If `L = 100 muH` and current changes by 1 A in 0.1 second. The emf produced is:

Options

A. 1 mV
B. 10 mV
C. 100 mV
D. 0.1 V

Explanation

To find the emf produced when the current changes in an inductor, we can use the formula:

\[ E = -L \frac{di}{dt} \]

Where:
- \( E \) is the electromotive force (emf),
- \( L \) is the inductance,
- \( di \) is the change in current,
- \( dt \) is the change in time.

### Step-by-Step Solution:

1. **Identify the given values**:
- Inductance, \( L = 100 \mu H = 100 \times 10^{-6} H \)
- Change in current, \( di = 1 A \)
- Change in time, \( dt = 0.1 s \)

2. **Substitute the values into the formula**:
\[ E = -L \frac{di}{dt} \]
\[ E = -100 \times 10^{-6} \frac{1}{0.1} \]

3. **Calculate \( \frac{di}{dt} \)**:
\[ \frac{di}{dt} = \frac{1 A}{0.1 s} = 10 A/s \]

4. **Substitute \( \frac{di}{dt} \) back into the equation**:
\[ E = -100 \times 10^{-6} \times 10 \]

5. **Calculate the emf**:
\[ E = -1000 \times 10^{-6} V = -1 \times 10^{-3} V \]
The negative sign indicates the direction of the emf according to Lenz's law, but we are interested in the magnitude:
\[ |E| = 1 \times 10^{-3} V = 1 mV \]

### Final Answer:
The emf produced is \( 1 mV \).