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*2m` and time period `pixx10^-7s`
B. a circle of radius `=0*1m` and time period `2pixx10^-7s`
C. a helix of radius `=0*1m` and time period `2pixx10^-7s`
D. a helix of radius `=0*2m` and time period `4pixx10^-7s`

Answer

C

Explanation

Since the proton is entering the magnetic field at some angle other than `90^@`, its path is helix,
Component velocity of proton along X-axis, figure.
`v_x=vcos60^@=2xx10^6xx1//2=10^6ms^-1`.

Due to component velocity `v_x`, th radius of the helix is described and is given by the relation
`r=m/q(v_x)/(B)=(1*67xx10^(-27)xx10^6)/(1*6xx10^(-19)xx0*104)=0*1m`
Now `T=(2pir)/(v_x)=(2pixx0*1)/(10^6)=2pixx10^-7s`