A monoenergetic electron beam with electron speed of `5.20 xx10^(6) "m s"^(-1)` is subject to a magnetic field of ` 1.30 xx 10^(-4)` T normal to the beam velocity . What is the radius of the circle traced by the beam , given e/m for electron equals ` 1.76 xx10^(11) " C kg"^(-1)`

Explanation

Given , `v = 5.20 xx 10^(6) ` m/s , `B = 1.30 xx 10^(-4) T , e/m = 1.76 xx 10^(11) ` C/kg
Since centripetal force is balanced by Force due to magnetic field , `(mv^(2))/r = Bv`
` [ :' (vec(v)xx vec (B))= e v B sin 90^(@) = Bev] `
` r = (mv)/(eB) = v /((e/m)xxm) = ( 5.20 xx10^(6))/(1.76 xx10^(11) xx 1.30 xx10^(-4)) = 0.227 m = 22.7 ` cm