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Biot-Savart law question

  1. Feb 18, 2015 #1
    1. The problem statement, all variables and given/known data
    there is no problem really this is not homework its just something I wanted to include in my lab report. the lab was working with the Helmholtz coil. I was trying to come up with the magnetic field at a distance, Z, from the center of a circular loop of radius, R, that has a steady current, I. I thought that I did it right... but I got a negative sign for some reason. where did I go wrong...

    2. Relevant equations
    see attached I think I got it all there

    3. The attempt at a solution
    see attached
     

    Attached Files:

  2. jcsd
  3. Feb 18, 2015 #2
    I updated my attached file. nothing big, just interchanged theta and phi to what I believe is a more conventional (at least in physics) use for theta and phi.
     
  4. Feb 18, 2015 #3
    any guidance on this would be great....
     
  5. Feb 18, 2015 #4
    ok this is my last update to attachment file. I just changed a few minor things with the ordering and included the Z hat in my final answer
     

    Attached Files:

  6. Feb 18, 2015 #5

    BvU

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    Good thing you added the ##\hat R## and ##\hat z## (even though now I have to edit my reply to remove the comments on those).

    I wondered how you got the minus sign in your cross product:
    In your figure 5.21 the z component of ##d{\bf B}## is clearly positive.

    Can't pinpoint where it goes wrong, but
    I suspect ##\ \vec \Re \equiv \vec r -\vec R \ ## has something to do with it ( ##\vec r = (0,0,Z)## ) .
     
  7. Feb 18, 2015 #6
    I don't think I did the cross product wrong.... but the values in the matrix might be off....
     
  8. Feb 18, 2015 #7
    ok ok I think I see where your going. so if (scriptR) ⃗ ≡r ⃗ −R ⃗ with ( r ⃗ =(0,0,Z)) then....

    R ⃗ simply equals (R,0,0)?

    making (scriptR) ⃗ = (-R,0,Z) and that's what I should put into my cross product?

    srry I don't know latex at all.
     
  9. Feb 18, 2015 #8
    well that definetly fixed the negative sign.... I am gonna go with it. thank you for fixing it for me. ill attach a the corrected document if anyones curious...

    and again thank you for your time on this
     

    Attached Files:

  10. Feb 18, 2015 #9

    BvU

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    But the problem still is that the r component of a vector in r,phi, z coordinates can't really be negative ....

    The guys here don't have the problem: their ##\alpha## is ##\pi/2 - \theta##
     
  11. Feb 18, 2015 #10
    well.. I am not understanding the issue.
     
  12. Feb 18, 2015 #11
    I wanna accept it just cause I guess. I am looking at pg 9 in my e&m book its Griffiths 3rd ed. it describes the r-r' formula, does not seem to have the limitation where the vector subtraction has to turn out positive. the problem 1.7 on the following page seems to illustrate this for me. is asks for (4,6,8) - (2,8,7) = (2,-2,1).

    so I don't see why the resulting vector cant have a negative component.
     
  13. Feb 18, 2015 #12

    BvU

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    ##\vec \Re## can have a negative component in cartesian coordinates, but not in the r component when working in cylindrical coordinates (then r is the distance to the z-axis and ##\phi## runs from 0 to ##\pi## -- the 2d polar coordinates).
     
  14. Feb 18, 2015 #13
    hmm phi is from 0 to 2pi. not that that voids your point. so is my derivation completely bogus? should I just throw it out and go with Cartesian coordinates?
     
  15. Feb 18, 2015 #14

    BvU

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    yes sorry.
     
  16. Feb 18, 2015 #15
    are we certain that r=(0,0,Z) and r' = (R,0,0).....

    could it be that r(-R,0,0) and r' = (0,0,Z) so I would start at the outside of the bottom circle and move to the center going a distance of -R then go up Z?
     
  17. Feb 18, 2015 #16
    is it ok for me to make my source point on the circle that dl is on?
     
  18. Feb 18, 2015 #17

    BvU

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    I still have a hard time finding out how to get a positive z component.

    I think we don't have the right ##\vec \Re## anymore: the r component is definitely not equal to -R.
    More like ##\vec \Re = (R, \phi+\pi, z)##. ##\phi+\pi## is multiplied with 0 and disappears (?).

    The determinant form looks correct, but gives us a minus sign, whereas ##\vec {dl} \times \vec\Re## definitely points as in Figure 5.21 with a positive z-component, so there's something wrong, but what ?

    And (I'm making things worse now): "The first integral equals 0 because of orthogonality" ?
    What exactly is orthogonal in the first integral (and not in the seceond) ?

    Bedtime for me..o0)

    Let's ask @haruspex :smile:
     
  19. Feb 18, 2015 #18
    that almost makes more sense to me. that the source point is on that circle. attached is an updated derivation.
     

    Attached Files:

  20. Feb 18, 2015 #19
    damn yea huh idk I guess that's not right either! soab I need that first integral to go away. it has to cancel itself out somehow....
     
  21. Feb 18, 2015 #20

    BvU

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    I like the Griffiths approach in the tekst under fig 5.21: "##d{\bf B}## sweeps out a cone. The horizontal components cancel..." (well, I don't like the term horizontal, but I understand what he means). But I still have a hard time expressing that in a decent integral of the vector product. Has to do with ##\phi## running for the current loop, but we don't want the ##\hat \phi## and the ##\hat r## of the point (0,?,z) where we want to know the field to run at all.

    Something screwy with using cylindrical coordinates for a running ##\phi## on the current loop wrt a point on the z-axis (for which ##\phi## is undefined) ?
     
    Last edited: Feb 18, 2015
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