Question

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 \( i_0 = 5 \, \text{A} \) and \( \omega = 50\pi \, \text{rad/s} \). The maximum value of emf in the second coil is \( \frac{\pi}{\alpha} \). The value of \( \alpha \) is ______.

Answer 1

Correct Answer: 2

The emf induced in the second coil is given by:

\[ \text{EMF} = -M \frac{di}{dt} \]

Substitute \( i = i_0 \sin \omega t \):

\[ \frac{di}{dt} = i_0 \omega \cos \omega t \]

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

\[ \text{EMF}_{\text{max}} = Mi_0 \omega = (0.002) \times (5) \times (50\pi) = \frac{\pi}{2} \, \text{V} \]

Therefore, \(\alpha = 2\).