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 splenoid is `4xx10^(-3)Wb` . The self-inductance of the solenoid is

Options

A. `1.0` henry
B. `4.0` henry
C. `2.5` henry
D. `2.0` henry

Explanation

To find the self-inductance of the solenoid, we can use the formula that relates the magnetic flux linked with the solenoid, the number of turns, the current, and the self-inductance. The relevant formula is:

\[ N \cdot \Phi = L \cdot I \]

Where:
- \( N \) = number of turns
- \( \Phi \) = magnetic flux per turn
- \( L \) = self-inductance
- \( I \) = current

### Step-by-step solution:

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

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

3. **Calculate the left side**:
\[ 500 \cdot 4 \times 10^{-3} = 2000 \times 10^{-3} = 2 \, \text{Wb} \]

4. **Set up the equation**:
\[ 2 = L \cdot 2 \]

5. **Solve for \( L \)**:
\[ L = \frac{2}{2} = 1 \, \text{H} \]

Thus, the self-inductance of the solenoid is \( L = 1 \, \text{Henry} \).