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Cyclotron Motion

  1. Feb 17, 2010 #1
    1. The problem statement, all variables and given/known data

    Given the equation of motion for charged particle in a magnetic field, with mass m and charge q moves in the y-z plane under the influence of a uniform magnetic field [tex]\hat{B}=B_{0}\hat{x}[/tex] is given by:

    [tex]m\frac{d\hat{v}}{dt}=q\hat{v}\times\hat{B}[/tex]

    Find the solution to the equation of motion, the cyclotron motion.

    (This isn't the full problem, I am stuck on this little part of it)

    2. Relevant equations



    3. The attempt at a solution

    so what i have done is treat the vector cross product as a multiplication sign. I have separated the variables and integrated:

    [tex]\int\frac{d\hat{v}}{\hat{v}}=\frac{qB_{0}}{m}dt[/tex]

    Upon integration, you get an exponential velocity, which makes no sense at all? I am not sure how else you'd find the cyclotron velocity solution?
     
    Last edited: Feb 17, 2010
  2. jcsd
  3. Feb 17, 2010 #2

    Physics Monkey

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    Hi ian2012,

    As you've discovered yourself, it is not correct to treat the cross product as simple multiplication. In fact, you should be able to show in this problem that the acceleration is orthogonal to the velocity. How familiar are you with the cross product? The most basic way to approach this problem is to write the equation of motion in coordinates rather than sticking with the vector notation.
     
  4. Feb 17, 2010 #3
    I am familiar with the cross product. If you compute the cross product with the direction of the field B and the velocity in the y-z plane, you will get:
    [tex]\hat{v}\times\hat{B}=\hat{v}B_{0}[/tex]
    But then I am again integrating the same expression.
     
  5. Feb 17, 2010 #4

    Physics Monkey

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    Sorry, I don't know your conventions, but if you take B to point in the x direction and v to lie in the y-z plane, then the cross product is not proportional to v. In fact, it is orthogonal to v, as I said earlier [tex] \vec{v} \cdot ( \vec{v} \times \vec{B} ) = 0 [/tex]. Physically, the magnetic force should be perpendicular to the velocity.
     
  6. Feb 17, 2010 #5
    Sorry, my hats represent vectors. I beg your pardon, if we say:
    [tex]\vec{v}=(0, v_{0}, v_{0})[/tex]

    then after taking the cross product, we get:
    [tex]m\frac{d\vec{v}}{dt}=qB_{0}(0, v_{0}, -v_{0})[/tex]

    I then did:
    [tex]m\frac{d}{dt}(0, v_{0}, v_{0})}=qB_{0}(0, v_{0}, -v_{0})[/tex]

    I split the problem up into the y, z component DE's:
    [tex]dv_{0y}=\frac{qB_{0}}{m}v_{0y}dt[/tex]
    [tex]dv_{0z}=-\frac{qB_{0}}{m}v_{0z}dt[/tex]

    Therefore the solution is:
    [tex]\vec{v}=(0, v_{0y}, v_{0z})=(0, C_{1}e^{\frac{qB_{0}t}{m}}, -C_{2}e^{-\frac{qB_{0}t}{m}})[/tex]

    Don't know if this is correct?
     
    Last edited: Feb 17, 2010
  7. Feb 17, 2010 #6

    Physics Monkey

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    You're almost there, you just have some confusion about the velocity components. Write [tex] \vec{v} = (0, v_y, v_z) [/tex] and then do the cross product. You should find that the equations of motion are slightly altered from what you wrote.
     
  8. Feb 17, 2010 #7
    But then how do you integrate the expressions:
    [tex]\frac{dv_{y}}{dt}=\frac{qB_{0}}{m}v_{z}[/tex]
    [tex]\frac{dv_{z}}{dt}=-\frac{qB_{0}}{m}v_{y}[/tex]

    you can't integrate v_{z} w.r.t dv_{y} and visa versa?
     
  9. Feb 17, 2010 #8

    Physics Monkey

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    Try differentiating the first equation with respect to time and using the second equation to produce an equation containing only one velocity component.
     
  10. Feb 17, 2010 #9
    I managed to get:
    [tex]v_{y}=c1sin\Omega t+c2cos\Omega t[/tex]
    [tex]v_{z}=c1cos\Omega t-c2sin\Omega t[/tex]

    [tex]\Omega = \frac{qB_{0}}{m}[/tex]
     
  11. Feb 17, 2010 #10

    Physics Monkey

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    Looks good to me.
     
  12. Feb 18, 2010 #11
    Oh, i just realised, i don't have to calculate the cross product. I found an easier way to solve my problem. Thanks for your help!
     
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