. A bar magnet is hung by a thin cotton thread in a uniform horizontal magnetic 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 60º is W, Now the torque required to keep the magnet in this new position is.

A. 2W/√3

B. W/√3

C. √3W

D. √3W/2

Best Answer

Explanation:

Torque acting on a bar magnet kept in a magnetic field B is given as,

τ = m B sin θ

work done in moving from θ = 0° to θ = 60° is

W = m B cos60° = 1/2mB

Torque action when magnet is placed at θ = 60° is

τ = m B sin60° = √3/2mB = √3W

Final answer:

Hence, the correct option is C, i.e. √3W

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