Question

When a charged particle moves in a magnetic field its kinetic energy is always constant explain?

Answer 1

hey, there's your answer.
This is one of the most elementary, at the same time the most interesting question.

The answer however is both simple and amusing, lets analyze!

Magnetic force is always perpendicular to its velocity, hence the B field does no work! Due to this being said the KE of the charged body remains constant.

F=ma=q(vxB)F=ma=q(vxB)

let us consider a system of an atom which is in a uniform magnetic field setup

now acceleration,a=dv/dta=dv/dt, lets manipulate this a bit…

dv/dt=(dv/dt)∗(dx/dx)dv/dt=(dv/dt)∗(dx/dx), now, rearranging we get

a=(dv/dx)∗(dx/dt)a=(dv/dx)∗(dx/dt), where the second term corresponds to the time rate of chance of position which is velocity.

hence we obtain: m(dv/dx)∗v=qBv(sinalpha)m(dv/dx)∗v=qBv(sinalpha)

we obtain upon integrating with limits for x ranging from the first bohr radius b(1) to b(2), v=Bq[b(2)−b(1)]/mv=Bq[b(2)−b(1)]/m

Hence, KE=m(v2)/2willturnouttobeKE=m(v2)/2willturnouttobe

KE= Bq[b(2)-b(1)]/2m= constant

So the kinetic energy of a charged body in a uniform B field is constant.
hope you got your answer.....