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 $B C$ and $A B$ will be same in magnitude.

$F_{B C}=F_{B A}=\frac{\mu_{0} I^{2}}{2 \pi d}$
$F=\sqrt{2} F_{B C}$
$=\sqrt{2} \frac{\mu_{0}}{2 \pi} \frac{l^{2}}{d}$
$F=\frac{\mu_{0} I^{2}}{\sqrt{2} \pi d}$