A copper wire is wire is wound on a wooden frame, whose shape is that of an equilateral triangle.If the number of turns of the coil per units length of the frame the same, then the self inductance of the coil:

Options

A. decreases by a factor of ` 9 sqrt3`
B. decreases by a factor of 9
C. increases by a factor of 27
D. increases by a factor of 3

Explanation

To solve the problem, we need to analyze the self-inductance of a coil wound around a wooden frame shaped like an equilateral triangle. We will derive the self-inductance for both cases and compare them.

### Step-by-Step Solution:

1. **Understanding the Geometry**:
- The wire is wound around a wooden frame shaped like an equilateral triangle. Let the length of each side of the triangle be \( L \).
- The perimeter \( P \) of the equilateral triangle is given by:
\[ P = 3L \]

2. **Number of Turns per Unit Length**:
- Let \( n_1 \) be the number of turns per unit length when the wire is wound around the triangle.
- The total length of the wire used is equal to the perimeter of the triangle, which is \( 3L \).
- Therefore, the number of turns per unit length can be expressed as:
\[ n_1 = \frac{N}{3L} \]
where \( N \) is the total number of turns.

3. **Self-Inductance Formula**:
- The self-inductance \( L \) of a solenoid is given by the formula:
\[ L = \mu_0 n^2 \pi r^2 l \]
where:
- \( \mu_0 \) = permeability of free space,
- \( n \) = number of turns per unit length,
- \( r \) = radius of the wire,
- \( l \) = length of the solenoid.

4. **Calculating Self-Inductance for the Triangle**:
- For the triangle, the effective length \( l \) is the perimeter \( P = 3L \).
- Therefore, the self-inductance \( L_1 \) can be expressed as:
\[ L_1 = \mu_0 n_1^2 \pi r^2 (3L) \]

5. **Considering the New Configuration**:
- Now, if we consider a new configuration where the wire is wound in a different manner but the number of turns per unit length remains the same, let’s denote the new number of turns per unit length as \( n_2 \).
- Since it is given that \( n_1 = n_2 \), we have:
\[ L_2 = \mu_0 n_2^2 \pi r^2 (9L) \]

6. **Relating the Two Self-Inductances**:
- Since \( n_1 = n_2 \), we can substitute \( n_1 \) into the equation for \( L_2 \):
\[ L_2 = \mu_0 n_1^2 \pi r^2 (9L) \]
- Comparing \( L_1 \) and \( L_2 \):
\[ L_2 = 3L_1 \]

7. **Conclusion**:
- The self-inductance of the coil increases by a factor of 3 when the configuration changes while keeping the number of turns per unit length constant.
- Therefore, the correct answer is that the self-inductance increases by a factor of 3.

### Final Answer:
The self-inductance of the coil increases by a factor of 3.