A magnetic dipole of moment M and moment of inertia I is initially at rest at an angular displacement 60° from its stable equilibrium position in a uniform magnetic field B.
When it passes through the angular position 30°, its angular speed is:

Question

A magnetic dipole of moment M and moment of inertia I is initially at rest at an angular displacement 60° from its stable equilibrium position in a uniform magnetic field B.

When it passes through the angular position 30°, its angular speed is:

G1 Physics · Accepted Answer

Use conservation of mechanical energy. Magnetic potential energy is U = -MB cosθ. Initially, at 60°, U i U_i U i ​ = -MB cos60° = -MB/2. The dipole starts from rest, so K i K_i K i ​ = 0. At 30°, U f U_f U f ​ = -MB cos30° = -√3MB/2. Let angular speed be ω. Then K f K_f K f ​ = ½Iω². Energy conservation gives U i U_i U i ​ = U f U_f U f ​ + ½Iω². Thus, -MB/2 = -√3MB/2 + ½Iω². Therefore, ½Iω² = MB(√3-1)/2. Multiplying by 2, Iω² = MB(√3-1). Hence, ω = √[MB(√3-1)/I].