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

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, $\alpha = 2$.