Radius of the circular path `R=(mv)/(qB)`
Angular velocity of the particle `omega=v/R=(qB)/m`
The speed of the particle will remain unchanged. The centre of circular path will be located on z-axis at point C. Let us take time interval t. during this time interval the particle will rotate an angel `theta=omegat=(qBt)/m`.
Let us draw the front view of the circular path and locate the instantaneous position of the particle.
Writing velocity vector at time t.
`vecv=v cos theta hatj-v sin theta hatk`
`vecv(t)=vcos ((qBt)/(m))hatj-v sin ((qBt)/(m))hatk`
Now writing y and z coordinates of the particle.
y-coordinate of `P=R sin theta`
z-coordinate of `P=-(R-R cos theta)`
`vecr(t)=r sin.(qBt)/(m) j-R(1-cos.(qBt)/(m))hatk`
[where `R=(mv)/(qB)`].