To find the inductance of a solenoid, we can use the formula:
\[
L = \frac{\mu_0 n^2 A}{l}
\]
where:
- \(L\) is the inductance,
- \(\mu_0\) is the permeability of free space (\(4\pi \times 10^{-7} \, \text{H/m}\)),
- \(n\) 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: Identify the given values
- Mean diameter \(D = 1 \, \text{cm} = 0.01 \, \text{m}\)
- Length of the solenoid \(l = 1 \, \text{m}\)
- Number of turns \(N = 1000\)
### Step 2: Calculate the radius
The radius \(r\) of the solenoid can be calculated as:
\[
r = \frac{D}{2} = \frac{0.01 \, \text{m}}{2} = 0.005 \, \text{m}
\]
### Step 3: Calculate the cross-sectional area \(A\)
The cross-sectional area \(A\) of the solenoid is given by:
\[
A = \pi r^2 = \pi (0.005)^2 = \pi \times 25 \times 10^{-6} \, \text{m}^2 = \frac{\pi}{4} \times 10^{-4} \, \text{m}^2
\]
### Step 4: Calculate the number of turns per unit length \(n\)
The number of turns per unit length \(n\) is given by:
\[
n = \frac{N}{l} = \frac{1000}{1} = 1000 \, \text{turns/m}
\]
### Step 5: Substitute the values into the inductance formula
Now we can substitute the values into the inductance formula:
\[
L = \frac{\mu_0 n^2 A}{l}
\]
Substituting the known values:
\[
L = \frac{(4\pi \times 10^{-7}) \times (1000)^2 \times \left(\frac{\pi}{4} \times 10^{-4}\right)}{1}
\]
### Step 6: Simplify the expression
Calculating the components:
- \(n^2 = 1000^2 = 10^6\)
- Therefore:
\[
L = \frac{(4\pi \times 10^{-7}) \times (10^6) \times \left(\frac{\pi}{4} \times 10^{-4}\right)}{1}
\]
\[
L = \frac{4\pi^2 \times 10^{-7} \times 10^6 \times 10^{-4}}{4}
\]
\[
L = \pi^2 \times 10^{-5} \, \text{H}
\]
### Step 7: Convert to Millihenry
To convert to Millihenry:
\[
L = \pi^2 \times 10^{-5} \, \text{H} \approx 0.1 \, \text{mH}
\]
### Final Answer
Thus, the inductance of the solenoid is approximately:
\[
L \approx 0.1 \, \text{mH}
\]
