45. Two coil have a mutual inductance $0.005 H$. The current changes in first coil according to equation $I = I_0 \sin \omega t$ where $I_0 = 2 A$ and $\omega = 100pi rad/sec$. The maximum value of emf in second coil is: (1998)
Solution:
Induced emf in the second coil is $e = -M \frac{dI}{dt} = -M \frac{d}{dt}(I_0 \sin \omega t) = -M I_0 \omega \cos \omega t$.
The maximum (peak) value of induced emf is $e_{max} = M I_0 \omega$.
$e_{max} = 0.005 \times 2 \times 100\pi = 0.01 \times 100\pi = \pi V$.
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