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Wire cutting magnetic field lines

  1. Jan 2, 2014 #1
    I have read in my schoolbook that when a wire of length l moves at velocity v in a magnetic field cutting the flux lines an emf is induced which it is calculated from emf = blv, I understand that emf is induced at time variation of magnetic flux across the wire, but I Can't see any change in the magnetic flux here, the magnetic flux should stay the same anywhere the wire moves in the field as long as it not getting nearer to or farther from the magnet. But It seems non logical that the flux changes just because the wire covers a certain area during its movement in the field !!!!!!! Please, this point is bothering me very much I would be very thankful if anyone could explain it to me, thanks in advance
     
  2. jcsd
  3. Jan 2, 2014 #2
    There is no magnetic field and there is no electrical field, they are both electromagnetic fields.

    If a charge is close to a magnet, the charge will not sense any force, but if the charge is moving next to a magnet it will sense an electrical force, because there is an electrical field.

    If a magnet is close to a charge, the magnet will not sense any force, but if the magnet is moving, it will sense a magnetic force, because there is a magnetic field.

    And this is all I can help.
     
  4. Jan 2, 2014 #3

    tiny-tim

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    Hi ElmorshedyDr! :smile:
    (We'll assume the magnetic field is uniform.)

    It depends what you imagine the wire is connected to, to complete a circuit.

    If the whole circuit is moving (in the field), and staying the same shape, then yes there's no change in flux.

    But we always imagine a rectangular circuit with one edge fixed, and two edges getting longer and longer, so that the flux through the circuit is increasing! :wink:

    (Because a potential always has to be measured relative to a fixed point of zero potential!)
     
  5. Jan 2, 2014 #4

    Thanks Tim, but I'm unable to visualize what you say please make it clearer
     
  6. Jan 2, 2014 #5

    tiny-tim

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    You need a fixed point of zero potential, and that fixed point has to be part of the circuit.

    Since the wire is moving, that means the whole circuit has to be getting large (or smaller).

    To simplify things, we make the circuit a rectangle: one edge is the wire, the opposite edge is fixed, and the two "side" edges are in the direction of motion of the wire. :wink:
     
  7. Jan 2, 2014 #6

    Yeah I got it, but it contradicts the bases of some problems in my school book, where the problems ask us to calculate the emf induced in the the second clock arrow cutting the magnetic field of the earth, according to what you say there shouldn't be a induced emf in the clock arrow
     
  8. Jan 2, 2014 #7

    tiny-tim

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    If I'm understanding correctly, the circuit here is a piece-of-pie with one edge fixed and the other getting further and further round the circle …

    so the area is increasing, and the flux is increasing. :smile:
     
  9. Jan 2, 2014 #8

    The arrow doesn't even make a closed circuit
     
  10. Jan 2, 2014 #9

    davenn

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    I think we are all having difficulty trying to visualise what you are talking about

    how about scanning the relevent images and text from your book and showing us
    exactly what you are talking about

    otherwise perople are just going to waste their time guessing about the setup

    cheers
    Dave
     
  11. Jan 3, 2014 #10

    tiny-tim

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    (just got up :zzz:)
    I'm assuming a circuit around the whole piece-of-pie, ie a sector of a cirucle, comprising one fixed radius, one moving radius (with the arrow), and the arc joining them.
     
  12. Jan 3, 2014 #11

    Philip Wood

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    Does this help? See thumbnail ...
     

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  13. Jan 3, 2014 #12

    Dave its, what I'm talking about is a clock arrow cutting the magnetic field it ask me to calculate the Emf in the arrow I cant understand how can an emf get induced in the arrow?
     
  14. Jan 3, 2014 #13

    Actually I don't understand why is a emf induced when a wire cuts a magnetic field although the shape of the wire didn't change besides the wire doesn't go nearer to or farther from the wire so delta BA should be zero
     
  15. Jan 3, 2014 #14

    Philip Wood

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    BA is the flux LINKED WITH (going through) the circuit. Because, in my diagram, the area of the circuit gets less as the wire moves to the right, BA (that is the flux linked) gets less.

    It's convenient to calculate the emf from the rate of change of flux linked, because the equation can be applied to both types of e-m induction (roughly speaking those due to A changing and to B changing). But it doesn't, imo, shed much light on WHY the emf is induced. One way to understand why, in the case of moving conductors, is to realise that there are motor effect (Bev) forces on the charge carriers in the wires, when these charge carriers move because the wire itself is moved.
     
  16. Jan 3, 2014 #15

    Philip Wood

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    I agree that the case of the clock-hand (I take it that's what you mean by 'clock arrow') is a difficult one. It's not easy to apply the concept of change of flux linkage to it. [Indeed, the closely related rotating disc is one of the awkward cases cited in the Wiki article on electromagnetic induction that vanHees2 justifiably recommends.]

    Nonetheless one can still apply the idea of flux CUTTING by the hand, or go back to basics and use the BEv method I mentioned in my previous post.

    Note that to MEASURE the voltage produced one would have to connect a voltmeter between the inner end (X) of the hand and its outer end (Y). It matters how you do this…

    If you could strap a mini-voltmeter to the hand itself, and connect wires from it going along the hand to X and Y, zero voltage would be measured. One way of looking at this is that the voltages induced in the hand and in the wires will be equal and in the same direction in space as each other, so in terms of the complete circuit, will cancel each other, being in opposite senses.

    To measure a voltage you'd have to let Y brush against a stationary conductor and connect the voltmeter between this and X, so the voltmeter and the wires are stationary (in the laboratory frame of reference), while just the hand itself moves. [Principle of Faraday disc generator]
     
  17. Jan 3, 2014 #16

    Thanks very much for your GREAT explanation, So you mean that the moving wire is connected to a circuit, and what happens when I move the wire, is that the area of the circuit increases or decreases, which certainly causes a Delta A x B
     
  18. Jan 3, 2014 #17

    I'm unable to visualize that when a hand of a clock is rotating the area of a circuit increases? Where is the circuit in the clock?
     
  19. Jan 3, 2014 #18

    Philip Wood

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    Re-read my last paragraph! You have to incorporate the hand in a circuit of your own making. And as I acknowledged in my first paragraph, even if you do this, applying to the circuit the idea of flux linkage changing is not straightforward and may lead to contradictions. But you can apply the concept of the hand CUTTING flux.
     
  20. Jan 3, 2014 #19

    Can I An emf be induced without a flowing current if the circuit is open ?
     
  21. Jan 3, 2014 #20

    Philip Wood

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    Yes. There will be no current if the circuit is open, but there will be an emf.[ Strictly you don't even have to have a physical circuit!, just a defined path - but I wouldn't start worrying about this!]
     
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