In a coil when current changes from 10 A to 2A in time 0.1 s, induced emf is 3.28 V. What is the self-inductance of coil?

Options

A. 4H
B. 0.4H
C. 0.04H
D. 5H

Answer

To solve the problem, we need to find the self-inductance (L) of the coil using the given information about the change in current, time, and induced emf.

### Step-by-Step Solution:

1. **Identify the given values**:
- Initial current (I₁) = 10 A
- Final current (I₂) = 2 A
- Time interval (Δt) = 0.1 s
- Induced emf (E) = 3.28 V

2. **Calculate the change in current (ΔI)**:
\[ \Delta I = I_2 - I_1 = 2 \, \text{A} - 10 \, \text{A} = -8 \, \text{A} \]

3. **Calculate the rate of change of current (di/dt)**:
\[ \frac{di}{dt} = \frac{\Delta I}{\Delta t} = \frac{-8 \, \text{A}}{0.1 \, \text{s}} = -80 \, \text{A/s} \]

4. **Use the formula for induced emf (E)**:
The formula for induced emf in terms of self-inductance (L) is:
\[ E = -L \frac{di}{dt} \]
Since we are interested in the magnitude, we can drop the negative sign:
\[ E = L \left| \frac{di}{dt} \right| \]

5. **Rearrange the formula to solve for L**:
\[ L = \frac{E}{\left| \frac{di}{dt} \right|} = \frac{3.28 \, \text{V}}{80 \, \text{A/s}} \]

6. **Calculate L**:
\[ L = \frac{3.28}{80} = 0.041 \, \text{H} \]

7. **Conclusion**:
The self-inductance of the coil is \( L = 0.041 \, \text{H} \) or \( 41 \, \text{mH} \).