The self inductance of a coil is L . Keeping the length and area same, the number of turns in the coil is increased to four times. The self inductance of the coil will now be

Options

A. `(1)/(4)L`
B. L
C. 4 L
D. 16 L

Explanation

To solve the problem, we need to understand the relationship between the self-inductance of a coil and the number of turns in the coil. The self-inductance \( L \) of a coil is given by the formula:

\[ L = \mu \frac{N^2 A}{l} \]

where:
- \( L \) is the self-inductance,
- \( \mu \) is the permeability of the core material,
- \( N \) is the number of turns,
- \( A \) is the cross-sectional area of the coil,
- \( l \) is the length of the coil.

### Step-by-Step Solution:

1. **Initial Self-Inductance**:
Let's denote the initial number of turns as \( N \). The initial self-inductance of the coil is given by:
\[ L = \mu \frac{N^2 A}{l} \]

2. **Increasing the Number of Turns**:
According to the problem, the number of turns is increased to four times the original number. Therefore, the new number of turns is:
\[ N' = 4N \]

3. **New Self-Inductance**:
We can now calculate the new self-inductance \( L' \) with the increased number of turns:
\[ L' = \mu \frac{(N')^2 A}{l} \]
Substituting \( N' = 4N \):
\[ L' = \mu \frac{(4N)^2 A}{l} \]

4. **Simplifying the Expression**:
Now, simplify the expression:
\[ L' = \mu \frac{16N^2 A}{l} \]
This can be rewritten as:
\[ L' = 16 \left( \mu \frac{N^2 A}{l} \right) \]
Thus:
\[ L' = 16L \]

5. **Conclusion**:
Therefore, the new self-inductance of the coil when the number of turns is increased to four times is:
\[ L' = 16L \]

### Final Answer:
The self-inductance of the coil will now be \( 16L \).

---