A New Look at Momentum — Question 5

Question 5

A magnetic orbit measures momentum through curvature

A particle with charge magnitude q moves perpendicular to a uniform magnetic field B on a circular orbit of radius R. Radiation losses are negligible. The magnetic force magnitude is qvB. Use $\mathbf p=\gamma m\mathbf v$, $\gamma=(1-v^2/c^2)^{-1/2}$, and $\mathbf F=d\mathbf p/dt$. Here m is invariant mass.

1. Derive the momentum-radius relation and the orbital period.

2. At $v=0.60c$, compare the period with $2\pi m/(qB)$.

Question 5 - Solution

A magnetic orbit measures momentum through curvature

(a) Method and calculation

$$\left|\frac{d\mathbf p}{dt}\right|=\frac{pv}{R}=qvB\quad\Rightarrow\quad\boxed{p=qBR}.$$$$T=2\pi R/v=\boxed{\frac{2\pi\gamma m}{qB}}.$$

(b) Interpretation and further calculation

$$v=.60c:\quad\gamma=1.25.$$$$\boxed{T=1.25\,\frac{2\pi m}{qB}}.$$

Physical check

The field changes momentum direction while keeping its magnitude fixed. The period grows with gamma.