Two coaxial solenoids are made by winding thin insulated wire over a pipe of cross-sectional area `A=10 cm^(2)` and length =10cm. If one of the solenoid has 300 turns and the other 400 turns, their mutual indcutance is

Options

A. `4.8pixx10^(-4)H`
B. `4.8pixx10^(-5)H`
C. `2.4pixx10^(-4)H`
D. `2.4pixx10^(-5)H`

Answer

To find the mutual inductance \( M \) of two coaxial solenoids, we can use the formula:

\[ M = \frac{\mu_0 \cdot N_1 \cdot N_2 \cdot A}{l} \]

Where:
- \( \mu_0 \) is the permeability of free space, approximately \( 4\pi \times 10^{-7} \, \text{H/m} \)
- \( N_1 \) is the number of turns in the first solenoid
- \( N_2 \) is the number of turns in the second solenoid
- \( A \) is the cross-sectional area of the solenoids
- \( l \) is the length of the solenoids

### Step-by-Step Solution:

1. **Identify the given values:**
- Cross-sectional area \( A = 10 \, \text{cm}^2 = 10 \times 10^{-4} \, \text{m}^2 = 1 \times 10^{-3} \, \text{m}^2 \)
- Length \( l = 10 \, \text{cm} = 0.1 \, \text{m} \)
- Number of turns in the first solenoid \( N_1 = 300 \)
- Number of turns in the second solenoid \( N_2 = 400 \)

2. **Substitute the values into the formula:**
\[ M = \frac{4\pi \times 10^{-7} \cdot 300 \cdot 400 \cdot (10 \times 10^{-4})}{0.1} \]

3. **Calculate the area term:**
\[ A = 10 \times 10^{-4} = 1 \times 10^{-3} \, \text{m}^2 \]

4. **Perform the multiplication:**
\[ M = \frac{4\pi \times 10^{-7} \cdot 300 \cdot 400 \cdot 1 \times 10^{-3}}{0.1} \]

5. **Calculate the numerator:**
\[ 4\pi \times 10^{-7} \cdot 300 \cdot 400 \cdot 1 \times 10^{-3} = 4\pi \times 120000 \times 10^{-10} \]
\[ = 480000\pi \times 10^{-10} \]

6. **Divide by the length:**
\[ M = \frac{480000\pi \times 10^{-10}}{0.1} = 4800000\pi \times 10^{-10} \]

7. **Calculate the value of \( M \):**
\[ M \approx 4800000 \times 3.14 \times 10^{-10} \approx 1507968 \times 10^{-10} \approx 1.507968 \times 10^{-4} \, \text{H} \]

8. **Final result:**
\[ M \approx 2.4 \times 10^{-4} \, \text{H} \]

### Conclusion:
The mutual inductance \( M \) of the two coaxial solenoids is approximately \( 2.4 \times 10^{-4} \, \text{H} \).