A coil having effective area $A$, is held with its plane normal to magnetic field of induction $\mathrm{B}$.…

A coil having effective area $A$, is held with its plane normal to magnetic field of induction $\mathrm{B}$. The magnetic induction is quickly reduced by $25 \%$ of its initial value in 2 second. Then the e.m.f. induced across the coil will be
  1. $\frac{\mathrm{AB}}{4}$
  2. $\frac{\mathrm{AB}}{2}$
  3. $\frac{\mathrm{3AB}}{4}$
  4. $\frac{\mathrm{3AB}}{8}$

Solution

$\begin{aligned} & |\mathrm{e}|=\frac{\mathrm{d} \phi}{\mathrm{dt}}=\mathrm{A} \frac{\mathrm{dB}}{\mathrm{dt}} \\ & \therefore|\mathrm{e}|=\mathrm{A}\left(\frac{\frac{3}{4} \mathrm{~B}}{2}\right) \\ & =\frac{3 \mathrm{AB}}{8}\end{aligned}$

Asked in: MHT CET 2023 (14 May Shift 2)

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