JEE Main 2020PhysicsMagnetic Effects of CurrentHardMCQ

JEE Main 2020 — Magnetic Effects of Current Question with Solution

JEE Main 2020 (08 Jan Shift 1)

Question

Proton with kinetic energy of 1 MeV moves from south to north. It gets an acceleration of 1012 m/s2 by an applied magnetic field (west to east). The value of magnetic field: (Rest mass of proton is 1.6×10-27kg )

Choose an option

Show full solutionCorrect option: A
Correct answer
A0.71mT

Step-by-step explanation


∴K.E.=1.6×10-13=12×1.6×10-27v2
v=2×107m/s
∴Bqv=ma
B=1.6×10-27×10121.6×10-19×2×107
=0.71×10-3T
So,

B=0.71 mT

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About this question

This is a previous-year question from JEE Main 2020, covering the Magnetic Effects of Current 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.