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?

Question

nottim · Answer

To speed things up, the equation is $$v=qBr/m$$

nottim · Answer

My only difficulty with this situation is the matter of "q". I don't know where the value comes from.

jim_thompson5910 · Answer

q looks like the charge (in Coulombs perhaps?) of the proton, but i'm not entirely sure

nottim · Answer

I think you are correct. Q is the charge.

nottim · Answer

IF anyone ever comes: value of q still undetermined.

jim_thompson5910 · Answer

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

jamesj · Answer

Yes, that's the charge of a proton.  You can solve it now.  Just make sure you use SI units throughout.

nottim · Answer

wait, so Q is a constant?

jamesj · Answer

q is the charge of object you're interested in, which is a proton.  And a proton most definitely has a fixed charge.

nottim · Answer

So this is knowledge I should have know already?

jamesj · Answer

it doesn't seem like an unreasonable thing that you should.

jim_thompson5910 · Answer

yeah it seems like they were assuming that you would have or should have known that

jim_thompson5910 · Answer

but if you didn't, you can easily look it up

nottim · Answer

Ok. Thank you very much.