Two long wires have uniform charge density $\lambda$ per unit length each. The wires are non$-$coplaner and mutually perpendicular. Shortest distance between them is $d$. Calculate interaction force between them.

Interaction force between two wires near the force exerted by one on the other. If our line of sight $x-$axis of one of these two wires, then that wire will appear like a point while the other in the form of straight line as shown in Fig.

Let foot of perpendicular drawn from lower wire to the upper wire be $N$. Consider two equal elemental lengths $dx$ each of upper wire on either side of $N$ as shown in Fig.

First, we have to calculate electric field at the position of each elemental length due to charge on lower wire, then we will calculate resultant of the forces acting on these two elemental lengths.

Distance of each elemental length from lower wire is$r=\sqrt{d²+x²}$

Electric field strength at position of each elemental length due to charge on lower wire is

$E=\frac{\lambda }{2{\mathrm{\pi \epsilon }}_{\circ }\mathrm{r}}$

Charge on each elemental length,$dq=\lambda dx$ 

Force on each elemental length$=E\lambda dx$

Resultant of these two forces is normal to the upper wire. Its magnitude, $dF=\left(E\lambda dx\right)2.\cos \theta$

or,

$dF=\frac{{\lambda }^2}{{\mathrm{\pi \epsilon }}_{\circ }\mathrm{r}}\cos \theta dx=\frac{{\lambda }^{2}d}{{\mathrm{\pi \epsilon }}_{\circ }\mathrm{r}}\cos \theta \frac{dx}{\left(d^2+x^2\right)}$

Total force on the wire is $F=\frac{{\lambda }^{2}d}{{\mathrm{\pi \epsilon }}_{\circ }}{\int }_{x=\circ }^{x=\infty }\frac{dx}{\left(d^2+x^2\right)}\frac{{\lambda }^2}{2{\epsilon }_{\circ }}$