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`

Explanation

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

\[ \text{e.m.f.} = -L \frac{di}{dt} \]

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

### Step 1: Calculate the change in current (\( di \))
The current changes from \( 8 \) A to \( 2 \) A:

\[ di = 2 - 8 = -6 \, \text{A} \]

### Step 2: Calculate the change in time (\( dt \))
The time interval given is:

\[ dt = 3 \times 10^{-2} \, \text{s} \]

### Step 3: Substitute the values into the e.m.f. formula
We know the induced e.m.f. is \( 2 \) V. We can substitute the values into the formula:

\[ 2 = -L \frac{-6}{3 \times 10^{-2}} \]

### Step 4: Simplify the equation
Rearranging the equation gives:

\[ 2 = L \frac{6}{3 \times 10^{-2}} \]

Calculating \( \frac{6}{3 \times 10^{-2}} \):

\[ \frac{6}{3 \times 10^{-2}} = \frac{6}{0.03} = 200 \, \text{A/s} \]

So the equation becomes:

\[ 2 = 200L \]

### Step 5: Solve for \( L \)
Now, we can solve for \( L \):

\[ L = \frac{2}{200} = 0.01 \, \text{H} \]

### Step 6: Convert \( L \) to millihenries
Since \( 1 \, \text{H} = 1000 \, \text{mH} \):

\[ L = 0.01 \, \text{H} = 0.01 \times 1000 \, \text{mH} = 10 \, \text{mH} \]

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

---