Mutual Inductance question from JEE Main 2024 (5 Apr, Shift 1)

Answer: (b)

1. From the figure, inner loop A has radius $a$, outer loop B has radius $b \gg a$.
2. Current $I$ in B gives at the centre $B_0 = \dfrac{\mu_0 I}{2b}$, nearly uniform over A
3. Flux through A: $\Phi = \dfrac{\mu_0 I}{2b}\pi a^2 \Rightarrow M = \dfrac{\Phi}{I} = \dfrac{\mu_0\pi a^2}{2b}$
4. So the answer is (b) $\dfrac{\mu_0\pi a^2}{2b}$.