What is the radius of the path of an electron (mass `9 xx 10^(-31)` kg and charge `1.6 xx 10^(-19)C)` moving at a speed of `3 xx 10^(7)m//s` in a magnetic field of `6 xx 10^(-4)T` perpendicular to it? What is its frequency? Calculate its energy in `keV.(1eV = 1.6 xx 10^(-19)J)`.

Explanation

Using Eq. (4.5) we find
`r = mv//(qB) = 9 xx 10^(-31) kg xx 3 xx 10^(7)ms^(-1)//(1.6 xx 10^(-19)C xx 6 xx 10^(-4)T)`
`= 26 xx 10^(-2)m = 26 cm`
`v = v//(2pir) = 2 xx 10^(6)s^(-1) = 2xx 10^(6)Hz = 2 MHz`.
`E = (½) mv^(2) = (½) 9 xx 10^(-31)kg xx 9 xx 10^(14)m^(2)//s^(2) = 40.5 xx 10^(-17)J`
`~~4 xx 10^(-16)J = 2.5 keV`.