A coil of wire of a certain radius has 100 turns and a self inductance of 15 mH. The self inductance of a second similar coil of 500 turns will be

Options

A. 75 mH
B. 375 mH
C. 15 mH
D. none of these

Explanation

To solve the problem of finding the self-inductance of a second coil with 500 turns, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Given Values:**
- For the first coil:
- Number of turns, \( n_1 = 100 \)
- Self-inductance, \( L_1 = 15 \, \text{mH} = 15 \times 10^{-3} \, \text{H} \)

- For the second coil:
- Number of turns, \( n_2 = 500 \)

2. **Understand the Relationship:**
- The self-inductance \( L \) of a coil is directly proportional to the square of the number of turns \( n \):
\[ L \propto n^2 \]
- This can be expressed as:
\[ \frac{L_2}{L_1} = \frac{n_2^2}{n_1^2} \]

3. **Set Up the Equation:**
- Rearranging the above relationship gives:
\[ L_2 = L_1 \cdot \frac{n_2^2}{n_1^2} \]

4. **Substitute the Known Values:**
- Substitute \( L_1 \), \( n_1 \), and \( n_2 \) into the equation:
\[ L_2 = 15 \times 10^{-3} \cdot \frac{500^2}{100^2} \]

5. **Calculate the Squares:**
- Calculate \( 500^2 = 250000 \) and \( 100^2 = 10000 \).
- Substitute these values back into the equation:
\[ L_2 = 15 \times 10^{-3} \cdot \frac{250000}{10000} \]

6. **Simplify the Fraction:**
- Simplifying \( \frac{250000}{10000} = 25 \):
\[ L_2 = 15 \times 10^{-3} \cdot 25 \]

7. **Final Calculation:**
- Calculate \( 15 \times 25 = 375 \):
\[ L_2 = 375 \times 10^{-3} \, \text{H} \]

8. **Convert to Millihenry:**
- Convert \( 375 \times 10^{-3} \, \text{H} \) to millihenry:
\[ L_2 = 375 \, \text{mH} \]

### Conclusion:
The self-inductance of the second similar coil is \( 375 \, \text{mH} \).