JEE Main 2020 · PhysicsMultiple choiceSingle correctEasyCalculation

JEE Main 3 September 2020, Shift 1, Physics Q2

Question 2 of 75 in this shift, Physics question 2 of 25.

A charged particle carrying charge 1

\mu

C is moving with velocity

(2\hat{i} + 3\hat{j} + 4\hat{k})

^{-1}

. If an external magnetic field of

(5\hat{i} + 3\hat{j} - 6\hat{k}) \times 10^{-3}

T exists in the region where the particle is moving then the force on the particle is

\overrightarrow{F} \times 10^{-9}

N. The vector

\overrightarrow{F}

is :

(1) $-0.30\hat{i} + 0.32\hat{j} - 0.09\hat{k}$
(2) $-3.0\hat{i} + 3.2\hat{j} - 0.9\hat{k}$
(3) $-30\hat{i} + 32\hat{j} - 9\hat{k}$ Official answer
(4) $-300\hat{i} + 320\hat{j} - 90\hat{k}$

Official answer

Option 3

NTA final key (Sep 2020).

Chapter: Moving Charges and Magnetism
Topic: Lorentz Force
Idea tested: Magnetic Force Vector Calculation

Same idea in other shifts

Asked 7× in all

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.