A beam of proton with a velocity of `4xx10^(5)ms^(-1)` enters a uniform magnetic field of 0.3 T at an angle of `60^(@)` to the magnetic field/The radius of helical path taken by proton beam is

Explanation

Velocity component along the field `v_(||)=4xx10^(5)xxcos60^(@)`
`=2xx10^(5)m//s`
and `v_(_|_)=(4xx10^(5))sin60^(@)=2sqrt(3)xx10^(5)`m/s
Proton will describe a circle in plane perpendicular to magnetic field with radius
`r=(mv_(_|_))/(qB)=((1.67xx10^(-27)kg)xx(2sqrt(3)xx10^(5)m//s))/((1.6xx10^(-19)C)xx(0.3T))`
`=1.2cm`
Time taken to complete one revolution is
`T=(2pir)/(v_(_|_))=(2xx3.14xx0.012)/(2sqrt(3)xx10^(5))`
Because of `v_(||)` protons will also move in the direction of magnetic field.
Pitch of helix `v_(||)xxT`
`=(2xx10^(5)xx2xx3.14xx0.012)/(2sqrt(3)xx10^(5))m`
`=0.044m`
`=4.4cm`