Question

Two long current carrying wires. separated by a 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 them is increased to two times and its direction is reversed. The distance is also Increased to $3d$ The new value of the force between them is

• A. $\frac{-F}{3}$
• B. $\frac{F}{3}$
• C. $\frac{2F}{3}$
• D. $\frac{-2F}{3}$

Answer 1

The correct option is D $\frac{-2F}{3}$

Force per unit length between two long current carrying wires is given by,

$F_1=\frac{{\mu }_0{I}_1{I}_2}{2\pi d}$

When $I_1$ is changed to $2I_1$ and $d$ is changed to $3d$

$\therefore F_2=\frac{{\mu }_0\left(2I_1\right)\left(I_2\right)}{2\pi \left(3d\right)}$

$=\frac{{\mu }_0{I}_1{I}_2}{2\pi d}\times \frac{2}{3}=\frac{2F}{3}$

As direction of current is reversed, the direction of the force will also get reversed.
$\therefore F_2=\frac{-2F}{3}$

Here, negative sign indicates that the force is now repulsive in nature.

Why this Question? Note: Force per unit length between two long current carrying wires is given by, $F_1=\frac{{\mu }_0{I}_1{I}_2}{2\pi d}$