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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
Explanation
Two equal perpendicular forces combine to $\sqrt{2}$ times one.
Detailed Solution
The forces on B due to C and due to A are equal in magnitude and at $90^\circ$ to each other.
$F_{BC} = F_{BA} = \frac{\mu_0I^2}{2\pi d}$
$F = \sqrt{2}F_{BC} = \sqrt{2}\frac{\mu_0I^2}{2\pi d} = \frac{\mu_0I^2}{\sqrt{2}\pi d}$
$F_{BC} = F_{BA} = \frac{\mu_0I^2}{2\pi d}$
$F = \sqrt{2}F_{BC} = \sqrt{2}\frac{\mu_0I^2}{2\pi d} = \frac{\mu_0I^2}{\sqrt{2}\pi d}$