When the current in a coil changeg from `8` amperes to `2` amperes in `3xx10^(-2)` seconds, the e.m.f. induced in the coil is `2` volt. The self-inductance of the coil (in millihenry) is

Options

A. 1
B. 5
C. 20
D. 10

Answer

To find the self-inductance of the coil, we can use the formula for the induced electromotive force (e.m.f.) in terms of self-inductance:

\[ E = L \frac{\Delta I}{\Delta t} \]

Where:
- \(E\) is the induced e.m.f. (in volts),
- \(L\) is the self-inductance (in henries),
- \(\Delta I\) is the change in current (in amperes),
- \(\Delta t\) is the change in time (in seconds).

### Step-by-Step Solution:

1. **Identify the given values:**
- Initial current, \(I_1 = 8 \, \text{A}\)
- Final current, \(I_2 = 2 \, \text{A}\)
- Change in time, \(\Delta t = 3 \times 10^{-2} \, \text{s}\)
- Induced e.m.f., \(E = 2 \, \text{V}\)

2. **Calculate the change in current (\(\Delta I\)):**
\[ \Delta I = I_2 - I_1 = 2 \, \text{A} - 8 \, \text{A} = -6 \, \text{A} \]

3. **Substitute the values into the e.m.f. formula:**
\[ E = L \frac{\Delta I}{\Delta t} \]
Rearranging gives:
\[ L = \frac{E \cdot \Delta t}{\Delta I} \]

4. **Substituting the known values:**
\[ L = \frac{2 \, \text{V} \cdot (3 \times 10^{-2} \, \text{s})}{-6 \, \text{A}} \]

5. **Calculate \(L\):**
\[ L = \frac{2 \cdot 3 \times 10^{-2}}{-6} = \frac{6 \times 10^{-2}}{-6} = -10^{-2} \, \text{H} \]
Since self-inductance is a positive quantity, we take the absolute value:
\[ L = 10^{-2} \, \text{H} \]

6. **Convert \(L\) to millihenries:**
\[ L = 10^{-2} \, \text{H} = 10 \, \text{mH} \]

### Final Answer:
The self-inductance of the coil is \(10 \, \text{mH}\).