A solenoid wound over a rectangular frame. If all the linear dimensions of the frame are increased by a factor 3 and he number of turns per unit length remains the same, the self inductance increased by a factor of

Options

A. 3
B. 9
C. 27
D. 63

Answer

To solve the problem, we need to analyze how the self-inductance of a solenoid changes when the dimensions of the rectangular frame it is wound around are increased by a factor of 3, while keeping the number of turns per unit length constant.

### Step-by-Step Solution:

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

2. **Effect of Increasing Dimensions**:
If all linear dimensions of the rectangular frame are increased by a factor of 3, then:
- The new length \( l' = 3l \),
- The new width \( w' = 3w \) (if applicable),
- The new height \( h' = 3h \) (if applicable).

3. **Calculating the New Area**:
The area \( A \) of the rectangular frame is given by:
\[ A = l \times w \]
When the dimensions are increased by a factor of 3:
\[ A' = l' \times w' = (3l) \times (3w) = 9lw = 9A \]

4. **Substituting into the Self-Inductance Formula**:
Now substituting the new area \( A' \) and the new length \( l' \) into the self-inductance formula:
\[ L' = \frac{\mu_0 n^2 A'}{l'} = \frac{\mu_0 n^2 (9A)}{3l} = \frac{9\mu_0 n^2 A}{3l} = 3 \cdot \frac{\mu_0 n^2 A}{l} = 3L \]

5. **Conclusion**:
Therefore, the self-inductance \( L' \) increases by a factor of 3 when all linear dimensions of the frame are increased by a factor of 3, while keeping the number of turns per unit length constant.

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

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