JEE Main 2020 · PhysicsMultiple choiceSingle correctMediumMulti-step

JEE Main 4 September 2020, Shift 1, Physics Q14

Question 14 of 75 in this shift, Physics question 14 of 25.

A small bar magnet placed with its axis at 30° with an external field of 0.06 T experiences a torque of 0.018 Nm. The minimum work required to rotate it from its stable to unstable equilibrium position is :

(1) $6.4\times10^{-2}$ J
(2) $7.2\times10^{-2}$ JOfficial answer
(3) $9.2\times10^{-3}$ J
(4) $11.7\times10^{-3}$ J

Official answer

Option 2

NTA final key (Sep 2020).

Chapter: Magnetism and Matter
Topic: Magnetic Dipole Moment
Idea tested: Work done in rotating a dipole

Same idea in other shifts

Asked 5× in all

4 Apr 2024, Shift 2 · Q43The magnetic moment of a bar magnet is 0.5 Am $^2$ . It is suspended in a uniform magnetic field of $8\times10^{-2}$ T. The work done in…EasySingle correct
6 Apr 2024, Shift 1 · Q57A circular coil having 200 turns, $2.5\times10^{-4}$ m $^2$ area and carrying 100 $\mu$ A current is placed in a uniform magnetic field of…EasyNumerical value
6 Apr 2024, Shift 2 · Q58A coil having 100 turns, area of $5\times10^{-3}\ \mathrm{m^2}$ , carrying current of 1 mA is placed in uniform magnetic field of 0.20 T…EasyNumerical value
26 Jun 2022, Shift 2 · Q42A bar magnet having a magnetic moment of $2.0 \times 10^{5}\ JT^{-1}$ , is placed along the direction of uniform magnetic field of…EasySingle correct

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.