Question

The force of repulsion between two parallel wires is $f$ when each one of them carries a certain current $I$. If the current in each is doubled, the force between them would be,

• A. $\raisebox{1ex}{$4$}\!\left/ \!\raisebox{-1ex}{$f$}\right.$
• B. $4f$
• C. $2f$
• D. $f$

Answer 1

The correct option is B

$4f$

Explanation for correct solution:

1. The force between two parallel wires is given by, $F=\frac{{\mu }_0{I}_1{I}_{2}L}{2\mathrm{\pi r}}$where , $I_1$ and $I_2$ are the currents in both wires, $L$ is the length of wire, $r$ is the distance between two parallel wires.
2. Given that $I_1$ and $I_2$ are equal i.e $I$ , we have $F=f$
3. When current is doubled, $I_1=I_2=2I$, $F=\frac{{\mu }_0{\left(2I\right)}^{2}L}{2\mathrm{\pi r}}=\frac{4{\mu }_0{I}^{2}L}{2\mathrm{\pi r}}$
4. Thus, $F=4f$

Hence, option B is the correct answer.