Electron in a magnetic field and angular momentum

In summary, the conversation discusses an electron moving in a circular path perpendicular to a constant magnetic field. The angular momentum of the electron is given and the goal is to solve for the velocity and radius. The equation qvB=mv^2/r is used and the square root of qB(vr)/m is found to get the velocity, from which the radius can be easily calculated.
  • #1
Munky
8
0

Homework Statement


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


Homework Equations



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


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
 
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  • #2
Welcome to PF!

Hi Munky ! Welcome to PF!
Munky said:
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:
 
  • #3
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
 
  • #4
Munky said:
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:
 

1. What is the direction of the angular momentum of an electron in a magnetic field?

The direction of the angular momentum of an electron in a magnetic field is perpendicular to both the direction of the electron's motion and the direction of the magnetic field.

2. How does a magnetic field affect the path of an electron?

A magnetic field can cause an electron to follow a curved path, as the force exerted by the magnetic field acts as a centripetal force, causing the electron to move in a circular motion.

3. How does the strength of the magnetic field affect the angular momentum of an electron?

The strength of the magnetic field directly affects the magnitude of the electron's angular momentum. A stronger magnetic field will result in a larger angular momentum for the electron.

4. Can the angular momentum of an electron in a magnetic field change?

Yes, the angular momentum of an electron in a magnetic field can change if the strength or direction of the magnetic field changes, or if the electron's speed or direction of motion changes.

5. What is the relationship between the magnetic moment and the angular momentum of an electron?

The magnetic moment of an electron is directly proportional to its angular momentum, with the constant of proportionality being the gyromagnetic ratio. This means that an increase in angular momentum will result in an increase in magnetic moment, and vice versa.

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