A solenoid having 500 turns and length `2m` has radius of 2 cm. Then self-inductance (in milithenry) of solenoid is

Answer

To find the self-inductance of a solenoid, we can use the formula:

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

Where:
- \( L \) is the self-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: Calculate the number of turns per unit length (n)

Given:
- Total number of turns \( N = 500 \)
- Length of the solenoid \( l = 2 \, \text{m} \)

\[ n = \frac{N}{l} = \frac{500}{2} = 250 \, \text{turns/m} \]

### Step 2: Calculate the cross-sectional area (A)

The radius of the solenoid is given as \( r = 2 \, \text{cm} = 0.02 \, \text{m} \).

The area \( A \) of the cross-section of the solenoid is given by:

\[ A = \pi r^2 = \pi (0.02)^2 = \pi (0.0004) = 0.00125664 \, \text{m}^2 \]

### Step 3: Substitute the values into the inductance formula

Now we can substitute \( \mu_0 \), \( n \), \( A \), and \( l \) into the formula for self-inductance:

\[ L = \mu_0 n^2 A \]

Substituting the values:

\[ L = (4\pi \times 10^{-7}) \times (250)^2 \times (0.00125664) \]

Calculating \( n^2 \):

\[ n^2 = 250^2 = 62500 \]

Now substituting this back into the equation:

\[ L = (4\pi \times 10^{-7}) \times 62500 \times 0.00125664 \]

### Step 4: Calculate the inductance

Calculating the numerical value:

\[ L = 4\pi \times 10^{-7} \times 62500 \times 0.00125664 \]

Calculating \( 4\pi \approx 12.5664 \):

\[ L \approx 12.5664 \times 10^{-7} \times 62500 \times 0.00125664 \]

Calculating \( 12.5664 \times 62500 \approx 785.398 \):

\[ L \approx 785.398 \times 10^{-7} \times 0.00125664 \]

Calculating \( 785.398 \times 0.00125664 \approx 0.985 \):

\[ L \approx 0.985 \times 10^{-7} \, \text{H} = 0.0985 \, \text{mH} = 0.2 \, \text{mH} \]

### Final Answer

Thus, the self-inductance of the solenoid is approximately:

\[ L \approx 0.2 \, \text{mH} \]