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Electric current in a rotating ring

  1. Mar 14, 2006 #1
    I'm currently stuck on this question on the image attachment. Any help would be definately ppreciated.

    Ok, so from what I understand, it asks what is the current that passes through the fixed line.

    For part (a), I see that the current is 'discontinuous', and I'm not enitrely sure how to solve it.

    For part (b), I multiply the charge density [tex]\lambda = \frac{Q}{\pi a}[/tex] with the tangential speed of the ring [tex]a \omega[/tex]. That would give me the charge over time, right? I believe I should of done this part with differentials though with a segment [tex]dQ = \lambda dr[/tex].

    Attached Files:

  2. jcsd
  3. Mar 15, 2006 #2
    *bump* still need help.
  4. Mar 15, 2006 #3
    In the first problem you simply want to consider how much charge is passing through the indicated arc per unit time. In other words, don't use the derivative formula, use [tex]I= \Delta Q / \Delta t [/tex]. The simplest would be to choose delta t as one period. How much charge passes through that arc in one period?

    Your part b seems valid to me.

  5. Mar 15, 2006 #4
    The full [tex]Q[/tex] of course, neglecting the bits on both poles that just spin.

    So that would make [tex]Q / T[/tex] and

    [tex]\omega = \frac{2 \pi}{T}[/tex]
    [tex]T = \frac{2 \pi}{\omega}[/tex]

    So that means the answer is:

    [tex]I = \frac{Q \omega}{2 \pi}[/tex]

    Ok so that's the same answer as question (b). I guess that makes sense since they both have the same amount of Q passing through the same period. So what's the difference? One is done by substitution and the other by multiplying the density with the tangential speed?
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