The coefficients of self induction of two inductance coil are 0.01 H and 0.03H respectively . When they are conneted in series so as to support each other, then the resultant self inductance becomes 0.06 Henry. The value of coefficient of mutual induction be __

Options

A. 0.02H
B. 0.05H
C. 0.01 H
D. zero

Answer

To solve the problem, we need to find the coefficient of mutual induction (M) between two coils when they are connected in series. The self-inductance values of the coils are given as L1 = 0.01 H and L2 = 0.03 H. The total self-inductance when connected in series is given as L_total = 0.06 H.

### Step-by-Step Solution:

1. **Write down the formula for total self-inductance in series:**
When two inductors are connected in series, the total self-inductance (L_total) can be expressed as:
\[ L_{total} = L_1 + L_2 + 2M \]

2. **Substitute the known values into the formula:**
We know:
- \( L_1 = 0.01 \, \text{H} \)
- \( L_2 = 0.03 \, \text{H} \)
- \( L_{total} = 0.06 \, \text{H} \)

Substituting these values into the formula gives:
\[ 0.06 = 0.01 + 0.03 + 2M \]

3. **Simplify the equation:**
Combine the self-inductance values:
\[ 0.06 = 0.04 + 2M \]

4. **Isolate the term with M:**
Subtract 0.04 from both sides:
\[ 0.06 - 0.04 = 2M \]
\[ 0.02 = 2M \]

5. **Solve for M:**
Divide both sides by 2:
\[ M = \frac{0.02}{2} = 0.01 \, \text{H} \]

6. **Conclusion:**
The coefficient of mutual induction (M) is:
\[ M = 0.01 \, \text{H} \]

### Final Answer:
The value of the coefficient of mutual induction is **0.01 Henry**.