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AC Generator With Repelling Magnets

  1. Aug 2, 2015 #1
    Hi Everyone,

    I have 4 scenarios in question . Will The EMF(Voltage) be the same in all of the scenarios?
    - the difference between the two photos are the polls: one repel one attract .
    - i understand that the current(amps) will change due to the wire length assuming same gauge

    1) Photo 1: d1 = 1cm, d2=20cm, wire length=240cm, Wire Exposed to field=80cm, each magnet is 1 Tesla(at surface), coil moves at 6meter/second
    2) Photo 1: d1 = 1cm, d2=10cm, wire length=200cm, Wire Exposed to field=80cm, each magnet is 1 Tesla(at surface), coil moves at 6meter/second
    3) Photo 2: d1 = 1cm, d2=20cm, wire length=240cm, Wire Exposed to field=80cm, each magnet is 1 Tesla(at surface), coil moves at 6meter/second
    4) Photo 2: d1 = 1cm, d2=10cm, wire length=200cm, Wire Exposed to field=80cm, each magnet is 1 Tesla(at surface), coil moves at 6meter/second


    Photo1:
    D1_2.png

    Photo2
    D1_3.png

    Thank you
     
  2. jcsd
  3. Aug 3, 2015 #2

    BvU

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    Hello Ben, welcome to PF :smile: !

    Do you have an expression for the EMF ? The F in EMF stands for force; forces are vectors: they have a magnitude and a direction, so you'll need to think about both of these.
     
  4. Aug 3, 2015 #3
    hi Bvu, yea I'm aware of it . the EMF(V) will be generated in the oppose direction . but still I'm looking for an answer to the above question. and when i ask if the EMF(Voltage) be the same i mean in magnitude
     
  5. Aug 3, 2015 #4

    CWatters

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    Do you have an expression for the magnitude of the EMF ?
     
  6. Aug 3, 2015 #5
    In your drawings #1, no current is generated in the loops.
     
  7. Aug 3, 2015 #6
    what do you mean ?
     
  8. Aug 3, 2015 #7
    Please explain why you think that ?
     
    Last edited by a moderator: Aug 3, 2015
  9. Aug 3, 2015 #8
    Use the right hand rule on the horizontal top section of a loop, and again, on the bottom section. The EMF of one section opposes the other. Symmetry says they are equal and opposite.
     
    Last edited by a moderator: Aug 3, 2015
  10. Aug 3, 2015 #9
    then how is it any difference from this ? acp14a.gif
     
  11. Aug 3, 2015 #10

    BvU

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    Blue and red move in opposite directions
     
  12. Aug 5, 2015 #11
    So essentially option 1,2 would not work as they are canceling each other and 3,4 would work?
     
  13. Aug 5, 2015 #12
    So essentially option 1,2 would not work as they are canceling each other and 3,4 would work?
     
  14. Aug 16, 2015 #13
    any thoughts ?
     
  15. Aug 16, 2015 #14
    any thoughts ?
     
  16. Aug 16, 2015 #15

    BvU

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    Plenty of thoughts. One of them: the situation isn't all that clear to you yet. What can be done to improve the situation ?

    Are you familiar with the Lorentz force on charge carriers ? $$
    \vec F = q\, \left (\vec E + \vec v \times \vec B \right ) $$
    There's a picture (bottom one: motion ##\Rightarrow ## emf) in post #16 in this thread
     
  17. Aug 16, 2015 #16
    all new to me. making sure I am getting it correctly .
    D1_3.png
     
  18. Aug 16, 2015 #17

    BvU

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    In you picture, e.g. at the lower end, you have M pointing to the left.
    M is the magnetic field.
    At the lower "horizontal" section of the wire frame, the magnetic field is pointing up.
    The velocity vector is pointing to the left. So the Lorentz force on the charge carriers in the wire is pointing away from the viewer.

    Your turn to try and find the direction of the emf for the upper "horizontal" section of the wire frame :smile: !
     
  19. Aug 16, 2015 #18
    Hold on, when it say magnetic field "right hand rule" little picture I added on top of the drawing. it is referring to the permanent magnet field? and the force is referring to in my case the direction of the coil.

    if that is the case then i believe this is the correct drawing .

    D1_3.png
     
    Last edited: Aug 16, 2015
  20. Aug 16, 2015 #19

    BvU

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    One out of three. Still for the bottom "horizontal" section of the wire frame:

    The magnetic field thumb is indeed the field from the magnet.
    So yes, the middle finger is up (no pun intended :rolleyes: ).

    The current in this situation is the motion of the charge carriers in the frame. The ##q\vec v## in the Lorentz force expression.
    So index finger points to the left.

    That leaves the thumb pointing into the screen. Thus representing the direction of the force that these charge carriers "feel". Corresponding to an emf that wants to move these charges.

    Your turn to try and find the direction of the emf for the upper "horizontal" section of the wire frame :smile: !

    And photo 1, where the emf allegedly oppose each other.
     
  21. Aug 16, 2015 #20
    But, if i look at "fleming's right hand rule" then:
    1) the thumb is pointing to the movement(motion of the coil in my case and we can also call it force)
    2) First finger (also called Index Finger) is the magnetic field (in my case the magnetic field due to the permeant magnet)
    3) second finger (also called middle Finger) is the direction of the current (EMF,V)

    no pun was taken

    so where am I getting it wrong in photo #2?

    eminduction3.png
     
  22. Aug 16, 2015 #21

    CWatters

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    Magnetic field is vertical so first finger should point vertically.
    Motion is to the left so thumb should point to the left.
    Last one left is current.
     
  23. Aug 16, 2015 #22

    BvU

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    Yeah, it works: in the scenario I described: thumb left, index up and middle points way from the viewer.

    But if you follow the wiki picture the result of the cross product goes with the thumb.
    Must say I think I remember (from 40+ years ago) that in ##\vec c = \vec a \times \vec b## the righthand thumb is a, index is b and result c is middle finger.

    Something to do with the properties of the cross product with respect to permutations: swap two -- the sign changes. Swap twice and then it's a plus sign again (even permutations). ​

    Learn yourself one and only one right hand rule for the cross product. Otherwise you'll end up with a whole life of only 50% chance you're indeed right ! For over 21's the corkscrew rule is perhaps even better...​


    But in your picture, at the bottom I see an m pointing up, and an f to the left. That's not good. You can't "also call it force". That's confusing.
    And i is not the current. The "cause" of the emf is the motion of the charge carriers (because the wire frame moves). That there is a current is a consequence of the F . There doesn't have to be a current (see e.g. photo 1!). But there is an emf.
     
  24. Aug 16, 2015 #23

    sophiecentaur

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    On the face of it, I don't think you are. You are reversing the field direction and, with the motion in the same direction, I should expect the resultant emf to be in the opposite direction at top and bottom of the coil - giving you a current circulating round the coil. Your logic certainly makes sense to me.
    This isn't a situation that's commonly met and I can't see it as a basis for a 'good' generator design. There is the problem that the field will not be as strong as you could get with the normal configuration ( with the iron core that you normally find in alternators) and the fact that you would need a reciprocating motion for the coil, which is not as convenient as a rotation in the normal style of alternator. If you wanted a (DC) Dynamo, the commutator would be pretty hard to implement, too.
     
  25. Aug 16, 2015 #24
    Can we simplify your answer . Yes there will be EMF i photo #2 and yes there will be current flowing . Please confirm .
     
  26. Aug 16, 2015 #25
    HI Cwatters, thank you for your reply . can we agree that Photo number 2 will generate current and EMF . plus that my last right hand rule drawing in the last picture is correct ?
     
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