Relativistic particle in magnetic field

The equation is valid, but the relativistic mass is multiplied by the speed of light squared to get the corrected mass.f
  • #1

Homework Statement

This is just a short response problem on my homework. It asks:
"If a particle is moving at a relativistic velocity, is the following equation still valid?
|q|vB = mv2/r
If the particle is an electron, what is the value of the mass?" (electron mass = 9.1094x10-31)

Homework Equations

|q|vB = mv2/r

The Attempt at a Solution

I would think that the equation still applies. Velocity is relative so it does not change when the frame of reference changes. Is that correct? Would I also just use the given electron mass for the mass? Maybe I'm thinking too much... Would .511MeV/c2 work too?

One side question. Would the radius be a contracted length? How would that be calculated? Any help is greatly appreciated. Thank you so much!
  • #2
This is a good question, I am not sure of the answer but I can tell you what I think. I think that the radius would be contracted, but the faster the particle goes the larger the radius.
  • #3
It's been a while, Noon, but I would say the equation is valid for relativistic velocity as long as the "m" in the equation is the relativistic mass. m = m0 + m0/sqrt[(1-vsquared/csquared)].
  • #4
I've been thinking. Does this make sense?

|q|vB = mv2/r
|q|vB = pv/r ...because mv = p
|q|vB = (ymc2)v/r ...p = ymc2 (y = gamma)
r = ymc2/(|q|B)

Would that work? Does that make sense?
  • #5
doesn't seem so. I don't see where you explicitely call out the rest mass when expanding the relativistic mass. try it with without using the momentum and get the equation in terms of q, v, B, m0, c and r. then solve for "r" if that is what you have to do. I have solved for r but will not add it so as not to "give you the answer"...also, I could be wrong since it's been a while...haha
  • #6
I also wonder what the other side of the equation will be for very high V. I know that relativistic effects can affect how charges "see" electric and magnetic fields, but I really can't address that since all my old e&m and advanced e&m texts are at home...good luck, Noon.