Question:

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?

Updated On: Jul 6, 2022
  • $4\,s$
  • $3\,s$
  • $2\,s$
  • $1\,s$
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The Correct Option is C

Solution and Explanation

$L=2\,mH=2\times10^{-3}\,H$ $I=t^{2}e^{-t}$ $\frac{dI}{dt}=t^{2}e^{-1}(-1)+e^{-t}(2t)=te^{-t}(-t+2)$ Emf $=L\frac{dI}{dt}=2\times10^{-3}\,t\,e^{-t}(-t+2)$ Now, emf = $0$, when $(-t+2)=0$ or $t=2\,s$
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Concepts Used:

Electromagnetic Induction

Electromagnetic Induction is a current produced by the voltage production due to a changing magnetic field. This happens in one of the two conditions:-

  1. When we place the conductor in a changing magnetic field.
  2. When the conductor constantly moves in a stationary field.

Formula:

The electromagnetic induction is mathematically represented as:-

e=N × d∅.dt

Where

  • e = induced voltage
  • N = number of turns in the coil
  • Φ = Magnetic flux (This is the amount of magnetic field present on the surface)
  • t = time

Applications of Electromagnetic Induction

  1. Electromagnetic induction in AC generator
  2. Electrical Transformers
  3. Magnetic Flow Meter