Q: A magnetic needle lying parallel to a magnetic field requires W units of work to turn it through 60° . What is the torque needed to maintain the needle in this position ?

Sol: In case of a dipole in a magnetic field,

W = MB(cosθ_{1}-cosθ_{2}) and τ = MB sinθ

Here, θ_{1} = 0° and θ_{2} = 60°

So, W = MB (1 – cosθ) = 2 MB sin^{2}θ/2

and, τ = MB sinθ = 2 MB sin(θ/2) cos(θ/2)

So, τ/W = cot( θ/2), i.e. τ = W cot 30° = √3 W.