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

Options

A. `2.5 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) = 500
- Current (I) = 2 A
- Magnetic flux linked with each turn (φ) = \(4 \times 10^{-3}\) Wb

### Step 2: Calculate the total magnetic flux (Φ) linked with the solenoid
The total magnetic flux linked with the solenoid can be calculated using the formula:
\[ \Phi = N \cdot \phi \]
Substituting the values:
\[ \Phi = 500 \cdot (4 \times 10^{-3}) = 2000 \times 10^{-3} = 2 \text{ Wb} \]

### Step 3: Use the formula for self-inductance (L)
The self-inductance (L) of the solenoid can be calculated using the formula:
\[ L = \frac{\Phi}{I} \]
Substituting the values:
\[ L = \frac{2 \text{ Wb}}{2 \text{ A}} = 1 \text{ H} \]

### Final Answer
The self-inductance of the solenoid is \(1 \text{ H}\).

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