A charged particle of unit mass and unit charge moves with velocity `vecv=(8hati+6hatj)ms^-1` in magnetic field of `vecB=2hatkT`. Choose the correct alternative (s).

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

Here magnitude of velocity, `|vec(v)| = sqrt(8^(2)+ 6^(2))= 10 ms^(-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.
So, time period of circular path, `T= (2pi r)/(v)`
`=(2pi xx 5)/(10) = pi` seconds