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(D) The magnetic field $B$ due to the straight wire at a distance $r$ is $B = \frac{\mu_0 I_2}{2 \pi r}$.For side $AB$ (distance $r_1 = L/3$),the forc

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

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(D) The magnetic field $B$ due to the straight wire at a distance $r$ is $B = \frac{\mu_0 I_2}{2 \pi r}$.For side $AB$ (distance $r_1 = L/3$),the force is $F_{AB} = I_1 L B_1 = I_1 L \left( \frac{\mu_0 I_2}{2 \pi (L/3)} \right) = \frac{3 \mu_0 I_1 I_2}{2 \pi}$ (Attractive,towards the wire).For side $CD$ (distance $r_2 = L/3 + L = 4L/3$),the force is $F_{CD} = I_1 L B_2 = I_1 L \left( \frac{\mu_0 I_2}{2 \pi (4L/3)} \right) = \frac{3 \mu_0 I_1 I_2}{8 \pi}$ (Repulsive,away from the wire).The sides $BC$ and $AD$ are perpendicular to the wire,and the forces on them cancel each other out due to symmetry.The net force $F_{net} = F_{AB} - F_{CD} = \frac{3 \mu_0 I_1 I_2}{2 \pi} - \frac{3 \mu_0 I_1 I_2}{8 \pi} = \frac{12 \mu_0 I_1 I_2 - 3 \mu_0 I_1 I_2}{8 \pi} = \frac{9 \mu_0 I_1 I_2}{8 \pi}$.

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