A loop made of straight edges has six corners at `A(0,0,0), B(L, O,0), C(L,L,0), D(0,L,0), E(0,L,L)` and `F(0,0,L)`. Where `L` is in meter. A magnetic field `B = B_(0)(hat(i) + hat(k))T` is present in the region. The flux passing through the loop `ABCDEFA` (in that order) is

Options

A. `B_0L^2Wb`
B. `2B_0L^2Wb`
C. `sqrt2B_0L^2Wb`
D. `sqrt4B_0L^2Wb`

Explanation

To solve the problem of finding the magnetic flux passing through the loop ABCDEFA, we will follow these steps:

### Step 1: Identify the Area of the Loop
The loop ABCDEFA consists of two parts: the base (ABCD) on the XY plane and the vertical sides (EF) connecting to point F. The area of the loop can be calculated as follows:

1. **Base Area (ABCD)**: The base is a square with side length L.
\[ A_{base} = L \times L = L^2 \]

2. **Vertical Area (ADEFA)**: The vertical area is also a square with side length L.
\[ A_{vertical} = L \times L = L^2 \]

3. **Total Area (A)**: Since the magnetic field is uniform and acts on both areas, we can consider the total area as:
\[ A = A_{base} + A_{vertical} = L^2 + L^2 = 2L^2 \]

### Step 2: Determine the Magnetic Field
The magnetic field is given as:
\[ \mathbf{B} = B_0 (\hat{i} + \hat{k}) \, \text{T} \]
This indicates that the magnetic field has components in both the x-direction and z-direction.

### Step 3: Calculate the Area Vector
The area vector \(\mathbf{A}\) for the loop will depend on the orientation of the loop. The area vector for the base (ABCD) will point in the positive z-direction, while the vertical area (ADEFA) will point in the positive x-direction.

1. **Area Vector for Base (ABCD)**:
\[ \mathbf{A}_{base} = L^2 \hat{k} \]

2. **Area Vector for Vertical (ADEFA)**:
\[ \mathbf{A}_{vertical} = L^2 \hat{i} \]

3. **Total Area Vector**:
\[ \mathbf{A} = \mathbf{A}_{base} + \mathbf{A}_{vertical} = L^2 \hat{i} + L^2 \hat{k} \]

### Step 4: Calculate the Magnetic Flux
The magnetic flux \(\Phi\) through the loop is given by the dot product of the magnetic field and the area vector:
\[ \Phi = \mathbf{B} \cdot \mathbf{A} \]

Substituting the values:
\[ \Phi = (B_0 (\hat{i} + \hat{k})) \cdot (L^2 \hat{i} + L^2 \hat{k}) \]

Calculating the dot product:
\[ \Phi = B_0 L^2 (\hat{i} \cdot \hat{i}) + B_0 L^2 (\hat{k} \cdot \hat{k}) = B_0 L^2 (1 + 1) = 2B_0 L^2 \]

### Final Result
The magnetic flux passing through the loop ABCDEFA is:
\[ \Phi = 2B_0 L^2 \, \text{Weber} \]