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

Explanation

Consider a charged particle of mass m and charge q entering a uniform magnetic field `vecB` with a velocity `vecv` along an angle `theta` with the direction of magnetic field in the plane of peper .
Velocity `vecv` may be resolved into two components(i) `v cos theta` along the magnetic field, and (ii) `v sin theta` perpendicular to the magnetic field as shown in fig. Obviously due to normal component `v sin theta` the charged particle will ecperience a force `F = q(v sin theta) B` along a direction perpendicular to `vecB` as well as `(v sin theta)` i.e, force F is along the z-axis. Under its effect, the charged particle describe a circular path of radius r, such that
`B q v sin theta = (m (v sin theta)^2)/(r)`
`implies r = (m v sin theta)/(B q) " " ......(i)`
The time period of revolution will be given by
`T = (2 pi r)/((v sin theta)) = (2 pi m)/(B q) " " .......(ii)`
or Frequency of revolution `v = 1/T = (Bq)/(2 pi m) " "........(iii)`
Due to component `v cos theta` the charged particle does not experience force due to magnetic field and tends to move linearly with a constant speed. Thus, under the combined effect of both the velocity componetns we can say that the charged particle will describe a helical path having its axis parallel to magnetic field.
In one complete revolution of its circular path the charged particle covers a linear distance along `vecB`, which is called as "pitch" of helix and is denoted by p. Obviously
`p = (v cos theta) T = (2 pi m v cos theta)/(Bq)`