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

Answer

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

### Step 1: Understand the relationship between magnetic flux, current, and self-inductance
The self-inductance \( L \) of a solenoid can be defined using the formula:
\[ L = \frac{\Phi}{I} \]
where:
- \( \Phi \) is the total magnetic flux linked with the solenoid,
- \( I \) is the current flowing through the solenoid.

### Step 2: Calculate the total magnetic flux linked with the solenoid
The total magnetic flux \( \Phi \) linked with the solenoid can be calculated as:
\[ \Phi = n \cdot \Phi_0 \]
where:
- \( n \) is the number of turns of the solenoid,
- \( \Phi_0 \) is the magnetic flux linked with each turn.

Given:
- \( n = 1000 \) turns,
- \( \Phi_0 = 4 \times 10^{-3} \, \text{Wb} \).

Substituting the values:
\[ \Phi = 1000 \cdot (4 \times 10^{-3}) = 4 \, \text{Wb} \]

### Step 3: Substitute the values into the self-inductance formula
Now, we can substitute the total magnetic flux \( \Phi \) and the current \( I \) into the self-inductance formula:
\[ L = \frac{\Phi}{I} \]
Given that:
- \( I = 4 \, \text{A} \),
- \( \Phi = 4 \, \text{Wb} \).

Substituting these values:
\[ L = \frac{4 \, \text{Wb}}{4 \, \text{A}} = 1 \, \text{H} \]

### Step 4: Conclusion
The self-inductance of the solenoid is:
\[ L = 1 \, \text{Henry} \]