Magnetic Properties of Matter

JEE Main level, single correct, medium

18 of 57

Question 18 of 57

A small bar magnet is placed with its axis at 30o{30}^{o} with an external magnetic field of 0.06T0.06T experiences a torque of 0.018Nm0.018Nm. The minimum work required to rotate it from its stable to unstable equilibrium position is:

Show answer and solution

Answer: (C) 7.2×10−2J7.2\times {10}^{-2}\text{J}

τ=MBsinθ=0.018\tau =MBsin\theta =0.018
M=0.018Bsin⁡θ=0.0180.06×0.5=0.64Am2M=\dfrac{0.018}{B\sin \theta }=\dfrac{0.018}{0.06\times 0.5}=0.64A{m}^{2}
W=ΔU=Uf−UiW=\Delta U={U}_{f}-{U}_{i}
=−MBcos⁡180∘−(−MBcos⁡0∘)=-MB\cos {180}^{^{\circ}}-\left(-MB\cos {0}^{^{\circ}}\right)
=2MB=2MB
=2×0.6×0.06=2\times 0.6\times 0.06
=0.072J=0.072J

JEE Main question paper PDF

Open a shift to read the full paper on this page, or download it.

JEE Main 2025 Apr 2 Shift 1 question paper

Open Download PDF

Opening paper…

Explain it as a whiteboard videoStep-by-step, in Hinglish or English. Your first video needs no login.Solved it on paper but got it wrong?Upload a photo of your working. Mistake Doctor marks the exact line that broke.

More Physics practice

JEE Main Physics practice question on Magnetic Properties of Matter (Magnetism and Properties of Magnet), with the answer and a step-by-step solution. Scan a question from your own book or paper to get a whiteboard video.