A charged particle of unit mass and unit charge moves with velocity of `v=(8hati+6hatj)m//s` in a magnetis field of `B=2hatKT. Then

Options

A. The path of the particle may be `x^2+y^2-4x-21=0`
B. The path of the particle may be `x^2+y^2=25`
C. The path of the particle may be `y^2+z^2=25`
D. The time period of the particle will be 3.14s

Answer

(a,b,d)

Explanation

`|vecv|=sqrt(8^2+6^2)=10ms^-1`
Radius of circular path `r=(mv)/(qB)=10/2=5m`
As the magnetic field is in z-direction, hence the path of the
charged particle will be circular of radius 5m in x-y plane, hence
option (a) and (b) may be possible.
Time period of circular path `T=(2pir)/v=(2pixx5)/10=pi` seconds.