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Two coils $\mathrm{P}$ and S have a mutual inductance of $3 \times 10^{-3} \mathrm{H}$. If the current in the coil, P is $I=20 \sin (50 \pi t) \mathrm{A}$, then the maximum value of the e.m.f. induced in coil $\mathrm{S}$ is
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The correct answer is:
$9.42 \mathrm{~V}$
The correct option is (B).
Concept: Flux associated among the coils is $\phi=\mathrm{MI}$ and the induced emf is given by $\mathrm{E}=-\frac{\mathrm{d} \phi}{\mathrm{dt}}$.
Therefore, $\mathrm{E}=-\mathrm{M} \frac{\mathrm{di}}{\mathrm{dt}}$.
Given, $\mathrm{I}=20 \sin (50 \pi \mathrm{t})$ and $\mathrm{M}=3 \times 10^{-3} \mathrm{H}$,
Therefore, $\mathrm{E}=-\mathrm{M} \frac{\mathrm{di}}{\mathrm{dt}}$.
Given, $\mathrm{I}=20 \sin (50 \pi \mathrm{t})$ and $\mathrm{M}=3 \times 10^{-3} \mathrm{H}$, therefore
$\mathrm{E}=-\mathrm{M} \frac{\mathrm{di}}{\mathrm{dt}}$.
Given, $\mathrm{I}=20 \sin (50 \pi \mathrm{t})$ and $\mathrm{M}=3 \times 10^{-3} \mathrm{H}$, therefore
$E=\left(-3 \times 10^{-3} \mathrm{H} \times 50 \pi \times 20\right) \cos (50 \pi \mathrm{t}) \mathrm{A}$
The maximum emf is given by
$\left|E_0\right|=3 \times 10^{-3} \times 50 \pi \times 20 \mathrm{~V}=9.42 \mathrm{~V}$
Concept: Flux associated among the coils is $\phi=\mathrm{MI}$ and the induced emf is given by $\mathrm{E}=-\frac{\mathrm{d} \phi}{\mathrm{dt}}$.
Therefore, $\mathrm{E}=-\mathrm{M} \frac{\mathrm{di}}{\mathrm{dt}}$.
Given, $\mathrm{I}=20 \sin (50 \pi \mathrm{t})$ and $\mathrm{M}=3 \times 10^{-3} \mathrm{H}$,
Therefore, $\mathrm{E}=-\mathrm{M} \frac{\mathrm{di}}{\mathrm{dt}}$.
Given, $\mathrm{I}=20 \sin (50 \pi \mathrm{t})$ and $\mathrm{M}=3 \times 10^{-3} \mathrm{H}$, therefore
$\mathrm{E}=-\mathrm{M} \frac{\mathrm{di}}{\mathrm{dt}}$.
Given, $\mathrm{I}=20 \sin (50 \pi \mathrm{t})$ and $\mathrm{M}=3 \times 10^{-3} \mathrm{H}$, therefore
$E=\left(-3 \times 10^{-3} \mathrm{H} \times 50 \pi \times 20\right) \cos (50 \pi \mathrm{t}) \mathrm{A}$
The maximum emf is given by
$\left|E_0\right|=3 \times 10^{-3} \times 50 \pi \times 20 \mathrm{~V}=9.42 \mathrm{~V}$
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