JEE Mains · Physics · STD 12 -Electromagnetic induction
- JEE Mains 2024
- Hard
- Numerical
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 \(\mathrm{emf}\) in the second coil is \(\frac{\pi}{\alpha} \mathrm{V}\). The value of \(\alpha\) is_______.
Pick an option to check your answer.
Answer & Solution
Correct Answer
(C) \(2\)
Step-by-step Solution
Detailed explanation
\( \phi=\mathrm{Mi}=\mathrm{Mi}_0 \sin \omega \mathrm{t} \)
\( \mathrm{EMF}=-\mathrm{M} \frac{\mathrm{di}}{\mathrm{dt}}=-0.002\left(\mathrm{i}_0 \omega \cos \omega \mathrm{t}\right) \)
\( \mathrm{EMF}_{\max }=\mathrm{i}_0 \omega(0.002)=(5)(50 \pi)(0.002) \)
\( \mathrm{EMF}_{\max }=\frac{\pi}{2} \mathrm{~V}\)
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