A monoenergetic electron beam with electron speed of `5.20xx10^(6) ms^(-1)` is subject to a magnetic field of `1.30xx10^(-4)T`normal to the beam velocity. What is the radius of the circle traced by the beam? Given `e//m = 1.76 xx 10^(11) kg^(-1)`.

Explanation

`(mcancel(v^(2)))/R = Be cancel(v) sin 90^(@)`
`R = (mv)/(Be) = (9.1xx10^(-31)xx5.2xx10^(6))/(1.3xx10^(-4)xx1.6xx10^(-19))`
i.e., `R = 22.75 xx 10^(-2)m`
`R = 22.75 ` cm
(ii) `R = (mv)/(Besqrt(1-v^(2)/c^(2)))=(22.75xx10^(-2))/sqrt(1-v^(2)/c^(2))` where `v/c le 1` with relativistic mass correction.
`v/c = (5.2xx10^(6))/(3xx10^(8))=1.73xx10^(-2) and sqrt(1-v^(2)/c^(2))=sqrt(1-0.003) = 0.99`
`R=(22.75xx10^(-2))/0.99=22.9797xx10^(-2)m.`