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 magnetic field of `6 xx 10^(-4)` T perpendicular to it ? What is its frequency ? Calculate its energy in keV. `(1 eV = 1.6 xx 10^(-19) J)`

Explanation

Using Eq. (5) we find
`r=(mv)/((qB)) = (9 xx 10^(-3) kg xx 3 xx 10^(7) ms ^(-1))/((1.6 xx 1-^(19) C xx 6 xx 10^(-4)T))=26 xx 10^2 m = 26 m`
`v=(v)/((2pir))=2 xx 10^6 s^(-1)= 2 xx 10^6 Hz = 2 MHz`
`E= (1/2)mv^2 = (1/2) 9 xx 10^(31) kg xx 9 xx 10^(14) m^2//s^2 = 40.5 xx 10^(-17) J`
`~~ 4xx 10^(-16) J = 2.5 KeV`