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B-Field between two ribbons problem

  1. Dec 7, 2009 #1
    1. The problem statement, all variables and given/known data

    Find the B-field between two ribbons with the current flowing in opposite directions on each.


    2. Relevant equations

    Biot-Savart equation


    3. The attempt at a solution

    In attempting to solve this I have found the B-field due to an infinite wire is

    B= (mu)I/2(pi)r

    Were r is the distance from the wire to the point at which B-field exists.

    Can someone tell me if this equation is directly applicable to the B-field from two parallel plates?
     
  2. jcsd
  3. Dec 7, 2009 #2

    kuruman

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    It is and you don't need the law of Biot-Savart. First you need to find an equation for the B field due to one ribbon. To do this consider the ribbon as made up from many wires of width dy, find the field due to it at some arbitrary point, then integrate over the width of the ribbon to find the total field.
     
  4. Dec 7, 2009 #3

    gabbagabbahey

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    Are the "ribbons" flat like parallel plates?
     
  5. Dec 8, 2009 #4
    It is a parallel plate problem.

    In integrating over the width, W, would I just use

    B=[tex]\int[/tex]((mu)I/2(pi)r)dx

    and is r dependent on x. If r is not dependent on x this would give

    B=(mu)IW/2(pi)r
     
    Last edited: Dec 8, 2009
  6. Dec 8, 2009 #5

    kuruman

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    Yes, r depends on x so you need to express it in terms of x in view of the geometry, i.e. the width of the ribbons and the separation between them.

    You cannot put the dx in just because you want to integrate over x. You have to justify where it comes from. Note that the strip of width dx carries only a fraction of the total current I which should be written as

    [tex]dI=I\frac{dx}{w}[/tex]

    where w is the width of the strip. Then the contribution of the strip to the magnetic field is

    [tex]dB=\frac{\mu_0}{2 \pi r}dI=\frac{\mu_0}{2 \pi r w}dx[/tex]
     
  7. Dec 8, 2009 #6

    gabbagabbahey

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    You'll also need to take the direction of dB into account....is it constant over the integration?
     
  8. Dec 9, 2009 #7
    I'm pretty sure the B-field is along the x-axis.

    The resulting equation to

    [tex]
    dB=\frac{\mu_0}{2 \pi r}dI=\frac{\mu_0}{2 \pi r w}dx
    [/tex]

    is

    [tex]B=\frac{\mu_0 I (W^2+L^2)^\frac{1}{2}}{\pi W^2}[/tex]

    after integrating over the width. With [tex]r=(x^2+y^2)^\frac{1}{2}[/tex] and [tex]y=\frac{L}{2}[/tex], were L is the distance between the plates. I have put the origin at the centre of both plates.

    However this does not allow for a description of the B-field outside either edge of the plates, i.e. past W. Since the only variable outside of W is x. Can you help with putting the outside explanation into this equation or do I have derive a different equation due to the change of conditions?
     
    Last edited: Dec 9, 2009
  9. Dec 9, 2009 #8

    kuruman

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    I am not sure I completely understand the geometry of the question. Imagine one of these ribbons in the xy plane with its long axis parallel to y and with its edges at x = -w/2 and x = +w/2. The other ribbon will be parallel to it at z = L. Using this picture, what are the coordinates of the point where you wish to measure the B field?
     
  10. Dec 9, 2009 #9
    The problem is to find the B-field along the x-axis. Which is parallel to the infinitely long parallel plates of width W, these are in the x-z plane, bit continues out of the parallel plates after a distance W/2.

    So it appears that either the initial equation needs to be modified or a new one formed for the field beyond the edges of the plates/ribbon...
     
  11. Dec 10, 2009 #10

    kuruman

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    So both ribbons are in the xz plane, one above the x-axis and the other below with the nearest edge of each ribbon at distance L/2 from the x-axis?
     
  12. Dec 10, 2009 #11
    Actually both ribbons extend to infinity in the xz plane, so the infinite wire equation is applicable. Yes they are at a distance of L/2 from the origin running parallel to the xz plane. As the current is in opposite directions in each ribbon the B fields add (I think...).
     
  13. Dec 10, 2009 #12

    kuruman

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    So the plane of one ribbon faces the plane of the other and the point of interest is on the x-axis which you might as well call the origin. Correct?
     
  14. Dec 10, 2009 #13

    gabbagabbahey

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    No, the field circles around each wire....the field from the wire (the infinitesimally thin slice of the ribbon) at say [itex]z=-w/2[/itex] will point in a different direction that the field from thew wire at [itex]z=0[/itex]....you need to find a way to express the direction for each of the pieces you are integrating over in a way that allows you to compute the integral.



    since the ribbons run parallel to the x-axis, you actually want to integrate over [itex]z[/itex] (the width of each ribbon), so that you are just adding up the fields of a whole bunch of infinite wires.

    [tex]d\textbf{B}=\pm\frac{\mu_0 I}{2\pi r w}\hat{\mathbf{\phi}'}dz'[/tex]


    Where the plus/minus depends on which ribbon you are looking at (remember, the current flows in the positive x-direction in one ribbon, and the negative x-direction in the other), [itex]r[/itex] is the distance from the wire located at [itex]z=z'[/itex] and [tex]\hat{\mathbf{\phi}'}[/itex] is the unit vector that circles around said wire....try expressing [tex]\hat{\mathbf{\phi}'}[/itex] in terms of Cartesian coordinates and unit vectors (you should find that it depends on [itex]z'[/itex])


    Isn't the distance from a wire (running parallel to the x-axis) at say [itex]y=\pm L/2[/itex] (upper or lower ribbon) and [itex]z=z'[/itex] to any point along the x-axis, given by [tex]r=\left(\frac{L^2}{4}+z'^2\right)^{1/2}[/itex]?
     
    Last edited: Dec 10, 2009
  15. Dec 10, 2009 #14
    What I mean is that the field that is circling the wires is directed along the positive direction of the x-axis and the equation

    [tex]

    dB=\frac{\mu_0 }{2 \pi r}dI=\frac{\mu_0}{2 \pi r w}dx

    [/tex]

    with the current flowing along the z axis, the direction of the ribbons/plates.

    [tex]
    r=\left(\frac{L^2}{4}+x^2\right)^{1/2}
    [/tex]

    Since the point of interest in on the x-axis and the plate is distance L/2 along the y-axis.

    This gives

    [tex]
    B=\frac{\mu_0 I (W^2+L^2)^\frac{1}{2}}{\pi W^2}
    [/tex]

    when integrating from 0 to W/2 along dx. What I am not sure of is how to account for the field when there is no plate either side, i.e. the field emanating from the plates along to x-axis past W/2.
     
  16. Dec 10, 2009 #15

    gabbagabbahey

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    No, draw yourself a picture of a wire parallel to the x-axis (and obviously not at y=z=0)...and draw the field lines that circle around the wire...do any of them point along the x-axis?

    I thought your setup had the current flowing parallel/antiparallel to the [itex]x[/itex]-axis, not the [itex]z[/itex] -axis?
     
  17. Dec 10, 2009 #16
    The current is flowing parallel to the z-axis.
     
  18. Dec 10, 2009 #17

    gabbagabbahey

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    Surely you can see how one might get confused here?

    Does this mean the plates are at [itex]y=\pm L/2[/itex] and extend from [itex]x=-w/2[/itex] to [itex]x=w/2[/itex] and from [itex]z=-\infty[/itex] to [itex]z=\infty[/itex]?....Is the current in the upper plate flowing in the positive z-direction and the current in the lower plate flowing in the negative z-direction?
     
  19. Dec 10, 2009 #18

    kuruman

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    This should be a simple problem, but I am still confused about the geometry. Is it possible to have a picture showing the ribbons and the point at which the field is to be calculated?
     
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