If emf induced in a coil is 2V by changing the current in it from 8 A to 6 A in `2 xx 10^(3)` s . Then , the coefficient of self -induction is

Options

A. `2 xx 10^(-3) H`
B. `10^(-3)` H
C. `0.5 xx 10^(-3)` H
D. `4 xx 10^(-3)` H

Explanation

To find the coefficient of self-induction (L) of a coil when an emf is induced, we can use the formula:

\[ \text{emf} = L \frac{\Delta I}{\Delta t} \]

Where:
- emf is the induced electromotive force (in volts)
- \(L\) is the coefficient of self-induction (in henries)
- \(\Delta I\) is the change in current (in amperes)
- \(\Delta t\) is the change in time (in seconds)

### Step 1: Identify the given values
- Induced emf (\(e\)) = 2 V
- Initial current (\(I_1\)) = 8 A
- Final current (\(I_2\)) = 6 A
- Time interval (\(\Delta t\)) = \(2 \times 10^{-3}\) s

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

### Step 3: Substitute the values into the formula
Using the formula for induced emf:
\[ e = L \frac{\Delta I}{\Delta t} \]
We can rearrange it to solve for \(L\):
\[ L = \frac{e \cdot \Delta t}{\Delta I} \]

### Step 4: Substitute the known values into the equation
\[ L = \frac{2 \, \text{V} \cdot (2 \times 10^{-3} \, \text{s})}{-2 \, \text{A}} \]

### Step 5: Calculate \(L\)
\[ L = \frac{2 \cdot 2 \times 10^{-3}}{-2} = \frac{4 \times 10^{-3}}{-2} = -2 \times 10^{-3} \, \text{H} \]

### Step 6: Interpret the result
The negative sign indicates the direction of induced emf opposing the change in current, which is consistent with Lenz's law. The magnitude of the coefficient of self-induction is:
\[ L = 2 \times 10^{-3} \, \text{H} \]

### Final Answer
The coefficient of self-induction \(L\) is \(2 \times 10^{-3} \, \text{H}\).

---