A Circulating Charged Particle — Question 1

Question 1

The orbit radius changes with speed while the nonrelativistic period does not

A proton of mass $1.67\times10^{-27}$ kg and charge $1.60\times10^{-19}$ C enters a .400 T uniform field perpendicular to its velocity of $3.00\times10^6$ m/s. Neglect radiation and relativity.

1. Find its circular radius and orbital period.

2. If the speed doubles in a separate trial, find the new radius, period and kinetic-energy ratio.

Question 1 - Solution

The orbit radius changes with speed while the nonrelativistic period does not

(a) Method and calculation

$$qvB=mv^2/r\Rightarrow r=\frac{mv}{qB}=.0782813\,\mathrm m.$$$$T=\frac{2\pi r}{v}=\frac{2\pi m}{qB}=1.63952\times10^{-7}\,\mathrm s.$$

(b) Interpretation and further calculation

$$v\prime=2v\Rightarrow r\prime=2r=.156563\,\mathrm m,\quad T\prime=T.$$$$K\prime/K=(v\prime/v)^2=4.$$

Physical check

The larger orbit and faster motion increase path length and speed in the same proportion.