A beam of protons with a velocity `4 xx 10^(5) m//sec` enters a uniform magnetic field of `0.4 T` at an angle of `37^(@)` to the magnetic field. Find the radius of the helium path taken by proton beam. Also find the pitch of helix.
`sin 37^(@) = 3//5 cos 37^(@) = 4//5`. `m_(p) ~= 1.6 xx 10^(-27) kg`.

Explanation

`R = (m v sin theta)/(B q)`
`= ((1.6 xx 10^(-27)) (4 xx 10^(5))(sin 37^(@)))/((0.4)(1.6 xx 10^(-19)))`
`= 10^(-2) xx 0.6 m`
`= 0.6 cm`
`T = (2 pi m)/(Bq) = (2pi (1.6 xx 10^(-27)))/(0.4(1.6 xx 10^(-19))) = 5pi xx 10^(-8) sec`
Pitch of helium path
`p = v cos theta T`
`= (4 xx 10^(5))(cos 37^(@))(5pi xx 10^(-8))`
`= 20pi xx 0.8 xx 10^(-3) = 1.6pi xx 10^(-2) m`
`= 1.6 pi cm`