Force on Current-Carrying Conductors question from NEET 2016 (Phase 1)

Answer: (c)

1. Forces on BC and DA cancel. AB is at $L/2$ and CD at $3L/2$, carrying opposite currents, so their forces are opposite.
2. $F_{AB} = \frac{\mu_0 I i L}{2\pi (L/2)} = \frac{\mu_0 I i}{\pi}$, $F_{CD} = \frac{\mu_0 I i L}{2\pi (3L/2)} = \frac{\mu_0 I i}{3\pi}$
3. $F_{net} = \frac{\mu_0 I i}{\pi} - \frac{\mu_0 I i}{3\pi} = \frac{2\mu_0 I i}{3\pi}$
4. So the answer is (c) $\frac{2\mu_0 Ii}{3\pi}$.