JEE Main 2024 · PhysicsNumerical answerNumerical valueEasyCalculation

JEE Main 6 April 2024, Shift 1, Physics Q57

Question 57 of 90 in this shift, Physics question 27 of 30, Section B.

A circular coil having 200 turns,

2.5\times10^{-4}

^2

area and carrying 100

\mu

A current is placed in a uniform magnetic field of 1T. Initially the magnetic dipole moment (

\vec{M}

) was directed along

\vec{B}

. Amount of work, required to rotate the coil through 90° from its initial orientation such that

\vec{M}

becomes perpendicular to

\vec{B}

, is ___________

\mu

Official answer

NTA final key.

Chapter: Moving Charges and Magnetism
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 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
4 Sep 2020, Shift 1 · Q14A 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…MediumSingle 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.