Question:

Two coils have mutual inductance 0.002 H. The current changes in the first coil according to the relation i=i0sinωt i = i_0 \sin \omega t , where i0=5A i_0 = 5 \, \text{A} and ω=50πrad/s \omega = 50\pi \, \text{rad/s} . The maximum value of emf in the second coil is πα \frac{\pi}{\alpha} . The value of α \alpha is ______.

Updated On: Jan 11, 2025
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Correct Answer: 2

Solution and Explanation

The emf induced in the second coil is given by:

EMF=Mdidt \text{EMF} = -M \frac{di}{dt}

Substitute i=i0sinωt i = i_0 \sin \omega t :

didt=i0ωcosωt \frac{di}{dt} = i_0 \omega \cos \omega t

Thus, the maximum emf (when cosωt=1 \cos \omega t = 1 ) is:

EMFmax=Mi0ω=(0.002)×(5)×(50π)=π2V \text{EMF}_{\text{max}} = Mi_0 \omega = (0.002) \times (5) \times (50\pi) = \frac{\pi}{2} \, \text{V}

Therefore, α=2\alpha = 2.

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