Self-inductance of a coil is 50mH. A current of 1 A passing through the coil reduces to zero at steady rate in 0.1 s, the self-induced emf is

Options

A. 5 V
B. 0.05 V
C. 50 V
D. 0.5 V

Explanation

To solve the problem of finding the self-induced emf in a coil with given parameters, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Given Values**:
- Self-inductance (L) = 50 mH = 50 × 10^(-3) H
- Initial current (I_initial) = 1 A
- Final current (I_final) = 0 A
- Time (t) = 0.1 s

2. **Calculate Change in Current (di)**:
- Change in current (di) = I_final - I_initial
- di = 0 A - 1 A = -1 A

3. **Calculate Rate of Change of Current (di/dt)**:
- di/dt = Change in current / Change in time
- di/dt = (-1 A) / (0.1 s) = -10 A/s

4. **Use the Formula for Self-Induced emf (ε)**:
- The formula for self-induced emf is given by:
\[ ε = -L \frac{di}{dt} \]
- Substitute the values:
\[ ε = - (50 × 10^{-3} H) × (-10 A/s) \]

5. **Calculate the Induced emf**:
- ε = 50 × 10^(-3) × 10 = 0.5 V

6. **Conclusion**:
- The self-induced emf is 0.5 volts.

### Final Answer:
The self-induced emf is **0.5 V**.

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