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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. |
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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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