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Lenz' law

  1. Feb 4, 2015 #1
    So I've been told that the direction that you wind a coil will determine the polarity of the induced voltage across the coil. I drew out two coils with clockwise and counterclockwise winding direction and with using the right hand rule and saying that the positive polarity of the induced voltage is in the direction in which current flows according to lenz law, I determined that the polarity is the same. Can someone please explain this further?
     
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  3. Feb 4, 2015 #2

    Andrew Mason

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    The direction of the current around any of the loops depends only on the rate of change of magnetic flux enclosed by the loop. Take the simplest case: a single loop. It is just a loop. Whether it is wound clockwise or counter-clockwise has no meaning. The induced emf and resulting current will be the same direction around the loop for a given rate of change of magnetic flux.

    All a coil is is a series of loops progressing in a certain direction. But the direction of the current in those loops, for a given rate of change of flux, will be the same regardless of which direction the loops progress.

    AM
     
  4. Feb 5, 2015 #3
    I already know the basics. E=L di/dt. So you're telling me that the direction in which a loop is wound has no meaning. Then why do transformers and potential and current coils all have polarity markings on them? What determines the polarity?
     
  5. Feb 5, 2015 #4

    Andrew Mason

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    The polarity is determined by which side of the loop the leads come from. The essential difference between a clockwise wound coil and a counter-clockwise coil is just the direction in which the coils progress. A counter-clockwise coil is identical to the clock-wise coil in all respects except that direction of coil progression. Essentially, the difference is that the leads are switched. So to make them electro-magnetically equivalent you just have to switch the leads on one.

    AM
     
  6. Feb 5, 2015 #5

    collinsmark

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    The [itex] \varepsilon= L\frac{di}{dt}[/itex] won't make any difference in this particular relationship for a single inductor. If you close a switch to introduce a battery into an RL circuit (using your handwound coil), and measure the relationship between the current and voltage across the coil, it will be the same if use the other clockwise vs. counterclockwise wound coil. The equation [itex] \varepsilon = L\frac{di}{dt}[/itex] doesn't depend on the coil's polarity.

    On the other hand, the application of Lenz' Law, [itex] \varepsilon = -\frac{d \Phi}{dt}[/itex] does make a difference. Changing the polarity of the coil (clockwise wound or counterclockwise wound) will change the direction of the magnetic field within the coil, all else being equal. In Lenz law, the magnetic flux, [itex] \Phi [/itex] is a scalar quantity, but it can take on both positive and negative values. All else being the same, it is to the polarity of the coil that has an effect on whether [itex] \Phi [/itex] is positive or negative, and thus whether [itex] \frac{d \Phi}{dt} [/itex] is positive or negative (again, all else being the same).

    Set up a RL circuit for your clockwise wound coil and place a compass near one end (and make note of which end). Notice how the coil deflects the compass needle (take careful notes, as the compass needle deflection may change with time). Now switch out the clockwise coil with the counterclockwise coil and set everything else up the same. The compass needle should deflect in the opposite direction than before.

    In summary, in the experiment, the circuit's voltage and current were the same, but the difference was in the way the compass needle was deflected. The circuit behaved the same either way, but the direction of the magnetic field within the coil changed direction.

    In ending, your choice of the winding clockwise vs counterclockwise determines which end of the coil is the north pole and which end is the south pole at a given time in a given circuit.

    *(You don't really need to swap out coils though, you could change the orientation of the coil that you already have, without changing its electrical connections in the circuit. Whichever way flips your thumb's direction when using right-hand-rule.)
     
    Last edited: Feb 5, 2015
  7. Feb 5, 2015 #6

    collinsmark

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    By the way, transformers are a little different. They have markings because they involve two, coupled coils. For example, swapping the leads in the secondary coil in the circuit without changing the primary coil connections, and you will reverse the polarity of the secondary emf. So in the case of transformers it can change circuit behavior.

    Some other coils/transformers are shielded, such that the external housing is electrically connected to one of the leads. When building the circuit it might be advantageous to know which lead is connected to the housing.
     
    Last edited: Feb 5, 2015
  8. Feb 5, 2015 #7
    Collin, what you've typed I already know. I know that the direction of the magnetic field changes but what I was asking is how to determine the polarity of the induced emf. The specific application I am confused about is a wattmeter. it does matter how you hook up coils in a wattmeter configuration and all that a watt meter is, is one current coil and one potential coil. so obviously the polarity of the induced emf is different depending on how you hook up those coils. I stated in my last message that I know all the basics because Andrew's message sort of implied that I don't know basic electrical relationships. I know what faradays law of induction is and I also know what lenz law is. I can reason my way through the polarity markings on a transformer using lenz law and the right hand rule. The right hand rule will tell you which way current flows and consequently the induced emf but when I apply the same logic to a single coil excited by an ac source I cant seem to figure it out. because the induced emf is always the same polarity.
     
  9. Feb 5, 2015 #8

    collinsmark

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    The induced emf is such that it would create a flux that would oppose the change of the original flux.

    The word "change" in the latter part of that sentence is vital.

    I'm not saying that that induced emf creates a flux that opposes the original flux. But rather that the induced flux opposes the change of the original flux.
    • Original flux is increasing: Induced flux is negative.
    • Original current is decreasing: Induced flux is positive.
    Keep in mind, I'm not saying anything about whether the original flux is positive or negative. It could be positive, negative, or even zero at any instant; it doesn't matter. All that matters is its direction of change.

    Then just use your knowledge of the right hand rule to determine the direction of the induced current, and thus the direction of the induced emf.

    Well, the polarity if the induced emf is ultimately independent of whether the coil is wound clockwise or counterclockwise; it doesn't matter here. That I agree. Edit: Sorry, that was really poorly worded and a mess up, Let me try that again: The polarity of the induced flux is ultimately independent of whether the coil is wound clockwise or counterclockwise for a given, time varying original flux.

    But the induced emf can still be either positive or negative depending on whether the flux through the inductor is decreasing or increasing.

    I think I understand what you are saying though: The final result in terms of voltage and current relationships do not matter on the "polarity" (counterclockwise vs. clockwise wound) of the inductor/coil. That I also agree with.
     
    Last edited: Feb 5, 2015
  10. Feb 5, 2015 #9

    collinsmark

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    Oh! I might have misunderstood. You may wish to ignore my last post then.

    What I though we were talking about in my last post was about the situation where you have a coil within some time-varying magnetic field that is generated externally.

    But that's not necessarily relevant if the coil inductor is just sitting there, connected in a circuit.

    For that situation, you do need to use [itex] \varepsilon = L \frac{di}{dt} [/itex] And that's probably all you need.

    You can switch back and forth between the voltage across the inductor and the current using that equation alone.

    The term "induced emf" doesn't really apply here, because in this situation we are not applying an external magnetic field.
     
  11. Feb 5, 2015 #10

    collinsmark

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    If you want the flux in an inductor in the absence of an external magnetic field, although with an applied voltage, skip the negative sign, and just use [itex] \varepsilon = \frac{d \Phi}{dt} [/itex].

    But again, when analyzing an inductor in an AC circuit, probably don't need to worry about the flux polarity, in which case [itex] v = L \frac{di}{dt} [/itex] is probably all you need.

    Sorry if I was confusing earlier. It's getting late. :)
     
  12. Feb 5, 2015 #11

    Andrew Mason

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    A problem clearly stated is a problem half solved. The question you are asking is not clear.

    If a magnet is passed through the centre of the coil along the axis of the coil, from right to left say, the current in the coil will produce a magnetic field that opposes the motion of the magnet ie. pushes the magnet from left to right - regardless of how the coil is wound. So the direction of the current around each loop of the coil has to be the same regardless of which way the coil is wound. Whether the current goes from right to left or left to right, however, depends on how the coil is is wound. For us to help you resolve the difficulty you are having you will have to describe more precisely what you did and what you observed.

    AM
     
    Last edited: Feb 6, 2015
  13. Feb 7, 2015 #12

    Andrew Mason

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    What do you mean when you refer to a "single coil" excited by an AC source? Are you are talking about an AC transformer to which there is an AC source connect to the primary? With an AC source on the primary, the secondary emf would be constantly changing direction. Perhaps you could explain what you mean by the "polarity" of the induced emf. Polarity would only have a meaning relative to the input AC source (ie in-phase or opposite phase).

    AM
     
    Last edited: Feb 7, 2015
  14. Feb 7, 2015 #13
    I thought I was clear in my problem statement but I guess not. Ok the specific application I am confused about is a watt meter. For those of you that don't know what a wattmeter is: it measures power. A current coil is placed in the circuit and a potential coil is placed across the load. Now all watt meters have polarity markings on them meaning you have to hook up these coils in a specific way or else you will get a negative voltage reading and the angle between current and voltage will be wrong. So this leads me to believe that the direction in which you wind the coil matters because the only thing you can really do is wind them clockwise or counterclockwise. Are ya with me so far??? I am not referring to mutual induction!!!! I am NOT referring to moving a magnet near a coil!!! So just to reiterate: the polarity markings on a wattmeter lead me to believe that the induced voltage (opposing an increase in current) in a coil is dependant on the direction of winding a coil.... otherwise if this is not correct please tell me why wattmeters have polarity markings?
     
  15. Feb 7, 2015 #14

    Andrew Mason

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    You have almost answered your question. To measure power you need two coils (1. a current transformer in series with the circuit having a very low resistance primary with few windings and many windings in the secondary 2. a voltage transformer which has a high resistance primary with many coils. Actual power is the product of the voltage x current using RMS values for each.) If the current and voltage are 180 degrees out of phase you will get a negative power reading. So when connecting the two coils into the circuit to measure power you have to make sure that the leads are not reversed. It is even more important if you are measuring 3 phase power. You will need a set of current and voltage coils for each phase.

    AM
     
  16. Feb 7, 2015 #15
    I don't think that's how it works. You only need two watt meters to measure 3 phase power. It's pretty simple to measure power. Vrms*Irms*cos theta, where theta is the angle between V and I. Now you still haven't answered my original question. How does the direction of winding (clockwise or counterclockwise) affect the polarity and if you say it doesn't, why do watt meters have polarity markings on them?
     
  17. Feb 7, 2015 #16

    Andrew Mason

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    You can do it with one meter. But the meter has to be able to determine the voltage and current in each phase. You may be able to do this by relating the third phase to the measurements of the other two.

    ?? What matters is the direction of the current in the coils (ie the direction of the 4 fingers using the right hand rule with the thumb pointing to the instantaneous north pole). That direction is independent of the whether the coil is wound clockwise or counter-clockwise. It depends only on the rate of change of flux in the area enclosed by the coil.

    If you were to have two secondary coils, one wound clockwise and the other wound counter-clockwise starting from the same place and ending in the same place, the leads at each end would always have opposite polarity (ie. would be 180 degrees out of phase) at any given instant. The current flowing in those coils would be in the same direction around the coil but opposite directions with respect to the leads.

    If the current is 180 degrees out of phase with voltage, you get -1 for cos theta. That is why the configuration of the leads from the voltage coil and the current coil matters. The correct polarity would be where the voltage and current are in phase. If you were to then switch one set of leads, the current measurement would be 180 degrees out of phase with the voltage. To avoid that, the meters have polarity markings. (Since there is no constant + or - in an AC circuit, the polarity of the leads from a coil only has meaning in relation to the phase of another coil).

    AM
     
  18. Feb 7, 2015 #17
    you cannot measure 3 phase power with one meter, its called the "two watt meter method" for a reason. You haven't really explained anything. Your second paragraph contradicts your first. First you say that the direction of current is independent coil winding. then in the second paragraph you go on to say that you would have two different polarities from two coils with different winding directions.
     
  19. Feb 8, 2015 #18

    Andrew Mason

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    Just a suggestion: try and think through what I have said. There is no contradiction. Draw the coils and work it out using the right hand rule for coils. I think you will realize why the right hand rule does not require you to take into account which way the coil is wound. When you have a clear question, let us know.

    As far as the "meter" is concerned, you will have to define what you mean by a meter. A single physical device with multiple elements can be used to measure 3 phase power.

    AM
     
  20. Feb 8, 2015 #19
    I did workout everything with the right hand rule and I did come to the conclusion that the direction of winding doesn't matter. But WHY do meters have polarity markings on them. I just figured out using the right hand rule that the induced emf is always going to be the same no matter how the coils are wound. As I have said before a watt meter is a device that measures power it has one current coil and 1 potential coil. so if the induced emf is always the same then there isn't going to be a cos(180) if you flip around the coils.
     
  21. Feb 8, 2015 #20
    I think I've come to the realization I wont get a satisfactory explanation over the internet so no need to reply. I think I am going to have to pester some electrical engineering profs.
     
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