JEE Main 2024PhysicsElectromagnetic InductionMediumMCQ

JEE Main 2024 — Electromagnetic Induction Question with Solution

JEE Main 2024 (31 Jan Shift 1)

Question

A coil is placed perpendicular to a magnetic field of 5000 T. When the field is changed to 3000 T in 2 s, an induced emf of 22 V is produced in the coil. If the diameter of the coil is 0.02 m, then the number of turns in the coil is:

Choose an option

Show full solutionCorrect option: B
Correct answer
B70

Step-by-step explanation

Average induced EMF is given by, ε=N∆ϕt

Also, change in flux ∆ϕ=(∆B)A

Given: Bi=5000 T,

Bf=3000 T

d=0.02 m ⇒r=d2=0.01 m

Hence, 

∆ϕ=(∆B)A

=2000π0.012=0.2π

EMF, ε=N∆ϕt⇒22=N0.2π2

⇒N=70

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About this question

This is a previous-year question from JEE Main 2024, 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.