# Motion of a charged particle in uniform magnetic field

In this topic Motion of a charged particle in uniform magnetic field we will discuss about how a charged particle moves in uniform magnetic field, when it is projected with some velocity at an angle with magnetic field.

Motion of a charged particle in uniform magnetic field

Suppose a charged particle of mass ‘m’ and charge ‘q’ is projected  with velocity ‘v’ at angle ‘ϴ’ with the magnetic field  ‘B’ as shown in figure (a) . Here v has two components  (v1= vcosϴ ) along x -axis and ( v2= v sinϴ) along y-axis.

Due to v1 it moves along x-axis and due to v2 a force F acts which is given as F = qv2 B = qvBsinϴ. Since this force F acts perpendicular to the velocity and magnitude of v2 not changes hence it moves on a circular path.  As shown in fig(b).

let r is the radius of the circular path , due to these two components of v1 and v2 particle follows a helical path as shown in fig(a) .

As we know when body moves on a circular path the centripetal forc F = mv22/r = mv2sin2ϴ/r ;

Which is provided by magnetic force F= qvBsinϴ

So we can write  mv2sin2ϴ/r = qvBsinϴ;

So, r = mvsinϴ/qB ……………….(1)

or vsinϴ=qBr/m …………………….(2)

angular velocity  ω= vsinϴ/r = Bqr/mr = Bq/m ……………..(3)

here frequency  f=ω/2∏ = Bq/2∏m  ( here frequency is independent of velocity)………….(4)

and time period T = 1/f = 2∏m/Bq ……………………………(5).

The pitch of the helix = vcosϴ x T = v cosϴ 2∏m/Bq .

Special cases –

Case (i) -If ϴ=00 i.e. v1=v and v2=0

So there will be no force and particle will move in the direction of B .

Case (ii) – if ϴ=900 then v1=0 and v2=v  i.e. body will move only in the circular path

And radius of the circular path  r= mv/qB  or  r= v/(q/m)B .