Electromagnetic Induction Question 4
Question 4 - 2024 (27 Jan Shift 1)
Two coils have mutual inductance $0.002 \mathrm{H}$. The current changes in the first coil according to the relation $\mathrm{i}=\mathrm{i}{0} \sin \omega \mathrm{t}$, where $\mathrm{i}{0}=5 \mathrm{~A}$ and $\omega=50 \pi \mathrm{rad} / \mathrm{s}$. The maximum value of emf in the second coil is $\frac{\pi}{\alpha} \mathrm{V}$. The value of $\alpha$ is
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Answer: (2)
Solution:
$\phi=\mathrm{Mi}=\mathrm{Mi}_{0} \sin \omega \mathrm{t}$
$\mathrm{EMF}=-\mathrm{M} \frac{\mathrm{di}}{\mathrm{dt}}=-0.002\left(\mathrm{i}_{0} \omega \cos \omega \mathrm{t}\right)$
$\mathrm{EMF}{\text {max }}=\mathrm{i}{0} \omega(0.002)=(5)(50 \pi)(0.002)$
$\mathrm{EMF}_{\max }=\frac{\pi}{2} \mathrm{~V}$
