When current in coil is uniformely reduced from 2A to 1A in 1 ms , the induced emf is 5V. The inductance of coil is :

Options

A. 5 H
B. 5000 H
C. 5 mH
D. 50 H

Answer

To find the inductance of the coil given the change in current and the induced emf, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Initial and Final Currents**:
- The initial current \( I_i = 2 \, \text{A} \)
- The final current \( I_f = 1 \, \text{A} \)

2. **Calculate the Change in Current**:
- The change in current \( \Delta I = I_f - I_i = 1 \, \text{A} - 2 \, \text{A} = -1 \, \text{A} \)

3. **Determine the Time Interval**:
- The time interval \( \Delta t = 1 \, \text{ms} = 1 \times 10^{-3} \, \text{s} \)

4. **Use the Formula for Induced EMF**:
- The induced emf \( \mathcal{E} \) is given by the formula:
\[ \mathcal{E} = -L \frac{\Delta I}{\Delta t} \]
- We know \( \mathcal{E} = 5 \, \text{V} \).

5. **Rearranging the Formula to Find Inductance**:
- Rearranging the formula gives:
\[ L = -\mathcal{E} \frac{\Delta t}{\Delta I} \]

6. **Substituting the Known Values**:
- Substitute \( \mathcal{E} = 5 \, \text{V} \), \( \Delta t = 1 \times 10^{-3} \, \text{s} \), and \( \Delta I = -1 \, \text{A} \):
\[ L = -5 \, \text{V} \times \frac{1 \times 10^{-3} \, \text{s}}{-1 \, \text{A}} \]
- This simplifies to:
\[ L = 5 \times 10^{-3} \, \text{H} \]

7. **Convert to Millihenries**:
- Since \( 1 \, \text{H} = 1000 \, \text{mH} \):
\[ L = 5 \, \text{mH} \]

### Final Answer:
The inductance of the coil is \( 5 \, \text{mH} \).

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