The magnetic flux in a closed circuit of resistance `10Omega` varies with time according to equation , `phi=6t^(2)-5t+1`. What is the magnitude of the induced current at `t=0.25s`?

Options

A. `1.2A`
B. `0.8A`
C. `0.6A`
D. `0.2A`

Answer

To find the magnitude of the induced current at \( t = 0.25 \, s \) in a closed circuit with a resistance of \( 10 \, \Omega \) and a magnetic flux given by the equation \( \phi = 6t^2 - 5t + 1 \), we can follow these steps:

### Step 1: Calculate the induced EMF
The induced electromotive force (EMF) in a circuit is given by Faraday's law of electromagnetic induction, which states:

\[ \text{EMF} = -\frac{d\phi}{dt} \]

First, we need to differentiate the magnetic flux \( \phi \) with respect to time \( t \).

Given:
\[ \phi = 6t^2 - 5t + 1 \]

Now, we differentiate \( \phi \):

\[ \frac{d\phi}{dt} = \frac{d}{dt}(6t^2 - 5t + 1) = 12t - 5 \]

Thus, the induced EMF is:

\[ \text{EMF} = -\frac{d\phi}{dt} = -(12t - 5) = -12t + 5 \]

### Step 2: Substitute \( t = 0.25 \, s \)
Now, we will calculate the EMF at \( t = 0.25 \, s \):

\[ \text{EMF} = -12(0.25) + 5 \]

Calculating this gives:

\[ \text{EMF} = -3 + 5 = 2 \, \text{V} \]

### Step 3: Calculate the induced current
Using Ohm's law, the current \( I \) can be calculated using the formula:

\[ I = \frac{\text{EMF}}{R} \]

Where \( R \) is the resistance of the circuit. Given \( R = 10 \, \Omega \):

\[ I = \frac{2 \, \text{V}}{10 \, \Omega} = 0.2 \, \text{A} \]

### Final Answer
The magnitude of the induced current at \( t = 0.25 \, s \) is:

\[ \boxed{0.2 \, \text{A}} \]