Trajectory of charged particles in a uniform magnetic field

  1. 1) An electron and a proton are injected into a uniform magnetic field at right angles to the direction of the field with the same Kinetic Energy. Then:
    a)the electron trajectory will be less curved than the proton trajectory
    b)the proton trajectory will be less curved than the electron trajectory
    c)both the trajectories will be equally curved
    d)both the trajectories will be straight

    My reasoning is as follows:
    Let –e be the charge of electron, +e be the charge of proton and B be the uniform magnetic field. Let mass of proton be Mp and mass of electron be Me. Let v1 and v2 be the velocity of electron and proton respectively.
    We know that Mp > Me
    We know that a charged particle describes a circular path in a perpendicular magnetic field.
    So, Bqv = (mv^2)/r
    r = (mv^2)/(Bqv)
    = {(1/2)mv^2}/{(1/2)Bqv}
    Since kinetic energy, magnetic field and charge are same for electron and proton,
    r proportional to 1/v
    Now, K.E of electron = K.E of proton
    (1/2)(Me)(v1^2) = (1/2)(Mp)(v2^2)
    (v1/v2)^2 = Mp/Me
    Since Mp > Me, v1>v2
    Now, r1/r2 = v2/v1
    Since v2<v1, r1<r2
    So, radius of circular path of electron is less than radius of circular path of proton. So the electron path is more curved that proton path because curvature is the reciprocal of radius of curvature. But the book answer is a)the electron trajectory will be less curved than the proton trajectory. Please help!
  2. jcsd
  3. siddharth

    siddharth 1,197
    Homework Helper
    Gold Member

    I can find nothing wrong with what you did.

    Your book once again seems to be wrong. I still think it's high time you got a better book.
  4. Hootenanny

    Hootenanny 9,681
    Staff Emeritus
    Science Advisor
    Gold Member

    I will agree with siddharth, the answer you gave is correct; as I have said before, your text does not give one the utmost confidence when it is riddled with simple errors.
  5. I think the book answer meant the radius of curvature. Thanks for the assistance.
    Last edited: Jul 25, 2006
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