The magnetic flux in a closed circuit of resistance `20 Omega`, varies with time (t) according to equation `phi=8t^(2)-6t+5`. What is the magnitude of induced current at time `t=1` sec ?

Options

A. 0.25 A
B. 0.5 A
C. 0.75 A
D. 0.1 A

Explanation

To find the magnitude of the induced current at time \( t = 1 \) second, we can follow these steps:

### Step 1: Determine the magnetic flux function
The magnetic flux \( \phi \) is given by the equation:
\[ \phi = 8t^2 - 6t + 5 \]

### Step 2: Calculate the induced electromotive force (emf)
The induced emf (\( \mathcal{E} \)) can be calculated using Faraday's law of electromagnetic induction, which states that the induced emf is equal to the rate of change of magnetic flux:
\[ \mathcal{E} = -\frac{d\phi}{dt} \]

### Step 3: Differentiate the magnetic flux function
Now, we differentiate \( \phi \) with respect to \( t \):
\[ \frac{d\phi}{dt} = \frac{d}{dt}(8t^2 - 6t + 5) \]
Using the power rule for differentiation:
\[ \frac{d\phi}{dt} = 16t - 6 \]

### Step 4: Substitute \( t = 1 \) second into the derivative
Now we substitute \( t = 1 \) second into the derivative to find the induced emf:
\[ \mathcal{E} = - (16(1) - 6) = - (16 - 6) = -10 \text{ volts} \]
The negative sign indicates the direction of the induced emf, but we are interested in the magnitude, which is:
\[ |\mathcal{E}| = 10 \text{ volts} \]

### Step 5: Calculate the induced current using Ohm's Law
Using Ohm's Law, the induced current \( I \) can be calculated as:
\[ I = \frac{\mathcal{E}}{R} \]
where \( R \) is the resistance of the circuit. Given \( R = 20 \, \Omega \):
\[ I = \frac{10 \text{ volts}}{20 \, \Omega} = 0.5 \text{ amperes} \]

### Final Answer
The magnitude of the induced current at \( t = 1 \) second is:
\[ \boxed{0.5 \text{ A}} \]

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