Q138 / 155 · WBJEE 2026 · Physics
A square of side $L$ lies in the $x - y$ plane, where the magnetic field is given by $\vec{B} = B_0\left(2\hat{i} + 3\hat{j} + 4\hat{k}\right)$ where $B_0$ is constant. The magnetic flux passing through the square is
- A$5\,B_0 L^2$
- B$2\,B_0 L^2$
- C$3\,B_0 L^2$
- Correct answerD$4\,B_0 L^2$
Solution
The square lies in the $x - y$ plane, so its area vector is perpendicular to the $x - y$ plane, which is along the $z$-axis.
The area vector is given by $\vec{A} = L^2 \hat{k}$
The magnetic field is given by $\vec{B} = B_0(2\hat{i} + 3\hat{j} + 4\hat{k})$
The magnetic flux $\Phi$ passing through the square is the dot product of the magnetic field and the area vector:
$\Phi = \vec{B} \cdot \vec{A}$
$\Phi = B_0(2\hat{i} + 3\hat{j} + 4\hat{k}) \cdot (L^2 \hat{k})$
$\Phi = 4 B_0 L^2$
Answer: $4\,B_0 L^2$
The area vector is given by $\vec{A} = L^2 \hat{k}$
The magnetic field is given by $\vec{B} = B_0(2\hat{i} + 3\hat{j} + 4\hat{k})$
The magnetic flux $\Phi$ passing through the square is the dot product of the magnetic field and the area vector:
$\Phi = \vec{B} \cdot \vec{A}$
$\Phi = B_0(2\hat{i} + 3\hat{j} + 4\hat{k}) \cdot (L^2 \hat{k})$
$\Phi = 4 B_0 L^2$
Answer: $4\,B_0 L^2$
Hence, the correct answer is option: 'D'.