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Homework Help: Induced potential difference for disk spinning in magnetic field

  1. Dec 17, 2013 #1
    1. The problem statement, all variables and given/known data

    A solid conducting disk of radius a rotates about its symmetry axis with
    angular speed ω rads/s. If there is a uniform magnetic field [itex]\mathbf{B}[/itex] perpendicular to the disk derive an expression for the potential difference induced between the centre of the disk and its rim.

    2. Relevant equations

    [tex]\epsilon=\oint \mathbf{f}\dot d\mathbf{l}[/tex]
    [tex]\Delta V=\int^{\mathbf{b}}_{\mathbf{a}}\mathbf{E}\cdot d\mathbf{l}[/tex]

    3. The attempt at a solution

    I found the force per unit charge

    [tex]f_{\text{mag}}=\mathbf{v}\times\mathbf{r}[/tex][tex]\mathbf{v}=\mathbf{\omega}\times\mathbf{r}=\omega r\hat{\phi}[/tex]
    [tex]f_{\text{mag}}=\omega r\hat{\phi}\times\mathbf{r}=\omega r B \hat{r}[/tex]

    Then I thought I could use

    [tex]\epsilon=\oint \mathbf{f} \cdot d\mathbf{l}[/tex]

    and get [tex]\epsilon=2\pi B\omega r^2[/tex] which would give a potential difference of [tex]2\pi B\omega a^2[/tex]

    but this is wrong. I think I have [itex]f_{\text{mag}}[/itex] correct. So now how do I find the potential difference?
  2. jcsd
  3. Dec 18, 2013 #2
    I am not too sure about your last steps. Did you put the emf equal to the electric field to find the potential difference? They aren't the same.

    Also it seems to me, like you omitted the electric field, when you found the force. You are inducing a current in the disk, and thus an electric field. So I think you should include it, and then isolate for it, when you have an expression for the emf.

    F = q(E+vB) as vB are perpendicular. Just keep the q, as you can get rid of it later, when you need to find the emf.

    You could probably even see the disk as a "wire", when you only compare the rim and the center. This way you can find an equation for it, as if it is a wire rotating in an electric field.
  4. Dec 18, 2013 #3
    What is the most basic relationship between "force" and "potential energy"?
  5. Dec 18, 2013 #4
    [tex]\mathbf{F}=-\nabla V[/tex]
  6. Dec 18, 2013 #5
    And you know the force is strictly radial - can you obtain ##V## from that?

    A word of caution. ##V## usually means "electric potential". You used that for "potential energy". "Potential energy" is usually denoted by ##U## (or ##\Pi## in some texts on mechanics). There is a relationship between ##U## and ##V##, but they are not the same thing. Be careful here.
  7. Dec 18, 2013 #6
    yeah I don't use that notation either. I meant to say E, which is F/charge right? so would I literally just use the third integral in my "relevant equations" then?
  8. Dec 18, 2013 #7
    That should work.
  9. Dec 18, 2013 #8

    rude man

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    Since the direction of emf is radial (emf = (B x l ) * v), you can also think of the disc as comprising many radial strips or wires, each separated by a thin insulator.
  10. Dec 18, 2013 #9
    ok this might sound stupid then, but what is the difference between [itex]\mathbf{E}[/itex] and [itex]\mathbf{f}_{\text{mag}}[/itex]. Like f is the force per unit charge. So isn't it the same thing as E?
  11. Dec 18, 2013 #10
    Yes indeed, the force per unit charge would be the same.
  12. Dec 18, 2013 #11

    rude man

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    If this is response to my post #8:

    Yes, but to get emf you need to integrate the force over the distance of the wire.

    My main point was that the disc can be modeled by a number of radial wires.

    BTW the polarity of my way of finding emf is not indicated by the expression I gave. To get the formula, think of the direction of force on a positive charge in the wire. The charge will bunch up in one end of the wire, from which the polarity then is obvious.
  13. Dec 18, 2013 #12
    ok so one final question. I have the formula for emf[tex]\epsilon=\oint \mathbf{f} \cdot d\mathbf{l}[/tex]So initially I was using the path length as the circumference, but now that I see that f is in the radial direction, this integral is just along r. Mathematically this is the same as what I did. So now I am forgetting the difference between EMF and potential difference. Also, this is a conductor, so why would this potential difference not lead to a radial current, which would balance out the charges and make the potential difference zero.
  14. Dec 18, 2013 #13
    The potential difference re-arranges elementary charges (electrons) in the disk radially; this in turn creates a counter-acting electric field, which will eventually compensate for the E.M.F. If there is a return path outside the disk however, there will be current through it. This arrangement is known as the homopolar generator.
  15. Dec 18, 2013 #14

    rude man

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    If the disc were pure copper there would be zero current since the emf from the front & back sides would cancel.

    What is usually meant is a Cu layer on one side of an insulated disc. The there would be an emf but a current only if you placed a resistive load between the center and the edge.

    Emf and potential difference are not mutually exclusive! Emf is a voltage generated by a source of energy other than electrical.

    Fact: the voltage across an inductor is an emf. Why?
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