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Consider a non conducting plate of radius a and mass m which has a charge q distributed uniformly over it, The plate is rotated about its own axis with an angular speed omega. Show that the magnetic moment M and the angular momentum L of the plate are related as M/L=q/(2m). |
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Answer» Solution :If `sigma` is the surface CHARGE density, then `q = sigmapia^(2)` Current i `= sigmaomegardr` The magnetic moment of the element ring `dM=i(dA)=sigmaomegar dr(pir^(2))=pisigmaomegar^(3) dr` and `M=pisigmaomegaint_(0)^(a)R^(3)dr=(pisigmaomegaa^(4))/4=(PIA^(2)sigma)(omegaa^(2))/4` The angular momentum of the disc about its axis `L=(ma^(2))/2omega`. The ratio `M/L=((qomegaa^(2))/4)/((omegama^(2))/2)=q/(2m)` |
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