1. sqrt3W 2. sqrt3W/2 3. 2W/sqrt3 4. W/sqrt3

Answer: 1. sqrt3W Work to rotate a magnet from equilibrium ( θ=0 ) is W=MB(1-cosθ) ; torque at angle θ is τ=MBsinθ . W=MB(1-cos60^°)=MB/2 , so MB=2W . τ=MBsin60^°=2W·sqrt3/2=sqrt3 W . Option (1).

Topic: Torque on magnetic dipole in uniform field· Magnetic Effects of Current and Magnetism

A bar magnet is hung by a thin cotton thread in a uniform horizontal magnetic field and is in equilibrium state. The energy required to rotate it by 60° is W. Now the torque required to keep the magnet in this new position is

1. \(\sqrt3W\)

2. \(\frac{\sqrt3W}{2}\)

3. \(\frac{2W}{\sqrt3}\)

4. \(\frac{W}{\sqrt3}\)

Doubtly Asked 30 Oct 2026 NEET UG 2016 Phase 2 Practice test neet Save WhatsApp Report

1 answer

Best answer

Answer: 1. \(\sqrt3W\)

Work to rotate a magnet from equilibrium (\(\theta=0\)) is \(W=MB(1-\cos\theta)\); torque at angle \(\theta\) is \(\tau=MB\sin\theta\).

\(W=MB(1-\cos60^\circ)=\dfrac{MB}{2}\), so \(MB=2W\).
\(\tau=MB\sin60^\circ=2W\cdot\dfrac{\sqrt3}{2}=\sqrt3\,W\).

Option (1).

Doubtly AI · 30 Oct 2026 AI-generated · not yet verified Report

Was this answer helpful?

A bar magnet is hung by a thin cotton thread in a uniform horizontal magnetic field and is in equilibrium state. The energy required to rotate it by 60° is W. Now the torque required to keep the magnet in this new position is

1 answer

Your answer