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. 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)`
` = 28 xx 10^(-2) m = 28 cm`
v `= v//(2pi r) = 17 xx 10^6 s^(-1) = 17 xx 10^6 Hz = 17 MHz`
`E = (1//2) mv^2 = (1//2) 9 xx 10^(-31) kg xx 9 xx 10^14 m^2 // s62 = 40.5 xx 10^(-17) J`
`= 4 xx 10^(-16) J = 2.5 keV. `