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

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$\frac{μ_{0} I^{2}}{\sqrt{2} π d}$

$\frac{μ_{0} I^{2}}{2 π d}$

$\frac{2 μ_{0} I^{2}}{π d}$

$\frac{\sqrt{2} μ_{0} I^{2}}{π d}$

answer is B.

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Detailed Solution

Force between wires A and B = Force between wires B and C

$F_{B c} = F_{A B} = \frac{μ_{0} I^{2} l}{2 π d}$

$A s \bar{F_{A B}} ⊥ \bar{F_{B C}}$

Net force on wire B, $F_{n e t} = \sqrt{2} F_{B C} = \sqrt{2} \frac{μ_{0} I^{2} l}{2 π d}$

$\frac{F_{n e t}}{l} = \frac{μ_{0} I^{2}}{\sqrt{2} π d}$

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