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

Explanation

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
- \( L \) is the length of the solenoids

### Step 1: Identify the given values
- \( N_1 = 300 \) turns (for the first solenoid)
- \( N_2 = 400 \) turns (for the second solenoid)
- \( A = 10 \, \text{cm}^2 = 10 \times 10^{-4} \, \text{m}^2 = 10^{-3} \, \text{m}^2 \) (convert cm² to m²)
- \( L = 20 \, \text{cm} = 0.2 \, \text{m} \) (convert cm to m)

### Step 2: Substitute the values into the formula
Now we substitute the values into the mutual inductance formula:

\[ M = \frac{(4\pi \times 10^{-7}) \times (300) \times (400) \times (10 \times 10^{-4})}{0.2} \]

### Step 3: Calculate the numerator
Calculate the numerator:

\[ 4\pi \times 10^{-7} \times 300 \times 400 \times 10 \times 10^{-4} \]

Calculating step by step:
1. \( 300 \times 400 = 120000 \)
2. \( 4\pi \times 10^{-7} \times 120000 = 4\pi \times 1.2 \times 10^{-3} = 4.8\pi \times 10^{-3} \)
3. \( 4.8\pi \times 10^{-3} \times 10^{-4} = 4.8\pi \times 10^{-7} \)

### Step 4: Calculate the final value of M
Now divide by \( L \):

\[ M = \frac{4.8\pi \times 10^{-7}}{0.2} = 24\pi \times 10^{-7} \]

### Step 5: Calculate the numerical value
Using \( \pi \approx 3.14 \):

\[ M \approx 24 \times 3.14 \times 10^{-7} \approx 75.36 \times 10^{-7} \, \text{H} = 7.536 \times 10^{-6} \, \text{H} \]

### Final Answer
Thus, the mutual inductance \( M \) is approximately:

\[ M \approx 7.54 \times 10^{-6} \, \text{H} \text{ or } 7.54 \, \mu\text{H} \]