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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A uniform time-varying magnetic field exists in a circular region of radius $R$, directed perpendicular into the plane of the paper, increasing at a constant rate $\alpha$. A straight conducting rod of length $2R$ is placed exactly along the diameter of the circular region (passing through the centre). Find the induced emf across the rod.

There is a ring of radius $r$ having linear charge density $\lambda$ and rotating with a uniform angular velocity $\omega$. The magnitude of the magnetic field produced by this ring at its own centre would be
($\mu_0 = \text{permeability of air}$)