A coil having an inductance of 0.5 H carries a current which is uniformly varying from zero to 10 ampere in 2 second. The e.m.f. (in volts) generated in the coil is

Options

A. 10
B. 5
C. 2.5
D. 1.25

Answer

To solve the problem, we need to calculate the electromotive force (e.m.f.) generated in a coil with an inductance of 0.5 H when the current varies uniformly from 0 to 10 A in 2 seconds.

### Step-by-Step Solution:

1. **Identify the given values:**
- Inductance (L) = 0.5 H
- Initial current (I_initial) = 0 A
- Final current (I_final) = 10 A
- Time interval (Δt) = 2 seconds

2. **Calculate the change in current (ΔI):**
\[ \Delta I = I_{\text{final}} - I_{\text{initial}} = 10 \, \text{A} - 0 \, \text{A} = 10 \, \text{A} \]

3. **Calculate the rate of change of current (di/dt):**
\[ \frac{di}{dt} = \frac{\Delta I}{\Delta t} = \frac{10 \, \text{A}}{2 \, \text{s}} = 5 \, \text{A/s} \]

4. **Use the formula for induced e.m.f. (E):**
The formula for the induced e.m.f. in a coil is given by:
\[ E = -L \frac{di}{dt} \]
Since we are interested in the magnitude, we can ignore the negative sign:
\[ |E| = L \frac{di}{dt} \]

5. **Substitute the values into the formula:**
\[ |E| = 0.5 \, \text{H} \times 5 \, \text{A/s} = 2.5 \, \text{V} \]

6. **Conclusion:**
The e.m.f. generated in the coil is 2.5 volts.

### Final Answer:
The e.m.f. generated in the coil is **2.5 volts**.

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