrm{dt}}$ where $\mathrm{A}$ is the area of the loop.
$$
\begin{array}{l}
\mathrm{e}=-\mathrm{B} \cdot \frac{\mathrm{d}}{\mathrm{dt}}\left(\pi \mathrm{r}^{2}\right)=-\mathrm{B} \pi 2 \mathrm{r} \frac{\mathrm{dr}}{\mathrm{dt}} \\
\mathrm{r}=2 \mathrm{~cm}=2 \times 10^{-2} \mathrm{~m} \\
\mathrm{dr}=2 \mathrm{~mm}=2 \times 10^{-3} \mathrm{~m} \\
\mathrm{dt}=1 \mathrm{~s} \\
\mathrm{e}=-0.04 \times 3.14 \times 2 \times 2 \times 10^{-2} \times \frac{2 \times 10^{-3}}{1} \mathrm{~V} \\
=0.32 \pi \times 10^{-5} \mathrm{~V} \\
=3.2 \pi \times 10^{-6} \mathrm{~V} \\
=3.2 \pi \mu \mathrm{V}
\end{array}
$$
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