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Force on a current loop near a straight current
Concepts tested here
- square-loop-near-wire-force
All Questions
2016 1 question
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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
Forces on the two parallel sides are unequal because they are at different distances.
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)