The magnetic flux through a coil perpendicluar to its plane and directed into paper is varying according to the relation `phi=(2t^(2)+4t+6)mWb`. The emf induced in the loop at t=4 s is

Options

A. 0.12 V
B. 2.4 V
C. 0.02 V
D. 1.2 V

Explanation

To find the induced EMF in the coil at \( t = 4 \) seconds, we will follow these steps:

### Step 1: Write down the expression for magnetic flux
The magnetic flux \( \phi \) through the coil is given by:
\[ \phi(t) = 2t^2 + 4t + 6 \, \text{mWb} \]

### Step 2: Convert the units of magnetic flux
Since we need to find the induced EMF in volts, we should convert milliwebers (mWb) to webers (Wb):
\[ \phi(t) = (2t^2 + 4t + 6) \times 10^{-3} \, \text{Wb} \]

### Step 3: Use Faraday's law of electromagnetic induction
The induced EMF \( \epsilon \) is given by the negative rate of change of magnetic flux with respect to time:
\[ \epsilon = -\frac{d\phi}{dt} \]

### Step 4: Differentiate the magnetic flux with respect to time
Now, we differentiate \( \phi(t) \):
\[ \frac{d\phi}{dt} = \frac{d}{dt}[(2t^2 + 4t + 6) \times 10^{-3}] \]
Using the power rule of differentiation:
\[ \frac{d\phi}{dt} = (4t + 4) \times 10^{-3} \]

### Step 5: Substitute \( t = 4 \) seconds into the derivative
Now, we substitute \( t = 4 \) seconds into the derivative:
\[ \epsilon = -\frac{d\phi}{dt} = -(4(4) + 4) \times 10^{-3} \]
Calculating the expression:
\[ \epsilon = -(16 + 4) \times 10^{-3} = -20 \times 10^{-3} \, \text{V} \]
Since EMF is considered as a magnitude, we take the absolute value:
\[ \epsilon = 20 \times 10^{-3} \, \text{V} = 0.02 \, \text{V} \]

### Final Answer
The induced EMF at \( t = 4 \) seconds is:
\[ \epsilon = 0.02 \, \text{V} \]

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