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

The radius of helical path `r=(mv_(bot))/(Bq)`
`r=(mv sin theta)/(Bq)=((1.67xx10^-27)(4xx10^5)(sin60^@))/((0.3)(1.6xx10^-19))`
`=1.2xx10^-2m`
Pitch of helical path `p=v_(||)xxT`
`p=((2pim)/(Bq))(v cos theta)`
`=((2pi)(1.67xx10^-27)(4xx10^5)(sin60^@))/((0.3)(1.6xx10^-19))=4.37xx10^-2m`