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

Question

Aicharya · Accepted Answer

Force per unit length between two parallel long current carrying wires is $= \frac{μ_{0} I_{1} I_{2}}{2πd}$
Here, force between BC and AB will be same in magnitude.

$F_{B C} = F_{B A} = \frac{μ_{0} I^{2}}{2 π d}$
$F_{r e s} = \sqrt{F_{B C}^{2} + F_{B A}^{2}}$
$⟹ F_{r e s} = \frac{\sqrt{2} μ_{0} I^{2}}{2 π d} = \frac{μ_{0} I^{2}}{\sqrt{2} π d}$

An arrangement of three parallel straight wires placed perpendicular to plane of paper carrying same current…

Solution