In circular coil, when no. of turns is doubled and resistance becomes `(1)/(4)th` of initial, then inductance becomes

Options

A. 4 times
B. 2 times
C. 8 times
D. no change

Explanation

To solve the problem, we need to analyze how the inductance of a circular coil changes when the number of turns is doubled and the resistance is reduced to one-fourth of its initial value.

### Step-by-Step Solution:

1. **Understanding Inductance**:
The self-inductance \( L \) of a circular coil is given by the formula:
\[ L = \frac{\mu_0 n^2 A}{2r} \]
where:
- \( \mu_0 \) is the permeability of free space,
- \( n \) is the number of turns,
- \( A \) is the area of the coil,
- \( r \) is the radius of the coil.

2. **Doubling the Number of Turns**:
If the number of turns \( n \) is doubled, we can express the new number of turns as \( n' = 2n \).

3. **Calculating New Inductance**:
Substitute \( n' \) into the inductance formula:
\[ L' = \frac{\mu_0 (n')^2 A}{2r} = \frac{\mu_0 (2n)^2 A}{2r} = \frac{\mu_0 \cdot 4n^2 A}{2r} = 2 \cdot \frac{\mu_0 n^2 A}{r} \]
Thus, we can see that the new inductance \( L' \) is:
\[ L' = 4L \]
This shows that the inductance becomes four times the initial inductance when the number of turns is doubled.

4. **Effect of Resistance**:
The problem states that the resistance \( R \) becomes \( \frac{1}{4} \) of its initial value. However, from the formula for inductance, we see that inductance \( L \) is independent of resistance. Therefore, changing the resistance does not affect the inductance.

5. **Final Conclusion**:
Since the inductance is directly proportional to the square of the number of turns and is independent of resistance, we conclude that the inductance becomes four times the initial inductance when the number of turns is doubled.

### Final Answer:
\[ \text{Inductance becomes } 4L \text{ (four times the initial inductance)} \]

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