A loop of wire is placed in a magnetic field `vec(B)=0.02 hat(i) T`. Then the flux through the loop if its area vector
`vec(A)=30hat(i)+16hat(j)+23hat(k) cm^(2)` is

Options

A. `60 mu Wb`
B. `32 mu Wb`
C. `46 mu Wb`
D. `138 mu Wb`

Answer

To find the magnetic flux through the loop, we can use the formula for magnetic flux, which is given by:

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

Where:
- \(\Phi\) is the magnetic flux,
- \(\vec{B}\) is the magnetic field vector,
- \(\vec{A}\) is the area vector.

### Step 1: Identify the vectors
The magnetic field vector is given as:
\[ \vec{B} = 0.02 \hat{i} \, \text{T} \]

The area vector is given as:
\[ \vec{A} = 30 \hat{i} + 16 \hat{j} + 23 \hat{k} \, \text{cm}^2 \]

### Step 2: Convert the area vector to meters squared
Since the area is given in cm², we need to convert it to m². We know that:
\[ 1 \, \text{cm}^2 = 10^{-4} \, \text{m}^2 \]
Thus,
\[ \vec{A} = (30 \hat{i} + 16 \hat{j} + 23 \hat{k}) \times 10^{-4} \, \text{m}^2 \]
This simplifies to:
\[ \vec{A} = 0.0030 \hat{i} + 0.0016 \hat{j} + 0.0023 \hat{k} \, \text{m}^2 \]

### Step 3: Calculate the dot product \(\vec{B} \cdot \vec{A}\)
The dot product of two vectors \(\vec{B} = B_x \hat{i} + B_y \hat{j} + B_z \hat{k}\) and \(\vec{A} = A_x \hat{i} + A_y \hat{j} + A_z \hat{k}\) is calculated as:
\[ \vec{B} \cdot \vec{A} = B_x A_x + B_y A_y + B_z A_z \]

In our case:
- \(B_x = 0.02\), \(B_y = 0\), \(B_z = 0\)
- \(A_x = 0.0030\), \(A_y = 0.0016\), \(A_z = 0.0023\)

Calculating the dot product:
\[ \vec{B} \cdot \vec{A} = (0.02)(0.0030) + (0)(0.0016) + (0)(0.0023) \]
\[ = 0.00006 \, \text{T m}^2 \]

### Step 4: Final result for magnetic flux
Thus, the magnetic flux \(\Phi\) through the loop is:
\[ \Phi = 0.00006 \, \text{Wb} = 60 \, \mu\text{Wb} \]

### Conclusion
The magnetic flux through the loop is \(60 \, \mu\text{Wb}\).

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