## NotTim 3 years ago 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?

1. NotTim

To speed things up, the equation is \[v=qBr/m\]

2. NotTim

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

3. jim_thompson5910

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

4. NotTim

I think you are correct. Q is the charge.

5. NotTim

IF anyone ever comes: value of q still undetermined.

6. jim_thompson5910

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

7. JamesJ

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

8. NotTim

wait, so Q is a constant?

9. JamesJ

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

10. NotTim

So this is knowledge I should have know already?

11. JamesJ

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

12. jim_thompson5910

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

13. jim_thompson5910

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

14. NotTim

Ok. Thank you very much.