Magnetic flux `phi` (in weber) linked with a closed circuit of resistance `10 ohm` varies with time `t` (in seconds) as
`phi=5t^(2)-4t+1`
The induced electromotive force in the circuit at `t=0.2` sec. is

Options

A. `0.4` volts
B. `-0.4` volts
C. `-2.0` volts
D. `2.0` volts

Explanation

To find the induced electromotive force (emf) in the circuit at \( t = 0.2 \) seconds, we will follow these steps:

### Step 1: Write down the expression for magnetic flux
The magnetic flux \( \phi \) linked with the circuit is given by:
\[ \phi(t) = 5t^2 - 4t + 1 \]

### Step 2: Differentiate the magnetic flux with respect to time
To find the induced emf, we need to calculate the rate of change of magnetic flux with respect to time, which is given by:
\[ \frac{d\phi}{dt} \]
We differentiate \( \phi(t) \):
\[ \frac{d\phi}{dt} = \frac{d}{dt}(5t^2 - 4t + 1) \]
Using the power rule of differentiation:
\[ \frac{d\phi}{dt} = 10t - 4 \]

### Step 3: Substitute \( t = 0.2 \) seconds into the derivative
Now, we substitute \( t = 0.2 \) into the expression for \( \frac{d\phi}{dt} \):
\[ \frac{d\phi}{dt} = 10(0.2) - 4 \]
Calculating this gives:
\[ \frac{d\phi}{dt} = 2 - 4 = -2 \, \text{weber/second} \]

### Step 4: Calculate the induced emf
The induced emf \( \mathcal{E} \) in the circuit can be calculated using Faraday's law of electromagnetic induction, which states:
\[ \mathcal{E} = -\frac{d\phi}{dt} \]
Substituting the value we found:
\[ \mathcal{E} = -(-2) = 2 \, \text{volts} \]

### Final Answer
The induced electromotive force in the circuit at \( t = 0.2 \) seconds is:
\[ \mathcal{E} = 2 \, \text{volts} \]

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