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Question in current equation in small loop.

  1. Jun 30, 2013 #1
    This is part of the derivation of vector magnetic potential of a small current loop center at origin and on xy plane. The current is stated to assumed to be constant along the loop. Attach is the scanned copy of the page in question.



    [PLAIN]http://i43.tinypic.com/5f2q7d.jpg[/PLAIN]


    I cannot understand why the author equate the current like this. It is constant along the loop, every other book I've seen just use I0 as constant and take it out of the integration.

    Furthermore, in equation (5-5), the xyz components are represented in cylindrical co and using [itex]\phi '[/itex] to show it's the SOURCE coordinates. Then equation ( 5-6) shows the general transformation of Spherical coordinates to xyz coordinates. (5-6) does not have [itex]\phi ',\;\theta '[/itex]. But then the author combine (5-5) and (5-6) into (5-3) and the result is (5-7) that contain both [itex]\phi ',\;\theta '[/itex] together with [itex]\phi ,\;\theta [/itex]!!! This is not correct to me as this is only concerning the current along the loop, there is no involvement of any FIELD POINT in this equation. I don't even see why the author put source and field angles into the current equation.

    to me

    [tex] I_{(x',y',z')}=I_{(x',y')}=\hat \phi I_0= -\hat x I_x \sin\phi '+\hat y I_y\cos \phi '[/tex]


    Where all angles are source angle with '. My point is there should never be any FIELD angles in the current loop equation as this is only about the current.

    Please help.

    [EDIT]: I read it over and over, the only way that can even make sense is if the current is observed AT the FIELD POINT. But this is very weak!!!
     
    Last edited: Jun 30, 2013
  2. jcsd
  3. Jun 30, 2013 #2
    I doesn't make sense to me either, can you give as the book's title and author name?
     
  4. Jun 30, 2013 #3
    It is antenna theory by Balanis 3rd edition. p233. Or p204 to 206 of the 2nd edition. It's a very famous book for antenna and he wrote an advanced electromagnetics text book also.

    Only thing I can even think of is if the author consider the differential current vector viewed at the OBSERVATION or FIELD point.

    Thanks
     
    Last edited: Jun 30, 2013
  5. Jun 30, 2013 #4
    How did you get your current equation? can you show, because it isn't obvious to me :/
     
  6. Jun 30, 2013 #5
    Started with (5.3) in xyz coordinates. Then transformed into cylindrical coordinates by (5.4). Then the radiated fields are in spherical coordinates using (5.6) to transform ##\hat x,\;\hat y,\;\hat z## into spherical unit vectors.

    Then appley both back to (5-3) and get (5.7)!!! There is nothing complicated with the procedures. It just does not make sense at all!!!
     
  7. Jun 30, 2013 #6
    I don't get what doesn't make sense?
     
  8. Jun 30, 2013 #7
    I know!!! I would chuck it to mistake if it were not such a famous book!!!
     
  9. Jul 1, 2013 #8
    Here is the next part of the the book, it just use some reasoning ( excuses) to shrink the formula down from (5-7) to (5-8) then to (5-13). You can see this is just continuation of (5-7) from my original post.


    [PLAIN]http://i42.tinypic.com/sb3akw.jpg[/PLAIN]


    Now you have just as much information as me, this is the whole thing!!!
     
  10. Jul 1, 2013 #9
    I tell you that I can easily obtein equation (5-14) bu still don't have a clue of what he is doing. I will try to down load the book a take a look at it, in the meantime, if you get to understand what he is doing please let me know.
     
  11. Jul 1, 2013 #10

    vanhees71

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    Obviously he simply solves the Helmholtz equation for the vector potential in Lorenz gauge for a current distribution,
    [tex](\Delta+k^2) \vec{A}=-\mu \vec{J}.[/tex]
    To specify the solution you further need that at infinity the waves should be outgoing. Obviously the book uses the sign convention [itex]\exp(+\mathrm{j} \omega t)[/itex]. So the Green's function reads
    [tex]G(\vec{x},\vec{x}')=\frac{1}{4 \pi |\vec{x}-\vec{x}'|} \exp[-\mathrm{j} k |\vec{x}-\vec{x}'|][/tex]
    for the outgoing solution. Then you get
    [tex]\vec{A}(\vec{x})=\frac{\mu}{4 \pi} \int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x}' G(\vec{x},\vec{x}') \vec{J}(\vec{x}').[/tex]
    The rest is just simplifying this expression for the case of a infinitesimally thin loop. There you have
    [tex]\mathrm{d}^3 \vec{x}' \vec{J}(\vec{x'})=I \mathrm{d} \vec{l}',[/tex]
    where [itex]I[/itex] is the current through the wire and [itex]\mathrm{d} \vec{l}'[/itex] is the infinitesimal line element along that wire. I think the rest is just the calculation of this integral for a circular loop in the xy plane using cylinder coordinates.

    CAVEAT: You have to write everything still in Cartesian components, do the integral first and then project to the appropriate cylinder-coordinate components!
     
  12. Jul 1, 2013 #11
    By the way that is one of the things that don't make sense, since, the book seems to integrate in curvilinear coordenates. Of course we know the book must be right so the problem is we don't understand what the book does and the question is centered in this issue:"to understand what the book is doing".
     
  13. Jul 1, 2013 #12
    I have another book that explained very clearly how to arrive to the final current equation. Basically the ##Id\vec {l}'=\hat \phi dl'=-\hat x \sin\phi'+\hat y \cos\phi'##.

    Since the current is ##\phi## independent, we choose the observation point along y axis. By drawing the vector components out, all the y components CANCEL out, leave only the x component. Therefore
    [tex]\hat \phi'=-\hat x sin\phi'[/tex]
    when observe along y axis.

    I am just totally puzzled by this book. No, books do make mistakes, mostly typos. But this is a very well respected book and is the bible for antenna engineers.
     
  14. Jul 1, 2013 #13

    vanhees71

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    I only know the here copied pages of this book, but frankly I'd not recommend to use it to learn about electromagnetic radiation. First of all it seems to make this elementary mistake with integrating naively over curvilinear components, but I haven't checked whether there is a mistake in this result. What I find worse is this awkward notation, not writing clearly out the integration variables (primed in my last posting) and the free variables parametrizing the point, at which you evaluate the field (unprimed in my last posting). There are plenty of very good books on classical electromagnetism. My favorites are


    J. Schwinger, Classical Electrodynamics
    J. D. Jackson, Classical Electrodynamics
    A. Sommerfeld, Lectures on Theoretical Physics, Vol. III (Electrodynamics)
    Landau/Lifgarbagez, Vols. II and VIII
    Feynman Lectures Vol. II
     
  15. Jul 1, 2013 #14
    Yeh, I think I spent enough time on this, more being puzzled rather than getting the answer. I already have the derivation using another book.

    The result is the same, that I verified before I even posted. I just thought I might missed something as this one looks quite out there!!!

    This is an Antenna Theory, not EM book............though he did write a book on advanced electromagnetics. That's the reason I never rely on one or even two books. I do have J D Jackson and I did look through it.

    This book is a post grad book that is not a common subject. Like other advanced level books that are not commonly used. They can be full of mistakes. I have two books on Phase Lock Loop by William Egan and Roland Best. Both have mistakes particular the one by Best. I contacted Egan and he gave me a site that I can get the revision on the mistakes. I posted my specific mistakes of Roland Best's book on Amazon, he actually responded and offered me a copy of the manuscript!!!

    From my experience, the more popular books like Electrodynamics by Griffiths, and lower division calculus etc, I don't find mistakes in those. Even in one of the PDE book by Asmar, there are sections that are quite out there. I am a self studier, I always have at least 4 books on every subject I studied and to cross check.
     
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