If the length and area of cross-section of an inductor remain same but the number of turns is doubled, its self-inductance will become-

Options

A. half
B. four times
C. double
D. one-fourth

Explanation

To solve the problem, we need to understand the relationship between the self-inductance of an inductor and the number of turns it has. The self-inductance (L) is directly proportional to the square of the number of turns (n) in the coil.

### Step-by-Step Solution:

1. **Understanding the Formula**:
The self-inductance (L) of an inductor is given by the formula:
\[ L \propto n^2 \]
where \( n \) is the number of turns.

2. **Initial Condition**:
Let's assume the initial number of turns is \( n \) and the initial self-inductance is \( L \). We can express this relationship as:
\[ L = k \cdot n^2 \]
where \( k \) is a constant that depends on the physical characteristics of the inductor (length, area of cross-section, and permeability of the core).

3. **Doubling the Number of Turns**:
If the number of turns is doubled, the new number of turns becomes \( 2n \). We can express the new self-inductance \( L' \) as:
\[ L' = k \cdot (2n)^2 \]

4. **Calculating the New Self-Inductance**:
Expanding the equation for \( L' \):
\[ L' = k \cdot (4n^2) = 4k \cdot n^2 \]
Since \( L = k \cdot n^2 \), we can substitute this into the equation:
\[ L' = 4L \]

5. **Conclusion**:
Therefore, the new self-inductance \( L' \) is four times the initial self-inductance \( L \):
\[ L' = 4L \]

### Final Answer:
The self-inductance will become **four times** the initial self-inductance.

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