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?

Torque acting on a bar magnet kept in a magnetic field →B is τ=→m×→B=mBsinθHence work done in moving from θ=0° to θ=60° is W=∫₀⁶⁰ mBsinθdθ = mB[-cosθ]₀⁶⁰ = mB(1-cos60°) = 1/2 mB Given that W = 1/2 mB = W, therefore mB = 2W Hence torque acting when magnet is placed at 60° = mBsin60° = 2Wsin60° = 2W(√3/2) = √3W