Inductance of a solenoid is 3H and it consist of 500 truns . If number of turn make twice , then the value of self inductance becomes :

Options

A. (a)`1.5` H
B. (b)3 H
C. (c)9 H
D. (d)12 H

Answer

To solve the problem, we need to understand how the self-inductance of a solenoid changes when the number of turns is altered. The self-inductance \( L \) of a solenoid is given by the formula:

\[ L = \frac{\mu N^2 A}{l} \]

where:
- \( L \) is the inductance,
- \( \mu \) is the permeability of the core,
- \( N \) is the number of turns,
- \( A \) is the cross-sectional area of the solenoid,
- \( l \) is the length of the solenoid.

### Step-by-Step Solution:

1. **Identify Initial Conditions**:
- Given: \( L = 3 \, \text{H} \) (initial inductance),
- \( N = 500 \) (initial number of turns).

2. **Determine New Number of Turns**:
- The number of turns is doubled:
\[ N' = 2N = 2 \times 500 = 1000. \]

3. **Apply the Inductance Formula**:
- The new inductance \( L' \) when the number of turns is doubled can be expressed as:
\[ L' = \frac{\mu (N')^2 A}{l}. \]

4. **Substituting the New Number of Turns**:
- Substitute \( N' \) into the inductance formula:
\[ L' = \frac{\mu (1000)^2 A}{l}. \]

5. **Relate New Inductance to Initial Inductance**:
- Since \( N' = 2N \), we can express \( L' \) in terms of \( L \):
\[ L' = \frac{\mu (2N)^2 A}{l} = \frac{\mu (4N^2) A}{l} = 4 \cdot \frac{\mu N^2 A}{l} = 4L. \]

6. **Calculate the New Inductance**:
- Substitute the value of \( L \):
\[ L' = 4 \times 3 \, \text{H} = 12 \, \text{H}. \]

### Final Answer:
The new self-inductance \( L' \) when the number of turns is doubled is:
\[ L' = 12 \, \text{H}. \]