1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Special relativity, circular motion

  1. Aug 16, 2011 #1
    1. The problem statement, all variables and given/known data
    A charged particle (mass [itex]m[/itex], charge [itex]q[/itex]) is moving with constant speed [itex]v[/itex]. A magnetic field [itex]\vec{B}[/itex] is perpendicular to the velocity of the particle. Find the strength of the field required to hold the particle on a circular orbit of radius [itex]R[/itex].


    2. Relevant equations
    [itex]\vec{F} = q\vec{v} \times \vec{B}[/itex]
    [itex]\vec{F} = m\vec{a}_c[/itex]


    3. The attempt at a solution
    Well, I know that in the "classical" case this is fairly easy. One just sets

    [itex]qvB = ma[/itex],

    and since [itex]a = \frac{v^2}{R}[/itex], one gets

    [itex]qvB = m \frac{v^2}{R}[/itex]
    [itex] \Rightarrow B = \frac{mv}{qR}[/itex]

    However, I am not sure if I can use this here, because the particle is assumed to be travelling at close to the speed of light. I have read somewhere that I should use the relativistic mass in the calculation of the centripetal force, i.e.

    [itex]F = \frac{\gamma m v^2}{R}[/itex],

    but I am not sure why this is the case. Could anyone help?
     
    Last edited: Aug 16, 2011
  2. jcsd
  3. Aug 17, 2011 #2
    One way to see this is go into the momentarily comoving frame (not an inertial frame, obviously) of the particle. In this frame, the particle is at rest so the Lorentz force only comes from the electric field in this frame, which is: [itex]q E^{\prime} = \gamma q \mathbf{v} \times \mathbf{B}[/itex]
    So it experiences a force which is perpendicular to both [itex]\mathbf{v}[/itex] and [itex]\mathbf{B}[/itex] with the [itex]\gamma[/itex] as promised.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Special relativity, circular motion
Loading...