What is the self-inductance of an air core solenoid 1 m long, diameter 0.05 m, if it has 700 turns?

Answer

To find the self-inductance of an air core solenoid, we can use the formula for the self-inductance \( L \) of a solenoid:

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

Where:
- \( L \) = self-inductance in Henrys (H)
- \( \mu_0 \) = permeability of free space \( = 4\pi \times 10^{-7} \, \text{H/m} \)
- \( n \) = number of turns per unit length (turns/m)
- \( A \) = cross-sectional area of the solenoid (m²)
- \( l \) = length of the solenoid (m)

### Step 1: Calculate the radius of the solenoid
Given the diameter \( d = 0.05 \, \text{m} \), the radius \( r \) is:

\[ r = \frac{d}{2} = \frac{0.05}{2} = 0.025 \, \text{m} \]

### Step 2: Calculate the cross-sectional area \( A \)
The cross-sectional area \( A \) of the solenoid can be calculated using the formula for the area of a circle:

\[ A = \pi r^2 = \pi (0.025)^2 \]

Calculating this gives:

\[ A = \pi \times 0.000625 = 0.001963495 \, \text{m}^2 \]

### Step 3: Calculate the number of turns per unit length \( n \)
The total number of turns \( N = 700 \) and the length \( l = 1 \, \text{m} \), so:

\[ n = \frac{N}{l} = \frac{700}{1} = 700 \, \text{turns/m} \]

### Step 4: Substitute values into the self-inductance formula
Now we can substitute \( \mu_0 \), \( n \), \( A \), and \( l \) into the self-inductance formula:

\[ L = \frac{(4\pi \times 10^{-7}) \times (700)^2 \times (0.001963495)}{1} \]

### Step 5: Calculate \( L \)
Calculating the components:

1. \( 700^2 = 490000 \)
2. \( 4\pi \times 10^{-7} \approx 1.256637 \times 10^{-6} \)
3. Now substituting:

\[ L = \frac{(1.256637 \times 10^{-6}) \times (490000) \times (0.001963495)}{1} \]

Calculating this gives:

\[ L \approx 1.256637 \times 10^{-6} \times 490000 \times 0.001963495 \approx 0.12078 \, \text{H} \]

### Final Result
Thus, the self-inductance \( L \) of the air core solenoid is approximately:

\[ L \approx 0.12078 \, \text{H} \, \text{or} \, 12078.01 \times 10^{-7} \, \text{H} \]