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Conservation of angular momentum (electromagnetism)
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[QUOTE="TSny, post: 4554697, member: 229090"] OK, feel free to shoot me.:redface: But now I'm getting [itex]\omega = 2\pi B\lambda/(8\pi^2 \rho R - \mu_{0}\lambda^2) [/itex]. So, there's a factor of R in just the first term of the denominator and I believe the dimensions of the two terms in the denominator are now consistent. This is what you will get if you take into account mfb's remarks. The correct comparison is that ##\rho R>> \mu_{0}\lambda^2## in any realistic setup. Then you can neglect the second term in the denominator which is equivalent to not worrying about the magnetic field produced by the rotating cylinder. [/QUOTE]
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Conservation of angular momentum (electromagnetism)
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