A proton of mass `1 . 67 xx 10^(-27)` kg and charge `1 . 6 xx 10^(-19)` C is projected with a speed of `2 xx 10^(6)` m/s at an anlge 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 `pi xx 10^(-7)` s.
B. A circle of radius = 0.1 m and time period `2pi xx 10^(-7)` s.
C. A helix of radius =0.1 m and time period `2 pi xx 10^(-7)` s
D. A helix of radius =0.2 m and time period `4 pi xx 10^(-7)` s

Answer

C

Explanation

Path of the proton will be a helix of radius
` r = (mv sin theta)/(qB)`
[where `theta ` = Angle between `vec(B) and vec(v)` ]
`rArr r = (1.67 xx 10^(-27) xx 2 xx 10^(6) xx sin 30^(@))/(1.6 xx 10^(-19) xx 0.104)`
= 0.1 m
Time period T ` = (2 pi m)/(qB) = (2 pi xx 1.67 xx 10^(-27))/( 1.6 xx 10^(-19) xx 0.104)`
`= 2 pi xx 10^(-7) s`