Three very long parallel wires each carry current I in the directions shown in Fig. If the separation between adjacent wires is d, calculate the magnitude and direction of the net magnetic force per unit length on mid wire.

The force per unit length between two current carrying wires is given by

$\frac{\mathrm{F}}{\mathrm{L}}=\frac{{\mu }_0{\mathrm{d}}_2}{2\pi d}$

Where,${\mu }_0$ is the permeability of vaccum,$l_1$ and$l_2$ is the current through the two wires and d is the distance between wire.

Attraction and repulsion between two parallel current carrying wires

When the current is flowing in same direction in both parallel wires then attraction occurs between two wires.

When the current is flowing in opposite direction in both parallel wires then repulsion occurs between two wires.

Given: The separation between wires is .

The current in three wires is.

The force per unit length on the middle wire is

$\begin{array}{r}{\left(\frac{\mathrm{F}}{\mathrm{L}}\right)}_{\text{net}}=\frac{{\mu }_0{l}_1{I}_2}{2\pi d}-\frac{{\mu }_0{l}_1}{2\pi d}\\ {\left(\frac{\mathrm{F}}{\mathrm{L}}\right)}_{\text{net}}=\frac{{\mu }_{0}l\mathrm{I}}{2\pi d}-\frac{{\mu }_0{I}_1}{2\pi d}\\ {\left(\frac{\mathrm{F}}{\mathrm{L}}\right)}_{\text{net}}=0\mathrm{N}/\mathrm{m}\end{array}$

Here, the force acting on the middle wire by both wires are equal in magnitude and opposite in direction.

Therefore, the magnitude of magnetic force per unit length on middle wire is zero.