The magnetic flux $\Phi_B$ through a surface is defined as the scalar product of the magnetic field vector $\mathbf{B}$ and the area vector $\mathbf{A}$: $\Phi_B = \mathbf{B} \cdot \mathbf{A} = BA \cos \theta$

1. Identify Given Values: Magnetic Field ($B$) = $0.5 \text{ T}$. Side length of square ($l$) = $1 \text{ m}$. Area of loop ($A$) = $l^2 = (1 \text{ m})^2 = 1 \text{ m}^2$. Resistance ($R$) = $1 \Omega$ (This is distractor information not needed for flux calculation).

2. Determine Angle ($\theta$):

• The problem states the plane of the loop is perpendicular to the magnetic field. The area vector ($\mathbf{A}$) is always normal (perpendicular) to the plane of the loop. Therefore, the angle $\theta$ between the magnetic field $\mathbf{B}$ and the area vector $\mathbf{A}$ is $0^\circ$.

3. Calculate Flux: $\Phi_B = (0.5 \text{ T})(1 \text{ m}^2) \cos(0^\circ)$ $\Phi_B = 0.5 \times 1 \times 1 = 0.5 \text{ Wb}$

Thus, the magnetic flux through the loop is $0.5 \text{ Wb}$.