Question

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

• A. $\frac{2{\mu }_0{I}^2}{\pi d}$
• B. $\frac{\sqrt{2}{\mu }_0{I}^2}{\pi d}$
• C. $\frac{{\mu }_0{I}^2}{\sqrt{2}\pi d}$
• D. $\frac{{\mu }_0{I}^2}{2\pi d}$

Answer 1

Answer: (C)

Force between $BC$ and $AB$ will be same in magnitude.

$F_{BC}=F_{BA}=\frac{{\mu }_0{I}^2}{2\pi d}$
$F=\sqrt{2}{F}_{BC}$
$=\sqrt{2}\frac{{\mu }_0}{2\pi }\frac{l^2}{d}$
$F=\frac{{\mu }_0{I}^2}{\sqrt{2}\pi d}$