To find the magnetic flux through the loop ABCDEFA due to the magnetic field \( \vec{B} = 3 \hat{i} + 4 \hat{k} \), we will follow these steps:
### Step 1: Identify the coordinates of the loop
The coordinates of the loop ABCDEFA are:
- A(0, 0, 0)
- B(5, 0, 0)
- C(5, 5, 0)
- D(0, 5, 0)
- E(0, 5, 5)
- F(0, 0, 5)
### Step 2: Determine the area of the loop
The loop can be divided into two parts:
1. Loop ABCDA (in the xy-plane)
2. Loop ADEFA (in the yz-plane)
#### Area of loop ABCDA:
- This loop is a square with side length 5 m.
- Area \( A_{ABCD} = \text{side}^2 = 5 \times 5 = 25 \, \text{m}^2 \).
- The area vector \( \vec{A}_{ABCD} \) is perpendicular to the plane of the loop, which is along the z-axis:
\[
\vec{A}_{ABCD} = 25 \hat{k} \, \text{m}^2
\]
#### Area of loop ADEFA:
- This loop is also a square with side length 5 m.
- Area \( A_{ADEFA} = \text{side}^2 = 5 \times 5 = 25 \, \text{m}^2 \).
- The area vector \( \vec{A}_{ADEFA} \) is perpendicular to the plane of the loop, which is along the x-axis:
\[
\vec{A}_{ADEFA} = 25 \hat{i} \, \text{m}^2
\]
### Step 3: Calculate the total area vector
The total area vector \( \vec{A} \) of the loop ABCDEFA is the sum of the area vectors of both loops:
\[
\vec{A} = \vec{A}_{ABCD} + \vec{A}_{ADEFA} = 25 \hat{k} + 25 \hat{i} = 25 \hat{i} + 25 \hat{k} \, \text{m}^2
\]
### Step 4: Calculate the magnetic flux
Magnetic flux \( \Phi \) through the loop is given by:
\[
\Phi = \vec{B} \cdot \vec{A}
\]
Substituting the values of \( \vec{B} \) and \( \vec{A} \):
\[
\vec{B} = 3 \hat{i} + 4 \hat{k}
\]
\[
\vec{A} = 25 \hat{i} + 25 \hat{k}
\]
Now, calculating the dot product:
\[
\Phi = (3 \hat{i} + 4 \hat{k}) \cdot (25 \hat{i} + 25 \hat{k}) = (3 \times 25) + (4 \times 25) = 75 + 100 = 175 \, \text{Weber}
\]
### Final Answer
The magnetic flux through the loop ABCDEFA due to the magnetic field \( \vec{B} \) is \( 175 \, \text{Weber} \).
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