Velocity and acceleration vector of a charged particle moving in a magnetic field at some instant are `vecv=3hati+4hatj and veca=2hati+xhatj`. Select the correct options.

Options

A. `x=-1*5`
B. `x=3`
C. magnetic field is along z-direction
D. K.E. of particle is constant

Answer

A, C, D

Explanation

(a) Since magnetic force `(F_m)` is always perpendicular to the velocity vector `vecv`, i.e.,
`vecF_m_|_vecv` or `veca_|_vecv` or `veca.vecv=0`
or `(2hati+xhatj).(3hati+4hatj)=0`, `x=-1*5`.
(c) Since `vecB` is `_|_` to the plane containing the velocity vector, in this case (XY-plane, `vecB` is along Z-direction)
(d) Since work done by magnetic force is zero, So K.E. of particle remains constant.