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Small angle approximation

  1. Aug 7, 2017 #1
    1. The problem statement, all variables and given/known data
    I want to solve the motion equation

    ## m \frac {dv_z} {dt} = - μ \frac {∂B_z} {∂z} ##
    with small angle approximation

    2. Relevant equations
    ## B_z(z) = B_0 -bCos(\frac {zπ} {2L}) ## is the magnetic field in the z-direction
    3. The attempt at a solution
    Started by derive the magnetic field equation with resp. z
    ## m \frac {∂B_z} {∂z} = \frac {bπ} {2L} * μ * sin(\frac {zπ} {2L}) ##
    and put it into eq above <=>

    ## m \frac {dv_z} {dt} = -μ \frac {bπ} {2L} * sin(\frac {zπ} {2L}) ##
    <=>
    small angles sin(x) ~ x
    ## \frac {dv_z} {dt} = - \frac μ m * \frac {bπ} {2L} * z ##

    ## v_z = \frac {dz} {dt} ## <=>

    ## \frac {d^2v_z} { dt^2} = - \frac {μb} {m} * \frac {μ} {2L} * z ##

    Im not sure if this is correct
     
  2. jcsd
  3. Aug 7, 2017 #2

    Charles Link

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    Homework Helper

    You dropped a factor of ## 2 L ## in the denominator, as well as ## \pi ## in the numerator, when you took the small angle approximation. (Also, your first equation should read ## \mu \frac{\partial{B}}{\partial{z}} ##, (instead of ## m ##), but that is clearly a "typo"). Your last equation needs to read ## \frac{d^2 z}{dt^2} ##, and please check your algebra on this. I believe you left off the ## \pi ##,(which should actually be ## \pi^2 ##), and you have an extra ## \mu ##. ## \\ ## Note: ## \sin(\frac{z \pi}{2 L})=\frac{z \pi}{2 L } ## for small angles. ## \\ ## Editing: Did they give you any initial conditions?=initial position and velocity? Otherwise, I guess you assume ## z=0 ## and ## v_z=0 ## at time ## t=0 ##. These initial conditions, though, would lead to the trivial solution ## z=0 ## for all ## t ## if I'm not mistaken.
     
    Last edited: Aug 7, 2017
  4. Aug 7, 2017 #3

    Ray Vickson

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    Science Advisor
    Homework Helper

    No, it is not. You dropped some factors, and your DE should be for ##d^2 z/dt^2##, not ##d^2 v_z / dt^2##.
     
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