If the self inductance of 500 turns coil is 125 mH, then the self inductance of the similar coil of 800 turns is

Options

A. `48.8 mH`
B. `200 mH`
C. `290 mH`
D. `320 mH`

Explanation

To find the self-inductance of a coil with 800 turns given that the self-inductance of a 500-turn coil is 125 mH, we can use the relationship that the self-inductance \( L \) is directly proportional to the square of the number of turns \( N \).

### Step-by-Step Solution:

1. **Write the relationship for self-inductance**:
\[ L \propto N^2 \]
This means that if \( L_1 \) is the self-inductance for \( N_1 \) turns, and \( L_2 \) is the self-inductance for \( N_2 \) turns, we can write:
\[ \frac{L_1}{L_2} = \frac{N_1^2}{N_2^2} \]

2. **Assign known values**:
Let:
- \( L_1 = 125 \, \text{mH} \) (self-inductance of the 500-turn coil)
- \( N_1 = 500 \) (number of turns in the first coil)
- \( N_2 = 800 \) (number of turns in the second coil)
- \( L_2 \) = unknown (self-inductance of the 800-turn coil)

3. **Set up the equation**:
Using the relationship we established:
\[ \frac{125 \, \text{mH}}{L_2} = \frac{500^2}{800^2} \]

4. **Calculate \( \frac{500^2}{800^2} \)**:
\[ \frac{500^2}{800^2} = \frac{250000}{640000} = \frac{25}{64} \]

5. **Substitute back into the equation**:
\[ \frac{125 \, \text{mH}}{L_2} = \frac{25}{64} \]

6. **Cross-multiply to solve for \( L_2 \)**:
\[ 125 \, \text{mH} \cdot 64 = 25 \cdot L_2 \]
\[ 8000 \, \text{mH} = 25 \cdot L_2 \]

7. **Divide both sides by 25**:
\[ L_2 = \frac{8000 \, \text{mH}}{25} = 320 \, \text{mH} \]

### Final Answer:
The self-inductance of the similar coil with 800 turns is **320 mH**.

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