1.

A thin uniform ring of radius R carrying charge q and mass m rotates about its axis with angular velocity omega. Find the ratio of its magnetic moment and angular momentum.

Answer»

Solution :The EQUIVALENT current in the RING is
`i=q/T=q/((2pi)/omega)=(qomega)/(2pi)`
MAGNETIC moment
`M=ia=(qomega)/(2pi)xxpiR^(2)=(qomegaR^(2))/2`
Angular momentum `L=Iomega(MR^(2))omegathereforeM/L=q/(2m)`


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