Skip to main content

A velocity selector consists of electric field E=E k and magnetic field B=B j with B=12 mT. The value E required for an electron of energy 728 eV…

JEE Main 2022 — Physics Electromagnetism

2022mcqeasy

A velocity selector consists of electric field E⃗=Ek^\vec{E}=E\hat{k}E=Ek^ and magnetic field B⃗=Bj^\vec{B}=B\hat{j}B=Bj^​ with B=12mTB=12\mathrm{mT}B=12mT. The value EEE required for an electron of energy 728eV728\mathrm{eV}728eV moving along the positive xxx-axis to pass undeflected is

(Given, mass of electron =9.1×10−31kg=9.1\times {10}^{-31}\mathrm{kg}=9.1×10−31kg)

Official previous-year question

Held on 26 Jul 2022 · Verified 6 Jul 2026.

Options

  1. A

    192kVm−1192\mathrm{kV}{m}^{-1}192kVm−1

  2. B

    192mVm−1192\mathrm{mV}{m}^{-1}192mVm−1

  3. C

    9600kVm−19600\mathrm{kV}{m}^{-1}9600kVm−1

  4. D

    16kVm−116\mathrm{kV}{m}^{-1}16kVm−1

Did you get this right?

Sign in to track your attempts and accuracy.

Your note

Sign in to keep a private note on this question. Nothing you write is ever public.

JEE Main Electromagnetism in other years

More JEE Main Electromagnetism questions

Other Physics topics for JEE Main

Keep practicing Electromagnetism

Work through every JEE Main Electromagnetism PYQ, year by year.

Browse all Physics topics