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Particle motion in a magnetic field

  1. Apr 7, 2013 #1
    1. The problem statement, all variables and given/known data
    Assume the earth's magnetic field is almost homogeneous with direction along the z-axis, with a small inhomogeneous modification which make the field lines converge towards the z-axis. Also ignore relativistic and gravitational effects.
    .
    First assume the magnetic field, B = B[itex]_{0}[/itex] = B[itex]_{0}[/itex]k, to be time independent and homogeneous, with
    k as a unit vector in the z-direction. A particle with charge q and mass m is moving in this field.
    Initially, at time t = 0 the particle has velocity v[itex]_{0}[/itex], with u[itex]_{0}[/itex] as the component in the z-direction and
    w0 as the component in the x; y-plane.
    a) Write the vector form of the equation of motion of the particle and show that it has solutions of
    the form
    r(t) = ρ[itex]_{0}[/itex](cos [itex]\omega[/itex][itex]_{0}[/itex]ti + sin [itex]\omega[/itex][itex]_{0}[/itex]0tj) + v[itex]_{z}[/itex]t k
    Determine the constants ρ[itex]_{0}[/itex], ω[itex]_{0}[/itex] and v[itex]_{0}[/itex] in terms of the initial velocity and magnetic field strength B0.

    2. Relevant equations
    F=q(E+v[itex]\times[/itex]B)


    3. The attempt at a solution
    E=0
    a=(q/m)(v[itex]_{0}[/itex][itex]\times[/itex]B[itex]_{0}[/itex])

    v[itex]_{0}[/itex]=(w[itex]_{0}[/itex],u[itex]_{0}[/itex])
    where w[itex]_{0}[/itex]=|w[itex]_{0}[/itex]|cos[itex]\vartheta[/itex]i+w[itex]_{0}[/itex]sin[itex]\vartheta[/itex]j)

    therefore v[itex]_{0}[/itex][itex]\times[/itex]B[itex]_{0}[/itex] = B[itex]_{0}[/itex]|w[itex]_{0}[/itex]|sin[itex]\vartheta[/itex]i-B[itex]_{0}[/itex]w[itex]_{0}[/itex]cos[itex]\vartheta[/itex]j
     
  2. jcsd
  3. Apr 7, 2013 #2

    mfb

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    What is ϑ?
    For (a), all you have to do is to calculate the acceleration vector based on the given formula for r(t), and show that it satisfies the equation for the Lorentz force.
     
  4. Apr 8, 2013 #3
    The [itex]\vartheta[/itex] was the angle used to separate w[itex]_{0}[/itex] into x and y components.

    But I'm sorry, can you please elaborate more? Do you mean all I need to do is differentiate r(t) twice and equate coefficients?
     
  5. Apr 8, 2013 #4

    mfb

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    "show that it has solutions of the form" -> show that [equation] is a solution -> show that [equation] has the Lorentz force as acceleration.

    With the calculated v0 x B0, right.
     
  6. Apr 8, 2013 #5
    Ok, but how do I calculate v[itex]_{0}[/itex][itex]\times[/itex]B[itex]_{0}[/itex] when v[itex]_{0}[/itex]=(w[itex]_{0}[/itex],u[itex]_{0}[/itex]) ? Don't I need to separate w[itex]_{0}[/itex] into x and y components?
     
  7. Apr 8, 2013 #6

    mfb

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    I would convert everything to polar coordinates, but splitting them in components in Cartesian coordinates is possible, too.
     
  8. Apr 11, 2013 #7
    I'm sorry, I still don't quite get it
     
  9. Apr 11, 2013 #8

    mfb

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    Did you do that? What did you get?
     
  10. Apr 11, 2013 #9
    I'm not going to lie, I couldn't do it.
     
  11. Apr 11, 2013 #10

    mfb

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    What did you get? Where did you run into problems?
    Those are obvious follow-up questions, you could have answered them without an extra post from me...
     
  12. Apr 11, 2013 #11
    I'm still not entirely sure what I am supposed to do. Is this right?

    d^2r/dt^2 = -[itex]\omega_{0}[/itex][itex]^{2}[/itex][itex]\rho_{0}[/itex](cos[itex]\omega_{0}[/itex]t i + sin[itex]\omega_{0}[/itex] j)

    Now we can say that, the above acceleration expression is equal to the Lorentz acceleration of (q/m)*(v[itex]_{0}[/itex][itex]\times[/itex]B[itex]_{0}[/itex]?

    But now how do I split v[itex]_{0}[/itex][itex] into i,j,k components? I tried splitting it into components before with the angle theta, but that was obviously wrong.
     
  13. Apr 11, 2013 #12

    mfb

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    $$\vec{v} \times \vec{B} = \begin{pmatrix}
    v_j B_0 \\
    -v_i B_0 \\
    0
    \end{pmatrix}$$ where I used that the B-field has a component in k-direction only.
    v_i and v_j are just the i- and j-component of ##\frac{d\vec{r}}{dt}##
     
  14. Apr 11, 2013 #13
    Ok, so then v[itex]_{j}[/itex]=cos([itex]\omega[/itex][itex]_{0}[/itex]t) and -v[itex]_{i}[/itex]=sin([itex]\omega[/itex][itex]_{0}[/itex]t)

    And then the constants -[itex]\omega[/itex][itex]_{0}[/itex][itex]^{2}[/itex][itex]\rho[/itex][itex]_{0}[/itex]=(q/m)*B[itex]_{0}[/itex]
     
  15. Apr 11, 2013 #14

    mfb

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    I think there are prefactors ρ_0 ω_0 missing, but the concept is right.
     
  16. Apr 11, 2013 #15
    So once I find the prefactors, I am given a 3 set of relations from which I can solve for the 3 constants?
     
  17. Apr 11, 2013 #16

    mfb

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    Right.
     
  18. Apr 12, 2013 #17
    Ok, thanks for all your help. How do I find them?
     
  19. Apr 12, 2013 #18

    mfb

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    Find what?
    It is all in the thread now, you just have to combine it.
     
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