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A current carrying square loop is placed near an infinitely long current carrying wire. The torque acting on the loop is

$(a)\;\frac{\mu_0}{2 \pi} \bigg( \frac{i_1i_2}{2}a \bigg) \\(b)\;\frac{\mu_0 ia}{2 \pi} \\ (c)\;\frac{\mu_o i_2 a}{2 \pi} \\ (d)\;zero $

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The forces acting on the limbs of the square due to $i_1$ all line in the plane of the coil.
Therefore there is no net torque.
Hence d is the correct answer.
answered Mar 27, 2014 by meena.p

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