The inductance of a solenoid `0.5m` long of cross-sectional area `20cm^(2)` and with `500` turns is

Options

A. `12.5 mH`
B. `1.25mH`
C. `15.0mH`
D. `0.12mH`

Explanation

To find the inductance of a solenoid, we can use the formula for the self-inductance of a solenoid, which is given by:

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

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

### Step 1: Convert the given values to appropriate units
- Length of the solenoid, \(l = 0.5 \, \text{m}\)
- Cross-sectional area, \(A = 20 \, \text{cm}^2 = 20 \times 10^{-4} \, \text{m}^2 = 2 \times 10^{-3} \, \text{m}^2\)
- Number of turns, \(N = 500\)

### Step 2: Calculate the number of turns per unit length
The number of turns per unit length \(n\) can be calculated as:

\[ n = \frac{N}{l} = \frac{500}{0.5} = 1000 \, \text{turns/m} \]

### Step 3: Substitute the values into the inductance formula
Now we can substitute the values into the inductance formula:

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

Substituting the values:

\[ L = \frac{(1000)^2 \cdot (4\pi \times 10^{-7}) \cdot (2 \times 10^{-3})}{0.5} \]

### Step 4: Calculate the inductance
Calculating the numerator:

\[ (1000)^2 = 1000000 \]

Now substituting:

\[ L = \frac{1000000 \cdot (4\pi \times 10^{-7}) \cdot (2 \times 10^{-3})}{0.5} \]

Calculating \(4\pi \times 10^{-7} \approx 1.25664 \times 10^{-6}\):

\[ L = \frac{1000000 \cdot 1.25664 \times 10^{-6} \cdot 2 \times 10^{-3}}{0.5} \]

Calculating the numerator:

\[ 1000000 \cdot 1.25664 \times 10^{-6} \cdot 2 \times 10^{-3} = 2.51328 \times 10^{-3} \]

Now dividing by \(0.5\):

\[ L = \frac{2.51328 \times 10^{-3}}{0.5} = 5.02656 \times 10^{-3} \, \text{H} = 5.02656 \, \text{mH} \]

### Final Result
The inductance of the solenoid is approximately:

\[ L \approx 5.03 \, \text{mH} \]