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Asked in GSEB Board July 2015 · Magnetic moment of magnets and loops
Given: two particles, each of mass m and charge q, at radius r (the rod is 2r long and turns about its centre), angular speed ω.
Idea: a charge going round a circle is a current loop. It passes any point once per period T=(2π)/ω, so the current it represents is I=q/T=(qω)/(2π).
Moment of one particle: μ₁=IA=(qω)/(2π)×π r²=(qω r²)/2 -- note the π cancels.
Both charges go round the same way, so the two moments add: μ=2μ₁=qω r².
Angular momentum: each particle contributes mr²ω, so L=2mr²ω.
μ/L=(qω r²)/(2mr²ω)=q/(2m).
Every trace of the geometry and the speed has cancelled: for any body whose charge and mass are distributed alike, this ratio is the gyromagnetic ratio q/(2m).
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