In an inductor of self-inductance L=2 mH, current changes with time according to relation `i=t^(2)e^(-t)`. At what time emf is zero ?

Options

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

Explanation

To solve the problem, we need to find the time at which the induced electromotive force (emf) in an inductor becomes zero. The formula for induced emf in an inductor is given by:

\[ \text{emf} = -L \frac{di}{dt} \]

where \( L \) is the self-inductance and \( i \) is the current.

### Step 1: Identify the given values
- Self-inductance \( L = 2 \, \text{mH} = 2 \times 10^{-3} \, \text{H} \)
- Current \( i(t) = t^2 e^{-t} \)

### Step 2: Differentiate the current with respect to time
To find \( \frac{di}{dt} \), we need to differentiate \( i(t) \):

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

Using the product rule for differentiation:

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

Calculating each part:

1. \( \frac{d}{dt}(t^2) = 2t \)
2. \( \frac{d}{dt}(e^{-t}) = -e^{-t} \)

Putting it together:

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

Factoring out \( e^{-t} \):

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

### Step 3: Substitute \( \frac{di}{dt} \) into the emf equation
Now substituting \( \frac{di}{dt} \) into the emf equation:

\[ \text{emf} = -L \frac{di}{dt} = -2 \times 10^{-3} e^{-t} (2t - t^2) \]

### Step 4: Set the emf to zero
To find when the emf is zero, we set the equation to zero:

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

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

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

### Step 5: Factor the equation
Factoring gives:

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

This gives us two solutions:

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

### Step 6: Identify the time when emf is zero
The time at which the emf is zero (other than the trivial solution \( t = 0 \)) is:

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

### Final Answer
The time at which the emf is zero is \( t = 2 \, \text{seconds} \).

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