Discussion
A relativistic particle of charge q and rest mass \(m_{0}\) moves with velocity \(\vec{v} = v_{x}\hat{x} + v_{y}\hat{y}\) in a uniform magnetic field, \(\vec{B} = B_{0}\hat{z}\). The relativistic Larmor radius R (the radius of the circular path the charged particle takes) is given by ..............
Answer: A
The radius is \(R = \frac{\gamma m_{0}\sqrt{v_{x}^{2} + v_{y}^{2}}}{qB_{0}}\).
The velocity lies in the plane perpendicular to \(\vec{B}\), so the particle moves on a circle. The magnetic force does no work, so the speed and \(\gamma\) stay constant, and the relativistic equation of motion is \(\gamma m_{0}\,d\vec{v}/dt = q\vec{v}\times\vec{B}\). Equating the force to the centripetal requirement:
\[ \begin{aligned} \frac{\gamma m_{0}v^{2}}{R} &= qvB_{0} \\ R &= \frac{\gamma m_{0}v}{qB_{0}} = \frac{p}{qB_{0}} \end{aligned} \]
with \(v = \sqrt{v_{x}^{2} + v_{y}^{2}}\). The radius is the relativistic momentum divided by \(qB_{0}\). Leaving out \(\gamma\) gives the non-relativistic radius; dividing by \(\gamma\) or using \(\gamma^{2}\) has no basis in the equation of motion.
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