Physics Circle home Student Login

Home›Chapters›Magnetic Properties of Materials PYQs›Field of a Magnet & Magnet in a Uniform Field›Question 19

NEET 2016 (Phase 2) · Field of a Magnet & Magnet in a Uniform Field

Field of a Magnet & Magnet in a Uniform Field question from NEET 2016 (Phase 2)

Q19NEET 2016 · PHASE 2Medium

Field of a Magnet & Magnet in a Uniform Field

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

Show answer and solution

Answer: (b)

  1. Work to rotate from equilibrium: W=MB(1−cos⁡60∘)=MB2W = MB(1-\cos 60^\circ) = \frac{MB}{2}, so MB=2WMB = 2W.
  2. Torque at 60∘60^\circ: τ=MBsin⁡60∘=2W⋅32\tau = MB\sin 60^\circ = 2W \cdot \frac{\sqrt{3}}{2}
  3. So the answer is (b) 3W\sqrt{3}W.
Report a mistake

More questions on Field of a Magnet & Magnet in a Uniform Field