# Electron in a magnetic field and angular momentum

#### Munky

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

Homework Helper
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! #### Munky

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

Homework Helper
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! 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. ### The Physics Forums Way

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