The current in self -inductance `L=40` mH is to be be increased uniformly from 1 A to 11 A is 4 millisecond . The emf induce in inductor during the process is

Options

A. 100 V
B. 0.4 V
C. 40 V
D. 440 V

Explanation

To find the induced electromotive force (emf) in a self-inductor when the current is increased uniformly, we can use the formula:

\[ E = L \frac{di}{dt} \]

Where:
- \(E\) is the induced emf,
- \(L\) is the self-inductance,
- \(\frac{di}{dt}\) is the rate of change of current.

### Step 1: Identify the given values
- Self-inductance \(L = 40 \, \text{mH} = 40 \times 10^{-3} \, \text{H}\)
- Initial current \(I_{\text{initial}} = 1 \, \text{A}\)
- Final current \(I_{\text{final}} = 11 \, \text{A}\)
- Time taken \(t = 4 \, \text{ms} = 4 \times 10^{-3} \, \text{s}\)

### Step 2: Calculate the change in current (\(di\))
\[ di = I_{\text{final}} - I_{\text{initial}} = 11 \, \text{A} - 1 \, \text{A} = 10 \, \text{A} \]

### Step 3: Calculate the rate of change of current (\(\frac{di}{dt}\))
\[ \frac{di}{dt} = \frac{di}{t} = \frac{10 \, \text{A}}{4 \times 10^{-3} \, \text{s}} = \frac{10}{0.004} = 2500 \, \text{A/s} \]

### Step 4: Substitute the values into the emf formula
\[ E = L \frac{di}{dt} = (40 \times 10^{-3} \, \text{H}) \times (2500 \, \text{A/s}) \]

### Step 5: Calculate the induced emf
\[ E = 40 \times 10^{-3} \times 2500 = 100 \, \text{V} \]

### Final Answer
The induced emf in the inductor during the process is \(100 \, \text{V}\).

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