The e.m.f. induced in a `1` milli henry inductor, in which the current changes from `5A` to `3A` in `10^(-3)` second is

Options

A. `2xx10^(-6)V`
B. `8xx10^(-6)V`
C. `2V`
D. `8V`

Answer

To solve the problem of finding the induced electromotive force (e.m.f.) in a 1 milli henry inductor when the current changes from 5 A to 3 A in \(10^{-3}\) seconds, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Values**:
- Inductance \(L = 1 \text{ mH} = 1 \times 10^{-3} \text{ H}\)
- Initial current \(I_1 = 5 \text{ A}\)
- Final current \(I_2 = 3 \text{ A}\)
- Time interval \(\Delta t = 10^{-3} \text{ s}\)

2. **Calculate the Change in Current (\(\Delta I\))**:
\[ \Delta I = I_2 - I_1 = 3 \text{ A} - 5 \text{ A} = -2 \text{ A} \]

3. **Use the Formula for Induced e.m.f.**:
The formula for the induced e.m.f. (\(\mathcal{E}\)) in an inductor is given by:
\[ \mathcal{E} = -L \frac{\Delta I}{\Delta t} \]
Here, \(\Delta I\) is the change in current and \(\Delta t\) is the change in time.

4. **Substitute the Values into the Formula**:
\[ \mathcal{E} = - (1 \times 10^{-3} \text{ H}) \frac{-2 \text{ A}}{10^{-3} \text{ s}} \]

5. **Simplify the Expression**:
\[ \mathcal{E} = (1 \times 10^{-3}) \times \frac{2}{10^{-3}} = 2 \text{ V} \]

6. **Final Result**:
The induced e.m.f. is:
\[ \mathcal{E} = 2 \text{ V} \]