A wire of fixed length is wound on a solenoid of length `l` and radius `r`. Its self-inductance is found to be `L`. Now, if the same wire is wound on a solenoid of length `l//2` and radius `r//2` then the self-inductance will be

Options

A. 2L
B. L
C. 4L
D. 8L

Answer

To solve the problem, we need to understand how the self-inductance of a solenoid changes when its dimensions are altered while keeping the length of the wire constant.

### Step-by-Step Solution:

1. **Understand the formula for self-inductance (L)**:
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 per unit length,
- \( A \) is the cross-sectional area of the solenoid,
- \( l \) is the length of the solenoid.

2. **Define the initial solenoid parameters**:
For the first solenoid, let:
- Length = \( l \)
- Radius = \( r \)
- Cross-sectional area \( A = \pi r^2 \)
- The total length of the wire = \( L_1 \)

3. **Calculate the number of turns (n)**:
The number of turns \( n \) can be expressed as:
\[ n = \frac{L_1}{2 \pi r} \]
where \( 2 \pi r \) is the circumference of the solenoid.

4. **Substituting \( n \) into the self-inductance formula**:
Substitute \( n \) into the self-inductance formula:
\[ L = \frac{\mu_0 \left(\frac{L_1}{2 \pi r}\right)^2 \pi r^2}{l} \]
Simplifying this gives:
\[ L = \frac{\mu_0 L_1^2}{4 \pi l} \]

5. **Change the dimensions of the solenoid**:
Now, we change the dimensions of the solenoid to:
- New length = \( \frac{l}{2} \)
- New radius = \( \frac{r}{2} \)
- New cross-sectional area \( A' = \pi \left(\frac{r}{2}\right)^2 = \frac{\pi r^2}{4} \)

6. **Calculate the new number of turns (n')**:
The new number of turns \( n' \) is:
\[ n' = \frac{L_1}{2 \pi \left(\frac{r}{2}\right)} = \frac{L_1}{\pi r} \]

7. **Substituting \( n' \) into the new self-inductance formula**:
Substitute \( n' \) into the self-inductance formula:
\[ L' = \frac{\mu_0 \left(\frac{L_1}{\pi r}\right)^2 \cdot \frac{\pi r^2}{4}}{\frac{l}{2}} \]
Simplifying this gives:
\[ L' = \frac{\mu_0 L_1^2}{\pi r^2} \cdot \frac{2}{l} = \frac{2 \mu_0 L_1^2}{4 \pi l} = 2L \]

8. **Conclusion**:
The self-inductance of the new solenoid is:
\[ L' = 2L \]

### Final Answer:
The self-inductance when the wire is wound on a solenoid of length \( \frac{l}{2} \) and radius \( \frac{r}{2} \) will be \( 2L \).