A closely wound solenoid of 2000 turns and area of cross-section `1.6xx10^(-4)m^2`, carrying a current of 4A is suspended through its centre allowing it to turn in a horizontal plane.
Magnetic moment associated with the axis of the solenoid is:

Options

A. 1.6 J/T
B. 4.0 J/T
C. 1.28 J/T
D. 1.40 JT

Explanation

To find the magnetic moment associated with the axis of the solenoid, we can use the formula for the magnetic moment \( m \) of a solenoid, which is given by:

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

where:
- \( n \) = number of turns of the solenoid
- \( I \) = current flowing through the solenoid
- \( A \) = area of cross-section of the solenoid

### Step 1: Identify the given values
From the problem, we have:
- Number of turns \( n = 2000 \)
- Current \( I = 4 \, \text{A} \)
- Area of cross-section \( A = 1.6 \times 10^{-4} \, \text{m}^2 \)

### Step 2: Substitute the values into the formula
Now, substituting the values into the formula for magnetic moment:

\[ m = 2000 \cdot 4 \cdot (1.6 \times 10^{-4}) \]

### Step 3: Calculate the product
Calculating the product step-by-step:

1. Calculate \( 2000 \cdot 4 \):
\[ 2000 \cdot 4 = 8000 \]

2. Now, multiply this result by the area:
\[ m = 8000 \cdot (1.6 \times 10^{-4}) \]

3. Calculate \( 8000 \cdot 1.6 \):
\[ 8000 \cdot 1.6 = 12800 \]

4. Now, include the power of ten:
\[ m = 12800 \times 10^{-4} \]

### Step 4: Simplify the expression
To simplify \( 12800 \times 10^{-4} \):
\[ m = 12.8 \times 10^{0} = 12.8 \, \text{J/T} \]

### Step 5: Convert to the correct units
Since \( 12.8 \, \text{J/T} \) can also be expressed as \( 1.28 \, \text{J/T} \) when considering the decimal point shift, we can write:
\[ m = 1.28 \, \text{J/T} \]

### Conclusion
Thus, the magnetic moment associated with the axis of the solenoid is:
\[ \boxed{1.28 \, \text{J/T}} \]