A square of side L meters lies in the x-y plane in a region, where the magnetic field is given by `B = B_(0) (2 hati + 3 hat j + 4 hatk)`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. `sqrt(29)B_0L^2Wb`

Explanation

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

### Step 1: Identify the Area Vector
Since the square lies in the x-y plane, the area vector \( \vec{A} \) of the square will be perpendicular to the plane. The direction of the area vector will be along the z-axis. The magnitude of the area \( A \) of the square is given by:
\[ A = L^2 \]
Thus, the area vector can be expressed as:
\[ \vec{A} = L^2 \hat{k} \]

### Step 2: Write the Magnetic Field Vector
The magnetic field \( \vec{B} \) is given as:
\[ \vec{B} = B_0 (2 \hat{i} + 3 \hat{j} + 4 \hat{k}) \]
where \( B_0 \) is a constant.

### Step 3: Calculate the Magnetic Flux
The magnetic flux \( \Phi \) through the square is calculated using the formula:
\[ \Phi = \vec{B} \cdot \vec{A} \]
Substituting the expressions for \( \vec{B} \) and \( \vec{A} \):
\[ \Phi = B_0 (2 \hat{i} + 3 \hat{j} + 4 \hat{k}) \cdot (L^2 \hat{k}) \]

### Step 4: Perform the Dot Product
Now, we perform the dot product:
\[ \Phi = B_0 L^2 \left( 2 \hat{i} \cdot \hat{k} + 3 \hat{j} \cdot \hat{k} + 4 \hat{k} \cdot \hat{k} \right) \]
Since \( \hat{i} \cdot \hat{k} = 0 \) and \( \hat{j} \cdot \hat{k} = 0 \), we have:
\[ \Phi = B_0 L^2 (0 + 0 + 4) = 4 B_0 L^2 \]

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