A long solenoid has self inductance. L. If its length is doubled keeping total number of turns constant then its new self inductance will be

Options

A. `(L)/(2)`
B. 2L
C. L
D. `(L)/(4)`

Answer

To solve the problem of finding the new self-inductance of a long solenoid when its length is doubled while keeping the total number of turns constant, we can follow these steps:

### Step-by-Step Solution:

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

2. **Identify Changes in Length**:
We are told that the length of the solenoid is doubled. Let the original length be \( l_1 \) and the new length be \( l_2 = 2l_1 \).

3. **Keeping Total Number of Turns Constant**:
Since the total number of turns \( N \) remains constant, we can express the number of turns per unit length \( n \) as:
\[ n = \frac{N}{l} \]
Thus, for the original solenoid:
\[ n_1 = \frac{N}{l_1} \]
and for the new solenoid:
\[ n_2 = \frac{N}{l_2} = \frac{N}{2l_1} = \frac{n_1}{2} \]

4. **Substituting into the Inductance Formula**:
The new self-inductance \( L_2 \) can be calculated using the modified values:
\[ L_2 = \mu_0 \frac{N^2 A}{l_2} \]
Substituting \( l_2 = 2l_1 \):
\[ L_2 = \mu_0 \frac{N^2 A}{2l_1} \]

5. **Relating New Inductance to Original Inductance**:
We can relate \( L_2 \) to \( L_1 \):
\[ L_1 = \mu_0 \frac{N^2 A}{l_1} \]
Therefore,
\[ L_2 = \frac{L_1}{2} \]

6. **Conclusion**:
The new self-inductance \( L_2 \) when the length of the solenoid is doubled while keeping the total number of turns constant is:
\[ L_2 = \frac{L_1}{2} \]

### Final Answer:
The new self-inductance will be \( \frac{L}{2} \).