A closely wound solenoid of 1000 turns and area of cross section `1.4xx10^(-4)m^(2)` carrying a current of 3 A is suspended through its centre allowing it to turn in a horizontal plane. The magnetic moment associated with this solenoid is

Options

A. `0.22JT^(-1)`
B. `0.32JT^(-1)`
C. `0.42JT^(-1)`
D. `0.52JT^(-1)`

Explanation

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

\[ M = NIA \]

where:
- \( M \) is the magnetic moment,
- \( N \) is the number of turns in 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, \( N = 1000 \)
- Current, \( I = 3 \, \text{A} \)
- Area of cross-section, \( A = 1.4 \times 10^{-4} \, \text{m}^2 \)

2. **Substitute the values into the formula**:
\[ M = NIA = 1000 \times 3 \times (1.4 \times 10^{-4}) \]

3. **Calculate the product**:
- First, calculate \( N \times I \):
\[ N \times I = 1000 \times 3 = 3000 \]
- Now, multiply this result by the area \( A \):
\[ M = 3000 \times (1.4 \times 10^{-4}) = 3000 \times 1.4 \times 10^{-4} \]

4. **Perform the multiplication**:
\[ 3000 \times 1.4 = 4200 \]
So,
\[ M = 4200 \times 10^{-4} \, \text{A m}^2 \]

5. **Convert to a more standard form**:
\[ M = 4.2 \times 10^{-1} \, \text{A m}^2 = 0.42 \, \text{A m}^2 \]

6. **Express in terms of Joules per Tesla**:
The unit of magnetic moment can also be expressed as Joules per Tesla:
\[ M = 0.42 \, \text{J/T} \]

### Final Answer:
The magnetic moment associated with the solenoid is \( 0.42 \, \text{J/T} \).