When current flowing in a coil changes from 3A to 2A in one millisecond, 5 volt emf is induced in it. The self-inductance of the coil will be-

Options

A. zero
B. 5kh
C. 5H
D. 5 mH

Explanation

To find the self-inductance of the coil, we can use the formula for induced electromotive force (emf) in a coil:

\[ \text{emf} = L \frac{di}{dt} \]

Where:
- \( L \) is the self-inductance,
- \( di \) is the change in current,
- \( dt \) is the change in time.

### Step-by-Step Solution:

1. **Identify the change in current (\( di \)):**
- The current changes from 3 A to 2 A.
- Therefore, \( di = 2 A - 3 A = -1 A \).

2. **Identify the change in time (\( dt \)):**
- The time interval is given as 1 millisecond.
- Convert milliseconds to seconds:
\[ dt = 1 \text{ ms} = 1 \times 10^{-3} \text{ s} = 0.001 \text{ s} \]

3. **Substitute the values into the emf formula:**
- The induced emf (\( \text{emf} \)) is given as 5 V.
- Substitute \( di \) and \( dt \) into the formula:
\[ 5 = L \frac{-1}{0.001} \]

4. **Rearranging the equation to solve for \( L \):**
- Rearranging gives:
\[ L = 5 \times \frac{0.001}{-1} \]
- This simplifies to:
\[ L = -5 \times 0.001 = -0.005 \text{ H} \]

5. **Convert to milliHenries:**
- Since \( 1 \text{ H} = 1000 \text{ mH} \), we convert:
\[ L = -5 \text{ mH} \]
- The negative sign indicates the direction of induced emf opposing the change in current, but we can express self-inductance as a positive quantity:
\[ L = 5 \text{ mH} \]

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

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