The magnetic flux through each turn of a 100 turn coil is `(t ^(3)– 2t) xx 10^(-3 )`Wb, where t is in second. The induced emf at t = 2 s is

Options

A. -4V
B. -1V
C. +1V
D. +4V.

Explanation

To find the induced emf at \( t = 2 \) seconds for a coil with a magnetic flux given by \( \Phi(t) = (t^3 - 2t) \times 10^{-3} \) Wb, we can follow these steps:

### Step 1: Identify the formula for induced emf
According to Faraday's law of electromagnetic induction, the induced emf (\( \mathcal{E} \)) in a coil is given by:
\[ \mathcal{E} = -n \frac{d\Phi}{dt} \]
where \( n \) is the number of turns in the coil and \( \Phi \) is the magnetic flux.

### Step 2: Substitute the given values
We know that the coil has \( n = 100 \) turns, and the magnetic flux is given as:
\[ \Phi(t) = (t^3 - 2t) \times 10^{-3} \]
So, we need to find \( \frac{d\Phi}{dt} \).

### Step 3: Differentiate the magnetic flux with respect to time
To find \( \frac{d\Phi}{dt} \), we differentiate \( \Phi(t) \):
\[ \frac{d\Phi}{dt} = \frac{d}{dt} \left( (t^3 - 2t) \times 10^{-3} \right) \]
Using the product rule and the power rule:
\[ \frac{d\Phi}{dt} = 10^{-3} \left( \frac{d}{dt}(t^3 - 2t) \right) = 10^{-3} (3t^2 - 2) \]

### Step 4: Substitute \( t = 2 \) seconds into the derivative
Now, we substitute \( t = 2 \) seconds into the derivative:
\[ \frac{d\Phi}{dt} \bigg|_{t=2} = 10^{-3} (3(2^2) - 2) = 10^{-3} (3 \cdot 4 - 2) = 10^{-3} (12 - 2) = 10^{-3} \cdot 10 = 10^{-2} \]

### Step 5: Calculate the induced emf
Now we can substitute \( \frac{d\Phi}{dt} \) back into the formula for induced emf:
\[ \mathcal{E} = -n \frac{d\Phi}{dt} = -100 \cdot 10^{-2} = -1 \text{ volts} \]

### Conclusion
Thus, the induced emf at \( t = 2 \) seconds is:
\[ \mathcal{E} = -1 \text{ volts} \]