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(A) For two concentric circular coils,the mutual inductance $M$ is given by $M = \frac{\mu_0 \pi a^2}{2b}$,where $a$ is the radius of the smaller inne

Vedclass · Accepted Answer

$\frac{\pi \mu_0 I_0}{2} \cdot \frac{a^2}{b} \omega \sin (\omega t)$

Vedclass · Answer

(A) For two concentric circular coils,the mutual inductance $M$ is given by $M = \frac{\mu_0 \pi a^2}{2b}$,where $a$ is the radius of the smaller inner loop and $b$ is the radius of the larger outer loop $(b \gg a)$.The current in the outer loop is $I = I_0 \cos (\omega t)$.According to Faraday's law of induction,the induced emf $e$ in the inner loop is given by $e = -M \frac{dI}{dt}$.Substituting the values:$e = -\left( \frac{\mu_0 \pi a^2}{2b} \right) \frac{d}{dt} [I_0 \cos (\omega t)]$$e = -\left( \frac{\mu_0 \pi a^2}{2b} \right) I_0 [-\omega \sin (\omega t)]$$e = \frac{\pi \mu_0 I_0}{2} \cdot \frac{a^2}{b} \omega \sin (\omega t)$.

$A$ 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

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