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Please explain why one of the integral vanishes

  1. Mar 25, 2013 #1

    This is a elemental magnetic dipole. The book shows how to find the vector magnetic potential at any distance from the origin. If you read the bottom, it said the second integral in 11.24 is obviously zero. It is not obvious to me.

    ##d\vec l'\;=\;\hat {\phi} bd\phi##. The second integral is

    ##\oint d\vec l'\;=\;\int_0^{2\pi} d\vec l'= \hat{\phi}2\pi b##. It is not zero.

    Please comment on this.


    Last edited: Mar 25, 2013
  2. jcsd
  3. Mar 25, 2013 #2
    This can't be right, the integrand is a vector, and the result you state is a scalar. The integral is like a vector sum- what do you get when you add up all the tangent vectors to a circle going around the loop? (think about the vector sum around a polygon)
  4. Mar 25, 2013 #3
    Well a curl free vector field integrated around any closed curve always gives 0.
    A constant vector field is of course curl-free
  5. Mar 25, 2013 #4


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    It's direction is changing with space so it obviously isn't a constant vector not to mention you are using heavy machinery to show something so simple. The poster above you already gave the most simple reason for why the integral vanishes.
  6. Mar 25, 2013 #5
    Thanks for the reply. I forgot to put in the ##\hat \phi##. So how do you integrate a vector? I am confused about this:

    If I translate ##\hat{\phi}=-\hat x \sin \phi+\hat y \cos \phi## and then integrate from 0 to ##2\pi##, then I get zero. But if I leave the vector as ##\hat {\phi}##, it is like my original post.
    Last edited: Mar 25, 2013
  7. Mar 25, 2013 #6
    Split it into a double integral in each component.
  8. Mar 25, 2013 #7
    What do you mean? There is only single integral with ##\hat{\phi}## as the component. Please read the post again as I added another part in a few seconds ago.
  9. Mar 25, 2013 #8


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    How about a simple appeal to geometry? For every dl vector, there is cancellation from an opposite-pointing vector on the opposite side of the circular loop. This is what Hechima was getting at back in Post #2, and I suspect what led the author to say the integral "obviously vanishes".

    But if you are determined to slog through the math:
    Yes, exactly.

    [STRIKE]Are you are taking the ##\hat {\phi}## factor out of the integral, as you would a constant factor? You can't do that, it's not constant owing to it's changing direction (it's a vector, remember). To clear this up, perhaps you could rewrite your integral including the ##\hat {\phi}## -- you say you left it out earlier, so it is not clear to me what you have done at this point. Then again, I am a little tired at the moment.[/STRIKE] [EDIT: I see you already made the change, never mind this striked-out part.]

    Okay, you had

    [tex]\oint d\vec l'\;=\;\int_0^{2\pi} d\vec l'[/tex]
    But that 2nd integral isn't quite right. dl is not going form 0 to 2π -- Φ is. As in Φ the scalar angle as one goes around the loop. So the variable of integration should be Φ (the scalar, not the unit vector) in that second integral.

    Since ##\vec{dl'}## has a magnitude of ##b \ d \phi## and direction ##\hat{\phi}##,
    [tex]\oint d\vec l'\;=\;\int_0^{2\pi} b \ \hat{\phi} \ d \phi[/tex]

    It's important to remember that ##\hat {\phi}## changes direction, therefore it is NOT a constant and can NOT be take outside the integral.

    I can't think of a better way to evaluate the integral than to write ##\hat {\phi}## in terms of its x and y components, which it appears you have already done.

    Hope that helps.
    Last edited: Mar 25, 2013
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