The resultant force on the current loop PQRS | vedclass

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(D) The force on segments $PS$ and $QR$ are equal and opposite,so their net force is zero.For the segments $PQ$ and $SR$ parallel to the long wire,the

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$5 	imes 10^{-4} \, N$

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(D) The force on segments $PS$ and $QR$ are equal and opposite,so their net force is zero.For the segments $PQ$ and $SR$ parallel to the long wire,the force is given by $F = \frac{\mu_0 I_1 I_2 L}{2 \pi r}$.Here,$I_1 = 20 \, A$ (long wire),$I_2 = 20 \, A$ (loop),and $L = 15 \, cm = 0.15 \, m$.Force on $PS$ (at distance $r_1 = 2 \, cm = 0.02 \, m$): $F_1 = \frac{2 	imes 10^{-7} 	imes 20 	imes 20 	imes 0.15}{0.02} = 6 	imes 10^{-4} \, N$ (Attractive).Force on $QR$ (at distance $r_2 = 2 + 10 = 12 \, cm = 0.12 \, m$): $F_2 = \frac{2 	imes 10^{-7} 	imes 20 	imes 20 	imes 0.15}{0.12} = 1 	imes 10^{-4} \, N$ (Repulsive).The resultant force is $F_{net} = F_1 - F_2 = 6 	imes 10^{-4} - 1 	imes 10^{-4} = 5 	imes 10^{-4} \, N$.

The resultant force on the current loop $PQRS$ due to a long current-carrying conductor is:

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