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(cosθ1-cosθ2) and τ = MB sinθ
Here, θ1 = 0° and θ2 = 60°
So, W = MB (1 – cosθ) = 2 MB sin2θ/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.