A square of side L metres lies in the xy-plane in a region, where the magnetic field is given by `B=B_0(2hati +4 hatj+4hatk)T`, where `B_0` is constant. The magnitude of flux passing through the square is :

Options

A. `2B_0L^2Wb`
B. `3B_0L^2Wb`
C. `4B_0L^2Wb`
D. `sqrt29B_0L^2Wb`

Answer

To find the magnitude of the magnetic flux passing through a square of side \( L \) metres lying in the xy-plane, we can follow these steps:

### Step 1: Understand the Magnetic Field
The magnetic field is given by:
\[ \mathbf{B} = B_0 (2 \hat{i} + 4 \hat{j} + 4 \hat{k}) \, \text{T} \]
where \( B_0 \) is a constant.

### Step 2: Identify the Area Vector
Since the square lies in the xy-plane, the area vector \( \mathbf{A} \) will be perpendicular to this plane. The direction of the area vector for a surface in the xy-plane is along the z-axis, which can be represented as:
\[ \mathbf{A} = L^2 \hat{k} \]
where \( L^2 \) is the area of the square.

### Step 3: Calculate the Magnetic Flux
The magnetic flux \( \Phi \) through the surface is given by the dot product of the magnetic field \( \mathbf{B} \) and the area vector \( \mathbf{A} \):
\[ \Phi = \mathbf{B} \cdot \mathbf{A} \]
Substituting the expressions for \( \mathbf{B} \) and \( \mathbf{A} \):
\[ \Phi = B_0 (2 \hat{i} + 4 \hat{j} + 4 \hat{k}) \cdot (L^2 \hat{k}) \]

### Step 4: Perform the Dot Product
Calculating the dot product:
\[ \Phi = B_0 L^2 \left( (2 \hat{i} \cdot \hat{k}) + (4 \hat{j} \cdot \hat{k}) + (4 \hat{k} \cdot \hat{k}) \right) \]
Since \( \hat{i} \cdot \hat{k} = 0 \) and \( \hat{j} \cdot \hat{k} = 0 \), we only consider the last term:
\[ \Phi = B_0 L^2 (0 + 0 + 4) = 4 B_0 L^2 \]

### Step 5: Write the Final Answer
Thus, the magnitude of the magnetic flux passing through the square is:
\[ \Phi = 4 B_0 L^2 \, \text{Weber} \]