Electron in a magnetic field and angular momentum

  • Thread starter Munky
  • Start date
1. The problem statement, all variables and given/known data
An electron moves in a circular path perpendicular to a constant magnetic field with a magnitude of 1 mT. The angular momentum of the electron about the center of the circle is 4.00E-25 kg m^2/s.

m= 9.109E-31 kg
q= 1.602E-19 C
B= 0.001 T

I also know that the answer is

r = 5.00 cm
v = 8.78E6 m/s


2. Relevant equations

F=qvB
or
qvB=mv^2/r
or
r=mv/qB


3. The attempt at a solution

My problem is that I know I need to get either F, v, or r from the angular momentum, but I cannot find any way to do that. If I were able to solve for one of those variables, the other two are simple to figure. I have another similar question where I need to know how to do this... and for the life of me I can't figure it out.

Thanks for any help you can be.

Munky
 

tiny-tim

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Welcome to PF!

Hi Munky ! Welcome to PF!
qvB=mv^2/r
ok … you know q B m and rv …

and your equation has q B m v and r …

just fiddle around with it! :smile:
 
Thanks for the response, cryptic as it seemed at first, it helped.

I don't know if I was clear, but I was supposed to solve for v and r, and wasn't sure how momentum fit in.

Your response was enough to make me look a little further, knowing that I had what I needed. I found that angular momentum "l" is the cross product of radius and mass times velocity. The cross product, in this case, is a non-issue because the sin theta = 1.

I did the resulting math and got the same (within rounding error) result as I got multiplying the answers (r times v).

Finally, the square root of qB(vr)/m gave me my velocity... from which getting radius is no problem.

Is that the easiest way to do this? It's been about 8 years since I took physics 1, and some of the stuff I should "know," I simply don't.

Thanks for your reply and the welcome.

Munky
 

tiny-tim

Science Advisor
Homework Helper
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Thanks for the response, cryptic as it seemed at first, it helped.
We try not to give away too much, so that you can get the answer yourself! :biggrin:
Finally, the square root of qB(vr)/m gave me my velocity... from which getting radius is no problem.

Is that the easiest way to do this?
Yup … √qB(vr)/m was what I meant … definitely the easiest way. :smile:
 

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