A square loop ABCD carrying a current i, is placed near and coplanar with a long straight conductor XY carrying a current I, the net force on the loop will be

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$\frac{2 μ_{0} I i}{3 π}$

$\frac{μ_{0} I i L}{2 π}$

$\frac{μ_{0} I i}{2 π}$

$\frac{2 μ_{0} I i L}{3 π}$

answer is C.

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Detailed Solution

$Force on arm A B due to current in conductor X Y is$

$F_{1} = \frac{μ_{0}}{4 π} \frac{2 I i L}{(L / 2)} = \frac{μ_{0} I i}{π}$

acting towards X Y in the plane of loop. Force on arm CD due to current in conductor X Y is

$F_{2} = \frac{μ_{0}}{4 π} \frac{2 I i L}{3 (L / 2)} = \frac{μ_{0} I i}{3 π}$

acting away from X Y in the plane of loop.
therefore Net force on the loop $= F_{1} - F_{2}$

$= \frac{μ_{0} I i}{π} [1 - \frac{1}{3}] = \frac{2}{3} \frac{μ_{0} I i}{π}$

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