The flux in a closed circuit of resistance `10 Omega` varies with time according to the equation `phi=3t^2-t+1`, where `phi` is in weber and t is in , s the value of induced current at t=1 s is

Options

A. 0.3 A
B. 0.5A
C. 0.7A
D. 1A

Answer

To solve the problem step by step, we will follow these instructions:

### Step 1: Understand the given information
We are given:
- The resistance \( R = 10 \, \Omega \)
- The magnetic flux \( \phi \) as a function of time:
\[ \phi(t) = 3t^2 - t + 1 \]

### Step 2: Calculate the induced electromotive force (emf)
The induced emf \( E \) in the circuit is given by Faraday's law of electromagnetic induction, which states that the induced emf is equal to the negative rate of change of magnetic flux:
\[ E = -\frac{d\phi}{dt} \]
We will differentiate \( \phi(t) \) with respect to \( t \).

### Step 3: Differentiate the flux function
Differentiate \( \phi(t) = 3t^2 - t + 1 \):
\[ \frac{d\phi}{dt} = \frac{d}{dt}(3t^2) - \frac{d}{dt}(t) + \frac{d}{dt}(1) \]
Calculating each term:
- The derivative of \( 3t^2 \) is \( 6t \)
- The derivative of \( -t \) is \( -1 \)
- The derivative of \( 1 \) is \( 0 \)

Thus, we have:
\[ \frac{d\phi}{dt} = 6t - 1 \]

### Step 4: Substitute \( t = 1 \) second into the derivative
Now, we will find the induced emf at \( t = 1 \) second:
\[ E = -\frac{d\phi}{dt} = -(6(1) - 1) = -(6 - 1) = -5 \, \text{V} \]
The magnitude of the induced emf is \( 5 \, \text{V} \).

### Step 5: Calculate the induced current
Using Ohm's law, the induced current \( I \) can be calculated as:
\[ I = \frac{E}{R} \]
Substituting the values:
\[ I = \frac{5 \, \text{V}}{10 \, \Omega} = 0.5 \, \text{A} \]

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

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