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 solve the problem, we need to find the induced emf (E) at \( t = 2 \) seconds for a coil with a given magnetic flux through each turn. The magnetic flux is given by the equation:

\[ \Phi(t) = (t^3 - 2t) \times 10^{-3} \text{ Wb} \]

where \( t \) is in seconds, and the number of turns \( N \) in the coil is 100.

### Step 1: Differentiate the magnetic flux with respect to time
The induced emf is given by Faraday's law of electromagnetic induction, which states:

\[ E = -N \frac{d\Phi}{dt} \]

First, we need to find \( \frac{d\Phi}{dt} \). We differentiate \( \Phi(t) \):

\[ \Phi(t) = (t^3 - 2t) \times 10^{-3} \]

Using the product rule, we differentiate:

\[ \frac{d\Phi}{dt} = \frac{d}{dt}[(t^3 - 2t) \times 10^{-3}] = \left(3t^2 - 2\right) \times 10^{-3} \]

### Step 2: Evaluate the derivative at \( t = 2 \) seconds
Now we substitute \( t = 2 \) seconds into the derivative:

\[ \frac{d\Phi}{dt} = (3(2^2) - 2) \times 10^{-3} \]

Calculating this:

\[ = (3 \times 4 - 2) \times 10^{-3} = (12 - 2) \times 10^{-3} = 10 \times 10^{-3} \text{ Wb/s} \]

### Step 3: Calculate the induced emf
Now we can substitute \( \frac{d\Phi}{dt} \) back into the equation for induced emf:

\[ E = -N \frac{d\Phi}{dt} = -100 \times (10 \times 10^{-3}) \]

Calculating this gives:

\[ E = -1000 \times 10^{-3} = -1 \text{ V} \]

### Final Answer
The induced emf at \( t = 2 \) seconds is:

\[ E = -1 \text{ V} \]