Force on Current-Carrying Conductors question from JEE Main 2019 (9 Jan, Shift 1)

Answer: (d)

1. The small loop is a magnetic dipole $m = I'\pi a^2$, and its moment is parallel to the wire's field (both perpendicular to the paper).
2. Force on a dipole in a non-uniform field: $F = m\frac{dB}{dx}$, with $B = \frac{\mu_0 I}{2\pi x}$, so $\frac{dB}{dx} = -\frac{\mu_0 I}{2\pi x^2}$.
3. $F = \frac{\mu_0 I I' \pi a^2}{2\pi d^2} \propto \frac{a^2}{d^2}$. By Newton's third law the force on the wire is equal.
4. So the answer is (d) $F \propto \left(\frac{a}{d}\right)^2$.