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Magnetic Flux from Infinite wire

  1. Nov 21, 2008 #1
    1. The problem statement, all variables and given/known data

    Infinite wire on Z axis
    Loop radius a lies in x,z plane, centered on positive x axis, with center a distance b away from origin.

    Find flux through loop.


    2. Relevant equations


    B field => from the infinite wire, dependent on 1/rho, in the (phi) direction, = (UoI/2pi)*(1/rho)
    flux = (B dot dA), where dA is normal to plane, so in the (Y) direction



    3. The attempt at a solution

    If am attaching my attempt. What I have done so far, is figured out my B (dot) dA, by taking (phi) from B field and changing it to cartesian, and dotting it with differential area of my loop, which is in (Y) direction, so my (X) component from the (phi) transformation zeroes out, and I am left with scalar flux.

    My question is this:
    I know that rho is the floating point, but the B field I am using is not the same for all values of rho, correct? It varies from (b-a) to (b+a) ?
    Do I integrate across rho from (b-a) to (b+a) ?? or has this difference already been accounted for?
    Should I be integrating across different form of B rather than just using the generic form from the infinite wire??
    and if so, do I do this before I take the dot product?
    It seems that rho is only dependent on an x coordinate because of the infinite wire, and not a z coordinate, so would there still a transformation of rho to cartesian for this integral?
    I am stuck here, any help varying this rho would be greatly appreciated.

    Thanks
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     

    Attached Files:

  2. jcsd
  3. Nov 22, 2008 #2
    Field B depends on rho as you wrote. Flux is:
    [itex] \int_{0}^{A} B \cdot dA [/itex]
    Since you know field B as the function of rho, you should also express differential area dA in terms of rho (and d(rho)), so you'll get dA = f(rho)d(rho). Then you integrate
    [itex] \int_{a-b}^{a+b} B(\rho)f(\rho)d\rho [/itex].

    Hope this would help
     
  4. Nov 22, 2008 #3

    gabbagabbahey

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    The loop is circular and hence has a constant radius so [itex]d \rho=0[/itex]....
     
  5. Nov 22, 2008 #4

    gabbagabbahey

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    The flux through a surface [itex]\mathcal{S}[/itex] is [tex]\int_{\mathcal{S}} \vec{B} \cdot \vec{dA}[/tex]....but you're not asked to find the flux through a surface; you're asked to find the flux through a loop, which is given by [tex]\oint_{\text{loop}} \vec{B} \cdot \vec{dl}[/tex] where [itex]\vec{dl}[/itex] is an infinitesimal length element for the loop.
     
  6. Nov 22, 2008 #5
    I think you're wrong, look at the picture in KleZMeR pdf. With rho I mean the distance from wire on z-axis, so rho is not constant through loop, if I understand the problem right.
     
  7. Nov 22, 2008 #6

    gabbagabbahey

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    The [itex]\rho[/itex] in his equation for the magnetic field represents the shortest distance from the wire on the z-axis to the field point. For any field point on the loop, that distance is just the radius of the loop; which is constant.
     
  8. Nov 22, 2008 #7
    The loop doesn't go around the wire, it has center moved for distance b from origin. Look at his sketch in attached pdf. Rho represents what you said, but the loop is not on constant rho, neither rho is the radius of the loop (which is a).
     
  9. Nov 22, 2008 #8

    gabbagabbahey

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    Okay, yes I just realized the loop is not centered at the origin, so rho will vary over the circumference of the loop.
     
  10. Nov 22, 2008 #9
    Do I really need the transformation of (phi) vector into (X) and (Y)? Can I just assume that the B field through the loop is in the (Y) direction, and that the differential area element is in the (Y) direction as well

    The integral of the cos term along the (Y) vector is giving me zeroes with limits 0->pi, and 0->2pi, and if I use limits of 0->pi/2, I get a factor of 2, and am still missing the (pi) factor in my area


    So in this case, can I assume that the element in terms of (rho) only is:
    f(rho) d(rho),

    and then is this just

    (pi)*(rho)^2 d(rho) ??

    Because wouldn't the pi term cover the earlier d(phi) that I am now eliminating?
     
  11. Nov 22, 2008 #10

    gabbagabbahey

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    Again, you are trying to find the flux through a loop, not through a surface....you don't have a differential area element; you have a differential length element. You don't have an area integral; you have a path integral.

    You need to start by finding an expression for [itex]\vec{dl}[/itex].
     
  12. Nov 22, 2008 #11
    Sorry, I have to disagree. The line integral around the loop is not a flux - Flux is always the number of field lines (or any other quantity) passing through a given area. In this case, the problem is to find the number of flux lines passing through the area of the loop.
     
  13. Nov 22, 2008 #12
    you don't really need any vectors, since area withing the loop and field are perpedicular. Differential area da is:

    [tex] dA = \sqrt{a^2+(\rho-b)^2} d\rho [/tex]

    Maybe I should sketch this, but I hope you understand that.
     
  14. Nov 22, 2008 #13
    I disagree with you gabbagabbahey also. He is looking for flux as naresh said.
     
  15. Nov 22, 2008 #14
    the field is not perpedicular to the normal of the surface! they have both the same direction. You should see that.
     
  16. Nov 22, 2008 #15
    your path integral is zero, since differential path element is perpendicular to magnetic field, of course, because there is no electric current through the loop, as states Ampere law. Your path integral gives total current through the loop which is zero of course. You don't get it right.
     
  17. Nov 22, 2008 #16

    gabbagabbahey

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    hmmmm.... After giving it more thought, I guess they must be looking for the flux through the surface enclosed by the loop....not the flux through the loop as I had interpreted this to mean...

     
  18. Nov 22, 2008 #17
    that your flux is called electric current :) glad we cleard that.
     
  19. Nov 22, 2008 #18
    ok, well because I am the one asking the question, I can not really make an educated comment on the discussion, but I have been looking through my notes and I did find a similar problem done.

    Is this a choice of English language?, where "through" can be chosen as through area enclosed, or through actual "surface" of loop, as in the surface of a line?, but the loop has no thickness, so how do field lines pass through line element? I don't know, possibly look at solution attached..

    I also made a false pass before, I realized that that area element is 2*(pi)*(rho)* d(rho)
    and with this attempt I integrate with b included, but it canceled out in the integral,

    and I did the problem on my own in a different way, and got the same units, but still did it a bit wrong, with a factor of (2a) rather than the correct factor of: (b - sqrt( b^2 - a^2 )

    so, in any case, I am attaching the notes I found
    see attached
     

    Attached Files:

  20. Nov 22, 2008 #19
    forget about that line integral, you need the surface one. From figure it should be clear why differential surface area (rectangle in figure) is as I wrote in one of previous posts.
     

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