JEE Main 2022 · PhysicsMultiple choiceSingle correctMediumMulti-step

JEE Main 26 July 2022, Shift 2, Physics Q47

Question 47 of 90 in this shift, Physics question 17 of 30, Section A.

A velocity selector consists of electric field

\vec{E} = E\hat{k}

and magnetic field

\vec{B} = B\hat{j}

with

B = 12\ mT

. The value of

E

required for an electron of energy

728\ eV

moving along the positive

x

-axis to pass undeflected is : (Given, mass of electron =

9.1 \times 10^{-31}

kg)

(1)192 $\text{kVm}^{-1}$ Official answer
(2)192 $\text{mVm}^{-1}$
(3)9600 $\text{kVm}^{-1}$
(4)16 $\text{kVm}^{-1}$

Official answer

Option 1

NTA final key.

Chapter: Moving Charges and Magnetism
Topic: Lorentz Force
Idea tested: Velocity Selector and Crossed Fields

Same idea in other shifts

Asked 4× in all

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Question text from the official JEE Main paper published by NTA; answer from the final answer key. Chapter, topic and idea tags, difficulty and skill are Lumi’s. Lumi analyses 108 of the 174 JEE Main shifts held from 2017 to 2026.