Physics - EMI/AC Question with Solution | TestHub

Let current is$I=I_0 sin \omega t,$

Where$I_0 =10, \omega =100 \pi$

Then induced e.m.f. is given by

$\omega =-\frac{MdI}{dt}=-\frac{Md}{dt}\left(I_0\right)$

$\omega =-M{I}_0\omega \cos \omega t$

$\therefore$Maximum e.m.f. is

$\omega \_\max =M{I}_0\omega$

$5\pi = M \times 10 \times 100\pi$

$M = 5 \mathrm{mH}$