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Conservation of angular momentum (electromagnetism)
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[QUOTE="TSny, post: 4554632, member: 229090"] This is what I got also, using the conservation of angular momentum. [EDIT: Actually, I see now that I got a factor of R that you don't have: [itex]\omega = 2\pi B\lambda/[R(8\pi^2 \rho - \mu_{0}\lambda^2)] [/itex]. I think the missing factor of R in your expression is due to the mistake that mfb has pointed to. [COLOR="DarkRed"][EDIT 2: After checking it over for the sixth time, I am no longer getting the factor of R in the denominator. So, I again agree with your answer. The dimensions seem to check, too. Sorry for the confusion and I hope I have it right now.] [/COLOR] (Nope, See correction in post below) In using Faraday's law, did you take into account the changing B field due to the increasing speed of rotation of the charged cylinder? In any realistic setup, you would have ##\rho >> \mu_o \lambda## and ##\omega## will be very small [see correction in post below]. So you could safely neglect the magnetic field of the spinning cylinder in both approaches. [/QUOTE]
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Conservation of angular momentum (electromagnetism)
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