The current in a coil varies with respect to time t as `I=3t^(2)+2t`. If the inductance of coil be 10 mH, the value of induced e.m.f. at` t=2s`will be-

Options

A. `0.14 V`
B. `0.12 V`
C. `0.11 V`
D. `0.13 V`

Answer

To solve the problem, we need to find the induced electromotive force (e.m.f.) in a coil when the current varies with time as given by the equation \( I = 3t^2 + 2t \). The inductance \( L \) of the coil is given as 10 mH (milliHenries). We will follow these steps:

### Step-by-Step Solution:

1. **Identify the formula for induced e.m.f.**:
The induced e.m.f. \( E \) in an inductor is given by the formula:
\[ E = -L \frac{dI}{dt} \]
where \( L \) is the inductance and \( \frac{dI}{dt} \) is the rate of change of current with respect to time.

2. **Differentiate the current function**:
We need to find \( \frac{dI}{dt} \) from the given current function:
\[ I = 3t^2 + 2t \]
Differentiating \( I \) with respect to \( t \):
\[ \frac{dI}{dt} = \frac{d}{dt}(3t^2 + 2t) = 6t + 2 \]

3. **Substitute the value of \( t \)**:
We need to evaluate \( \frac{dI}{dt} \) at \( t = 2 \) seconds:
\[ \frac{dI}{dt} \bigg|_{t=2} = 6(2) + 2 = 12 + 2 = 14 \]

4. **Substitute values into the e.m.f. formula**:
Now we can substitute \( L = 10 \) mH (which is \( 10 \times 10^{-3} \) H) and \( \frac{dI}{dt} = 14 \) into the e.m.f. formula:
\[ E = -L \frac{dI}{dt} = -10 \times 10^{-3} \times 14 \]

5. **Calculate the induced e.m.f.**:
\[ E = -0.14 \text{ volts} \]

6. **Find the magnitude of the induced e.m.f.**:
The magnitude of the induced e.m.f. is:
\[ |E| = 0.14 \text{ volts} \]

### Final Answer:
The value of the induced e.m.f. at \( t = 2 \) seconds is \( 0.14 \) volts.

---