A closely wound solenoid of 3000 turns and area of cross-section `2 xx 10^(-4)m^(2)`, carrying a current of `6 A`, is suspended through its center allowing it to turn in a horizontal plane. The magnetic moment (in `(J) (T)^(-1)` ) associated with the solenoid is

Answer

To find the magnetic moment associated with the solenoid, we can use the formula for the magnetic moment \( \mu \) of a solenoid:

\[ \mu = nIA \]

where:
- \( n \) is the number of turns per unit length (turns/m),
- \( I \) is the current in amperes (A),
- \( A \) is the area of cross-section in square meters (m²).

### Step 1: Calculate the number of turns per unit length (n)

Given:
- Total number of turns \( N = 3000 \)
- Length of the solenoid \( L \) is not given directly, but we can assume it is long enough for the turns to be closely wound.

For a closely wound solenoid, we can assume \( n \) is the total number of turns divided by the length of the solenoid. However, since we do not have the length, we will keep it in terms of \( L \):

\[ n = \frac{N}{L} = \frac{3000}{L} \]

### Step 2: Use the formula for magnetic moment

Now substituting \( n \) into the magnetic moment formula:

\[ \mu = nIA = \left(\frac{3000}{L}\right) \cdot (6) \cdot (2 \times 10^{-4}) \]

### Step 3: Simplify the expression

\[ \mu = \frac{3000 \cdot 6 \cdot 2 \times 10^{-4}}{L} \]

Calculating the constant part:

\[ 3000 \cdot 6 = 18000 \]
\[ 18000 \cdot 2 \times 10^{-4} = 36 \times 10^{-1} = 3.6 \]

Thus, we have:

\[ \mu = \frac{3.6}{L} \text{ (J)(T}^{-1}\text{)} \]

### Step 4: Conclusion

Since we do not have the length \( L \) of the solenoid, we cannot compute a numerical value for the magnetic moment without it. However, we can express the magnetic moment in terms of \( L \):

\[ \mu = \frac{3.6}{L} \text{ (J)(T}^{-1}\text{)} \]