An air core solenoid has `1000`turns and is one metre long. Its cross-sectional area is `10cm^(2)`. Its self-inductance is

Options

A. `0.1256 mH`
B. `12.56 mH`
C. `1.256 mH`
D. `125.6 mH`

Explanation

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

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

where:
- \( L \) = self-inductance (in henries)
- \( \mu_0 \) = permeability of free space \( = 4\pi \times 10^{-7} \, \text{H/m} \)
- \( N \) = number of turns (1000 turns)
- \( A \) = cross-sectional area (in square meters)
- \( l \) = length of the solenoid (1 meter)

### Step 1: Convert the cross-sectional area from cm² to m²
The cross-sectional area \( A \) is given as \( 10 \, \text{cm}^2 \). We need to convert this to square meters:

\[ A = 10 \, \text{cm}^2 = 10 \times 10^{-4} \, \text{m}^2 = 0.001 \, \text{m}^2 \]

### Step 2: Substitute the values into the self-inductance formula
Now we can substitute the values into the self-inductance formula:

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

Substituting the known values:

\[ L = (4\pi \times 10^{-7}) \frac{(1000)^2 (0.001)}{1} \]

### Step 3: Calculate \( L \)
Now we perform the calculations:

\[ L = (4\pi \times 10^{-7}) \frac{1000000 \times 0.001}{1} \]
\[ L = (4\pi \times 10^{-7}) \times 1000 \]
\[ L = 4\pi \times 10^{-4} \, \text{H} \]

Using \( \pi \approx 3.14 \):

\[ L \approx 4 \times 3.14 \times 10^{-4} \approx 1.256 \times 10^{-3} \, \text{H} \]

### Step 4: Convert to millihenries
To convert henries to millihenries:

\[ L \approx 1.256 \, \text{mH} \]

### Final Answer
The self-inductance of the air core solenoid is approximately:

\[ L \approx 1.256 \, \text{mH} \]

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