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Particle Motion in a Magnetic field

  1. Feb 17, 2017 #1
    1. The problem statement, all variables and given/known data
    Consider a charged particle entering a region of uniform magnetic field B - for example the earth's field. Determine it's subsequent motion when the y-axis is parallel with the magnetic field

    2. Relevant equations
    F=qv x B = ma
    vector components of velocity, acceleration, and magnetic field

    3. The attempt at a solution
    This isn't homework, it's an example from the book, but they skip many gory details that I can't seem to figure out, and the way we did it in class makes even less sense.

    To cut a long story short, eventually I get two equations:
    z'''=-α2z' and x'''=-α2x'
    Then the book says they use the technique's used in appendix C, which are the use of Auxiliary functions (or characteristic) and homogeneous equations, which we aren't using in class apparently, I'd like to know how they are used and the examples given in the back make sense, but they are all 2nd degree functions, not 3rd like mine.
    My attempt with the equations as they are:
    r32r=0
    which I can see 3 solutions from:
    r=0, r=±iα
    the second solution will yield the cos and sin functions I desire, but I'm missing a "t" that needs to be with the alpha in the trig functions and I have no idea how to get it.
    x=Acos(αt) + Bsin(αt) + x0
    z=A'cos(αt) + Bsin(αt) + z0 These two are the end results in the book
    I feel as if though I'm supposed to integrate the functions first which will give me the initial position constants, but I don't know where the "t" comes from.
     
  2. jcsd
  3. Feb 17, 2017 #2

    berkeman

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    Staff: Mentor

    Are those triple derivatives? Why?

    Can you just post a diagram of the situation, and show how the Lorentz force guides the motion of the charged particle?
     
  4. Feb 17, 2017 #3
    Here are the pics straight from the book, I thought this would be easier than typing it out, hopefully they turn out fine.
     

    Attached Files:

  5. Feb 17, 2017 #4

    berkeman

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    Staff: Mentor

    Yoiks. Can you scan them and Upload them?
     
  6. Feb 17, 2017 #5
    No sadly, I'll just type it out, give me a few mins
     
  7. Feb 17, 2017 #6

    berkeman

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    Staff: Mentor

    Thanks
     
  8. Feb 17, 2017 #7
    Alright so the cartesian coordinate system is chosen with the y-axis parallel to the magnetic field. If q is the charge on the particle, v it's velocity, "a" it's acceleration, and B the earth's magnetic field, then:
    v=x'i + y'j + z'k
    a = x''i + y''j + z''k
    B = Boj
    The magnetic force F=qv X B = ma, so
    m(x''i +y''j + z''k) = q(x'i + y'j +z'k) X Boj = qBo(x'k - z'i)​

    Equating like vector components gives:
    mx''=-qBoz'
    my''=0
    mz''=qBox'​
    Integrating the second of these equations, my''=0, yields:
    y'=yo'​
    where yois a constant and is the initial value of y'. Integrating a second time gives:
    y=yo't + yo
    where yois also a constant.

    To integrate the first and last equations (the x'' and z''), let α=qBo/m, so that
    x''=-αz'
    z''=αx'​
    These coupled, simultaneous differential equations can be easily uncoupled by differentiating one and substituting it into the other, giving:
    z'''=αx''=-α2z'
    x'''=-αz''=-α2x'​
    so that
    z'''=-α2z'
    x'''=-α2x'​

    Everything up to here makes complete sense, but it is the bridge between here and this next part that I can't figure out:

    Both of these differential equations have the same form of solution. Using auxiliary functions and homogeneous equations, we have:
    x=Acos(αt) + Bsin(αt) + xo
    z=A'cos(αt) + B'sin(αt) + zo
     
  9. Feb 17, 2017 #8
    Also I should add this isn't the final solution to the problem, it's just the step I'm having trouble understanding.
     
  10. Feb 17, 2017 #9

    James R

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    Science Advisor
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    Gold Member

    Conside the equation ##x'''=-\alpha^2 x'##. If we define ##u=x'## then we can re-write it as
    $$u''+\alpha^2 u=0.$$
    The general solution to this, as you're probably aware, is:
    $$u(t)=a\cos(\alpha t) + b\sin(\alpha t) = x'(t)$$
    Now, just integrate the solution with respect to time, and we get
    $$x(t) = A\cos(\alpha t)+B\sin(\alpha t) + x_0,$$
    where, for example, ##A=-b/\alpha##.

    Hope this helps.
     
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