Electromagnetic Induction Question 4
Question 4 - 2024 (27 Jan Shift 1)
Two coils have mutual inductance $0.002 H$. The current changes in the first coil according to the relation $i=i _0 \sin \omega t$, where $i _0=5 A$ and $\omega=50 \pi rad / s$. The maximum value of emf in the second coil is $\frac{\pi}{\alpha} V$. The value of $\alpha$ is
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Answer: (2)
Solution:
$\phi=Mi=Mi _0 \sin \omega t$
$EMF=-M \frac{di}{dt}=-0.002\left(i _0 \omega \cos \omega t\right)$
$EMF _{\text {max }}=i _0 \omega(0.002)=(5)(50 \pi)(0.002)$
$EMF _{\max }=\frac{\pi}{2} V$
