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Question

A square of side $L$ lies in the $x-y$ plane, where the magnetic field is given by $B = B_0 (2\hat{i} + 3\hat{j} + 4\hat{k})$ where $B_0$ is constant. The magnetic flux passing through the square is

This question was previously asked in
WBJEE 2026 Physics and Chemistry Question Paper (24-May-2026)
The correct answer is
$4 B_0 L^2$

To determine the magnetic flux passing through the square, we need to evaluate the contribution of the magnetic field component perpendicular to the square's surface.

The square is situated in the \(x-y\) plane, which means its normal vector is along the \(z\)-axis, i.e., \(\hat{k}\).

The given magnetic field is \(B = B_0 (2\hat{i} + 3\hat{j} + 4\hat{k})\). For calculating the magnetic flux through the square, we need only the component of the magnetic field along the normal to the surface. Thus, the relevant component of the magnetic field is the one along \(\hat{k}\) direction, which is \(4B_0\).

Magnetic flux \(\Phi\) through a surface is given by:

\(\Phi = \int \vec{B} \cdot d\vec{A}\)

For a flat surface with uniform magnetic field component perpendicular to it, this simplifies to:

\(\Phi = B_{\text{perpendicular}} \cdot A\)

Here, \(B_{\text{perpendicular}} = 4B_0\) and the area \(A\) of the square is \(L^2\).

Substituting these values, we get:

\(\Phi = 4B_0 \cdot L^2 = 4B_0L^2\)

The magnetic flux passing through the square is therefore \(4B_0L^2\), which matches the correct answer choice:

$4 B_0 L^2$

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