Question

A particle of charge q moving through a magnetic field B experiences the force where v is the velocity of the particle. Show that: (i) The force does no work in moving the particle along its path from t = 0 to t=T. (i1) The speed v(t) Iv(O)ll of the particle is constant in time e particle 1s constant in time

Answer 1

Answer:

       dx/Dt x B . x  =0

Explanation:

Let's calculate the work and the magnetic force, the expression for magnetic force is

        F = qv x B

Bold indicate vector quantities, the expression for the job is

         W = F. X

Let's replace in this equation

       W = q v x B . X

The definition of speed is

      v =  dX / dt

With what work is left

     W = q dX / dt x B . X

As we can see the vector product gives us a vector perpendicular to dX and its scalar product by X of zero

Second part

   The speed a vector and although the magnitude is constant the change of direction implies a change in the speed.

   Let's calculate the magnitudes of speed (speed)

       F = qv B sin θ

       F = ma

       q v B sin θ = ma

       a = qvB / m senT

     

This acceleration is perpendicular to the magnetic field and the velocity, so it does not change if magnitude but its direction, it is directed to the center of the circle.

   | v | = q vB/m sin θ