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

The key idea is that magnetic flux depends only on the component of $\mathbf{B}$ perpendicular to the surface. Since the square lies in the $x$-$y$ plane, its area vector is along $\hat{k}$.

1. Area vector: $\mathbf{A} = L^2 \hat{k} \ \text{m}^2$.
2. Flux: $\Phi = \mathbf{B} \cdot \mathbf{A} = \big(B_0(2\hat{i} + 3\hat{j} + 4\hat{k})\big) \cdot (L^2 \hat{k})$.
3. Only the $\hat{k}$ component contributes: $\Phi = B_0 (4) \cdot L^2 = 4 B_0 L^2 \ \text{Wb}$.

The key idea here is that magnetic flux depends only on the component of the magnetic field that is perpendicular to the surface. If the field has components parallel to the surface, those components slide along the surface and never actually "pierce" through it — so they contribute zero to the flux.

The square lies in the $x$-$y$ plane. That means its normal vector (the direction perpendicular to its surface) points along the $z$-axis. The area vector $\mathbf{A}$ is therefore $L^2 \hat{k}$ (taking the positive $z$ direction by convention, though the sign only affects the sign of the flux, not the magnitude).

Now let’s work through the calculation.

1. Write the magnetic field vector clearly

2. Define the area vector The square has side $L$, so area $= L^2$. Since it lies in the $x$-$y$ plane, the area vector is perpendicular to that plane:

3. Apply the definition of magnetic flux Magnetic flux $\Phi$ through a surface is given by the dot product of the field and the area vector:

4. Compute the dot product

The $\hat{i}$ and $\hat{j}$ components dot with $\hat{k}$ give zero — they are perpendicular to the area vector. Only the $\hat{k}$ component survives:

5. Magnitude of flux The question asks for the magnitude of flux. Since $4 B_0 L^2$ is already positive (assuming $B_0 > 0$), the magnitude is the same. If $B_0$ were negative, the magnitude would still be $4 |B_0| L^2$, but the problem states $B_0$ is constant — typically taken as positive unless specified otherwise.

Method: Computing Magnetic Flux Through a Flat Surface Given a Vector Field

Use this method whenever you're given $\vec B$ as a vector (with $\hat i,\hat j,\hat k$ components) and a flat surface lying in one of the coordinate planes.

Steps

Step 1: Write down the area vector of the surface

A flat surface's area vector $\vec A$ points along its normal (perpendicular to the surface) with magnitude equal to the surface's area. If the surface lies in the $x$-$y$ plane, its normal is along $\hat k$; in the $y$-$z$ plane, along $\hat i$; in the $x$-$z$ plane, along $\hat j$.

Step 2: Apply the definition of flux as a dot product

This is the key formula to remember — flux is not $|\vec B| \times \text{area}$; it is only the component of $\vec B$ along the surface's own normal, times the area.

Step 3: Expand the dot product component-by-component

Multiply matching components ($\hat i\cdot\hat i = \hat j\cdot\hat j=\hat k\cdot\hat k=1$) and discard the rest ($\hat i\cdot\hat j=\hat i\cdot\hat k=\hat j\cdot\hat k=0$, since the coordinate axes are mutually perpendicular). Only the component of $\vec B$ parallel to the area vector's own axis survives — every other component of $\vec B$ runs parallel to the surface and contributes nothing.

Step 4: Read off the magnitude

Once you have $\Phi_B$ as a signed number, its magnitude is simply $|\Phi_B|$ — the question is usually asking for this, not the signed value (the sign only tells you which face of the surface the flux is "coming out of").

The recurring trap this method guards against: never take $|\vec B|$ (the full magnitude, via $\sqrt{B_x^2+B_y^2+B_z^2}$) and multiply it by the area — that only gives the correct flux when $\vec B$ is entirely along the surface's normal.