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 a given self-inductance and a changing current, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Given Values:**
- Self-inductance \( L = 50 \, \text{mH} = 50 \times 10^{-3} \, \text{H} \)
- Initial current \( I_i = 1 \, \text{A} \)
- Final current \( I_f = 0 \, \text{A} \)
- Time interval \( dt = 0.1 \, \text{s} \)

2. **Calculate the Change in Current (\( dI \)):**
\[ dI = I_f - I_i = 0 - 1 = -1 \, \text{A} \]

3. **Calculate the Rate of Change of Current (\( \frac{dI}{dt} \)):**
\[ \frac{dI}{dt} = \frac{-1 \, \text{A}}{0.1 \, \text{s}} = -10 \, \text{A/s} \]

4. **Use the Formula for Self-Induced EMF (\( \mathcal{E} \)):**
The formula for self-induced EMF is given by:
\[ \mathcal{E} = -L \frac{dI}{dt} \]
Substituting the values:
\[ \mathcal{E} = - (50 \times 10^{-3} \, \text{H}) \times (-10 \, \text{A/s}) \]

5. **Calculate the Self-Induced EMF:**
\[ \mathcal{E} = 50 \times 10^{-3} \times 10 = 0.5 \, \text{V} \]

6. **Final Answer:**
The self-induced EMF is \( 0.5 \, \text{V} \).