A proton (charge `=1*6xx10^(-19)C`, mass `m=1*67xx10^(-27)kg`) is shot with a speed `8xx10^6ms^-1` at an angle of `30^@` with the x-axis. A uniform magnetic field `B=0*30T` exists along the x-axis. Show that the path of the proton is helix and find the radius of the helix.

Answer

`13*92cm`

Explanation

Component velocity of proton along x-axis will be
`v_x=v cos 30^@=8xx10^6xxsqrt3/2`
`=6*93xx10^6ms^-1`.
Component velocity of proton along y-axis will be
`v_y=vsin 30^@=8xx10^6xx1/2=4xx10^6ms^-1`
Since the angle between velocity component `v_x` and magnetic field B is `0^@`, therefore magnetic force on proton due to component velocity `v_x` will be
`F=qv_xBsin0^@=0`.
Thus the proton will move uniformly along x-axis. As the component velocity of proton along y-axis, `v_y` is perpendicular to the direction of magnetic field, therefore, the magnetic force on proton
`F=qv_yBsin90^@=qv_yB`
will acts as centripetal force and proton will describe a circular path due to this component velocity. As the proton covers linear distance as well as describes a circular path, hence the path of the proton will be helix. Radius of the helix will be given by
`qv_yB=mv_y^2//r`
or `r=(mv_y)/(qB)=(1*67xx10^(-27)xx4xx10^6)/(1*6xx10^(-19)xx0*30)`
`=13*92xx10^-2m=13*92cm`