Self-inductance of a coil is 2 mH and rate of current flow is `10^(3)` A/s. The induced emf is:

Options

A. 1V
B. 2 V
C. 3V
D. 4V

Explanation

To find the induced emf (electromotive force) in a coil due to self-inductance, we can use the formula:

\[ \text{Induced emf} (E) = -L \frac{di}{dt} \]

where:
- \( L \) is the self-inductance of the coil,
- \( \frac{di}{dt} \) is the rate of change of current.

### Step-by-Step Solution:

1. **Identify the given values**:
- Self-inductance \( L = 2 \, \text{mH} = 2 \times 10^{-3} \, \text{H} \)
- Rate of change of current \( \frac{di}{dt} = 10^{3} \, \text{A/s} \)

2. **Substitute the values into the formula**:
\[ E = -L \frac{di}{dt} \]
\[ E = - (2 \times 10^{-3} \, \text{H}) \times (10^{3} \, \text{A/s}) \]

3. **Calculate the induced emf**:
\[ E = - (2 \times 10^{-3}) \times (10^{3}) = -2 \, \text{V} \]

4. **Interpret the result**:
The negative sign indicates that the induced emf opposes the change in current according to Lenz's law. However, the magnitude of the induced emf is \( 2 \, \text{V} \).

### Final Answer:
The induced emf is \( 2 \, \text{V} \).