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If the coefficient of mutual induction of the primary and secondary coils of an induction coil is $5 \mathrm{H}$ and a current of $10 \mathrm{~A}$ is cut off in $5 \times 10^{-4}$ second, the $e m f$ induced(in volt) in the secondary coil is
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Verified Answer
The correct answer is:
$1 \times 10^{5}$
$\mathrm{M}=5 \mathrm{H}, \mathrm{I}=10 \mathrm{~A}, \mathrm{t}=5 \times 10^{-4} \mathrm{~s}$.
Now, emf induced in secondary is
$\begin{array}{l}
\mathrm{e}=-\mathrm{M} \frac{\mathrm{d}}{\mathrm{dt}}, \quad \text { (-ve sign shows direction) } \\
\Rightarrow \mathrm{e}=5 \times \frac{10}{5 \times 10^{-4}}=1 \times 10^{+5} \mathrm{~V}
\end{array}$
Now, emf induced in secondary is
$\begin{array}{l}
\mathrm{e}=-\mathrm{M} \frac{\mathrm{d}}{\mathrm{dt}}, \quad \text { (-ve sign shows direction) } \\
\Rightarrow \mathrm{e}=5 \times \frac{10}{5 \times 10^{-4}}=1 \times 10^{+5} \mathrm{~V}
\end{array}$
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