A long solenoid has `500` turns. When a current of `2A` is passed through it, the resulting magnetic flux linked with each turn of the solenoid is `4xx10^(-3)Wb` . The self-inductance of the solenoid is

Options

A. 1.0 H
B. 4.0 H
C. 2.5 H
D. 2.0 H

Explanation

To find the self-inductance of the solenoid, we can use the formula for self-inductance \( L \):

\[ L = \frac{n \Phi}{I} \]

where:
- \( n \) is the number of turns in the solenoid,
- \( \Phi \) is the magnetic flux linked with each turn,
- \( I \) is the current flowing through the solenoid.

### Step-by-step Solution:

1. **Identify the given values**:
- Number of turns, \( n = 500 \)
- Current, \( I = 2 \, \text{A} \)
- Magnetic flux per turn, \( \Phi = 4 \times 10^{-3} \, \text{Wb} \)

2. **Substitute the values into the formula**:
\[ L = \frac{n \Phi}{I} = \frac{500 \times (4 \times 10^{-3})}{2} \]

3. **Calculate the numerator**:
\[ 500 \times (4 \times 10^{-3}) = 2000 \times 10^{-3} = 2 \, \text{Wb} \]

4. **Now divide by the current**:
\[ L = \frac{2 \, \text{Wb}}{2 \, \text{A}} = 1 \, \text{H} \]

5. **Final result**:
The self-inductance of the solenoid is \( L = 1 \, \text{H} \).