A solenoid having a core of cross-section `4cm^(2)`, half air and half iron (relative permeability =500), is 22 cm long . If the number of turns on it is `10^(3)` its self inductance is __

Options

A. 570 H
B. 57 H
C. 5.7 H
D. 0.57 H

Answer

To find the self-inductance \( L \) of the solenoid, we will use the formula for the self-inductance of a solenoid with a core that has different materials. The formula is given by:

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

Where:
- \( L \) = self-inductance
- \( \mu_0 \) = permeability of free space = \( 4\pi \times 10^{-7} \, \text{H/m} \)
- \( \mu_r \) = relative permeability of the core material
- \( N \) = number of turns
- \( A \) = cross-sectional area
- \( l \) = length of the solenoid

### Step 1: Identify the parameters
- Cross-sectional area \( A = 4 \, \text{cm}^2 = 4 \times 10^{-4} \, \text{m}^2 \)
- Length \( l = 22 \, \text{cm} = 0.22 \, \text{m} \)
- Number of turns \( N = 10^3 = 1000 \)
- Relative permeability of air \( \mu_{r1} = 1 \)
- Relative permeability of iron \( \mu_{r2} = 500 \)

### Step 2: Calculate the effective relative permeability
Since the solenoid is half air and half iron, the effective relative permeability \( \mu_r \) can be calculated as the average:

\[ \mu_r = \frac{\mu_{r1} + \mu_{r2}}{2} = \frac{1 + 500}{2} = \frac{501}{2} = 250.5 \]

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

\[ L = \frac{(4\pi \times 10^{-7}) \times 250.5 \times (1000)^2 \times (4 \times 10^{-4})}{0.22} \]

### Step 4: Calculate \( L \)
Calculating the numerator:

\[ L = \frac{(4\pi \times 10^{-7}) \times 250.5 \times 1000000 \times 4 \times 10^{-4}}{0.22} \]

Calculating step by step:

1. \( 1000^2 = 1000000 \)
2. \( 4 \times 10^{-4} = 0.0004 \)
3. \( 4\pi \times 10^{-7} \approx 1.25664 \times 10^{-6} \)

Now, substituting these values:

\[ L = \frac{(1.25664 \times 10^{-6}) \times 250.5 \times 1000000 \times 0.0004}{0.22} \]

Calculating the numerator:

\[ = 1.25664 \times 250.5 \times 0.4 \approx 126.56 \times 10^{-6} \]

Now dividing by \( 0.22 \):

\[ L \approx \frac{126.56 \times 10^{-6}}{0.22} \approx 0.575 \, \text{H} \]

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

\[ L \approx 0.571 \, \text{H} \]