An arrangement of three parallel straight wires placed perpendicular to plane of paper carrying same current ‘I’ along the same direction as shown in figure. Magnitude of force per unit length on the middle wire ‘B’ is given by

 

 

Magnetic field produced by wire A at the place of B is, $\frac{{\mu }_{0}l}{2\pi d}$ along CB

Force per unit length on B is
$\frac{{\mathrm{F}}_1}{\ell }=1\frac{{\mu }_0\mathrm{I}}{2\pi \mathrm{d}}$ along BA.
Similarly, magnetic fi eld by wire C at the place of B is $\frac{{\mu }_{0}l}{2\pi d}$ along BA
And force per unit length on B, $\frac{F_2}{\ell }=\frac{\left|{\mu }_0\right|}{2\pi d}$ along BC
As forces are perpendicular and having equal magnitude, net force on B is $F_B=\sqrt{2}\frac{{\mu }_0{I}^2}{2\pi d}=\frac{{\mu }_0{l}^2}{\sqrt{2}\pi d}$