KCET 2025 Physics Paper

To find the magnetic flux through a square loop of side 2 meters that lies in the $Y-Z$ plane, given the magnetic field $\vec{B} = (5 \hat{i} + 3 \hat{j} - 4 \hat{k}) \text{ T}$, we proceed as follows: Determine the area vector:Since the loop is in the Y-Z plane, its area vector $\overline{A}$ is perpendicular to this plane and points along the Xaxis. The magnitude of the area vector is given by the area of the square, which is $2 \times 2 = 4$ square meters. The area vector is thus $\overline{A} = 4 \hat{i}$. Calculate the magnetic flux:The magnetic flux $\phi$ through the loop is given by the dot product of the magnetic field $\overline{B}$ and the area vector $\overline{A}$ :$\phi = \overline{B} \cdot \overline{A} = (5 \hat{i} + 3 \hat{j} - 4 \hat{k}) \cdot 4 \hat{i}$Performing the dot product, we have:$\phi = (5 \times 4) + (3 \times 0) + (-4 \times 0) = 20 \text{ Wb}$Thus, the magnitude of the magnetic flux through the square loop is 20 Wb .