- #1

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[itex]\gamma \frac{mv^2}{r} = qvB \implies \frac{v}{\sqrt{1-(v/c)^2}} = \frac{qBr}{m} [/itex]?

- Thread starter jdstokes
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- #1

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[itex]\gamma \frac{mv^2}{r} = qvB \implies \frac{v}{\sqrt{1-(v/c)^2}} = \frac{qBr}{m} [/itex]?

- #2

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How is the relativistic momentum?

- #3

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Yes. But just to be clear, using your symbols, the transverse mass = gamma*m.

[itex]\gamma \frac{mv^2}{r} = qvB \implies \frac{v}{\sqrt{1-(v/c)^2}} = \frac{qBr}{m} [/itex]?

Pete

- #4

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It's clearer in terms of momentum pv/r=qvB-->p=qBr.

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