When the current in a coil changes from `8A` to `2A` in `3xx10^(-2)s` , th eemf induced in the coil is `2V` . The self-induced of coil in `mH`

Options

A. `10`
B. `20`
C. `30`
D. `40`

Explanation

To find the self-inductance of the coil in millihenries (mH), we can follow these steps:

### Step 1: Identify the given values
- Initial current (I1) = 8 A
- Final current (I2) = 2 A
- Time interval (Δt) = 3 × 10^(-2) s
- Induced emf (ε) = 2 V

### Step 2: Calculate the change in current (ΔI)
ΔI = I2 - I1 = 2 A - 8 A = -6 A

### Step 3: Use the formula for induced emf
The formula relating the induced emf (ε) to self-inductance (L) is given by:
\[ \epsilon = -L \frac{\Delta I}{\Delta t} \]

### Step 4: Rearrange the formula to solve for L
Rearranging the formula gives us:
\[ L = -\frac{\epsilon \cdot \Delta t}{\Delta I} \]

### Step 5: Substitute the known values into the equation
Substituting the values we have:
\[ L = -\frac{2 \, \text{V} \cdot (3 \times 10^{-2} \, \text{s})}{-6 \, \text{A}} \]

### Step 6: Calculate L
\[ L = \frac{2 \cdot 3 \times 10^{-2}}{6} \]
\[ L = \frac{6 \times 10^{-2}}{6} \]
\[ L = 10^{-2} \, \text{H} \]

### Step 7: Convert L to millihenries
To convert henries to millihenries, we multiply by 1000:
\[ L = 10^{-2} \, \text{H} = 10 \, \text{mH} \]

### Final Answer
The self-inductance of the coil is **10 mH**.

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