A magnetic field in a certain region is given by `B = (40 hat(i) - 15 hat(k)) xx 10^(-4) T`. The magnetic flux passes through a loop of area `5.0 cm^(2)` is placed flat on `xy` plane is

Options

A. `750n Wb`
B. `-750n Wb`
C. `360n Wb`
D. `-360n Wb`

Explanation

To solve the problem of finding the magnetic flux through a loop placed in the xy-plane, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Magnetic Field (B)**:
The magnetic field is given as:
\[ \mathbf{B} = (40 \hat{i} - 15 \hat{k}) \times 10^{-4} \, \text{T} \]

2. **Determine the Area of the Loop (A)**:
The area of the loop is given as:
\[ A = 5.0 \, \text{cm}^2 = 5.0 \times 10^{-4} \, \text{m}^2 \]

3. **Determine the Area Vector (A)**:
Since the loop is placed flat on the xy-plane, the area vector will be perpendicular to the plane, which is in the z-direction:
\[ \mathbf{A} = 5.0 \times 10^{-4} \hat{k} \, \text{m}^2 \]

4. **Calculate the Magnetic Flux (Φ)**:
The magnetic flux 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:
\[ \Phi = (40 \hat{i} - 15 \hat{k}) \times 10^{-4} \cdot (5.0 \times 10^{-4} \hat{k}) \]

5. **Perform the Dot Product**:
The dot product can be calculated as follows:
\[ \Phi = (40 \hat{i} \cdot 5.0 \times 10^{-4} \hat{k}) + (-15 \hat{k} \cdot 5.0 \times 10^{-4} \hat{k}) \]
Since \(\hat{i} \cdot \hat{k} = 0\) and \(\hat{k} \cdot \hat{k} = 1\):
\[ \Phi = 0 + (-15) \cdot (5.0 \times 10^{-4}) \times 10^{-4} \]
\[ \Phi = -75 \times 10^{-8} \, \text{Wb} \]

6. **Convert to Nano Webers**:
\[ \Phi = -750 \, \text{nWb} \]

### Final Answer:
The magnetic flux passing through the loop is:
\[ \Phi = -750 \, \text{nWb} \]