When the number of turns and the length of the solenoid are doubled keeping the area of cross-section same, the inductance

Options

A. Four times
B. Doubled
C. Halved
D. Unchanged

Answer

To solve the problem, we need to analyze how the inductance of a solenoid changes when the number of turns and the length of the solenoid are both doubled, while keeping the area of cross-section constant.

### Step-by-Step Solution:

1. **Understand the Formula for Inductance**:
The self-inductance \( L \) of a solenoid is given by the formula:
\[ L = \frac{\mu_0 N^2 A}{l} \]
where:
- \( \mu_0 \) is the permeability of free space,
- \( N \) is the number of turns,
- \( A \) is the cross-sectional area,
- \( l \) is the length of the solenoid.

2. **Identify Initial Conditions**:
Let the initial number of turns be \( N \), the initial length be \( l \), and the cross-sectional area \( A \) remains constant.

3. **Determine New Conditions**:
When the number of turns is doubled, the new number of turns \( N' \) becomes:
\[ N' = 2N \]
When the length is doubled, the new length \( l' \) becomes:
\[ l' = 2l \]

4. **Substitute into the Inductance Formula**:
The new inductance \( L' \) can be expressed as:
\[ L' = \frac{\mu_0 (N')^2 A}{l'} \]
Substituting \( N' \) and \( l' \):
\[ L' = \frac{\mu_0 (2N)^2 A}{2l} \]

5. **Simplify the Expression**:
Simplifying the equation:
\[ L' = \frac{\mu_0 (4N^2) A}{2l} = \frac{4 \mu_0 N^2 A}{2l} = \frac{2 \mu_0 N^2 A}{l} \]
This shows that:
\[ L' = 2L \]
where \( L \) is the initial inductance.

6. **Conclusion**:
Therefore, the new inductance \( L' \) is twice the initial inductance \( L \):
\[ L' = 2L \]

### Final Answer:
The inductance is doubled.