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How to calculate magnetic vector potential?

  1. Apr 19, 2015 #1
    Source WIki:
    An axially symmetric toroidal inductor with no circumferential current totally confines the B field within the windings, the A field (magnetic vector potential) is not confined. Arrow #1 in the picture depicts the vector potential on the axis of symmetry. Radial current sections a and b are equal distances from the axis but pointed in opposite directions, so they will cancel. Likewise segments c and d cancel. In fact all the radial current segments cancel. The situation for axial currents is different. The axial current on the outside of the toroid is pointed down and the axial current on the inside of the toroid is pointed up. Each axial current segment on the outside of the toroid can be matched with an equal but oppositely directed segment on the inside of the toroid. The segments on the inside are closer than the segments on the outside to the axis, therefore there is a net upward component of the A field along the axis of symmetry.
    1.PNG
    2.PNG
    The E and B fields can be computed from the A and cd014731964c742c274df08d7cc238fb.png (scalar electric potential) fields

    6b767392e32fd1dc6b95c651b28cd324.png [8] and : 29bb4c095ea3dd6ddfa37f81a68ad953.png [8] and so even if the region outside the windings is devoid of B field, it is filled with non-zero E field.
    The quantity 7f32f1a07c684049f6f2bbd5929594bd.png is responsible for the desirable magnetic field coupling between primary and secondary while the quantity 34a78e8d1464a0318308fa582aaba3fc.png is responsible for the undesirable electric field coupling between primary and secondary.


    What I'm wondering is how do you actually calculate 'A' the mag vector potential in the first place?

    Thanks
     
    Last edited by a moderator: Apr 15, 2017
  2. jcsd
  3. Apr 19, 2015 #2

    Baluncore

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  4. Apr 19, 2015 #3
    Thanks for the reply Baluncore, do you think you could help me put that equation in perspective. Say you're putting a current through the red wire of some number N turns below, you've got B inside the core, so A is up through the centre of the toroid which we say is the z-axis. You'd have a 'J' current (density?) circling like in the previous picture but as far as that equation is concerned, is that the current (density) flowing through the wire? (which we can measure with an ammeter and calculate the J from the diamter of the wire)
    And how do or should I define r the position vector? Should the defined origin be along the z-axis and if so should z = 0 be in the middle of the toroid, so O(0,0,0)?
    oh.png
    Any chance you should show me a dummy example of evaluating that expression for A? I'm having a bit of trouble understanding it. What does that d3 mean, is it a triple integral?
    Thank you
     
  5. Apr 20, 2015 #4

    dlgoff

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    Yes, r is the position vector from a current density volume element to a vector potential point of interest; d3r is the current density volume element. A is not just along the Z-axis.

    vecpot3.gif
    image compliments of http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/magvec.html
     
  6. Apr 20, 2015 #5
    That is a helpful post, thanks. Glad to know I don't have to do a triple integral. Say I defined the point of origin halfway in the toroid on the centre axis as below (A is in the z direction):
    oh2.png
    So the 'current density volume element' in my case that's the (red) current carrying wire wraped around the toroid right? Not some currend flowing in the surface of the toroid? I''m still having trouble adapting my model to your equation because in mine the current is circeling the toroid vertically on the Z-Y plane and only makes one turn on the X-Y plane every 360o and in practice I'd cancel that out with a retrun wire so I want to ignore that. In yours the current is circling on the X-Y plane.So J would be the current 'I' in the wire divided by the cross secional area of the wire?
    If so what does this mean for dS and dl?
    Thanks!
     
  7. Apr 20, 2015 #6

    dlgoff

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    dS is the toroids cross-sectional area (see your first picture; the area contained in afce) and dl is the infinitesimal distance as you integrate around the toroid.
     
  8. Apr 20, 2015 #7
    Yeah but the difference is that toroid has the A vector tangential to the toroid and mine has the A vector pointing up inside the toroid and down outside. So it's kind of like the toroid in that picture on it's side rotated 360o, for each turn of the wire around another toroid core. but in my case the wire is a rectangle not a toroid as in your picture. That's why I thought dS might have been the cross sectional area of the wire.

    So what do I do for 'r' for some arbitary point on the z-axis? I assume r is all the same magnitude on everywhere on the z-axis? And does that mean in my case:
    A = μ0/4π * (j current density of wire) * (cross sectional area of toroid) * 2π*(mid toroid radius) / r
    I was thinking I don't need to integrate around dl because it's a circle, hence the 2π*radius?

    Should r be the same as the middle of the toroid radius? (in which case they'd cancel out) Or perhaps dl is 2h+2R in my diagram? (where R is the width of the toroid as in the diagram)
    In which case it would be: A = μ0/4π * (j current density of wire) * (cross sectional area of toroid) * (2h + 2R) / r

    Remember about those gold arrows in the first picture that it says "The axial current on the outside of the toroid is pointed down and the axial current on the inside of the toroid is pointed up. Each axial current segment on the outside of the toroid can be matched with an equal but oppositely directed segment on the inside of the toroid. The segments on the inside are closer than the segments on the outside to the axis, therefore there is a net upward component of the A field along the axis of symmetry."

    You can see why I'm still not sure what the equation will be. I guess its like the A I'm looking for is the sum of the A from your equation.
    Thanks a lot
     
  9. Apr 20, 2015 #8

    dlgoff

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    I think you are confusing the toroid with a current loop.

    1) The magnetic field from current in a straight wire:

    magcur.gif

    2) The magnetic field from current in a loop of wire:

    curloo.gif

    3) The magnetic field from current in a coil of wire:

    magcoi.gif

    Now if a coil is around a toroid, as you have in your situation, there's a magnetic field of the toroid.

    tor3.gif


    You should now spend some time on these links and try to understand how the magnetic vector potential is determined anywhere in space relative to your toroid.

    Perhaps there are other members here that can explain this better than I.

    Edit: I'm having trouble getting the images to show up but you can find them in the referenced links I gave from http://hyperphysics.phy-astr.gsu.edu/hbase/Indxback.html [Broken]
     
    Last edited by a moderator: May 7, 2017
  10. Apr 21, 2015 #9
    Yeah I'm already familiar with those examples, the problem is that those pictures are different than the instance I'm analylising, I think you're confusing what the toroid is in each case, as I said: 'the difference is that your toroid is the current wire, and has the A vector tangential to the toroid. Whereas mine has the A vector pointing up inside the toroid (as my toroid is the magnetic core) and A goes down outside the toroid. So it's kind of like the toroid in your picture is on it's side rotated 360deg (for each turn of the wire around another toroid core).'
    Thanks anyway
     
    Last edited by a moderator: May 7, 2017
  11. Apr 21, 2015 #10

    jim hardy

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    Vector Calculus was the course that showed me the limit of my abilities. It drove me to my knees, i passed only by memorizing techniques i didn't understand. So take following with a grain of salt.

    This image from Baluncore's link
    660px-Magnetic_Vector_Potential_Circular_Toroid.png
    is followed by this text
    It'd take me days to figure out that vector calculus.

    I'd start with this observation:
    Red circles, A, appear to represent the external magnetic field of a toroid, outside the spiral winding.
    It of course is way smaller than the field inside the spiral winding, along its axis.
    Let's just pick that purple one on right half.
    Mr Gauss tells us that closed loop integral of h dot dl = current enclosed.
    I can accept that because it's how a clamp-on ammeter works.
    So if i walk that purple circle have i enclosed any current ?
    Why yes, because the current in that winding has corkscrewed its way from in front of the paper to behind the paper.
    Viewed from the top the current in the winding would corkscrew its way completely around the toroid, counterclockwise.
    That amounts to one turn in the plane of the toroid.
    And that's why toroids can have a small external field.
    Some folks wind one turn around the toroid in the direction to cancel out that mmf from the corkscrew current's one turn in plane of core.

    When the formula that i derive reflects that observation, i would have confidence i'd done it correctly .
    But until then i'd know vector calculus was still mastering me , not the other way 'round.

    Does that help any ?

    That corkscrew current indeed is equivalent to a current flowing around toroid's surface or maybe even at its centroid in cross section . I'll guess the math would say surface or centroid give same result. Biot-Savart ?

    VC is my valley of death and i am sore afraid of it.

    old jim
     
    Last edited: Apr 21, 2015
  12. Apr 21, 2015 #11
    Thanks Jim,
    Yes, in that image you reference B is inside a core, so A is coming out like a corkscrew (right hand rule)
    Did you see the image:
    1-png.82237.png
    Just assuming that B is cancelled externally, and is confined internally, the B in my example in my 2nd and 3rd posts is caused by a current in a wire wrapped around the toroid (the gold arrows) and the wiki article states about that above picture "Arrow #1 in the picture depicts the vector potential on the axis of symmetry. Radial current sections a and b are equal distances from the axis but pointed in opposite directions, so they will cancel. Likewise segments c and d cancel. In fact all the radial current segments cancel. The situation for axial currents is different. The axial current on the outside of the toroid is pointed down and the axial current on the inside of the toroid is pointed up. Each axial current segment on the outside of the toroid can be matched with an equal but oppositely directed segment on the inside of the toroid. The segments on the inside are closer than the segments on the outside to the axis, therefore there is a net upward component of the A field along the axis of symmetry."
    So I don't know if full-on vector calculus to calculate A is necessary. (I'm interested in calculating A because I want to calculate the B field flowing inside the toroid from B = ∇xA)

    dlgoff gave a nice formula for A in his first post which I think I can apply, but as in my 3rd post I'm just not sure how the variables interchange. Because the difference between the formula in his current loop and mine is that his toroid IS the current wire, and has the A vector of that is tangential to his toroid (the current path). Whereas my toroid is a magnetic core and has the A vector pointing up inside the toroid and A goes down outside the toroid (as the above wiki quote explains). So it's kind of like the toroid in his formula is on it's side rotated 360deg around my magnetic core (that is his toroid is each turn of the wire around another toroid core, the golden arrows making a rectangle in the aboe picture).

    Does that clarify my point, or am I wrong about something?
    Thanks again for the input!
     
  13. Apr 21, 2015 #12

    jim hardy

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    Caution: this post was written under the mis-conception that A was some form of mmf , not "magnetic vector potential" which it turns out has quite a different meaning.
    sorry about that, old jim



    My bifurcated brain has two hangups that'll take me a few hours to resolve. I mention them so you'll understand my plight.

    1. In top picture post 10 we have current density j, in bottom picture post 11 we have current i. There are more square inches on the outside of a toroid than on the inside, so if amps per square inch is same both inside and out then how do i square that with Kirchoff?
    Chewing on that as we speak. Expect resolution when go out to work on my TV antenna for my alleged brain thinks best in background..

    2. This picture dlgoff linked shows current flowing around the donut core in a circle, in one plane.
    In fact a toroid has such a current because the winding progresses all the way around the core, just it's not a nice circular path but a spiral path that encircles the core. Imagine cutting the core and straightening it out - it's an electromagnet with winding along its whole length.

    vecpot3.gif
    Now when i look at the picture you posted in 11, my alleged brain jumps to this thought experiment:
    What if those two current loops afce and bhdg were not part of a spiral winding that circumnavigates the toroid , instead were single turns in the plane of the paper powered by individual batteries?
    Now there'd be no current traversing the circumference of the toroid, only those two loops n the same plane.
    So my closed loop integral(i assume that;s what the circle in middle of ∫ sign means) should traverse those current loops in plane of paper, not the loop made by toroid core which goes behind and in front of the paper.
    So the problem has changed. In right hand equation above, our dL is now rotated 90 degrees from what it was. Moving A out of the plane of the paper, into the plane of the toroid.

    When i first looked at this image i said "something doesn't seem right."
    1-png.82237.png

    Here's what it is.
    Apply right hand rule to ANY of those eight yellow arrow currents.
    All eight of them create mmf in toroid that's ccw viewed from above.
    I don't see a vertical mmf component along A from the vertical currents, because the current loops are in the same plane as A loops.

    and everywhere i look i see mmf only normal to plane of loop where it passes through that plane
    dipole.jpg
    (Sorry, image should be flipped so current loop is in plane of paper)
    and if this formula is any good
    http://www.netdenizen.com/emagnettest/offaxis/equation2.offaxisloop.img.png

    then there is no radial component in the plane of the loop because anywhere in the plane of the loop ϒ is zero.

    I'm condemned to this meager sort of understanding.

    If i'm wrong please point it out.

    So all that having been said ,

    i reject this sentence
    as bilgewater.

    stricken paragraph was based on misconception A is mmf. I don't know what is A or how it's calculated , i withdraw the assertion of "bilgewater" jh

    The upward mmf field results from the one effective turn in plane of core carrying current around that loop , aforementioned "corkscrew current".
    Extreme quality toroid transformers wrap a single turn circumferentially around the whole core in direction to cancel that corkscrew mmf.

    make sense ?

    old jim
     
    Last edited: Apr 22, 2015
  14. Apr 21, 2015 #13

    jim hardy

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    Sorry for typos above.

    Congratulations to you and dlgoff, between you i learned quite a bit.. my alleged brain is smoldering now.

    nice job all.
    see also this link for off axis field.
    https://www.wakari.io/sharing/bundle/ericdennison/magnets/Off Axis Field of a Current Loop.ipynb

    i shouldn't whine so about vector calculus for i have highest respect and admiration for those who are able, and i woudn't want anyone to think otherwise.. just please don't anybody think i'm a resource for it.

    old jim
     
  15. Apr 21, 2015 #14

    jim hardy

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    In this toroid current I enters at left wire and exits at right wire, having circumnavigated the core ounterclockwise , albeit like a corkscrew ,
    So by right hand rule there's I X one turn of mmf pushing flux up out of the page,
    tor.gif
    image at http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/toroid.html
    i shudder at the thought of writing the equation for that spiral path.

    that can be corrected so there's no net z axis mmf

    from http://physics.stackexchange.com/qu...ly-confined-within-the-core-of-a-toroid-but-n
    in this toroid, current entering on red wire progresses around the core clockwise making I X one turn of mmf as before, but this time into the page. Notice that red wire turns white and goes back around the core counterclockwise, that I X one turn cancels red wire's mmf.


    3nkTa.jpg

    That's why i think the wiki statement in post 1 about mmf resulting from different radii for inside vs outside of turns is wrong.

    my mistake, wiki was talking abut "A" not about mmf
    MMF appears because the direction of current flow has a circumferential component.
     
    Last edited: Apr 22, 2015
  16. Apr 21, 2015 #15
    I need to read your Post #12 again, I only just then skimmed through it, but I must admit I did have my doubts about the wiki statement, but keep in mind B is the Curl of A, (B inside the core). So even though I doubt it, I am still struggling to accept that it's "Bildgewater" yet. Maybe on reflecting on what you said again it'll click with me. Are you saying that there is no A on the axis of symmetry?
    Flipping this all on it's head, say instead of calculating B from A maybe you did A from B, (would) the only otherway you were calculating B inside the core would be:
    B = (Number of turns * Current / Reluctance ) / cross sectoinal area of core
    wouldn't it?
     
  17. Apr 21, 2015 #16

    jim hardy

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    1. What the heck is "A" ? Is it a mmf? If so, where are the amp-turns? They're in the circumferential component of current in the spiral winding.
    In this picture
    1-png.82237.png
    there is no mmf in direction of red arrow A because there's no circumferential component in either of those gold rectangles.

    In this picture (which was your link not baluncore's like i thought)
    wikimedia.org%2Fwikipedia%2Fcommons%2Fthumb%2F4%2F45%2FMagnetic_Vector_Potential_Circular_Toroid.png

    B is equated to zero, so no flux means no net mmf.... so what the heck is A ? Have i missed not only the boat but the whole pier ?
     
  18. Apr 21, 2015 #17

    jim hardy

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    that looks right. B X area = Φ = mmf/ℝ

    old jim
     
  19. Apr 21, 2015 #18
    Hee hee hee, I'm still only half way through re-reading your post #12.
    But A is the (vector) magnetic potential, I think of it as like when you imagine the current is your right thumb and the curl of your fingers is the B field circling around it. Same thing with A as far as I know, imagine A is your thumb and B is your fingers, or vice versa, as far as I know A is often parallel to current or like the tangent of B. I've just been going from this really:
    http://www.feynmanlectures.caltech.edu/II_15.html
     
  20. Apr 21, 2015 #19

    dlgoff

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  21. Apr 21, 2015 #20

    jim hardy

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    just glanced at your feynman lecture
    Missed the pier ? Heck, I'm in the wrong harbor....

    Seems to me A is one step removed from H. A surrounds mmf just as mmf surrounds current.

    Part of becoming wise is learning what it is that i don't know.
    I have to leave such math to the children of greater gods.

    I'm sorry. Hope i didn't distract you from your studies.


    old jim
     
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