A beam of electrons enters a uniform magnetic field of `0*3T` with a velocity of `4xx10^5ms^-1` at an angle of `60^@` to the field. Find the radius of the helical path taken by the beam. Also find the pitch of the helix (distance travelled by a proton parallel to the magnetic field during one period of rotation). Mass of proton is `1*67xx10^(-27)kg`.

Explanation

Here, `B=0*3T`, `v=4xx10^5ms^-1`,
`theta=60^@`
`q=1*6xx10^(-19)C` `m=1*67xx10^(-27)kg`
Components of the velocity of proton parallel and perpendicular to the direction of magnetic field are `v_(||)=v cos theta =4xx10^5xx cos 60^@`
`=(4xx10^5)xx(1//2)=2xx10^5ms^-1`
`v_|_=v sin theta=(4xx10^5)xx sin 60^@`
`=4xx10^5xx(sqrt3//2)=3*464xx10^5m//s`.
The proton moving with component velocity `v_(||)` will move along the magnetic field and with component velocity `v_(_|_)` will describe a circular path in the magnetic field.
Hence, the path of proton in the magnetic field is helical path.
The radius of the helical path is
`qv_(_|_)B=(mv_(_|_)^2)/(r)`
or `r=(mv_(_|_))/(qB)=((1*67xx10^(-27))xx(3*464xx10^5))/((1*6xx10^(-19))xx0*3)`
`=1*2xx10^-2m=1*2cm`
Period of revolution of electron is
`T=(2pir)/(v_(_|_))=(2xx3*14xx(1*2xx10^-2))/((3*464xx10^5))`
`=2*175xx10^-7s`
Pitch of the helix is
`p=v_(||)xxT=(2xx10^5)xx(2*175xx10^-7)`
`=4*35xx10^-2cm=4*35cm`