A uniform magnetic field B is set up along the positive x-axis. A particle of charge 'q' and mass 'm' moving with a velocity v enters the field at the origin in XY-plane such that it has velocity components both along and perpendicular to the magnetic field B. Trace, giving reason, the trajectory followed by the particle. Find out the expression for the distance moved by the particle along the magnetic field in one rotation.

Let angle made by the particle velocity with the magnetic field initially be θ.
Component of velocity perpendicular to the magnetic field will bev⊥=v sin(θ)
Component of velocity parallel to the magnetic field will bev∥=v cos(θ)
It will face a centripetal force due to the magnetic field which is perpendicular to both the magnetic field and v⊥. This will give the particle a circular motion.
However, the field has no effect on v∥ due to which this component will remain constant.
Hence, the trajectory of the particle in the field will be helical as shown in the attached figure.
Centripetal force is given by:
F=mv⊥²/R=qv⊥B
R=mv⊥/qB
Time period is given by:
T=2πR/v⊥
T=2πm/qB
Pitch is equal to the distance moved by the particle along the magnetic field in one time period.
Pitch=v∥T=2πmv cos(θ)/qB