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A coil having effective area ' $\mathrm{A}$ ' is held with its plane normal to a magnetic field of induction ' $\mathrm{B}$ '. The magnetic induction is quickly reduced to $25 \%$ of its initial value in 1 second. The e.m.f. induced in the coil (in volt) will be
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The correct answer is:
$\frac{3 \mathrm{AB}}{4}$
Induced emf is equal to change in flux per unit time:
$\begin{aligned} & \mathrm{e}=\frac{\mathrm{d} \phi}{\mathrm{dt}}=\mathrm{A} \frac{\mathrm{dB}}{\mathrm{dt}}=\mathrm{A} \frac{3}{4} \frac{\mathrm{B}}{1} \\ & \left(\because \mathrm{dB}=\frac{3}{4} \mathrm{~B}\right) \\ & \therefore \mathrm{E}=\frac{3 \mathrm{AB}}{4} \mathrm{~V}\end{aligned}$
$\begin{aligned} & \mathrm{e}=\frac{\mathrm{d} \phi}{\mathrm{dt}}=\mathrm{A} \frac{\mathrm{dB}}{\mathrm{dt}}=\mathrm{A} \frac{3}{4} \frac{\mathrm{B}}{1} \\ & \left(\because \mathrm{dB}=\frac{3}{4} \mathrm{~B}\right) \\ & \therefore \mathrm{E}=\frac{3 \mathrm{AB}}{4} \mathrm{~V}\end{aligned}$
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