Two coils are wound on the same iron rod so that the flux generated by one passes through the other . The primary coil has N turns in it and when a current 2 A flows through it the flux in it is `2.5 xx10^(-4)` Wb . If the secondary coil has 12 turns the mutual inductance of te coils is ( assume the secondary coil is in open circuit )

Options

A. `10xx10^(4)`H
B. `15xx10^(-4)`H
C. `20xx10^(-4)`H
D. `25xx10^(-4)` H

Answer

To find the mutual inductance \( M \) of the two coils, we can use the relationship between the mutual inductance, the number of turns in the secondary coil, and the magnetic flux. Here’s a step-by-step solution:

### Step 1: Understand the given values
- Current in the primary coil \( I = 2 \, \text{A} \)
- Flux in the primary coil \( \Phi_p = 2.5 \times 10^{-4} \, \text{Wb} \)
- Number of turns in the secondary coil \( N_s = 12 \)

### Step 2: Calculate the total flux linked with the secondary coil
The total flux \( \Phi_s \) linked with the secondary coil can be calculated as:
\[ \Phi_s = N_s \cdot \Phi_p \]
Substituting the values:
\[ \Phi_s = 12 \cdot (2.5 \times 10^{-4}) = 3.0 \times 10^{-3} \, \text{Wb} \]

### Step 3: Use the formula for mutual inductance
The mutual inductance \( M \) can be defined as:
\[ M = \frac{\Phi_s}{I} \]
Where \( I \) is the current in the primary coil. Substituting the values:
\[ M = \frac{3.0 \times 10^{-3}}{2} \]

### Step 4: Calculate the mutual inductance
\[ M = 1.5 \times 10^{-3} \, \text{H} = 15 \times 10^{-4} \, \text{H} \]

### Final Answer
The mutual inductance of the coils is:
\[ M = 15 \times 10^{-4} \, \text{H} \]

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