An e.m.f of 5 millivolt is induced in a coil when in a nearby placed another coil the current changes by 5 ampere in 0.1 second . The coefficient of mutual induction between the two coils will be :

Options

A. 1 Henry
B. 0.1 Henry
C. 0.1 millihenry
D. 0.001 millihenry

Answer

To find the coefficient of mutual induction between two coils, we can use the formula:

\[ M = -\frac{E}{\frac{dI}{dt}} \]

where:
- \( M \) is the coefficient of mutual induction,
- \( E \) is the induced electromotive force (e.m.f),
- \( \frac{dI}{dt} \) is the rate of change of current in the nearby coil.

### Step-by-Step Solution:

1. **Identify the given values:**
- Induced e.m.f (\( E \)) = 5 millivolts = \( 5 \times 10^{-3} \) volts
- Change in current (\( dI \)) = 5 amperes
- Change in time (\( dt \)) = 0.1 seconds

2. **Calculate the rate of change of current (\( \frac{dI}{dt} \)):**
\[ \frac{dI}{dt} = \frac{dI}{dt} = \frac{5 \, \text{A}}{0.1 \, \text{s}} = 50 \, \text{A/s} \]

3. **Substitute the values into the mutual induction formula:**
\[ M = -\frac{5 \times 10^{-3}}{50} \]

4. **Perform the calculation:**
\[ M = -\frac{5 \times 10^{-3}}{50} = -0.1 \times 10^{-3} = -0.1 \, \text{mH} \]

5. **Interpret the result:**
The negative sign indicates the direction of the induced e.m.f. However, we are interested in the magnitude of the coefficient of mutual induction:
\[ |M| = 0.1 \, \text{mH} \]

### Final Answer:
The coefficient of mutual induction between the two coils is \( 0.1 \, \text{mH} \).

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