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Friday, 9 October

JEE Main 2020 · PhysicsMultiple choiceSingle correctMediumMulti-step

JEE Main 8 January 2020, Shift 1, Physics Q12

Question 12 of 75 in this shift, Physics question 12 of 25, Section A.

Proton with kinetic energy of 1 MeV moves from south to north. It gets an acceleration of 101210^{12} m/s2^2 by an applied magnetic field (west to east). The value of magnetic field : (Rest mass of proton is 1.6×10−271.6\times10^{-27} kg)
  1. (1)71 mT
  2. (2)7.1 mT
  3. (3)0.71 mTOfficial answer
  4. (4)0.071 mT

Official answer

Option 3

NTA final key (Jan 2020).

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

Same idea in other shifts

Asked 3× in all
  1. 2 Apr 2026, Shift 1 · Q481 μ1\ \muC charge moving with velocity v⃗=(i^−2j^+3k^)\vec v=\left(\hat i-2\hat j+3\hat k\right) m/s in the region of magnetic field…EasyNumerical value
  2. 5 Apr 2026, Shift 1 · Q49The charged particle moving in a uniform magnetic field of (3i^+2j^) T(3\hat{i} + 2\hat{j})\,\mathrm{T} has an acceleration…MediumNumerical value

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 42 of the 174 JEE Main shifts held from 2017 to 2026.