A coil is wound as a transformer of rectangular cross section. If all the linear dimension of the transformer are increased by a factor `2` and the number of turns per unit length of the coil remain the same, the self-inductance increased by a factor of

Options

A. `16`
B. `12`
C. `8`
D. `4`

Explanation

To solve the problem, we need to analyze how the self-inductance of a transformer changes when the linear dimensions are increased by a factor of 2, 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 can be expressed in terms of the number of turns \( N \), the area \( A \), and the magnetic field \( B \). For a solenoid, the self-inductance is given by:
\[ L = \frac{\mu_0 N^2 A}{l} \]
where \( \mu_0 \) is the permeability of free space, \( N \) is the total number of turns, \( A \) is the cross-sectional area, and \( l \) is the length of the solenoid.

2. **Initial Dimensions**:
Let the initial length and breadth of the transformer be \( L \) and \( B \) respectively. The area \( A \) is given by:
\[ A = L \times B \]
The total number of turns \( N \) can be expressed as:
\[ N = n \times L \]
where \( n \) is the number of turns per unit length.

3. **Calculating Initial Self-Inductance**:
The initial self-inductance \( L_1 \) can be defined as:
\[ L_1 = \frac{\mu_0 (nL)^2 (LB)}{L} = \mu_0 n^2 L^2 B \]

4. **New Dimensions**:
When all linear dimensions are increased by a factor of 2:
- New length \( L' = 2L \)
- New breadth \( B' = 2B \)
- The area \( A' \) becomes:
\[ A' = L' \times B' = (2L) \times (2B) = 4LB \]

5. **Calculating New Total Number of Turns**:
Since the number of turns per unit length remains the same, the total number of turns \( N' \) is:
\[ N' = n \times L' = n \times (2L) = 2nL \]

6. **Calculating New Self-Inductance**:
The new self-inductance \( L_2 \) can be defined as:
\[ L_2 = \frac{\mu_0 (N')^2 A'}{l'} \]
Substituting the new values:
\[ L_2 = \frac{\mu_0 (2nL)^2 (4LB)}{2L} = \frac{\mu_0 (4n^2L^2)(4LB)}{2L} \]
Simplifying this gives:
\[ L_2 = \frac{\mu_0 \cdot 16n^2L^2B}{2} = 8\mu_0 n^2 L^2 B \]

7. **Finding the Factor of Increase**:
To find the factor by which the self-inductance has increased, we take the ratio of the new self-inductance to the old self-inductance:
\[ \text{Factor} = \frac{L_2}{L_1} = \frac{8\mu_0 n^2 L^2 B}{\mu_0 n^2 L^2 B} = 8 \]

### Conclusion:
The self-inductance of the transformer increases by a factor of **8** when all linear dimensions are doubled.