Two long parallel wires separated by distance ' \(d\) ' carry currents \(I_1\) and \(I_2\) in the same direction. They exert a force \(F\) on each other. Now the current in one of the wire is increased to three times and its direction is made opposite. The distance between the wires is doubled. The magnitude of force between them is

The correct option is (B).
Using Ampere's circuital law, the magnetic field due to a long current carrying wire at a distance \(r\) is: \(B=\frac{\mu_0 I}{2 \pi r}\)
Force on the wire carrying current in the parallel direction is, \(F=I_2 B l\)
Substituting the expression for \(B\) into the last equation and rearranging terms gives
\(F=\frac{\mu_0 I_1 I_2 l}{2 \pi r}\)
For long wires, it is convenient to think in terms of \(\frac{F}{l}\), the force per unit length.
\(\frac{F}{l}=\frac{\mu_0 I_1 I_2}{2 \pi r}\)
If \(I_2\) is increased 3 times and the separation \(r\) is doubled, then
\(\frac{F_{n e w}}{l}=\frac{3}{2}\left(\frac{\mu_0 I_1 I_2}{2 \pi r}\right)\)
Therefore, \(F_{\text {new }}=\frac{3}{2} F\).