A proton of mass `1.67xx10^(-27)` kg charge `1.6xx10^(-19) C` is projected in xy-plane with a speed of `2xx10^(6) m//s` at an angle of `60^(@)` to the X-axis. If a uniform magnetic field of 0.14 T is applies along the Y-axis, then the path of the 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.
Corresponding velocity of the proton along X-axis,
`v_x=vcos 60^@=2xx10^6xx1/2=10^6ms^-1`
Due to velocity component `v_x`, the radius of the helix is
described and is given by the relation
`r=(mv_x)/(qB) =(1.6xx10^-27xx10^6)/(1.6xx10^-19xx0.10)=0.1m`
Now, `T=(2pir)/(v_x) =(2pixx0.1)/(10^6)=2pixx10^-7s`