A closely wound solenoid of 750 turns and area of cross section of `5xx 10^(4)m^(2)` carries a current of 3.0 A. Its associated magnetic moment is

Options

A. `4.12JT^(-1)`
B. `3.12JT^(-1)`
C. `2.12JT^(-1)`
D. `1.13JT^(-1)`

Explanation

To find the magnetic moment of a closely wound solenoid, we can use the formula:

\[ M = N \cdot I \cdot A \]

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

### Step-by-Step Solution:

1. **Identify the given values:**
- Number of turns, \( N = 750 \)
- Current, \( I = 3.0 \, \text{A} \)
- Area of cross-section, \( A = 5 \times 10^{-4} \, \text{m}^2 \)

2. **Substitute the values into the formula:**
\[ M = N \cdot I \cdot A \]
\[ M = 750 \cdot 3.0 \cdot (5 \times 10^{-4}) \]

3. **Calculate the product:**
- First, calculate \( 750 \cdot 3.0 = 2250 \)
- Then, multiply by the area:
\[ M = 2250 \cdot (5 \times 10^{-4}) = 2250 \cdot 0.0005 \]

4. **Perform the multiplication:**
\[ M = 2250 \cdot 0.0005 = 1.125 \]

5. **Express the result in appropriate units:**
- The magnetic moment is expressed in Joules per Tesla (J/T):
\[ M = 1.125 \, \text{J/T} \]

6. **Round the result:**
- Rounding \( 1.125 \) gives approximately \( 1.13 \, \text{J/T} \).

### Final Answer:
The associated magnetic moment of the solenoid is approximately \( 1.13 \, \text{J/T} \).

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