A square of side `L` meters lies in the `x-y` plane in a region where tbe magnctic fleld is given by `vec(B)=B_(0)(2 hat(i)+3 hat(j)+4 hat(k))` tesla `_(3)` where `B_(0)` is a constant. If the magnitude of flux passing through the square is `y B_(a) L^(2)` weber, then find the value of `y .`

Answer

To solve the problem, we need to calculate the magnetic flux passing through a square of side \( L \) meters that lies in the \( x-y \) plane, given the magnetic field \( \vec{B} = B_0 (2 \hat{i} + 3 \hat{j} + 4 \hat{k}) \) tesla.

### Step-by-Step Solution:

1. **Identify the Area Vector**:
The square lies in the \( x-y \) plane. The area vector \( \vec{A} \) for a surface in the \( x-y \) plane is directed along the \( z \)-axis. Therefore, the area vector can be expressed as:
\[ \vec{A} = L^2 \hat{k} \]
where \( L^2 \) is the area of the square.

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

3. **Evaluate the Dot Product**:
The dot product will only consider the \( k \) component of the magnetic field:
\[ \Phi = B_0 (2 \cdot 0 + 3 \cdot 0 + 4 \cdot L^2) = 4 B_0 L^2 \]

4. **Relate to Given Flux**:
According to the problem, the magnitude of the flux is given as:
\[ \Phi = y B_0 L^2 \]
Setting the two expressions for flux equal to each other:
\[ 4 B_0 L^2 = y B_0 L^2 \]

5. **Solve for \( y \)**:
We can divide both sides by \( B_0 L^2 \) (assuming \( B_0 \) and \( L^2 \) are not zero):
\[ y = 4 \]

### Final Answer:
Thus, the value of \( y \) is:
\[ \boxed{4} \]