Two coils A and B have 200 and 400 turns respectively. A current of 1 A in coil A causes a flux per turn of `10^(-3)` Wb to link with A and a flux per turn of `0.8xx10^(-3)` Wb through B. the ratio of mutual inductance of A and B is

Options

A. 43955
B. 1/1.6
C. 1.6
D. 1

Answer

To find the ratio of the mutual inductance of coils A and B, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Number of turns in coil A, \( N_A = 200 \)
- Number of turns in coil B, \( N_B = 400 \)
- Flux per turn linked with coil A, \( \Phi_A = 10^{-3} \, \text{Wb} \)
- Flux per turn linked with coil B, \( \Phi_B = 0.8 \times 10^{-3} \, \text{Wb} \)
- Current in coil A, \( I_A = 1 \, \text{A} \)

2. **Calculate the Total Flux Linkage for Each Coil:**
- Total flux linkage for coil A, \( \Psi_A = N_A \cdot \Phi_A = 200 \cdot 10^{-3} = 0.2 \, \text{Wb} \)
- Total flux linkage for coil B, \( \Psi_B = N_B \cdot \Phi_B = 400 \cdot (0.8 \times 10^{-3}) = 0.32 \, \text{Wb} \)

3. **Determine Self-Inductance of Coil A:**
- Self-inductance \( L_A \) can be calculated using the formula:
\[ L_A = \frac{\Psi_A}{I_A} = \frac{0.2}{1} = 0.2 \, \text{H} \]

4. **Determine Mutual Inductance between Coils A and B:**
- The mutual inductance \( M_{AB} \) is given by:
\[ M_{AB} = \frac{\Psi_B}{I_A} = \frac{0.32}{1} = 0.32 \, \text{H} \]

5. **Calculate the Ratio of Mutual Inductance:**
- The ratio of the mutual inductance of A and B is given by:
\[ \frac{L_A}{M_{AB}} = \frac{0.2}{0.32} = \frac{20}{32} = \frac{5}{8} = 0.625 \]
- To express this in a more common form, we can also express it as:
\[ \frac{1}{1.6} \]

### Final Answer:
The ratio of mutual inductance of A and B is \( \frac{1}{1.6} \).