To solve this problem, we need to calculate the net force on the square loop ABCD due to the long straight conductor XY.$\\$The force between two parallel current-carrying conductors is given by:$\\ F = \frac{\mu_0}{2\pi} \cdot \frac{I_1 I_2}{d} \cdot L \\$where $\mu_0$ is the permeability of free space, $I_1$ and $I_2$ are the currents,$\\ d$ is the distance between the conductors, and $L$ is the length of the conductor.$\\$Consider each side of the square loop:$\\$1. Side AB:$\\$• Current $i$ flows from A to B.$\\$• Distance from XY is $\frac{L}{2}. \\ F_{AB} = \frac{\mu_0}{2\pi} \cdot \frac{I \cdot i}{\frac{L}{2}} \cdot L = \frac{\mu_0 IiL}{\pi L} = \frac{\mu_0 Ii}{\pi} \\$2. Side CD:$\\$• Current $i$ flows from C to D.$\\$• Distance from XY is $\frac{3L}{2}. \\ F_{CD} = \frac{\mu_0}{2\pi} \cdot \frac{I \cdot i}{\frac{3L}{2}} \cdot L = \frac{\mu_0 IiL}{3\pi L} = \frac{\mu_0 Ii}{3\pi} \\$3. Side BC and AD:$\\$• These sides are perpendicular to XY, so the force is zero.$\\$Net force on the loop:$\\$The forces on AB and CD are in opposite directions.$\\$Net force $F_{net} = F_{AB} - F_{CD} \\ F_{net} = \frac{\mu_0 Ii}{\pi} - \frac{\mu_0 Ii}{3\pi} \\ F_{net} = \frac{\mu_0 Ii}{\pi} \left(1 - \frac{1}{3}\right) \\ F_{net} = \frac{\mu_0 Ii}{\pi} \cdot \frac{2}{3} \\ F_{net} = \frac{2\mu_0 Ii}{3\pi} \\$Therefore, the net force on the loop is $\frac{2\mu_0 Ii}{3\pi}. \\$This corresponds to Option 2.