Force on Current-Carrying Conductors question from JEE Main 2026 (28 Jan, Shift 1)

Answer: (b)

1. $\frac{F}{l} = \frac{\mu_0 I_1 I_2}{2\pi r}$; like currents attract, opposite currents repel.
2. P (3 A up) and Q (1 A down) are opposite: repulsion, Q pushed towards R. $\frac{F}{l} = \frac{2\times10^{-7}\times3\times1}{0.03} = 2\times10^{-5}\text{ N/m}$
3. R (2 A down) and Q are in the same direction: attraction, Q pulled towards R. $\frac{F}{l} = \frac{2\times10^{-7}\times2\times1}{0.02} = 2\times10^{-5}\text{ N/m}$
4. Both act towards R: $F = 4\times10^{-5}\times0.15 = 6\times10^{-6}\text{ N}$
5. So the answer is (b) $6\times10^{-6}$ N towards R.