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JEE Main 2017 · Physics · Magnetic Effects of Current · medium

A long straight wire carries a current of 10 A. A square loop of side a carrying 5 A lies in the same plane with two sides parallel to the wire; the…

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A long straight wire carries a current of 10 A10\ \mathrm{A}. A square loop of side aa carrying 5 A5\ \mathrm{A} lies in the same plane with two sides parallel to the wire; the nearer of these sides is at a distance aa from the wire and has current in the same direction as the wire. The net force on the loop is
  1. A. 5×10−6 N5\times10^{-6}\ \mathrm{N} towards the wire
  2. B. 1×10−5 N1\times10^{-5}\ \mathrm{N} towards the wire
  3. C. 2.5×10−6 N2.5\times10^{-6}\ \mathrm{N} away from the wire
  4. D. 7.5×10−6 N7.5\times10^{-6}\ \mathrm{N} towards the wire
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Answer: (A) 5×10−6 N5\times10^{-6}\ \mathrm{N} towards the wire

Forces on the two sides perpendicular to the wire cancel. For the near side (distance aa, parallel current) the force is attractive: F1=μ0I1I2a2πa=2×10−7×50=10−5 NF_1 = \dfrac{\mu_0 I_1I_2a}{2\pi a} = 2\times10^{-7}\times50 = 10^{-5}\ \mathrm{N}. The far side is at 2a2a with opposite current, so the force is repulsive: F2=μ0I1I2a2π(2a)=5×10−6 NF_2 = \dfrac{\mu_0I_1I_2 a}{2\pi(2a)} = 5\times10^{-6}\ \mathrm{N}. Net force =10−5−5×10−6=5×10−6 N= 10^{-5} - 5\times10^{-6} = 5\times10^{-6}\ \mathrm{N} towards the wire. The side length aa cancels out of the result.
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