A 200 turn closely wound circular coil of radius 15 cm carries a current of 4 A. The magnetic moment of this coil is

Options

A. `36.5A m^(2)`
B. `56.5A m^(2)`
C. `66.5A m^(2)`
D. `108A m^(2)`

Explanation

To find the magnetic moment of the coil, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the given values**:
- Number of turns, \( n = 200 \)
- Radius of the coil, \( r = 15 \, \text{cm} = 15 \times 10^{-2} \, \text{m} \)
- Current in the coil, \( I = 4 \, \text{A} \)

2. **Calculate the area of the coil**:
The area \( A \) of a circular coil is given by the formula:
\[ A = \pi r^2 \]
Substituting the value of \( r \):
\[ A = \pi (15 \times 10^{-2})^2 \]
\[ A = \pi (0.15)^2 \]
\[ A = \pi \times 0.0225 \, \text{m}^2 \]
\[ A \approx 3.14 \times 0.0225 \approx 0.07065 \, \text{m}^2 \]

3. **Use the formula for magnetic moment**:
The magnetic moment \( \mu \) of the coil is given by:
\[ \mu = n \cdot I \cdot A \]
Substituting the values of \( n \), \( I \), and \( A \):
\[ \mu = 200 \cdot 4 \cdot 0.07065 \]
\[ \mu = 800 \cdot 0.07065 \]
\[ \mu \approx 56.52 \, \text{A m}^2 \]

4. **Round off the result**:
The magnetic moment can be rounded to:
\[ \mu \approx 56.5 \, \text{A m}^2 \]

5. **Conclusion**:
Therefore, the magnetic moment of the coil is approximately \( 56.5 \, \text{A m}^2 \).

### Final Answer:
The magnetic moment of the coil is \( 56.5 \, \text{A m}^2 \).