Two long straight wires $$P$$ and $$Q$$ carrying equal current 10 A each were kept parallel to each other at 5 cm distance. Magnitude of magnetic force experienced by 10 cm length of wire $$P$$ is $$F_1$$. If distance between wires is halved and currents on them are doubled, force $$F_2$$ on 10 cm length of wire $$P$$ will be:

The force between two parallel wires carrying currents is given by $$F = \frac{\mu_0 I_1 I_2 L}{2\pi d}$$. For the first wire, the current is $$I = 10$$ A and the separation is $$d = 5$$ cm, while for the second wire the current is $$I' = 20$$ A and the separation is $$d' = 2.5$$ cm.

The ratio of the forces is found by substituting these values into the formula for force, which yields $$\frac{F_2}{F_1} = \frac{I'^2}{I^2} \times \frac{d}{d'} = \frac{400}{100} \times \frac{5}{2.5} = 4 \times 2 = 8$$, and hence $$F_2 = 8F_1$$.