A 2 m long solenoil with diameter 2 cm and 2000 turns has a secondary coil of 1000 turns wound closely near its midpoint. The mutuat inductance between the two coils is.

Options

A. `2.4xx10^(-4)H`
B. `3.9xx10^(-4)H`
C. `1.28xx10^(-3)H`
D. `3.14xx10^(-3)H`

Explanation

To find the mutual inductance between the solenoid and the secondary coil, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Given Data:**
- Length of the solenoid (l) = 2 m
- Diameter of the solenoid = 2 cm = 0.02 m
- Number of turns in the solenoid (N1) = 2000 turns
- Number of turns in the secondary coil (N2) = 1000 turns

2. **Calculate the Radius of the Solenoid:**
- Radius (r) = Diameter / 2 = 0.02 m / 2 = 0.01 m

3. **Calculate the Area of Cross-section of the Solenoid:**
- Area (A) = πr²
- A = π(0.01 m)² = π(0.0001 m²) = π × 10⁻⁴ m²

4. **Use the Formula for Mutual Inductance (M):**
- The formula for mutual inductance is:
\[ M = \mu_0 \frac{N_1 N_2 A}{l} \]
- Where:
- \(\mu_0\) (permeability of free space) = \(4\pi \times 10^{-7} \, \text{H/m}\)

5. **Substitute the Values into the Formula:**
- M = \(4\pi \times 10^{-7} \, \text{H/m} \times \frac{2000 \times 1000 \times \pi \times 10^{-4}}{2}\)

6. **Calculate the Mutual Inductance:**
- First, calculate the numerator:
\[ 2000 \times 1000 \times \pi \times 10^{-4} = 2 \times 10^6 \times \pi \times 10^{-4} = 2\pi \times 10^{2} \]
- Now substitute back:
\[ M = 4\pi \times 10^{-7} \times \frac{2\pi \times 10^{2}}{2} \]
- Simplifying gives:
\[ M = 4\pi \times 10^{-7} \times \pi \times 10^{2} = 4\pi^2 \times 10^{-5} \, \text{H} \]
- Using \(\pi \approx 3.14\):
\[ M \approx 4 \times (3.14)^2 \times 10^{-5} \approx 4 \times 9.86 \times 10^{-5} \approx 3.94 \times 10^{-4} \, \text{H} \]

7. **Final Answer:**
- The mutual inductance \(M \approx 3.94 \times 10^{-4} \, \text{H}\), which corresponds to the option \(3.9 \times 10^{-4} \, \text{H}\).