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Homework Help: Motion of charged particle in magnitc field given by potential of magnetic dipole

  1. Jan 2, 2010 #1
    Find the fi rst integrals of motion for a particle of mass m and charge q in a magnetic field given by the vector potential (scalar potential [tex]\Phi[/tex]= 0)

    (i) of a constant magnetic dipole [tex]m_{d}[/tex]

    [tex]A=\frac{\mu_{0}}{4 pi}\frac{m_{d} \times r}{r^{3}}[/tex]

    Hint: Cylindrical coordinates are useful.


    I think what i should do is to compute A for cylindrical coordinate system and then use Lagrangian mechanics to get a equation of motion? Is this correct? (we have the charge q given, so we can use the kinetic engergy?)

    I tried to compute A but i dont really understand what to do with the magnetic dipole (as a vector)? Whats the story whith that scalar potential?

    Thanks for your help,
    Mumba
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Jan 2, 2010 #2

    diazona

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    Homework Helper

    That would probably be my first thought as well. But I wouldn't write A in cylindrical coordinates just yet. Try to go as far as you can without expressing it in any particular coordinate system.
    [itex]\vec{m}_d[/itex] is just a given vector. You don't really do anything with it, except carry it through the calculation. And I think they just tell you that the scalar potential is zero to clarify the conditions of the problem... I can't imagine why you would think that the scalar potential would be nonzero, unless you were specifically told so.
     
  4. Jan 3, 2010 #3
    Hi Thanks for the answer.

    But i cant go very far without a coordinate system, can I?
    I mean, using cyl. coordinates, my degrees of freedom would be just R and [tex]\Theta[/tex].

    So i can get the components of r=(Rcos[tex]\Theta[/tex], Rsin[tex]\Theta[/tex], z), where the z-axis is pointing upwards and [tex]\Theta[/tex] the angle between the x-axis (pointing towards you) and R.

    What can i do with [tex]m_{d}[/tex]? Should be a vector so i cant really just set it on the origin?! Can i give it an arbitrary direction, say [tex]m_{d}=(m_{d},0,0)[/tex]?

    Thanks,
    Mumba
     
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