JEE Main 2024PhysicsElectromagnetic InductionMediumNumerical

JEE Main 2024 — Electromagnetic Induction Question with Solution

JEE Main 2024 (01 Feb Shift 2)

Question

A coil of 200 turns and area 0.20 m2 is rotated at half a revolution per second and is placed in uniform magnetic field of 0.01 T perpendicular to axis of rotation of the coil. The maximum voltage generated in the coil is 2πβ volt. The value of β is ______.

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Show full solutionCorrect answer: 5
Correct answer
5

Step-by-step explanation

Time taken for one complete revolution = time period =2 s.

Therefore, ω=2πT=π rad s-1

Now flux, ϕ=NABcos(ωt)

Therefore, induced EMF ε=-dϕdt=NABωsin(ωt)

The maximum voltage generated, εmax=NABω

=200×0.2×0.01×π

=4π10=2π5 volt

Hence, β=5.

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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.