A proton of mass `1.67xx10^(-27)kg` and charge `1.6xx10^(-19)c` is projected with a speed of `2xx10^(6)m//s` at an angle of `60^(@)` to the x-axis. If a uniform magnetic field of `0.104` Tesla is applied along Y-axis, the path of proton is

Options

A. a circle of radius `~=0.1m` and time perio `pi xx 10^(-7)s`
B. a circle of radius `~=0.2m` and time period `2pixx10^(-7)s`
C. a helix of radius `~=0.1m` and time period `2pixx10^(-7)s`
D. a helix of radius `~=0.2m` and time period `4pixx10^(-7)s`

Answer

C

Explanation

`v_(_|_)=v cos 60`, find `r=(mv_(_|_))/(qB),r=0.1m`
Hence `t=(2pim)/(qB)=2pixx10^(-7)s`. Hence answer is `(c )` and path is helix.