JEE Main 2023PhysicsElectromagnetic InductionEasyMCQ

JEE Main 2023 — Electromagnetic Induction Question with Solution

JEE Main 2023 (25 Jan Shift 2)

Question

A wire of length 1 m moving with velocity 8 m s-1 at right angles to a magnetic field of 2 T. The magnitude of induced emf, between the ends of wire will be ___________.

Choose an option

Show full solutionCorrect option: D
Correct answer
D16 V

Step-by-step explanation

The expression of motional emf is e=BLvsinθ, where, θ is angle between magnetic field and direction of motion.

Here, θ=90°, 

So, induced emf across the ends of wire is e= BLvsin90°=BLv

= 2 × 1 × 8 = 16 V

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About this question

This is a previous-year question from JEE Main 2023, 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.