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Thornton and Marion, Chapter 2, Prob. 22

  1. Jan 26, 2010 #1
    1. The problem statement, all variables and given/known data

    This is a four part problem -- the only issue I have is on part (c) so I'll condense the question:

    If the equation of motion of a particle of charge [tex]q[/tex] in an electric field and magnetic field is

    [tex]qv_y B \hat{\mathbf{i}} + \left( qE_y - qv_xB \right) \hat{\mathbf{j}} + qE_z \hat{\mathbf{k}} = m \frac{d}{dt} \left( v_x \hat{\mathbf{i}} +v_y \hat{\mathbf{j}} + v_z \hat{\mathbf{k}}\right), [/tex]

    obtain expressions for [tex] v_x \left(t\right)[/tex] and [tex] v_y \left(t\right)[/tex]. Show that the time averages of these velocity components are

    [tex]\left\langle v_x\left(t\right)\right\rangle = \frac{E_y}{B}[/tex]


    [tex] \left\langle v_y\left(t\right)\right\rangle = 0[/tex]

    (Show that the motion is periodic and then average over one complete period.)

    2. Relevant equations

    3. The attempt at a solution

    Solutions to the DE are

    [tex]v_x \left(t\right) = \frac{E_y}{B} + C_1 \cos \left(\omega t\right) + C_2 \sin \left(\omega t\right)[/tex]


    [tex] v_y \left(t\right) = -C_1 \sin \left(\omega t\right) + C_2 \cos \left(\omega t\right)[/tex],

    where [tex]\omega[/tex] is the cyclotron frequency. Now, clearly the velocity functions are periodic with period [tex]2\pi/\omega[/tex]. What do they mean "time averages," and what does the [tex]\left\langle \right\rangle[/tex] notation mean?


    [tex] \left\langle v_x\left(t\right)\right\rangle = \frac{ v_x\left(0\right) + v_x \left(2\pi/\omega\right)}{2}[/tex],

    like an arithmetic mean, but I'm just guessing based on the hint (and that doesn't yield the correct answer). Do they mean average velocity?
  2. jcsd
  3. Jan 26, 2010 #2


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    Hi union68! :smile:
    "Time average" means the average value over time (and the <> notation just means average, or expectation value) …

    in this case, theoretically you'd integrate over a whole period, and then divide by the period.

    However, in this case it's a bit obvious that the average of a cos or sin over the period is 0, so that just leaves you with … ? :wink:
  4. Jan 26, 2010 #3
    Ah, yes. Integrating over the period and dividing by the period gives you the correct answer. The cosines and sines vanish.

    However, what exactly is this procedure though? I guess I've never seen integration related to an average.
  5. Jan 26, 2010 #4


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    uhh? :redface: … how else would you define an average of something that varies continuously? :smile:

    (and how do you think root-mean-square voltage etc is defined? :wink:)
  6. Jan 26, 2010 #5
    After a quick Google search I found some explanation. I have never seen continuous averaging before.

    I have no idea what RMS voltage is either -- I'm a math major dabbling in physics.

    Thank you for the help.
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