In a transformer , the coefficient of mutual inductance between the primary and the secondary coil is `0.2` henry. When the current changes by `5` ampere//second in the primary, the induced e.m.f. in the secondary will be

Options

A. 5V
B. 1 V
C. 25 V
D. 10 V

Explanation

To solve the problem, we will use the formula for the induced electromotive force (e.m.f.) in the secondary coil of a transformer due to a change in current in the primary coil. The formula is given by:

\[ \text{Induced e.m.f.} (E) = -M \frac{dI}{dt} \]

where:
- \( E \) is the induced e.m.f. in the secondary coil,
- \( M \) is the mutual inductance between the primary and secondary coils,
- \( \frac{dI}{dt} \) is the rate of change of current in the primary coil.

### Step-by-Step Solution:

1. **Identify the given values:**
- Mutual inductance \( M = 0.2 \, \text{H} \)
- Rate of change of current \( \frac{dI}{dt} = 5 \, \text{A/s} \)

2. **Substitute the values into the formula:**
\[ E = -M \frac{dI}{dt} \]
\[ E = -0.2 \, \text{H} \times 5 \, \text{A/s} \]

3. **Calculate the induced e.m.f.:**
\[ E = -1 \, \text{V} \]

4. **Interpret the result:**
The negative sign indicates the direction of the induced e.m.f. according to Lenz's law, but the magnitude of the induced e.m.f. is \( 1 \, \text{V} \).

5. **Final answer:**
The induced e.m.f. in the secondary coil is \( 1 \, \text{V} \).