A loop ABCA of straight edges has three corner points A (8, 0, 0), B (0, 8, 0 ) and C (0, 0, 8). The magnetic field in this region is `vecB=5(hati+hatj+hatk)T`0 . The quantity of flux through the loop ABCA (in Wb) is _____.

Explanation

To find the magnetic flux through the loop ABCA, we can follow these steps:

### Step 1: Identify the coordinates of points A, B, and C
- Point A: \( A(8, 0, 0) \)
- Point B: \( B(0, 8, 0) \)
- Point C: \( C(0, 0, 8) \)

### Step 2: Determine the area of the triangle ABC
The area \( A \) of triangle ABC can be calculated using the formula for the area of a triangle given its vertices:
\[ A = \frac{1}{2} \times \text{base} \times \text{height} \]
In this case, we can use the lengths of the sides of the triangle. The lengths of the sides can be calculated using the distance formula:
- Length AB = \( \sqrt{(8-0)^2 + (0-8)^2} = \sqrt{64 + 64} = 8\sqrt{2} \)
- Length BC = \( \sqrt{(0-0)^2 + (8-0)^2 + (0-8)^2} = 8\sqrt{2} \)
- Length AC = \( \sqrt{(8-0)^2 + (0-0)^2 + (0-8)^2} = 8\sqrt{2} \)

Since triangle ABC is equilateral, we can use the formula for the area of an equilateral triangle:
\[ A = \frac{\sqrt{3}}{4} s^2 \]
where \( s \) is the length of a side. Here, \( s = 8\sqrt{2} \):
\[ A = \frac{\sqrt{3}}{4} (8\sqrt{2})^2 = \frac{\sqrt{3}}{4} \times 128 = 32\sqrt{3} \]

### Step 3: Calculate the magnetic flux
The magnetic flux \( \Phi \) through the loop is given by:
\[ \Phi = \vec{B} \cdot \vec{A} \]
where \( \vec{B} = 5(\hat{i} + \hat{j} + \hat{k}) \) T and \( \vec{A} \) is the area vector of the triangle ABC. The area vector is perpendicular to the plane of the triangle and has a magnitude equal to the area of the triangle.

Assuming the area vector points in the direction of \( \hat{i} + \hat{j} + \hat{k} \):
\[ \vec{A} = A \hat{n} = 32\sqrt{3} \hat{n} \]
where \( \hat{n} \) is the unit vector in the direction of \( \hat{i} + \hat{j} + \hat{k} \). The magnitude of \( \hat{n} \) is:
\[ |\hat{n}| = \sqrt{1^2 + 1^2 + 1^2} = \sqrt{3} \]
Thus, the unit vector is:
\[ \hat{n} = \frac{1}{\sqrt{3}}(\hat{i} + \hat{j} + \hat{k}) \]
Therefore, the area vector becomes:
\[ \vec{A} = 32\sqrt{3} \cdot \frac{1}{\sqrt{3}}(\hat{i} + \hat{j} + \hat{k}) = 32(\hat{i} + \hat{j} + \hat{k}) \]

### Step 4: Calculate the dot product
Now we can calculate the dot product:
\[ \Phi = \vec{B} \cdot \vec{A} = 5(\hat{i} + \hat{j} + \hat{k}) \cdot 32(\hat{i} + \hat{j} + \hat{k}) = 5 \cdot 32 \cdot (1 + 1 + 1) = 5 \cdot 32 \cdot 3 = 480 \, \text{Wb} \]

### Final Answer
The quantity of flux through the loop ABCA is \( \Phi = 480 \, \text{Wb} \).

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