Question

An electron is accelerated through a potential difference of 5 kV and enters a uniform magnetic field of 0.02 `(wb)/(m^2)` acting normal to the direction of electron motion.
Determine the radius of path. 

Answer 1

`v = sqrt(2qV)/m = sqrt(2 xx 1.6 xx 10^(−19) xx 5000)/(9.1 xx 10^(−31)) = 4193 xx 10^4`

`r = (mv)/(qB) =( 9.1 xx 10^(−31) xx  4193 xx 10^4)/( 1.6 xx 10^(−19) xx 0.02)  = 0.0119 m`

Radius of path = 0.0119 m