A closely wound solenoid of 3000 turns and area of cross section `2xx 10^(-4)m^(2)`, carrying a current of 6 A is suspended through its centre allowing it to turn in a horizontal plane. The magnetic moment associated with this solenoid is

Options

A. `1.2JT^(-1)`
B. `2.4JT^(-1)`
C. `3.0JT^(-1)`
D. `3.6JT^(-1)`

Explanation

To find the magnetic moment associated with the solenoid, we can use the formula:

\[ m = n \cdot I \cdot A \]

where:
- \( m \) is the magnetic moment,
- \( n \) is the number of turns of the solenoid,
- \( I \) is the current flowing through the solenoid,
- \( A \) is the area of cross-section of the solenoid.

### Step-by-step Solution:

1. **Identify the given values**:
- Number of turns, \( N = 3000 \)
- Current, \( I = 6 \, \text{A} \)
- Area of cross-section, \( A = 2 \times 10^{-4} \, \text{m}^2 \)

2. **Substitute the values into the formula**:
\[ m = N \cdot I \cdot A \]
\[ m = 3000 \cdot 6 \cdot (2 \times 10^{-4}) \]

3. **Calculate the product**:
- First, calculate \( 3000 \cdot 6 \):
\[ 3000 \cdot 6 = 18000 \]

- Now multiply by the area:
\[ m = 18000 \cdot (2 \times 10^{-4}) = 18000 \cdot 0.0002 \]

4. **Perform the final multiplication**:
\[ m = 18000 \cdot 0.0002 = 3.6 \]

5. **State the unit of magnetic moment**:
The unit of magnetic moment is Joules per Tesla (J/T).

6. **Final answer**:
\[ m = 3.6 \, \text{J/T} \]

### Conclusion:
The magnetic moment associated with the solenoid is \( 3.6 \, \text{J/T} \).