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Question

For two concentric circular coil,

Mutual Inductance $M=\frac{\mu_{0} \pi \mathrm{N}_{1} \mathrm{N}_{2} \mathrm{a}^{2}}{2 \mathrm{b}}$

here, $\ma

Vedclass · Accepted Answer

$\frac{{\pi {\mu _0}{I_0}}}{2}.\frac{{{a^2}}}{b}\omega \,\sin \,\left( {\omega t} \right)$

Vedclass · Answer

For two concentric circular coil,

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. $(ii)$

$\mathrm{e}=\frac{-\mu_{0} \pi \mathrm{a}^{2}}{2 \mathrm{b}} \frac{\mathrm{d}}{\mathrm{dt}}\left(\mathrm{I}_{0} \cos \omega \mathrm{t}\right)$

$\mathrm{e}=\frac{\mu_{0} \pi \mathrm{a}^{2}}{2 \mathrm{b}} \mathrm{I}_{0} \sin \omega \mathrm{t}(\omega)$

$\mathrm{e}=\frac{\pi \mu_{0} \mathrm{I}_{0}}{2} \cdot \frac{\mathrm{a}^{2}}{\mathrm{b}} \omega \sin \omega \mathrm{t}$

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