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 will follow these steps:

### Step 1: Identify the coordinates of the points A, B, and C
The coordinates are given as:
- A (8, 0, 0)
- B (0, 8, 0)
- C (0, 0, 8)

### Step 2: Determine the area of triangle ABC
The triangle ABC is formed by points A, B, and C. We can calculate the area of triangle ABC using the formula for the area of a triangle given by vertices in 3D space:

\[ \text{Area} = \frac{1}{2} \left| \vec{AB} \times \vec{AC} \right| \]

Where:
- \(\vec{AB} = B - A = (0 - 8, 8 - 0, 0 - 0) = (-8, 8, 0)\)
- \(\vec{AC} = C - A = (0 - 8, 0 - 0, 8 - 0) = (-8, 0, 8)\)

Now, we compute the cross product \(\vec{AB} \times \vec{AC}\):

\[ \vec{AB} \times \vec{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -8 & 8 & 0 \\ -8 & 0 & 8 \end{vmatrix} \]

Calculating the determinant:

\[ = \hat{i}(8 \cdot 8 - 0 \cdot 0) - \hat{j}(-8 \cdot 8 - 0 \cdot -8) + \hat{k}(-8 \cdot 0 - 8 \cdot -8) \]
\[ = \hat{i}(64) - \hat{j}(-64) + \hat{k}(64) \]
\[ = 64\hat{i} + 64\hat{j} + 64\hat{k} \]

Now, find the magnitude of the cross product:

\[ \left| \vec{AB} \times \vec{AC} \right| = \sqrt{(64)^2 + (64)^2 + (64)^2} = \sqrt{3 \cdot (64)^2} = 64\sqrt{3} \]

Thus, the area of triangle ABC is:

\[ \text{Area} = \frac{1}{2} \cdot 64\sqrt{3} = 32\sqrt{3} \]

### Step 3: Determine the direction of the area vector
The area vector \(\vec{A}\) is perpendicular to the plane of the triangle ABC. Since the triangle is in the first octant and the coordinates of A, B, and C are positive, we can take the direction of the area vector as:

\[ \vec{A} = 32\sqrt{3} \hat{n} \]

Where \(\hat{n}\) is the unit vector in the direction of the normal to the triangle. Since \(\vec{AB}\) and \(\vec{AC}\) are both in the positive octant, we can take:

\[ \hat{n} = \frac{1}{\sqrt{3}}(\hat{i} + \hat{j} + \hat{k}) \]

Thus,

\[ \vec{A} = 32\sqrt{3} \cdot \frac{1}{\sqrt{3}}(\hat{i} + \hat{j} + \hat{k}) = 32(\hat{i} + \hat{j} + \hat{k}) \]

### Step 4: Calculate the magnetic flux
The magnetic flux \(\Phi\) through the loop is given by:

\[ \Phi = \vec{B} \cdot \vec{A} \]

Where \(\vec{B} = 5(\hat{i} + \hat{j} + \hat{k})\) and \(\vec{A} = 32(\hat{i} + \hat{j} + \hat{k})\).

Calculating the dot product:

\[ \Phi = 5(\hat{i} + \hat{j} + \hat{k}) \cdot 32(\hat{i} + \hat{j} + \hat{k}) = 5 \cdot 32 \cdot (1 + 1 + 1) = 160 \cdot 3 = 480 \]

### Final Answer
Thus, the magnetic flux through the loop ABCA is:

\[ \Phi = 480 \, \text{Wb} \]