At any instant t current I through a coil of self inductance 2 mH is given by `i=t^(2)e^(-t)`. The induced e.m.f. will be zero at time

Options

A. 1 sec
B. 2 sec
C. 3 sec
D. 4 sec

Explanation

To find the time at which the induced electromotive force (e.m.f.) is zero in a coil with a self-inductance of 2 mH, we start with the given current through the coil, which is expressed as:

\[ I(t) = t^2 e^{-t} \]

### Step 1: Write the formula for induced e.m.f.
The induced e.m.f. (ε) in an inductor is given by the formula:

\[ \epsilon = -L \frac{dI}{dt} \]

where \( L \) is the self-inductance of the coil and \( \frac{dI}{dt} \) is the rate of change of current with respect to time.

### Step 2: Differentiate the current function
We need to differentiate the current function \( I(t) = t^2 e^{-t} \) with respect to time \( t \). We will use the product rule for differentiation, which states that if you have two functions \( u(t) \) and \( v(t) \), then:

\[ \frac{d(uv)}{dt} = u \frac{dv}{dt} + v \frac{du}{dt} \]

Let:
- \( u(t) = t^2 \)
- \( v(t) = e^{-t} \)

Now, we find the derivatives:
- \( \frac{du}{dt} = 2t \)
- \( \frac{dv}{dt} = -e^{-t} \)

Using the product rule:

\[ \frac{dI}{dt} = u \frac{dv}{dt} + v \frac{du}{dt} = t^2 (-e^{-t}) + e^{-t} (2t) \]

This simplifies to:

\[ \frac{dI}{dt} = -t^2 e^{-t} + 2t e^{-t} = e^{-t} (2t - t^2) \]

### Step 3: Substitute into the e.m.f. formula
Now substitute \( \frac{dI}{dt} \) back into the e.m.f. formula:

\[ \epsilon = -L \frac{dI}{dt} = -2 \times 10^{-3} \cdot e^{-t} (2t - t^2) \]

### Step 4: Set the e.m.f. to zero
To find when the induced e.m.f. is zero, we set the equation to zero:

\[ -2 \times 10^{-3} \cdot e^{-t} (2t - t^2) = 0 \]

Since \( e^{-t} \) is never zero, we can simplify this to:

\[ 2t - t^2 = 0 \]

### Step 5: Factor the equation
Factoring out \( t \):

\[ t(2 - t) = 0 \]

### Step 6: Solve for \( t \)
Setting each factor to zero gives us:

1. \( t = 0 \)
2. \( 2 - t = 0 \) → \( t = 2 \)

Thus, the induced e.m.f. will be zero at:

\[ t = 0 \text{ seconds or } t = 2 \text{ seconds} \]

### Conclusion
Since we are interested in the time after the current starts flowing, the relevant solution is:

\[ t = 2 \text{ seconds} \]