NEET 2016 Phase 2 · PhysicsDipole energyPrevious Year Question
A bar magnet is hung by a thin cotton thread in a uniform horizontal magnetic field and is in equilibrium. The energy required to rotate it by 60° is W. The torque required to keep the magnet in this new (60°) position is:
Answer: (B) √3 W. Answer: (B) Solution: At equilibrium θ=0 (m ∥ B).
- A.W/√3
- B.√3 W✓
- C.√3 W/2
- D.2W/√3
Correct Answer
(B) √3 W
Solution & Explanation
Answer: (B) Solution: At equilibrium θ=0 (m ∥ B). Work to rotate to 60°: W = mB(1 − cos60°) = mB(1 − ½) = mB/2 ⟹ mB = 2W. Torque to hold at 60°: τ = mB sin60° = 2W·(√3/2) = √3 W.
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