Two coaxial solenoids are made by winding thin insulated wire over a pipe of cross-sectional area `A=10 cm^(2)` and length =20cm. If one of the solenoid has 300 turns and the other 400 turns, their mutual indcutance is

Options

A. `2.4 pi xx 10^(-5)H`
B. `4.8 pi xx 10^(-4)H`
C. `4.8 pi xx 10^(-5)H`
D. `2.4 pi xx 10^(-4)H`

Answer

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

\[ M = \frac{\mu_0 n_1 n_2 A}{l} \]

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 in square meters,
- \( l \) is the length of the solenoids in meters.

### Step-by-Step Solution:

1. **Identify the given values**:
- \( n_1 = 300 \) turns (for the first solenoid)
- \( n_2 = 400 \) turns (for the second solenoid)
- \( A = 10 \, \text{cm}^2 = 10 \times 10^{-4} \, \text{m}^2 = 10^{-3} \, \text{m}^2 \)
- \( l = 20 \, \text{cm} = 0.2 \, \text{m} \)

2. **Substitute the values into the formula**:
\[ M = \frac{(4\pi \times 10^{-7}) \times 300 \times 400 \times (10^{-3})}{0.2} \]

3. **Calculate the product of turns**:
\[ n_1 n_2 = 300 \times 400 = 120000 = 1.2 \times 10^5 \]

4. **Substitute this back into the equation**:
\[ M = \frac{(4\pi \times 10^{-7}) \times (1.2 \times 10^5) \times (10^{-3})}{0.2} \]

5. **Simplify the equation**:
\[ M = \frac{4\pi \times 1.2 \times 10^{-5}}{0.2} \]

6. **Calculate \( \frac{1.2}{0.2} \)**:
\[ \frac{1.2}{0.2} = 6 \]

7. **Now substitute back**:
\[ M = 4\pi \times 6 \times 10^{-5} = 24\pi \times 10^{-5} \]

8. **Convert to standard form**:
\[ M = 2.4\pi \times 10^{-4} \, \text{H} \]

### Final Answer:
\[ M = 2.4\pi \times 10^{-4} \, \text{H} \]