Question

Relation between angular momentum and magnetic moment

Answer 1

Relation between magnetic moment and angular momentum — classic theory. The angular momentum is →L=→r×→p or L=mrv and therefore ⇒μ=q2mL=γL.

Answer 2

μ=I⋅A=qtπr2=qv2πrmmπr2

(I=q/t: current, A: area of a loop, q charge, t=2πr/v: time of 1 rotation, v: velocity of particle, m: mass )

The angular momentum is L⃗ =r⃗ ×p⃗  or L=mrv and therefore

⇒μ=q2mL=γL.

The easiest way to see the equality is to use a more general formula for the magnetic dipole moment of a particle. For a flat planar loop of current, it's true that μ=IA, with the direction of the dipole normal to the loop. However, the more general case is that of a a volume current J⃗  in some finite region of space. In this case, the the general formula for the magnetic dipole moment of the configuration is

μ⃗ =12∫r⃗ ×J⃗ d3r.

(Showing that this reduces to the above formula for a flat planar loop is left as an exercise to the reader.) If we further assume that the current density is due to a number of particles with number density n, charge q, velocity v⃗ , and mass m, then we have current density J⃗ =nqv⃗ ; thus,

μ⃗ =12∫r⃗ ×(nqv⃗ )d3r=q2m∫r⃗ ×(nmv⃗ )d3r.

But nmv⃗ =ρv⃗ , where ρ is the mass density of the cloud; thus, the above integral can be rewritten as

μ⃗ =q2m∫ρr⃗ ×v⃗ d3r=q2mL⃗ .