JEE Main 2019PhysicsElectromagnetic InductionMediumMCQ

JEE Main 2019 — Electromagnetic Induction Question with Solution

JEE Main 2019 (10 Jan Shift 2)

Question

The self induced emf of a coil is 25 volts. When the current in it is changed at uniform rate from 10A to 25A in 1s , the change in the energy of the inductance is:

Choose an option

Show full solutionCorrect option: C
Correct answer
C437.5 J

Step-by-step explanation

Induced emf / back emf in an inductor coil is given by 

ε=Ldidt

⇒25=L×25-101

⇒L=2515=53H

Energy of inductor is given by

E=12Li2

Energy change △E=12LI22-I12

⇒∆E=12×53625-100 J

∆E=437.5 J

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About this question

This is a previous-year question from JEE Main 2019, covering the Electromagnetic Induction chapter of Physics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.