JEE Main 2021PhysicsElectromagnetic InductionEasyNumerical

JEE Main 2021 — Electromagnetic Induction Question with Solution

JEE Main 2021 (27 Jul Shift 2)

Question

In the given figure the magnetic flux through the loop increases according to the relation ϕB(t)=10t2+20t, where ϕB is in milliwebers and t is in seconds. The magnitude of current through R=2 Ω resistor at t=5 s is _____ mA.

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Show full solutionCorrect answer: 60
Correct answer
60

Step-by-step explanation

|ϵ|=dϕdt=20t+20 mV

|i|=|ϵ|R=10t+10 mA

At t=5

|i|=60 mA

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About this question

This is a previous-year question from JEE Main 2021, 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.