Question

The work done for rotating a magnet with magnetic dipole momentm, through 90° from its magnetic meridian is n times the work done to rotate it through 60°. Find the value of n.

Answer 1

Data : θ₀ = 0°, θ₁ = 90°, θ₂ = 60°, W₁ = nW₂

The work done by an external agent to rotate the magnet from θ₀ to θ is

W = MB (cos θ₀ – cos θ) 

∴ W₁ = MB(cos θ₀ – cosθ₁) 

= MB (cos 0° – cos 90°)

= MB (1 – 0) 

= MB

∴ W₂ = MB (cos 0°- cos 60°)