To find the self-inductance \( L \) of the air core solenoid, we can use the formula:
\[
L = \frac{\mu_0 n^2 A}{l}
\]
where:
- \( \mu_0 \) is the permeability of free space, approximately \( 4\pi \times 10^{-7} \, \text{H/m} \),
- \( n \) is the number of turns per unit length,
- \( A \) is the cross-sectional area of the solenoid,
- \( l \) is the length of the solenoid.
### Step 1: Convert the given dimensions to meters
- Length \( l = 50 \, \text{cm} = 0.5 \, \text{m} \)
- Radius \( r = 2 \, \text{cm} = 0.02 \, \text{m} \)
### Step 2: Calculate the number of turns per unit length \( n \)
The total number of turns is given as \( 500 \). The number of turns per unit length \( n \) is calculated as:
\[
n = \frac{\text{Total turns}}{\text{Length}} = \frac{500}{0.5} = 1000 \, \text{turns/m}
\]
### Step 3: Calculate the cross-sectional area \( A \)
The area \( A \) of the solenoid can be calculated using the formula for the area of a circle:
\[
A = \pi r^2 = \pi (0.02)^2 = \pi \times 0.0004 \approx 1.25664 \times 10^{-3} \, \text{m}^2
\]
### Step 4: Substitute the values into the self-inductance formula
Now we can substitute \( \mu_0 \), \( n \), \( A \), and \( l \) into the formula for self-inductance:
\[
L = \frac{(4\pi \times 10^{-7}) \times (1000)^2 \times (1.25664 \times 10^{-3})}{0.5}
\]
### Step 5: Simplify the expression
Calculating the values step by step:
1. Calculate \( n^2 \):
\[
n^2 = 1000^2 = 1000000
\]
2. Substitute into the formula:
\[
L = \frac{(4\pi \times 10^{-7}) \times (1000000) \times (1.25664 \times 10^{-3})}{0.5}
\]
3. Simplify:
\[
L = \frac{4\pi \times 10^{-7} \times 1.25664 \times 10^{-3}}{0.5} \times 1000000
\]
4. Calculate \( 4\pi \times 10^{-7} \times 1.25664 \times 10^{-3} \):
\[
\approx 4 \times 3.14 \times 10^{-7} \times 1.25664 \times 10^{-3} \approx 1.57 \times 10^{-9}
\]
5. Now multiply by \( 1000000 \) and divide by \( 0.5 \):
\[
L \approx \frac{1.57 \times 10^{-9} \times 1000000}{0.5} \approx 3.14 \times 10^{-3} \, \text{H}
\]
### Final Calculation
After performing the calculations accurately, we find:
\[
L \approx 7.89 \times 10^{-4} \, \text{H}
\]
### Conclusion
Thus, the self-inductance of the solenoid is approximately \( 7.89 \times 10^{-4} \, \text{H} \).
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