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(B) Let the square loop have vertices $P, Q, R, S$ such that $PQ$ is the side closest to the wire at distance $a$. The magnetic field $B$ due to the l

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Repulsive and equal to $\frac{\mu_0 I_1 I_2}{4\pi}$

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(B) Let the square loop have vertices $P, Q, R, S$ such that $PQ$ is the side closest to the wire at distance $a$. The magnetic field $B$ due to the long wire at a distance $r$ is $B = \frac{\mu_0 I_1}{2\pi r}$.$1$. For the side $PQ$ (length $a$,distance $a$): The current flows downwards. The magnetic field is directed into the page. By Fleming's Left-Hand Rule,the force $F_1$ is directed away from the wire (repulsive) with magnitude $F_1 = I_2 B_1 a = I_2 \left( \frac{\mu_0 I_1}{2\pi a} \right) a = \frac{\mu_0 I_1 I_2}{2\pi}$.$2$. For the side $RS$ (length $a$,distance $2a$): The current flows upwards. The magnetic field is directed into the page. By Fleming's Left-Hand Rule,the force $F_2$ is directed towards the wire (attractive) with magnitude $F_2 = I_2 B_2 a = I_2 \left( \frac{\mu_0 I_1}{2\pi (2a)} \right) a = \frac{\mu_0 I_1 I_2}{4\pi}$.$3$. For the top and bottom sides: The forces on these segments are equal and opposite,thus they cancel each other out.$4$. The net force is $F_{net} = F_1 - F_2 = \frac{\mu_0 I_1 I_2}{2\pi} - \frac{\mu_0 I_1 I_2}{4\pi} = \frac{\mu_0 I_1 I_2}{4\pi}$. Since $F_1 > F_2$,the net force is repulsive.

$A$ rigid square loop of side $a$ carrying current $I_2$ is lying on a horizontal surface near a long wire carrying current $I_1$ in the same plane as shown in the figure. The net force on the loop due to the wire will be:

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