JEE Main · 2019 · 09 Jan 2019 · Shift 1 · Magnetic Effects of Current

Treat the small loop as a magnetic dipole of moment $M=I\pi a^2$ sitting in the wire's non-uniform field $B\propto \frac{1}{d}$. The force on a dipole comes from the field gradient, $F=M\frac{dB}{dd}$, and since $\frac{dB}{dd}\propto \frac{1}{d^2}$ we get $F\propto a^2\cdot \frac{1}{d^2}={\left(\frac{a}{d}\right)}^2$. By Newton's third law the loop pushes on the wire exactly as hard as the wire pushes on the loop, so the force on the wire carries the same ${\left(\frac{a}{d}\right)}^2$ dependence.