The magnetic flux linked with a coil , in Webers, is given by the equation `phi=4t^2+3t+5` . Then the magnitude of induced emf at t=2s will be

Options

A. 12 V
B. 19 V
C. 18 V
D. 16 V

Explanation

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

### Step 1: Understand the formula for induced emf
The induced emf (\( \varepsilon \)) in a coil is given by the rate of change of magnetic flux (\( \Phi \)) linked with the coil. Mathematically, this is expressed as:
\[ \varepsilon = -\frac{d\Phi}{dt} \]
For the magnitude, we can ignore the negative sign:
\[ |\varepsilon| = \frac{d\Phi}{dt} \]

### Step 2: Differentiate the magnetic flux equation
The magnetic flux linked with the coil is given by the equation:
\[ \Phi = 4t^2 + 3t + 5 \]
We need to differentiate this equation with respect to time \( t \):
\[ \frac{d\Phi}{dt} = \frac{d}{dt}(4t^2 + 3t + 5) \]
Using the power rule of differentiation:
- The derivative of \( 4t^2 \) is \( 8t \).
- The derivative of \( 3t \) is \( 3 \).
- The derivative of a constant (5) is \( 0 \).

Thus, we have:
\[ \frac{d\Phi}{dt} = 8t + 3 \]

### Step 3: Substitute \( t = 2 \) seconds into the derivative
Now, we substitute \( t = 2 \) seconds into the expression we found for \( \frac{d\Phi}{dt} \):
\[ \frac{d\Phi}{dt} = 8(2) + 3 \]
Calculating this gives:
\[ \frac{d\Phi}{dt} = 16 + 3 = 19 \]

### Step 4: State the magnitude of the induced emf
Therefore, the magnitude of the induced emf at \( t = 2 \) seconds is:
\[ |\varepsilon| = 19 \text{ volts} \]

### Final Answer
The magnitude of the induced emf at \( t = 2 \) seconds is \( 19 \text{ V} \).

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