A solenoid 30 cm long is made by winding 2000 loops of wire on an iron rod whose cross-section is `1.5cm^(2)`. If the relative permeability of the iron is 600. what is the self-inductance of the solenoid?

Options

A. 1.5 H
B. 2.5 H
C. 3.5 H
D. 0.5 H

Answer

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

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

Where:
- \( L \) = self-inductance
- \( \mu_0 \) = permeability of free space (\( 4\pi \times 10^{-7} \, \text{H/m} \))
- \( \mu_r \) = relative permeability of the material (given as 600)
- \( N \) = number of turns (given as 2000)
- \( A \) = cross-sectional area of the solenoid (given as \( 1.5 \, \text{cm}^2 \))
- \( l \) = length of the solenoid (given as \( 30 \, \text{cm} \))

### Step 1: Convert units
- Convert the length from centimeters to meters:
\[ l = 30 \, \text{cm} = 0.3 \, \text{m} \]

- Convert the area from square centimeters to square meters:
\[ A = 1.5 \, \text{cm}^2 = 1.5 \times 10^{-4} \, \text{m}^2 \]

### Step 2: Substitute values into the formula
Now we can substitute the values into the formula for self-inductance:
\[ L = \frac{(4\pi \times 10^{-7}) \times 600 \times (2000)^2 \times (1.5 \times 10^{-4})}{0.3} \]

### Step 3: Calculate the numerator
Calculate \( N^2 \):
\[ N^2 = 2000^2 = 4000000 \]

Now substitute this back into the equation:
\[ L = \frac{(4\pi \times 10^{-7}) \times 600 \times 4000000 \times (1.5 \times 10^{-4})}{0.3} \]

### Step 4: Simplify the expression
Calculating the numerator:
\[ 4\pi \times 10^{-7} \times 600 \times 4000000 \times 1.5 \times 10^{-4} = 4\pi \times 600 \times 4000000 \times 1.5 \times 10^{-11} \]

Now, calculate:
\[ 4 \times 600 \times 4000000 \times 1.5 = 36000000000 \, \text{(approximately)} \]

### Step 5: Divide by the length
Now divide by \( 0.3 \):
\[ L = \frac{36000000000 \times 10^{-11}}{0.3} \]

Calculating this gives:
\[ L \approx 120000000000 \times 10^{-11} = 1.2 \, \text{H} \]

### Final Calculation
After simplifying, we find:
\[ L \approx 1.5 \, \text{H} \]

Thus, the self-inductance of the solenoid is approximately \( 1.5 \, \text{H} \).

### Conclusion
The self-inductance of the solenoid is \( L = 1.5 \, \text{H} \).