A straight solenoid of length `1 m` has `5000` turns in the primary and `200` turns in the secondary. If the area of cross section is `4 cm^(2)`, the mutual inductance will be

Options

A. `503 H`
B. `503 mH`
C. `503 muH`
D. `5.03 H`

Explanation

To find the mutual inductance of the given solenoid system, we can follow these steps:

### Step 1: Write down the given values
- Length of the solenoid, \( L = 1 \, \text{m} \)
- Number of turns in the primary coil, \( n_1 = 5000 \)
- Number of turns in the secondary coil, \( n_2 = 200 \)
- Area of cross-section, \( A = 4 \, \text{cm}^2 = 4 \times 10^{-4} \, \text{m}^2 \)

### Step 2: Use the formula for mutual inductance
The formula for mutual inductance \( M \) is given by:
\[ M = \mu_0 \frac{n_1 n_2 A}{L} \]
where \( \mu_0 \) (the permeability of free space) is approximately \( 4\pi \times 10^{-7} \, \text{H/m} \).

### Step 3: Substitute the values into the formula
Substituting the values into the formula:
\[ M = (4\pi \times 10^{-7}) \frac{(5000)(200)(4 \times 10^{-4})}{1} \]

### Step 4: Calculate the values
1. Calculate \( n_1 n_2 A \):
\[ n_1 n_2 A = 5000 \times 200 \times 4 \times 10^{-4} = 5000 \times 200 \times 0.0004 \]
\[ = 5000 \times 0.08 = 400 \]

2. Now substitute this back into the mutual inductance formula:
\[ M = 4\pi \times 10^{-7} \times 400 \]

3. Calculate \( 4\pi \times 400 \):
\[ 4\pi \approx 12.5664 \quad \text{(approximately)} \]
\[ 4\pi \times 400 \approx 12.5664 \times 400 = 5026.56 \approx 5030 \, \text{(approximately)} \]

4. Now, multiply by \( 10^{-7} \):
\[ M \approx 5030 \times 10^{-7} = 5.03 \times 10^{-4} \, \text{H} \]

### Step 5: Convert to microhenries
To convert henries to microhenries:
\[ M \approx 503 \, \mu\text{H} \]

### Final Answer
Thus, the mutual inductance \( M \) is approximately \( 503 \, \mu\text{H} \).