In a given solenoid if the number of turns and the length of the solenoid are doubled keeping the area of cross-section same, then it's inductance:

Options

A. a. halved
B. b. doubled
C. c. 1/4 times the original value
D. d. unaffected

Answer

To solve the problem, we need to understand how the inductance of a solenoid is calculated and how changes in the number of turns and length affect it.

### Step-by-Step Solution:

1. **Understanding the Formula for Inductance**:
The 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 per unit length,
- \( A \) is the cross-sectional area,
- \( l \) is the length of the solenoid.

2. **Identifying Initial Conditions**:
Let the initial number of turns be \( n_1 \), the initial length be \( l_1 \), and the area be \( A \). Thus, the initial inductance \( L_1 \) can be expressed as:
\[ L_1 = \frac{\mu_0 n_1^2 A}{l_1} \]

3. **Modifying the Number of Turns and Length**:
According to the problem, both the number of turns and the length of the solenoid are doubled:
- New number of turns \( n_2 = 2n_1 \)
- New length \( l_2 = 2l_1 \)

4. **Calculating New Inductance**:
Substitute \( n_2 \) and \( l_2 \) into the inductance formula:
\[ L_2 = \frac{\mu_0 n_2^2 A}{l_2} = \frac{\mu_0 (2n_1)^2 A}{2l_1} \]
Simplifying this gives:
\[ L_2 = \frac{\mu_0 \cdot 4n_1^2 A}{2l_1} = \frac{2\mu_0 n_1^2 A}{l_1} \]

5. **Relating New Inductance to Initial Inductance**:
Now, we can express \( L_2 \) in terms of \( L_1 \):
\[ L_2 = 2 \cdot \frac{\mu_0 n_1^2 A}{l_1} = 2L_1 \]

6. **Conclusion**:
Thus, the inductance \( L_2 \) after doubling the number of turns and the length of the solenoid is:
\[ L_2 = 2L_1 \]

### Final Answer:
The inductance of the solenoid is doubled.