A small circular loop of wire of radius \(a\) is located at the centre of a much larger circular wire loop of radius \(b\). The two loops are in the same plane. The outer loop of radius \(b\) carries an alternating current \(I = I_0\, cos\, (\omega t)\) . The emf induced in the smaller inner loop is nearly

Mutual Inductance \(M=\frac{\mu_{0} \pi \mathrm{N}_{1} \mathrm{N}_{2} \mathrm{a}^{2}}{2 \mathrm{b}}\)

here, \(\mathrm{N}_{1}=\mathrm{N}_{2}=1\)

Hence, \(M=\frac{\mu_{0} \pi a^{2}}{2 b}.........(i)\)

and given \(I=I_{0} \cos \omega t.........(ii)\)

Now according to Faraday's second law induced \(emf\)

\(\mathrm{e}=-\mathrm{M} \frac{\mathrm{d} \mathrm{I}}{\mathrm{dt}}\)

Fromeq. \((ii)\)

\(\mathrm{e}=\frac{-\mu_{0} \pi \mathrm{a}^{2}}{2 \mathrm{b}} \frac{\mathrm{d}}{\mathrm{dt}}\left(\mathrm{I}_{0} \cos \omega \mathrm{t}\right)\)

\(\mathrm{e}=\frac{\mu_{0} \pi \mathrm{a}^{2}}{2 \mathrm{b}} \mathrm{I}_{0} \sin \omega \mathrm{t}(\omega)\)

\(\mathrm{e}=\frac{\pi \mu_{0} \mathrm{I}_{0}}{2} \cdot \frac{\mathrm{a}^{2}}{\mathrm{b}} \omega \sin \omega \mathrm{t}\)