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 `= 0.2 m` & time period `pi xx 10^(-7) sec`
B. A circle of radius `= 0.1 m` & time period `2pi xx 10^(-7) sec`
C. A helix of radius `= 0.1 m` & time period `2pi xx 10^(-7) sec`
D. A helix of radius `= 0.2 m` & time period `4pi xx 10^(-7) sec`

Answer

C

Explanation

Due to `v cos theta`, particle moves in a circle of
Radius `R = (mv cos theta)/(Bq) = (1.67 xx 10^(-27) xx 2 xx 10^(6) xx cos 60^(@))/(0.014 xx 1.6 xx 10^(-19))`
With time period `T = (2pi m)/(Bq) = (2pi xx 1.67 xx 10^(-27))/(0.104 xx 1.6 xx 10^(-19))`
`= 2pi xx 10^(-7) sec`
Due to `v sin theta`, moves along the field, hence path will be helical.