Question

In an inductor of self inductance l = 2mH, current changes with time according to the relation $I=t^2{e}^{-t}$. The time at which the emf becomes zero is

• A. 4s
• B. 2s
• C. 1s
• D. 0.5s

Answer 1

The correct option is B

2s

Inductor

1. A coil, choke, or reactor is a type of inductor.
2. It is a two-terminal electrical component that stores energy in a magnetic field.
3. An inductor is made up of insulated wire twisted into a coil around a core.

Self-inductance

1. Electromagnetic induction in the form of self-inductance is a type of electromagnetic induction.
2. When the current in a current-carrying wire, solenoid, or conductor changes, it is characterized as the induction of a voltage in the wire.
3. In the instance of self-inductance, a voltage is induced by the magnetic field formed by fluctuating or altering the current in the circuit.
4. As a result, the voltage is known as self-induced voltage.

Step 1: Given

Inductance $L=2mHandI=t^2{e}^{-t}$

Step 2: Formula used

$E=-L\left(\frac{dl}{dt}\right)[whereE=emf,L=induc\tan ce]$

Step 3: Formula of induced emf in an inductor

$$\begin{gathered} E=-L\left(\frac{dl}{dt}\right) \\ E=-L\left(\frac{d}{dt}\left(t^2{e}^{-t}\right)\right) \\ E=-L\left|2t{e}^{-t}-t^2{e}^{-t}\right| \end{gathered}$$

Step 4: Calculating time at which emf becomes zero

If emf is zero, $E=0$

$$\begin{gathered} 2t{e}^{-t}=t^2{e}^{-t} \\ t=2s \end{gathered}$$

Therefore, the emf becomes zero at $2s$.

Hence, option B is the correct answer.