The current in drlf -inducatance `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. 4 v
D. 440 v

Explanation

To solve the problem of finding the induced EMF in an inductor when the current changes uniformly, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the given values:**
- Inductance \( L = 40 \, \text{mH} = 40 \times 10^{-3} \, \text{H} \)
- Initial current \( I_1 = 1 \, \text{A} \)
- Final current \( I_2 = 11 \, \text{A} \)
- Time duration \( \Delta t = 4 \, \text{ms} = 4 \times 10^{-3} \, \text{s} \)

2. **Calculate the change in current (\( \Delta I \)):**
\[ \Delta I = I_2 - I_1 = 11 \, \text{A} - 1 \, \text{A} = 10 \, \text{A} \]

3. **Calculate the rate of change of current (\( \frac{dI}{dt} \)):**
\[ \frac{dI}{dt} = \frac{\Delta I}{\Delta t} = \frac{10 \, \text{A}}{4 \times 10^{-3} \, \text{s}} = \frac{10}{0.004} = 2500 \, \text{A/s} \]

4. **Use the formula for induced EMF (\( \mathcal{E} \)):**
The induced EMF in an inductor is given by:
\[ \mathcal{E} = -L \frac{dI}{dt} \]
Since we are interested in the magnitude, we can ignore the negative sign:
\[ \mathcal{E} = L \frac{dI}{dt} \]

5. **Substitute the values into the formula:**
\[ \mathcal{E} = 40 \times 10^{-3} \, \text{H} \times 2500 \, \text{A/s} \]
\[ \mathcal{E} = 40 \times 2500 \times 10^{-3} = 100 \, \text{V} \]

6. **Conclusion:**
The induced EMF in the inductor during the process is \( \mathcal{E} = 100 \, \text{V} \).

### Final Answer:
The induced EMF is **100 volts**.

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