For two coils with number of turns 500 and 200 each of length 1 m and cross-sectional area `4 xx 10^(-4) m^2` , the mutual inductance is :

Options

A. `0.5 m mu H`
B. 0.5 H
C. `5 m mu H`
D. 0.05 `m H`

Answer

To find the mutual inductance \( M \) between two coils, we can use the formula:

\[ M = \frac{\mu_0 N_1 N_2 A}{l} \]

Where:
- \( M \) is the mutual inductance,
- \( \mu_0 \) is the permeability of free space (\( 4\pi \times 10^{-7} \, \text{H/m} \)),
- \( N_1 \) is the number of turns in the first coil,
- \( N_2 \) is the number of turns in the second coil,
- \( A \) is the cross-sectional area of the coils (in square meters),
- \( l \) is the length of the coils (in meters).

### Step-by-step Solution:

1. **Identify the given values:**
- \( N_1 = 500 \) (number of turns in coil 1)
- \( N_2 = 200 \) (number of turns in coil 2)
- \( A = 4 \times 10^{-4} \, \text{m}^2 \) (cross-sectional area)
- \( l = 1 \, \text{m} \) (length of the coils)

2. **Substitute the values into the formula:**
\[ M = \frac{(4\pi \times 10^{-7}) \times 500 \times 200 \times (4 \times 10^{-4})}{1} \]

3. **Calculate the numerator:**
- First, calculate \( 500 \times 200 = 100000 \).
- Next, calculate \( 100000 \times 4 \times 10^{-4} = 40 \).
- Finally, multiply by \( 4\pi \times 10^{-7} \):
\[ M = 40 \times (4\pi \times 10^{-7}) = 160\pi \times 10^{-7} \]

4. **Approximate \( \pi \):**
- Using \( \pi \approx 3.14 \):
\[ M \approx 160 \times 3.14 \times 10^{-7} = 502.4 \times 10^{-7} \, \text{H} \]

5. **Convert to milliHenries:**
- Since \( 1 \, \text{H} = 1000 \, \text{mH} \):
\[ M \approx 0.05024 \, \text{H} = 50.24 \, \text{mH} \]

6. **Final Answer:**
- Rounding to two decimal places, the mutual inductance \( M \) is approximately \( 0.05 \, \text{mH} \).