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

Answer

C

Explanation

path of the proton will be a helix of radius `r=(mvsin theta)/(qB)`
(where `theata`=Angle between `vecB` and `vecv`)

`implies r=(1.67xx10^(-27)xx2xx10^(6)xxsin30^(@))/(1.6xx10^(-19)xx0.104)`
`=0.1m`
Time period `T=(2pim)/(qB)=(2pixx1.67xx10^(-27))/(1.6xx10^(-19))xx0.104`
`=2pixx10^(-7)sec`