When the number of turns in a coil is doubled without any change in the length of the coil, its self-inductance becomes

Options

A. Half
B. Double
C. Four times
D. Eight times

Answer

To solve the problem of how the self-inductance of a coil changes when the number of turns is doubled, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Formula for Self-Inductance**:
The self-inductance \( L \) of a coil 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 of the coil,
- \( l \) is the length of the coil.

2. **Identify Constants**:
In this problem, we are told that the length of the coil \( l \) remains unchanged, and we assume that the permeability \( \mu_0 \) and the cross-sectional area \( A \) also remain constant.

3. **Determine the Effect of Doubling the Number of Turns**:
If the number of turns \( N \) is doubled, we can denote the initial number of turns as \( N \) and the new number of turns as \( 2N \).

4. **Calculate the New Self-Inductance**:
Substitute \( 2N \) into the self-inductance formula:
\[ L' = \frac{\mu_0 (2N)^2 A}{l} \]
Simplifying this gives:
\[ L' = \frac{\mu_0 (4N^2) A}{l} = 4 \left( \frac{\mu_0 N^2 A}{l} \right) = 4L \]
Therefore, the new self-inductance \( L' \) is four times the original self-inductance \( L \).

5. **Conclusion**:
The self-inductance becomes \( 4L \) when the number of turns is doubled.

### Final Answer:
The self-inductance becomes 4 times the initial self-inductance.

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