A beam of proton with velocity `4xx10^(5)m*s^(-1)` centres a uniform magnetic field of `0.4T` at an angle of `60^(@)` to the magnetic field. Find the radius of the helical path of the proton beam and the time period of revolution. Also find the pitch of helix. Mass of proton `=1.67xx10^(-27)kg`.

Explanation

Mass of the proton `m=1.67xx10^(-27)kg`, velocity of the proton `v=4xx10^(5)m*s^(-1)`, charge `q=1.6xx10^(-19)-C`.

`:.` Component of velocity perpendicular to field is `v sin theta`, the radius of the helical path,
`r=(m v sin theta)/(qB)=((1.67xx10^(-27))(4xx10^(5)) sqrt(3))/((1.6xx10^(-19))xx0.4xx2)`
`=9xx10^(-3) m =0.9m`
Time period of revolution.
`T=(2pir)/(v sin theta)=( 2xx3.14xx0.009xx2)/(4xx10^(5)xx sqrt(3))=1.63xx10^(-7)s`
Pitch of the helix
`p= v cos thetaxxT=4xx10^(5)xx(1)/(2)xx1.63xx10^(-7)`
`3.26xx10^(-2)m=3.26 cm`