To solve this problem, we need to calculate the magnetic flux through the square loop.$\\$The formula for magnetic flux $\Phi$ through a surface is given by:$\\ \Phi = B \cdot A \cdot \cos(\theta) \\$Where:$\\$• $B$ is the magnetic field strength.$\\$• $A$ is the area of the loop.$\\$• $\theta$ is the angle between the magnetic field and the normal to the surface.$\\$Given:$\\$• $B = 0.5 \,$T$\\$• The side of the square loop is $1 \,$m, so the area $A = 1 \times 1 = 1 \,$m^2.} \\ • The plane of the loop is perpendicular to the magnetic field, so $\theta = 0^\circ. \\$Since the plane is perpendicular, the angle between the magnetic field and the normal is $\theta = 0^\circ. \\$Thus, $\cos(\theta) = \cos(0^\circ) = 1. \\$Substitute the values into the formula:$\\ \Phi = 0.5 \cdot 1 \cdot 1 \\ \Phi = 0.5 \,$Weber$\\$Therefore, the magnetic flux through the loop is $0.5 \,$Weber$. \\$This corresponds to Option 2.