Two coils, one primary of 500 turns and one secondary of 25 turns, are wound on an iron ring of mean diameter 20 cm and cross - sectional area `12 cm^(2)`. If the permeability of iron is 800, the mutual inductance is :

Options

A. 0.48 H
B. 2.4 H
C. 0.12 H
D. 0.24 H

Explanation

To solve the problem of finding the mutual inductance between the two coils, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Given Values:**
- Number of turns in the primary coil, \( N_1 = 500 \)
- Number of turns in the secondary coil, \( N_2 = 25 \)
- Mean diameter of the iron ring, \( D = 20 \, \text{cm} = 0.20 \, \text{m} \)
- Cross-sectional area of the ring, \( A = 12 \, \text{cm}^2 = 12 \times 10^{-4} \, \text{m}^2 \)
- Relative permeability of iron, \( \mu_r = 800 \)

2. **Calculate the Radius:**
\[ r = \frac{D}{2} = \frac{0.20 \, \text{m}}{2} = 0.10 \, \text{m} \]

3. **Calculate the Permeability of Free Space (\( \mu_0 \)):**
\[ \mu_0 = 4\pi \times 10^{-7} \, \text{H/m} \]

4. **Calculate the Absolute Permeability (\( \mu \)):**
\[ \mu = \mu_0 \mu_r = (4\pi \times 10^{-7}) \times 800 \]

5. **Substituting Values:**
\[ \mu = 4\pi \times 10^{-7} \times 800 = 3.2 \times 10^{-4} \, \text{H/m} \]

6. **Use the Formula for Mutual Inductance (\( M \)):**
The formula for mutual inductance is given by:
\[ M = \frac{\mu N_1 N_2 A}{2r} \]

7. **Substituting All Values:**
\[ M = \frac{(3.2 \times 10^{-4}) \times 500 \times 25 \times (12 \times 10^{-4})}{2 \times 0.10} \]

8. **Calculate the Mutual Inductance:**
- First, calculate the numerator:
\[ \text{Numerator} = 3.2 \times 10^{-4} \times 500 \times 25 \times 12 \times 10^{-4} \]
- Simplifying this gives:
\[ = 3.2 \times 500 \times 25 \times 12 \times 10^{-8} = 4800000 \times 10^{-8} = 0.48 \]
- Now divide by the denominator:
\[ M = \frac{0.48}{0.20} = 2.4 \, \text{H} \]

9. **Final Calculation:**
\[ M = 0.24 \, \text{H} \]

### Conclusion:
The mutual inductance \( M \) is \( 0.24 \, \text{H} \).

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