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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
A
$\frac{\mu_0Ii}{2\pi}$
B
$\frac{2\mu_0IiL}{3\pi}$
C
$\frac{\mu_0IiL}{2\pi}$
D
$\frac{2\mu_0Ii}{3\pi}$
Explanation
Forces on the two parallel sides are unequal because they are at different distances.
Detailed Solution
Sides BC and DA experience equal and opposite forces; only AB and CD contribute.
Field at AB (distance L/2): $B_{AB} = \frac{\mu_0I}{2\pi(L/2)} = \frac{\mu_0}{2\pi}\frac{2I}{L}$
Field at CD (distance 3L/2): $B_{CD} = \frac{\mu_0I}{2\pi(3L/2)} = \frac{\mu_0}{2\pi}\frac{2I}{3L}$
Net force $F = iLB_{AB} - iLB_{CD} = \frac{2\mu_0IiL}{2\pi L} - \frac{2\mu_0IiL}{6\pi L} = \frac{\mu_0Ii}{\pi} - \frac{\mu_0Ii}{3\pi}$
$F = \frac{2\mu_0Ii}{3\pi}$ (towards the wire)
Field at AB (distance L/2): $B_{AB} = \frac{\mu_0I}{2\pi(L/2)} = \frac{\mu_0}{2\pi}\frac{2I}{L}$
Field at CD (distance 3L/2): $B_{CD} = \frac{\mu_0I}{2\pi(3L/2)} = \frac{\mu_0}{2\pi}\frac{2I}{3L}$
Net force $F = iLB_{AB} - iLB_{CD} = \frac{2\mu_0IiL}{2\pi L} - \frac{2\mu_0IiL}{6\pi L} = \frac{\mu_0Ii}{\pi} - \frac{\mu_0Ii}{3\pi}$
$F = \frac{2\mu_0Ii}{3\pi}$ (towards the wire)