Show that the time period (T) of oscillations of a freely suspended magnetic dipole of magnetic moment (m) in a uniform magnetic field (B) is given by T = 2π√(I/mB), where I is the moment of inertia of the magnetic dipole.
Solution:
τ=M→×B→=MBsinθ For small angle θ τ=MBθ α=(MBI)θ Comparing with angular SHM ω²=MBI T=2πω=2π√(I/mB) where, m=magnetic moment I=moment of inertia B=magnetic field intensity