A long solenoid has `1000` turns. When a current of `4A` flows through it, the magnetic flux linked with each turn of the solenoid is `4xx10^(-3)Wb`. The self-inductance of the solenoid is

Options

A. `3 H`
B. `2 H`
C. `1 H`
D. `4 H`

Explanation

To find the self-inductance of the solenoid, we can follow these steps:

### Step 1: Identify the given values
- Number of turns (n) = 1000
- Current (I) = 4 A
- Magnetic flux linked with each turn (Φ) = 4 x 10^(-3) Wb

### Step 2: Calculate the total magnetic flux linked with the solenoid
The total magnetic flux (Φ_net) linked with the solenoid can be calculated using the formula:
\[ \Phi_{\text{net}} = \Phi \times n \]
Substituting the values:
\[ \Phi_{\text{net}} = (4 \times 10^{-3} \, \text{Wb}) \times 1000 = 4 \, \text{Wb} \]

### Step 3: Use the formula for self-inductance
The self-inductance (L) of the solenoid can be calculated using the relationship:
\[ \Phi_{\text{net}} = L \times I \]
Rearranging this formula gives us:
\[ L = \frac{\Phi_{\text{net}}}{I} \]
Substituting the values we calculated:
\[ L = \frac{4 \, \text{Wb}}{4 \, \text{A}} = 1 \, \text{H} \]

### Conclusion
The self-inductance of the solenoid is \(1 \, \text{Henry}\).

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