A coil is wound on a frame of rectangular cross - section. If all the linear dimensions of the frame are increased by a factor x and the number of turns per unit length of the coil remains the same, self - inductance of the coil increases by a factor of

Options

A. `x^(2)`
B. `x^(3)`
C. `x^(4)`
D. `x^(5)`

Explanation

To solve the problem, we need to analyze how the self-inductance of the coil changes when the linear dimensions of the frame are increased by a factor of \( x \), while keeping the number of turns per unit length constant.

### Step-by-Step Solution:

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

2. **Expressing Number of Turns**:
The number of turns \( N \) can be expressed in terms of the number of turns per unit length \( n \) and the length \( l \):
\[ N = n \cdot l \]

3. **Substituting \( N \) in the Inductance Formula**:
Substituting \( N \) into the inductance formula gives:
\[ L = \mu_0 \frac{(n \cdot l)^2 A}{l} = \mu_0 n^2 l A \]

4. **Changing Dimensions**:
When all linear dimensions are increased by a factor of \( x \):
- The new length \( l' = x \cdot l \)
- The new cross-sectional area \( A' = x^2 \cdot A \) (since area is proportional to the square of the linear dimensions)

5. **Calculating New Inductance**:
The new inductance \( L' \) can be calculated as follows:
\[ L' = \mu_0 n^2 l' A' = \mu_0 n^2 (x \cdot l) (x^2 \cdot A) = \mu_0 n^2 x^3 l A \]

6. **Finding the Factor of Increase**:
Now, we can express the new inductance in terms of the old inductance:
\[ L' = x^3 \cdot (\mu_0 n^2 l A) = x^3 \cdot L \]
Therefore, the self-inductance increases by a factor of \( x^3 \).

### Conclusion:
The self-inductance of the coil increases by a factor of \( x^3 \).