Motion of Charged Particle in The Magnetic Field

  • As we have mentioned earlier magnetic force F=(vXB) does not do any work on the particle as it is perpendicular to the velocity.
  • Hence magnetic force does not cause any change in kinetic energy or speed of the particle.
  • Let us consider there is a uniform magnetic field B perpendicular to the plane of paper and directed in downward direction and is indicated by the symbol C in figure shown below.

  • Now a charge particle +q is projected with a velocity v to the magnetic field at point O with velocity v directed perpendicular to the magnetic field.
  • Magnetic force acting on the particle is

    F=q(v X B) = qvBsinθ
    Since v is perpendicular to B i.e., angle between v and B is θ=90 Thus charged particle at point O is acted upon by the force of magnitude
    and the direction of force would be perpendicular to both v and B
  • Since the force f is perpendicular to the velocity, it would not change the magnitude of the velocity and the peffect of this force is only to change the direction of the velocity.
  • Thus under the action of the magnetic force of the particle will more along the circle perpendicular to the field.
  • Therefore the charged particle describe an anticlockwise circular path with constant speed v and here magnetic force work as centripetal force. Thus

    where radius of the circular path traversed by the particle in the magnetic in field B is given as
    r=mv/qB                   ---(5)
    thus radius of the path is proportional to the momentum mv pof the charged particle.
  • 2πr is the distance traveled by the particle in one revolution and the period T of the complete revolution is

    T=2 πr /v
    From equation(5)
    time period T is
    T=2πm/qB                   (6)
    and the frequency of the particle is f=1/T=qB/2πm                   (7)
  • From equation (6) and (7) we see that both time period and frequency does not dependent on the velocity of the moving charged particle.
  • Increasing the speed of the charged particle would result in the increace in the radius of the circle. So that time taken to complete one revolution would remains same.
  • If the moving charged particle exerts the magnetic field in such a that velocity v of particle makes an angle θ with the magnetic field then we can resolve the velocity in two components

    vparallel : Compenents of the velocity parallel to field
    vperpendicular :component of velocity perpendicular to magnetic field B
  • The component vpar would remain unchanged as magnetic force is perpendicular to it.
  • In the plane perpendicular to the field the particle travels in a helical path. Radius of the circular path of the helex is

    r=mvperpendicular/qB=mvsinθ/qB                   (8)
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