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 use the formula:

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

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

### Step 1: Identify the given values
- Number of turns, \(N = 1000\)
- Current, \(I = 4 \, \text{A}\)
- Magnetic flux linked with each turn, \(\Phi = 4 \times 10^{-3} \, \text{Wb}\)

### Step 2: Substitute the values into the formula
Now, we can substitute the values into the formula for self-inductance:

\[ L = \frac{N \Phi}{I} = \frac{1000 \times (4 \times 10^{-3})}{4} \]

### Step 3: Simplify the expression
First, calculate the numerator:

\[ 1000 \times (4 \times 10^{-3}) = 4 \, \text{Wb} \]

Now substitute this back into the equation:

\[ L = \frac{4}{4} = 1 \, \text{H} \]

### Final Answer
The self-inductance of the solenoid is:

\[ L = 1 \, \text{H} \]

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