The flux in a closed circuit of resistance `20 Omega` varies with time according to the equation `phi = 6t^(2)-5t+1`. What is the induced current at time `t = 0.25` second?

Options

A. 0.5 A
B. 0.4 A
C. 0.1 A
D. 0.2 A

Explanation

To solve the problem, we need to find the induced current in a closed circuit with a given resistance and a time-varying magnetic flux. The steps are as follows:

### Step 1: Write down the flux equation
The magnetic flux is given by the equation:
\[ \phi(t) = 6t^2 - 5t + 1 \]

### Step 2: Differentiate the flux to find the induced EMF
The induced electromotive force (EMF) can be calculated using Faraday's law of electromagnetic induction, which states:
\[ \text{EMF} = -\frac{d\phi}{dt} \]
We need to differentiate the flux equation with respect to time \( t \):
\[ \frac{d\phi}{dt} = \frac{d}{dt}(6t^2 - 5t + 1) \]
Calculating the derivative:
\[ \frac{d\phi}{dt} = 12t - 5 \]
Thus, the induced EMF is:
\[ \text{EMF} = - (12t - 5) = -12t + 5 \]

### Step 3: Substitute \( t = 0.25 \) seconds into the EMF equation
Now, we will substitute \( t = 0.25 \) seconds into the EMF equation:
\[ \text{EMF} = -12(0.25) + 5 \]
Calculating this:
\[ \text{EMF} = -3 + 5 = 2 \text{ volts} \]

### Step 4: Use Ohm's Law to find the induced current
Ohm's Law states that:
\[ I = \frac{E}{R} \]
where \( I \) is the current, \( E \) is the EMF, and \( R \) is the resistance. Given that the resistance \( R = 20 \, \Omega \):
\[ I = \frac{2 \text{ volts}}{20 \, \Omega} \]
Calculating the current:
\[ I = 0.1 \text{ amperes} \]

### Final Answer
The induced current at \( t = 0.25 \) seconds is:
\[ I = 0.1 \text{ A} \]

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