Two coils have a mutual inductance of 0.005 H. The current changes in the first coil according to the equation \( I = I_0 \sin\omega t \), where \( I_0 = 10\,\mathrm{A} \) and \( \omega = 100\pi\,\mathrm{rad/s} \). The maximum value of emf in the second coil is :
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( 5\pi \)
By mutual induction, the emf induced in the second coil is proportional to the rate of change of current in the first coil.
\( \displaystyle \varepsilon = -M\,\frac{dI}{dt} = -M\,\frac{d}{dt}\left(I_0 \sin\omega t\right) = -M I_0 \omega \cos\omega t \)
The maximum value occurs when \( |\cos\omega t| = 1 \):
\( \displaystyle \varepsilon_{\max} = M I_0 \omega = 0.005\,\mathrm{H} \times 10\,\mathrm{A} \times 100\pi\,\mathrm{rad\,s^{-1}} = 5\pi\,\mathrm{V} \)
Note: The same question was also asked in CUET (UG) 2023 on 11-06-2023 (Slot 3).