A bar magnet of magnetic moment M and moment of inertia I is suspended freely in a uniform horizontal magnetic field B. It is displaced through a small angle from its stable equilibrium position and released.
Its time period of small angular oscillations is:

Question

A bar magnet of magnetic moment M and moment of inertia I is suspended freely in a uniform horizontal magnetic field B. It is displaced through a small angle from its stable equilibrium position and released.

Its time period of small angular oscillations is:

G1 Physics · Accepted Answer

For a magnetic dipole in a uniform magnetic field, τ = -MB sinθ. For small angular displacement, sinθ ≈ θ. Hence, τ ≈ -MBθ. Using rotational dynamics, I d²θ/dt² = -MBθ. Therefore, d²θ/dt² + (MB/I)θ = 0. This is SHM with angular frequency ω = √(MB/I). Hence, T = 2π/ω = 2π√(I/MB).