Area of cross section of plane circular coil makes twice keeping it number of turns same, then percantage increase in its self inductance :-

Options

A. `41.4%`
B. `141.4%`
C. `200%`
D. `100%`

Answer

To solve the problem of finding the percentage increase in self-inductance when the area of a circular coil is doubled while keeping the number of turns constant, we can follow these steps:

### Step-by-Step Solution:

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

2. **Identify Changes in Area and Length**:
- The area \( A \) is doubled, so the new area \( A' = 2A \).
- The number of turns \( N \) remains constant.
- We need to find out how the length \( l \) changes when the area is doubled.

3. **Relate Area to Radius**:
The area of a circular coil is given by:
\[ A = \pi r^2 \]
If the area is doubled:
\[ 2A = \pi r'^2 \]
Substituting for \( A \):
\[ 2\pi r^2 = \pi r'^2 \]
Dividing both sides by \( \pi \):
\[ 2r^2 = r'^2 \]
Taking the square root:
\[ r' = \sqrt{2} r \]

4. **Determine the New Length**:
The length \( l \) of the coil can be expressed in terms of the radius:
\[ l = N \cdot 2\pi r \]
With the new radius \( r' \):
\[ l' = N \cdot 2\pi r' = N \cdot 2\pi (\sqrt{2} r) = \sqrt{2} \cdot l \]

5. **Calculate the New Self-Inductance \( L' \)**:
Substitute the new area \( A' \) and new length \( l' \) into the inductance formula:
\[ L' = \frac{\mu_0 N^2 (2A)}{(\sqrt{2} l)} = \frac{2\mu_0 N^2 A}{\sqrt{2} l} = \frac{\sqrt{2} \cdot \mu_0 N^2 A}{l} = \sqrt{2} L \]

6. **Find the Percentage Increase**:
The percentage increase in self-inductance is given by:
\[ \text{Percentage Increase} = \left( \frac{L' - L}{L} \right) \times 100 \]
Substituting \( L' = \sqrt{2} L \):
\[ \text{Percentage Increase} = \left( \frac{\sqrt{2} L - L}{L} \right) \times 100 = \left( \sqrt{2} - 1 \right) \times 100 \]
Approximating \( \sqrt{2} \approx 1.414 \):
\[ \text{Percentage Increase} = (1.414 - 1) \times 100 = 0.414 \times 100 = 41.4\% \]

### Final Answer:
The percentage increase in self-inductance is **41.4%**.