The work done for rotating a magnet with magnetic dipole momentm, through 90° from its magnetic meridian is n

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°)
= MB(1 – \(\frac{1}{2}\))
= 0.5MB
∴ W₁ = 2W₂ = MB
Given W₁ = nW₂. Therefore n = 2.