JEE Main 2019PhysicsElectromagnetic InductionEasyMCQ

JEE Main 2019 — Electromagnetic Induction Question with Solution

JEE Main 2019 (10 Jan Shift 1)

Question

A solid metal cube of edge length 2 cm is moving in the positive y-direction, at a constant speed of 6 m s-1. There is a uniform magnetic field of 0.1 T in the positive z-direction. The potential difference between the two faces of the cube, perpendicular to the x-axis, is

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Show full solutionCorrect option: A
Correct answer
A12 mV

Step-by-step explanation

emf developed is given by, e=Bvl

∴e=0.12×10-2×6⇒e=12×10-3=12 mV

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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.