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Physics Q&A Game

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  1. May 25, 2003 #1

    chroot

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    Just like the Astronomy Q&A game in the Astronomy forum, let's play a physics Q&A game here. The astronomy game has certainly taught me some very neat facts -- maybe this one will do the same.

    I'm going to post one question posed by marcus:

    "A favorite version of the question chroot just answered is the one about the airplane flying the polar route. It is going 200 meters per second and its wingspan is 30 meters----what is the voltage difference between the two wingtips? I am not asking this question because it is a cousin of the one asked by Ivan, just recalling it. the plane is in a region where the earth's magn. field is roughly vertical and of such and such a strength etc etc."

    It seems that there should be no voltage. Faraday's law dictates that

    E = -N d[phi]/dt

    Since [phi] is defined as the flux: [phi] = B * A, where B is the magnitude of the magnetic field and A is the area enclosed by the loop. In this problem, neither A nor B is changing -- so the induced voltage should be zero.

    Am I right? Was it really just a trick question?

    - Warren
     
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  3. May 25, 2003 #2

    Ivan Seeking

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    -[30X200][50X10^-6][+-30]= +-9 volts
     
    Last edited: May 25, 2003
  4. May 25, 2003 #3
    My answer

    When the plane takes off a voltage will be induced across the wings, and it will be sustained in flight. But you can't measure it with a voltmeter across the wingtips because the very same induction will occur with the meter leads. However, if the plane can land with the charge on the wingtips remaining, it can be measured on the ground with an electrometer, which is a very sensitive voltmeter.
     
  5. May 25, 2003 #4

    Ivan Seeking

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    Re: My answer

    I bought perfect test leads for my perfect voltmeter. They have them down at the gedunken supply house.

    Re: "There are more things in heaven and earth, Horatio, Than are dreamt of in your philosophy."
    I like you already!
     
  6. May 25, 2003 #5

    marcus

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    30 meter wingspan x 200 meter/sec speed
    x 50 microTesla (the vertical component of the geomagn. field)

    let's see how that multiplies out

    it comes to 0.3 volts

    but you have an extra factor of 30 you multiplied in
    which brings it to 9.0

    You are permitted to edit your answer if it is a typo!

    If it is not a typo then explain please the factor of "[+-30]"
    You may certainly be right to have it. I am often forgetting
    details and there may be some obvious thing I am missing.
    I got this problem from an old edition of Halliday Resnick
    but dont remember the exact numbers
     
  7. May 26, 2003 #6

    Ivan Seeking

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    Shoot no. I was integrating the emf over the length of the wings. twas thinking E and not ε
     
  8. May 26, 2003 #7

    Ivan Seeking

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    canni still go next? canni canni please please
     
  9. May 26, 2003 #8

    marcus

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    Your turn!

    sorry I got preoccupied with something else and just got
    back

    please proceed

    (this game was partly your idea----the teaser problem---which chroot generalized, you and chroot make the rules I think)
     
  10. May 26, 2003 #9
    It's the Hall effect, Warren. :) Actually a real-life voltmeter should work fine, I think -- just stretch the leads out, doesn't matter if you even touch the wing or not.
     
  11. May 26, 2003 #10

    chroot

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    I'm still not sure how this answer was arrived at.

    - Warren
     
  12. May 26, 2003 #11

    chroot

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    Ohhh.. the Hall effect. Hmph.

    :wink:

    - Warren
     
  13. May 26, 2003 #12

    Ivan Seeking

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    yippie!

    The dA/dt was just the width of the wings multiplied by the speed of the plane.

    Ok. I don't know if this will be really easy, or really hard, or somewhere in between. A True Story.

    The answer to a very complex physics or mathematics problem [I don't remember the exact nature of the problems] was to be calculated on one of the world's top supercomputers of the time; circa 1975. Two approaches were considered. The first and preferred approach was to calculate an extremely precise answer, and next was to approximate the answer. The best answer would require about a year of CPU time. The approximation could be calculated in relatively short order: in a matter of days.

    The cost of CPU time caused this issue to be scrutinized very closely. It was decided that in spite of the money, the approximate answer was the best choice. What is the physical motivation [the physics] for their choice?
     
  14. May 26, 2003 #13
    Re: My answer

    Charge can't remain (separated when plane is on the ground) because plane is made of metal.
     
    Last edited by a moderator: May 26, 2003
  15. May 26, 2003 #14
    No, voltmeter won't show anything (because emf [vB]l is evenly distributed along a wire l).
     
  16. May 26, 2003 #15

    marcus

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    I often disagree with you Alexander or else am put off by your argumentatitive tone of voice (I do not remember which it is) but here I must commend you and voice agreement.

    This is not normally considered an example of the Hall effect.

    It is a standard textbook example that comes in right after the Faraday law. Halliday Resnick has it as an antenna of a car driving where the field is horizontal and the vertical antenna is cutting across field lines.
    Giancoli has it with airplane wings cutting across vertical field lines.

    The basic law (as in so many things) is the Lorentz force.
    (Which also explains the classical Hall effect! but the problem is elementary and appears long before the Hall effect is discussed.)

    A voltmeter could not be used, I reckon, to measure the voltage difference between the wingtips in flight (for the reason Alexander gave----no current would flow thru the meter altho the voltage difference would be there)
    A conductor is not necessarily a constant potential thing when it is moving thru a B field as someone said, maybe Alexander.

    Lorentz force is so basic, almost the F=ma of E/M
    Lorentz force law is really the definition of the E and B fields, it is part of the groundwork.

    dropping out the electrostatic part of the law it just says
    F = qv X B

    The B arrows near the north geomagnetic pole, which is a "south" pole---paradoxically---point DOWN.

    So conduction electrons in the wing feel a force to the right

    The voltage is higher on the left wingtip, where positive charge carriers would concentrate if there were some.

    In Giancoli the problem comes right after the example of the bar sliding along two parallel rails, in a uniform vertical field

    The rails merely permit the voltage difference between the ends of the bar to be measured (it is a Faraday law loop including the rails)

    The airplane wing is the bar with the two rails removed.

    You could put rails in the sky and have the airplane swoop down and fly along them with its wingtips touching them so that the voltage could be measured but the FAA and the Airline's Pilots Association would not allow this so one must trust in Faraday and believe on grounds of faith.
     
  17. May 26, 2003 #16
    ^^^ Well, you would need 'leads' with a great deal of capacitance; then charges will build up at the ends and cause a voltage difference across the voltmeter inputs. emfs in 'series' along each lead act just like one big emf, remember.
     
  18. May 26, 2003 #17

    marcus

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    not a bad idea. As I picture your idea it involves remote switches that can be used to isolate parts of a conductor running the length of the wing
    one might even "freeze" the voltage differerence in carefully insulated elements inside the airplane wing and keep
    the difference "on hold" so to speak until the plane could
    land and have it measured.

    the humble textbook writers did not bother to discuss how the voltage could be measured!
     
  19. May 26, 2003 #18
    Re: Re: My answer

    So lets make the plane out of fiberglass with metal wingtips and have a wire and a switch between them. We can close the switch when the plane takes off and open it before it lands, keeping the charge on the wingtips. And I see marcus thought of this too.
     
  20. May 26, 2003 #19
    Markus, why this long post? Any relevance to a voltmeter reading? Looking across this long post I could not find what you state about voltmeter. Can you be concrete in you posts, without mentioning all what textbooks have cover-to-cover?

    Voltmeter with its leads span as airplane wings will show zero reading despite that there may be quite high potential between tips of leads. (And I can prove that if needs be.)

    From you post I understood that you claim opposite. Well, then prove it.
     
    Last edited by a moderator: May 26, 2003
  21. May 26, 2003 #20

    marcus

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    Me too, this is what I was saying. You did not understand.

    Then you REALLY misunderstood me---180 degrees. Well that happens. Better luck next time.

    My post was opposed to using the Hall effect (which damgo referred to) as an explanation. Sorry it confused you!
     
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