Two coils `A` and `B` having turns `300` and `600` respectively are placed near each other, on passing a current of `3.0` ampere in `A`, the flux linked with `A` is `1.2xx10^(-4)` and with `B` it is `9.0xx10^(-5)` weber. The mutual inductance of the system is

Options

A. `2xx10^(-5)`henry
B. `3xx10^(-5)`henry
C. `4xx10^(-5)`henry
D. `6xx10^(-5)`henry

Explanation

To find the mutual inductance \( M \) of the system with two coils \( A \) and \( B \), we can use the relationship between the flux linked with coil \( B \) and the current flowing through coil \( A \). The formula for mutual inductance is given by:

\[ M = \frac{\Phi_B}{I_A} \]

where:
- \( \Phi_B \) is the flux linked with coil \( B \)
- \( I_A \) is the current flowing through coil \( A \)

### Step-by-Step Solution:

1. **Identify the Given Values**:
- Number of turns in coil \( A \) (\( N_A \)) = 300
- Number of turns in coil \( B \) (\( N_B \)) = 600
- Current in coil \( A \) (\( I_A \)) = 3.0 A
- Flux linked with coil \( A \) (\( \Phi_A \)) = \( 1.2 \times 10^{-4} \) Wb
- Flux linked with coil \( B \) (\( \Phi_B \)) = \( 9.0 \times 10^{-5} \) Wb

2. **Use the Formula for Mutual Inductance**:
- Substitute the values into the mutual inductance formula:
\[ M = \frac{\Phi_B}{I_A} \]
- Here, \( \Phi_B = 9.0 \times 10^{-5} \) Wb and \( I_A = 3.0 \) A.

3. **Calculate the Mutual Inductance**:
\[ M = \frac{9.0 \times 10^{-5}}{3.0} \]
- Performing the division:
\[ M = 3.0 \times 10^{-5} \text{ H} \]

4. **Final Result**:
- The mutual inductance of the system is:
\[ M = 3.0 \times 10^{-5} \text{ H} \]