Kerala KEAM 5 June 2024 Solved Paper

The force per unit length between two parallel wires carrying currents is given by Ampere's law:$$F = \frac{\mu_0 I_1 I_2}{2 \pi r}$$where:- $F$ is the force between the wires,- $\mu_0$ is the permeability of free space,- $I_1$ and $I_2$ are the currents in the wires,- $r$ is the distance between the wires.Initially, the currents in the two wires are equal but opposite, so the force between them is repulsive. The magnitude of the force is given by:$$F_1 = \frac{\mu_0 I^2}{2 \pi r}$$where $I$ is the current in each wire, and $r = 5 \, cm = 0.05 \, m$.Now, when the current in one of the wires is doubled and reversed, the new current in one wire is $-2I$ (the negative sign indicates the reversed direction). The new force between the wires will be:$$F_2 = \frac{\mu_0 (2I)(-I)}{2 \pi r} = -\frac{\mu_0 (2I^2)}{2 \pi r}$$Thus, the new force will be:$$F_2 = 2 \times F_1$$The direction of the force will now be attractive because the currents are now in opposite directions.Therefore, the force between the wires will be 2F and attractive.Thus, the correct answer is:$$\boxed{2F and attractive}$$ Quick Tip: When two currents in parallel wires are in opposite directions, they repel each other. Doubling the current in one wire and reversing its direction makes the force between the wires twice as large and attractive.