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 solve the problem of finding the mutual inductance between the solenoid and the secondary coil, we can follow these steps:

### Step 1: Identify the given values
- Length of the solenoid (L) = 2 m
- Diameter of the solenoid (D) = 2 cm = 0.02 m
- Radius of the solenoid (r) = D/2 = 0.01 m
- Number of turns in the primary coil (N1) = 2000 turns
- Number of turns in the secondary coil (N2) = 1000 turns

### Step 2: Calculate the cross-sectional area (A) of the solenoid
The area \( A \) of the solenoid can be calculated using the formula for the area of a circle:
\[ A = \pi r^2 \]
Substituting the radius:
\[ A = \pi (0.01)^2 = \pi \times 10^{-4} \text{ m}^2 \]

### Step 3: Use the formula for mutual inductance (M)
The formula for mutual inductance between two coils is given by:
\[ M = \frac{\mu_0 N_1 N_2 A}{L} \]
Where:
- \( \mu_0 \) (permeability of free space) = \( 4\pi \times 10^{-7} \text{ H/m} \)
- \( N_1 \) = number of turns in the primary coil
- \( N_2 \) = number of turns in the secondary coil
- \( A \) = cross-sectional area
- \( L \) = length of the solenoid

### Step 4: Substitute the values into the formula
Substituting the known values into the formula:
\[ M = \frac{(4\pi \times 10^{-7}) \times 2000 \times 1000 \times (\pi \times 10^{-4})}{2} \]

### Step 5: Simplify the expression
Calculating the above expression:
\[ M = \frac{(4\pi \times 10^{-7}) \times 2000 \times 1000 \times \pi \times 10^{-4}}{2} \]
\[ = \frac{4 \times 2000 \times 1000 \times \pi^2 \times 10^{-11}}{2} \]
\[ = 2 \times 2000 \times 1000 \times \pi^2 \times 10^{-11} \]
\[ = 3.9 \times 10^{-4} \text{ H} \]

### Final Answer
The mutual inductance \( M \) between the two coils is:
\[ M \approx 3.9 \times 10^{-4} \text{ Henry} \]