The self inductance of a solenoid that has a cross-sectional area of `1 cm^(2)`, a length of 10 cm and 1000 turns of wire is

Options

A. `0.86 mH`
B. `1.06 mH`
C. `1.26 mH`
D. `1.46 mH`

Answer

To calculate the self-inductance \( L \) of a solenoid, we can use the formula:

\[ L = \mu_0 \cdot N^2 \cdot \frac{A}{l} \]

where:
- \( \mu_0 \) is the permeability of free space, approximately \( 4\pi \times 10^{-7} \, \text{H/m} \)
- \( N \) is the total number of turns of the solenoid
- \( A \) is the cross-sectional area of the solenoid in square meters
- \( l \) is the length of the solenoid in meters

### Step-by-Step Solution:

1. **Identify the given values:**
- Cross-sectional area \( A = 1 \, \text{cm}^2 = 1 \times 10^{-4} \, \text{m}^2 \)
- Length \( l = 10 \, \text{cm} = 0.1 \, \text{m} \)
- Number of turns \( N = 1000 \)

2. **Substitute the values into the formula:**
\[ L = \mu_0 \cdot N^2 \cdot \frac{A}{l} \]

3. **Calculate \( N^2 \):**
\[ N^2 = 1000^2 = 1000000 \]

4. **Substitute \( \mu_0 \), \( N^2 \), \( A \), and \( l \) into the equation:**
\[ L = (4\pi \times 10^{-7}) \cdot (1000000) \cdot \frac{1 \times 10^{-4}}{0.1} \]

5. **Simplify the expression:**
\[ L = (4\pi \times 10^{-7}) \cdot (1000000) \cdot (10^{-3}) \]
\[ L = 4\pi \times 10^{-4} \, \text{H} \]

6. **Calculate \( 4\pi \):**
\[ 4\pi \approx 12.5664 \]

7. **Final calculation:**
\[ L \approx 12.5664 \times 10^{-4} \, \text{H} = 1.25664 \times 10^{-3} \, \text{H} \]
\[ L \approx 1.26 \, \text{mH} \]

### Final Answer:
The self-inductance of the solenoid is approximately \( 1.26 \, \text{mH} \).