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(C) The magnetic field $B$ at a distance $r$ from a long straight wire carrying current $I_2$ is given by $B = \frac{\mu_0 I_2}{2\pi r}$.For the squar

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$\frac{9\mu_0 I_1 I_2}{8\pi}$

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(C) The magnetic field $B$ at a distance $r$ from a long straight wire carrying current $I_2$ is given by $B = \frac{\mu_0 I_2}{2\pi r}$.For the square loop, the forces on the top segment $BC$ and bottom segment $AD$ are equal and opposite, so they cancel each other out.The force on the segment $AB$ (at distance $r_1 = L/3$) is $F_{AB} = \frac{\mu_0 I_1 I_2 L}{2\pi (L/3)} = \frac{3\mu_0 I_1 I_2}{2\pi}$ (attractive, towards the wire).The force on the segment $CD$ (at distance $r_2 = L/3 + L = 4L/3$) is $F_{CD} = \frac{\mu_0 I_1 I_2 L}{2\pi (4L/3)} = \frac{3\mu_0 I_1 I_2}{8\pi}$ (repulsive, away from the wire).The net force $F_{net} = F_{AB} - F_{CD} = \frac{3\mu_0 I_1 I_2}{2\pi} - \frac{3\mu_0 I_1 I_2}{8\pi} = \frac{12\mu_0 I_1 I_2 - 3\mu_0 I_1 I_2}{8\pi} = \frac{9\mu_0 I_1 I_2}{8\pi}$.

$A$ square loop $ABCD$ of side $L$ carrying a current $I_1$ is placed at a distance $(L/3)$ from a straight conductor $XY$ carrying current $I_2$. The loop and the conductor are coplanar. The net force on the loop will be ($\mu_0$ = magnetic permeability).

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