In a circular conducting coil, when current increases from `2A` to `18A` in `0.05` sec., the induced e.m.f. is `20V`. The self-inductance of the coil is

Options

A. `62.5mH`
B. `6.25mH`
C. `50mH`
D. none of these

Explanation

To find the self-inductance of the coil, we can use the formula for induced electromotive force (e.m.f.) due to a change in current in an inductor:

\[ \text{Induced e.m.f.} (E) = L \frac{di}{dt} \]

Where:
- \( E \) is the induced e.m.f.
- \( L \) is the self-inductance of the coil.
- \( di \) is the change in current.
- \( dt \) is the change in time.

### Step-by-Step Solution:

1. **Identify the given values:**
- Initial current, \( I_1 = 2 \, A \)
- Final current, \( I_2 = 18 \, A \)
- Change in time, \( dt = 0.05 \, s \)
- Induced e.m.f., \( E = 20 \, V \)

2. **Calculate the change in current (\( di \)):**
\[ di = I_2 - I_1 = 18 \, A - 2 \, A = 16 \, A \]

3. **Substitute the values into the induced e.m.f. formula:**
\[ 20 \, V = L \frac{16 \, A}{0.05 \, s} \]

4. **Calculate \( \frac{di}{dt} \):**
\[ \frac{di}{dt} = \frac{16 \, A}{0.05 \, s} = 320 \, A/s \]

5. **Rearranging the formula to solve for \( L \):**
\[ L = \frac{E}{\frac{di}{dt}} = \frac{20 \, V}{320 \, A/s} \]

6. **Calculate \( L \):**
\[ L = \frac{20}{320} = 0.0625 \, H \]

7. **Convert \( L \) to millihenries:**
\[ L = 0.0625 \, H = 62.5 \, mH \]

### Final Answer:
The self-inductance of the coil is \( 62.5 \, mH \).