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Calculating a Homopolar Generator's EMF

  1. Apr 14, 2016 #1
    1. The problem statement, all variables and given/known data
    Screen Shot 2016-04-14 at 1.21.10 PM.png Screen Shot 2016-04-14 at 1.21.18 PM.png
    2. Relevant equations
    ##\omega = v/r##
    ##q\vec v \times \vec B\ =\ \vec F_B##
    ##|\vec F|d\ =\ |\vec E|##
    ##q \Delta V\ =\ E##
    3. The attempt at a solution
    (So, my answer is twice and large as the correct answer. This is the second time this has happened on a question like this . . .)
    Since a negative charge would collect on the outside of the disk I began by analyzing one negative charge and treated it in equilibrium on the edge of the disk,
    q\vec v \times \vec B- \frac{|\vec E|}{d}\ &=\ 0 \\
    qvB \ &=\ \frac{q \Delta V}{r} \\
    vBr \ &=\ \Delta V
    And since ##v\ =\ \omega r##:
    (\omega r)Br \ &=\ \Delta V \\
    \Delta V \ =\ \epsilon \ &\approx\ 48.3\ V
    Any help would be wonderful c:
  2. jcsd
  3. Apr 15, 2016 #2

    rude man

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    Start with the differential form for emf:
    d(emf) = Bv(r)dr
    and show off your powers of integration! :-)
  4. Apr 16, 2016 #3
    I'm back tracking the work a little to make sure i understand where your getting that form of emf:
    F_B - F_E \ &=\ 0 \\
    F_B \ &=\ F_E\ =\ q|\vec E|\ =\ q\frac{d(\epsilon)}{dr} \\
    vB\ =\ \omega r B\ &=\ \frac{d(emf)}{dr} \\
    \int (\omega r B)dr\ &=\ \epsilon \\
    \frac{\omega B r^2}{2} \ &=\ \epsilon\ \approx 24.1\ V
    thanks so much!
  5. Apr 16, 2016 #4

    rude man

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    Looking good!

    d(emf) = B(r) v(r) dr is just the generalized (differential) form of emf = Bvr. Just basic calculus. It allows B and/or v to be functions of r. If B and v are not functions of r then it's just Bvr. But in your case B = constant but v = v(r) so then
    d(emf) = B v(r) dr = B ωr dr
    and when integrated over r=0 to r= R you get emf = ωBR2/2
    where R is the disk's radius.
    Last edited: Apr 16, 2016
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