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A proton of mass 1.67x10^27 kg, moves in a circle in the plane perpendicular to a uniform magnetic field of magnitude 1.8T. The radius of curvature is 3.0cm. what is the speed of the proton?

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To speed things up, the equation is \[v=qBr/m\]
My only difficulty with this situation is the matter of "q". I don't know where the value comes from.
q looks like the charge (in Coulombs perhaps?) of the proton, but i'm not entirely sure

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Other answers:

I think you are correct. Q is the charge.
IF anyone ever comes: value of q still undetermined.
If q is the charge, then in coulombs, the charge of a proton is q = 1.602 * 10^(-19) C where C is the abbreviation of a coulomb
Yes, that's the charge of a proton. You can solve it now. Just make sure you use SI units throughout.
wait, so Q is a constant?
q is the charge of object you're interested in, which is a proton. And a proton most definitely has a fixed charge.
So this is knowledge I should have know already?
it doesn't seem like an unreasonable thing that you should.
yeah it seems like they were assuming that you would have or should have known that
but if you didn't, you can easily look it up
Ok. Thank you very much.

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