The self-inductance L of a solenoid of area of cross-section A and length l, with N turns increase as:

Options

A. A and l decreases
B. A and I increases
C. A increases and l decreases
D. A decreases and l increase

Explanation

To determine how the self-inductance \( L \) of a solenoid changes with an increase in the number of turns \( N \), we can start with the formula for the self-inductance of a solenoid:

\[ L = \frac{\mu_0 N^2 A}{l} \]

where:
- \( L \) is the self-inductance,
- \( \mu_0 \) is the permeability of free space,
- \( N \) 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. **Understanding the Formula**:
The self-inductance \( L \) is directly proportional to the square of the number of turns \( N \) and the cross-sectional area \( A \), and inversely proportional to the length \( l \) of the solenoid.

2. **Increasing the Number of Turns**:
If we increase the number of turns \( N \), we can see from the formula that \( L \) will increase because \( N^2 \) is in the numerator.

3. **Effect of Cross-Sectional Area**:
If we keep the area \( A \) constant and increase \( N \), \( L \) will increase. However, if we were to decrease the area \( A \) while increasing \( N \), the effect on \( L \) would depend on the relative changes in \( N \) and \( A \).

4. **Keeping Length Constant**:
Assuming the length \( l \) remains constant, the increase in \( N \) will lead to a significant increase in \( L \).

5. **Conclusion**:
Therefore, as the number of turns \( N \) increases, the self-inductance \( L \) of the solenoid increases.

### Final Answer:
The self-inductance \( L \) of a solenoid increases as the number of turns \( N \) increases.