The magnetic flux linked with a coil varies with time as `phi=3t^(2)+4t+9` webers. The induced emf at `t=2s` is

Options

A. `10V`
B. `1V`
C. `6V`
D. `16V`

Answer

To find the induced emf at \( t = 2 \) seconds, we will follow these steps:

### Step 1: Write down the expression for magnetic flux
The magnetic flux linked with the coil is given by:
\[ \phi(t) = 3t^2 + 4t + 9 \]

### Step 2: Differentiate the magnetic flux with respect to time
To find the induced emf (\( \mathcal{E} \)), we need to differentiate the magnetic flux with respect to time \( t \):
\[ \mathcal{E} = -\frac{d\phi}{dt} \]
Calculating the derivative:
\[ \frac{d\phi}{dt} = \frac{d}{dt}(3t^2 + 4t + 9) \]
Using the power rule of differentiation:
\[ \frac{d\phi}{dt} = 6t + 4 \]

### Step 3: Substitute \( t = 2 \) seconds into the derivative
Now, we will substitute \( t = 2 \) seconds into the expression we found for \( \frac{d\phi}{dt} \):
\[ \frac{d\phi}{dt} \bigg|_{t=2} = 6(2) + 4 \]
Calculating this gives:
\[ \frac{d\phi}{dt} \bigg|_{t=2} = 12 + 4 = 16 \]

### Step 4: Calculate the induced emf
Now, substituting into the expression for induced emf:
\[ \mathcal{E} = -\frac{d\phi}{dt} = -16 \text{ volts} \]
However, since we are interested in the magnitude of the induced emf, we take the absolute value:
\[ \mathcal{E} = 16 \text{ volts} \]

### Final Answer
The induced emf at \( t = 2 \) seconds is \( 16 \) volts.

---