What is the net force on the square coil?

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(A) The force on sides $BC$ and $AD$ are equal in magnitude but opposite in direction,so their net force is zero.For side $AB$ (at distance $r_1 = 2 \

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$25 	imes 10^{-7} \, N$ moving towards wire

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(A) The force on sides $BC$ and $AD$ are equal in magnitude but opposite in direction,so their net force is zero.For side $AB$ (at distance $r_1 = 2 \, cm = 2 	imes 10^{-2} \, m$): $F_{AB} = \frac{\mu_0 I_1 I_2 L}{2 \pi r_1} = 2 	imes 10^{-7} 	imes \frac{2 	imes 1 	imes 0.15}{2 	imes 10^{-2}} = 3 	imes 10^{-6} \, N$ (attractive,towards the wire).For side $CD$ (at distance $r_2 = 2 \, cm + 10 \, cm = 12 \, cm = 12 	imes 10^{-2} \, m$): $F_{CD} = \frac{\mu_0 I_1 I_2 L}{2 \pi r_2} = 2 	imes 10^{-7} 	imes \frac{2 	imes 1 	imes 0.15}{12 	imes 10^{-2}} = 0.5 	imes 10^{-6} \, N$ (repulsive,away from the wire).The net force is $F_{net} = F_{AB} - F_{CD} = 3 	imes 10^{-6} - 0.5 	imes 10^{-6} = 2.5 	imes 10^{-6} \, N = 25 	imes 10^{-7} \, N$,directed towards the wire.

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