magnetic flux and mean path length

Discussion in 'Homework Help' started by PG1995, Mar 26, 2012.

  1. PG1995

    Thread Starter Active Member

    Apr 15, 2011
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    Hi

    Please have a look on the attachment. Please help me with the query. Thank you.

    Regards
    PG
     
  2. PG1995

    Thread Starter Active Member

    Apr 15, 2011
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    Hi

    It seems mean length is used for corner values. e.g. 30 + (2*15)/2 = 30. But why is mean value used for corners only, why not mean of entire length used? Please let me know. Thanks.
     
  3. steveb

    Senior Member

    Jul 3, 2008
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    Think about how the field will flow. Some of the field will be on the outer perimeter, while some of the field will be on the inner perimeter. Doing an exact field calculation is complex and time consuming, so people use the approximate method of considering the mean path length. The field is assumed to be spread out, but on average the path around is more in the middle. So, just imagine a line (as they show in the figure) that goes through the middle fo the core. The shape is rectangular and you know how to find the perimeter of a rectangle. Only, here you do not consider the entire perimeter as one piece because the cross sectional area is not the same over the whole perimeter. So, they break it into two pieces.
     
    Last edited: Mar 26, 2012
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  4. PG1995

    Thread Starter Active Member

    Apr 15, 2011
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    Thank you, Steve.

    Please help me with the queries in the attachment. If you like you can combine both questions. I have unsuccessfully tried to find some decent diagram on the net using the search words "voltage on conductor moving in magnetic field" Thanks a lot.

    Best wishes
    PG
     
  5. steveb

    Senior Member

    Jul 3, 2008
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    For Q1, it appears that the wire is laid flat on the page and is not perpendicular to it. This is clear in the picture and it is also consistent with the wording you highlighted.

    For Q2 (which relates back to Q1) I think your confusion is that you think that the vector vXB is perpendicular to the page. This is not true. The vector vXB must be perpendicular to the both v and B. The vector B is into the page and the velocity is in the page to the right. Hence, vxB must be in the page and facing toward the top of the page as shown in the diagram.
     
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  6. PG1995

    Thread Starter Active Member

    Apr 15, 2011
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    Thanks a lot, Steve.

    Actually, I just came back to let you know that I understand it now but you had already replied. You are right that I was thinking that direction of vXB is out of page.

    Once again, thank you for the help.

    Best regards
    PG
     
  7. PG1995

    Thread Starter Active Member

    Apr 15, 2011
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    Hi :)

    Please have a look on the attachment and please help me with the query. Thank you.

    Regards
    PG
     
  8. steveb

    Senior Member

    Jul 3, 2008
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    You should keep two facts in your mind. First, the method suggested here is approximate; and, second, the two answers you mentioned are very close to each other in value (< 3 % error).

    What this means is that your alternate method is not really wrong because is tries to approximately consider the path length. However, the method suggested is considered the proper way and it is based on the logical idea that the cross sectional area that is perpendicular to the path is the best estimate to use. So, if you want to apply that method, with the understanding that it also is not a perfect method, you should obey the rules setforth.
     
  9. PG1995

    Thread Starter Active Member

    Apr 15, 2011
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    Thanks a lot.

    Q1: Do you mean magnetic flux path, "perpendicular to the path"? Could you please elaborate on this? Thank you.


    Q2: Please have a look on the attachment and please help me with the query there. Many thanks.

    Regards
    PG
     
  10. steveb

    Senior Member

    Jul 3, 2008
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    Yes, the flux path.

    Consider any rectangular path. Take each side. It has a length and a cross sectional area associated with it. Don't mix and match the lengths and the cross sectional areas. Take each side-length entirely and associate one cross sectional area to it.



    He is calculating the reluctance of each air gap separately, and he calls this Ra. Then, he includes two separate air gaps in the schematic drawing Ra1 and Ra2.
     
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  11. PG1995

    Thread Starter Active Member

    Apr 15, 2011
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    Thank you very much, Steve.

    Best wishes
    PG
     
  12. PG1995

    Thread Starter Active Member

    Apr 15, 2011
    753
    5
    Hi

    Please have a look on the attachment and kindly help me with the query. Thanks a lot.

    Regards
    PG
     
  13. steveb

    Senior Member

    Jul 3, 2008
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    A Weber is a unit that measures the magnetic field density integrated over an area. The entity being quantified is a property of the magnetic field, and a region of space where you define a closed loop to which you may assign any surface area.

    When you apply Faraday's law to find the total voltage over the encircling path, you are interested in the flux that is linked by the path. If you were only allowed to encircle the magnetic field once, then there would be no ambiguity, but there is nothing to stop you from going around more than once. If you were to go around twice, you would get twice the voltage compared to going around once, yet the field was the same in the same region of space in both cases.

    The above case of two encirclements allows two possible interpretations. You can think of one circle and one surface with two turns, or can you try to imagine one path of double the length that defines a spiraled surface that cuts the flux twice.

    So, the unit of Weber turns allows both interpretations without changing the fundamental unit of Weber. Remember, "number of turns" which must be an integer has no real units.

    I'm sure this is still confusing, because it confuses me too. So, let's walk through it again.

    Imagine a wire that wraps a changing flux 2 times. There are two ways to calculate the voltage across the wire.

    1. Consider the wire as wrapping an area twice. Calculate the flux change in the area in units of Webers/s and then multiply by the two turns. Hence, you have F Webers/s times 2 turns for 2F Weber-turns/s.

    2. Consider the wire as defining a spiraled surface that cuts the flux twice. Calculate the flux change over the new area (which is now twice as large) and you get 2F Webers/s. However, since you defined this path to be one total path (not two turns of the same path), the number of turns is one. Hence you have 2F Webers/s times 1 turn for 2F Weber-turns/s.

    Note that units of volts is the same as Webers/s which is the same as Weber-turns/s.

    In a nutshell, the unit Weber-turn is invented as a way to allow people to think of one area of flux and number of encirclements. Without this allowance, we would be forced to imagine these spiraled surfaces cutting the flux multiple times, rather than number of turns. It is a little difficult to visualize the spirals and the way they define areas that cut the flux more than once, but perhaps this is the better way to think of it if you have trouble understanding Weber-turns. Using spiraled surfaces forces the number of turns to always be one.
     
    Last edited: Apr 3, 2012
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  14. PG1995

    Thread Starter Active Member

    Apr 15, 2011
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    Thank you, Steve, for the previous post. I do have some follow on questions about it but it's better to ask them later, and I was happy to know that my confusion was right.

    Please have a look on the attachment. Could you please help me with the query there? Thanks a lot.

    Best regards
    PG
     
    Last edited: Apr 6, 2012
  15. Kermit2

    AAC Fanatic!

    Feb 5, 2010
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    AVERAGE VALUE - The average value of a sine wave of voltage or current is the average of all the instantaneous values during one alternation. The average value is equal to 0.636 of the peak value.




    EFFECTIVE VALUE - The effective value of an alternating current or voltage is the value of alternating current or voltage that produces the same amount of heat in a resistive component that would be produced in the same component by a direct current or voltage of the same value. The effective value of a sine wave is equal to 0.707 times the peak value. The effective value is also called the root mean square or rms value.


    I don't know if this will spark anything for you or not. I just quickly read the post and thought it might.

    The 1.11 comes from the two values above. .707 divided by .636 is 1.11
     
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  16. steveb

    Senior Member

    Jul 3, 2008
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    I agree with Kermit.

    PG, I also see your point about the average being zero over the span indicated by the author. This is misleading as you say. Normally average value means the average value of one half cycle, but it is the average of the positive half cycle, not the half cycle that has values both positive and negative.
     
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  17. PG1995

    Thread Starter Active Member

    Apr 15, 2011
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    Thank you, Kermit, Steve.

    1: Average value of a sinusoidal wave: Does multiplying by 0.636 give average of one half cycle or complete cycle? I understand that RMS or effective value is average of one complete cycle.

    2:
    The author also says "But for a sinusoidal wave the r.m.s. or effective value is 1.11 times average value,...". Why is so? If the answer involves some kind of derivation, then just tell me and I will memorize the fact. Thanks a lot for the help.

    Regards
    PG
     
  18. steveb

    Senior Member

    Jul 3, 2008
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    What don't you understand about Kermit2's answer? The RMS value is 0.707*A, where A is the amplitude, and the average value is 0.636*A. Hence, the ratio of RMS to average value is 1.11.

    Are you wondering where the 0.707 and 0.636 numbers come from? The RMS is the square root of the mean of the square of the wave. The average is the mean of the absolute value of the wave.
     
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  19. PG1995

    Thread Starter Active Member

    Apr 15, 2011
    753
    5
    Hi

    Please have a look on the attachment and please help me with the query there. I understand that my query is not as clear as it should be but I have a test tomorrow and still have so much to do. I hope you would understand my question as you mostly do. Thanks a lot.

    Regards
    PG
     
  20. steveb

    Senior Member

    Jul 3, 2008
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    You said "... the voltage supplying the current would need to push the current with more effort to generate more flux ... you can't simply say that you can double the magnetic flux density by just doubling the number of turns."

    You seem to be inspired to invent your own laws of physics even though your views are not supported by experiment. :eek:

    What if you had a superconductor? Wouldn't the current flow without loss or effort? Would you even need a voltage, once the current loop was established?

    Maxwell's equations provide a concise description of how electromagnetic fields behave. One of those equations is Ampere's law, which in the static case looks as follows.

    \oint \vec{H} \cdot d\vec{l}=\int\int \vec{J}\cdot d\vec{s}=I_{pen}

    , where H is the magnetic field, J is the current density and Ipen is the current that penetrates the area that J is distributed over.

    This is saying that the line integral of H around a circuit inside the core, is equal to the current that penetrates the area that is bounded by that circuit path. It is clear that more loops of wire will result in more current that penetrates the area. This is a law of nature and you can't really argue with it unless you develop and experiment that contradicts it. What you view as intuitively correct has no bearing on reality.

    I think what your mind is doing is considering conservation of energy, and you are correct that the magnetic field contains energy. However, it is potential energy or stored energy that it contains. So, there is a dynamic process involving voltage and current that helps establish the field, but once it is established, no voltage is required (in principle) to maintain it. The fact that a real system has voltage driving the coil is due to losses such as wire resistance.
     
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