Two coils A and B have mutual inductance 0.008 $\mathrm{H}$. The current changes in the coil $\mathrm{A}$,…

Two coils A and B have mutual inductance 0.008 $\mathrm{H}$. The current changes in the coil $\mathrm{A}$, according to the equation $\mathrm{I}=\mathrm{I}_{\mathrm{m}} \sin \omega \mathrm{t}$, where $I_m=5 A$ and $\omega=200 \pi \mathrm{rad} \mathrm{s}^{-1}$. The maximum value of the e.m.f. induced in the coil B in volt is
  1. $4 \pi$
  2. $8 \pi$
  3. $10 \pi$
  4. $16 \pi$

Solution

$\mathrm{e}=\mathrm{M} \frac{\mathrm{dI}}{\mathrm{dt}}$ Given: $\mathrm{M}=0.008, \mathrm{I}_{\mathrm{m}}=5 \mathrm{~A}, \omega=200 \mathrm{rad} / \mathrm{s}$ $\therefore \quad \mathrm{e}=0.008 \times \mathrm{I}_{\mathrm{m}} \omega \mathrm{e} \mathrm{os} \omega \mathrm{t}$ For $\mathrm{e}=\mathrm{e}_{\max }, \cos \omega \mathrm{t}=1$ $\begin{aligned} \therefore \quad \mathrm{e}_{\max } & =0.008 \times \mathrm{I}_{\mathrm{m}} \times \omega \\ & =0.008 \times 5 \times 200 \pi \\ & =8 \pi \end{aligned}$

Asked in: MHT CET 2023 (09 May Shift 2)

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