Q 12-06-148JEE MainJEE Main 2017 (8 Apr)Medium
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:
Answer: (D) $\dfrac{\pi\mu_0I_0}{2}\cdot\dfrac{a^2}{b}\omega\sin(\omega t)$
Since $a \ll b$, the field over the small loop is nearly the field at the centre of the large loop, $B = \dfrac{\mu_0I}{2b}$:
$$\phi = \frac{\mu_0I}{2b}\pi a^2 = \frac{\pi\mu_0a^2}{2b}I_0\cos\omega t$$
$$\varepsilon = -\frac{d\phi}{dt} = \frac{\pi\mu_0I_0}{2}\cdot\frac{a^2}{b}\,\omega\sin(\omega t)$$
Solution by Sreeraj P, M.Sc Physics