1. A coil of area `2 cm^2` with turns 500 is placed in a magnetic field of 0.5 T with its plane parallel to field. The flux linked with the coil is

Options

A. 50mWb
B. Zero
C. 25mWb
D. 5mWb

Answer

To solve the problem of finding the magnetic flux linked with a coil placed in a magnetic field, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Values**:
- Area of the coil, \( A = 2 \, \text{cm}^2 \)
- Number of turns, \( n = 500 \)
- Magnetic field strength, \( B = 0.5 \, \text{T} \)

2. **Convert the Area to Standard Units**:
- Since the area is given in cm², we need to convert it to m².
- \( 1 \, \text{cm}^2 = 10^{-4} \, \text{m}^2 \)
- Therefore, \( A = 2 \, \text{cm}^2 = 2 \times 10^{-4} \, \text{m}^2 \)

3. **Understand the Orientation of the Coil**:
- The problem states that the plane of the coil is parallel to the magnetic field.
- When the plane of the coil is parallel to the magnetic field, the angle \( \theta \) between the magnetic field and the normal to the coil's surface is \( 90^\circ \).

4. **Use the Formula for Magnetic Flux**:
- The magnetic flux \( \Phi \) linked with the coil is given by the formula:
\[ \Phi = n \cdot B \cdot A \cdot \cos(\theta) \]
- Since \( \theta = 90^\circ \), we have \( \cos(90^\circ) = 0 \).

5. **Calculate the Magnetic Flux**:
- Substituting the values into the formula:
\[ \Phi = 500 \cdot 0.5 \cdot (2 \times 10^{-4}) \cdot \cos(90^\circ) \]
- Since \( \cos(90^\circ) = 0 \):
\[ \Phi = 500 \cdot 0.5 \cdot (2 \times 10^{-4}) \cdot 0 = 0 \]

6. **Conclusion**:
- The magnetic flux linked with the coil is \( \Phi = 0 \, \text{Wb} \).

### Final Answer:
- The flux linked with the coil is **0 Weber**.