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Anti-Helmholtz coil configuration

  1. Jun 20, 2012 #1
    Hi

    Say I have to coils in the anti-Helmholtz configuration as in the attached picture. It is pretty easy to find the field along the z-direction, as this is introduced in introductory EM. However say that I would like to know the field along the x-direction. This I don't know how to find.

    What I *do* know is that the Maxwell Equations (div B = 0) tell me that
    [tex]
    \frac{dB}{dx} = \frac{dB}{dy} = -\frac{1}{2}\frac{dB}{dz}
    [/tex]
    But does this imply that the field along x, B(x), is simply -B(z), the negated B-field along the z-direction?

    Best,
    Niles.
     

    Attached Files:

  2. jcsd
  3. Jun 21, 2012 #2

    mfb

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    Staff: Mentor

    div B = 0 is equal to [itex]\frac{\partial B_x}{\partial x} + \frac{\partial B_y}{\partial y} = - \frac{\partial B_z}{\partial z}[/itex]
    Using symmetry, x and y must be the same, therefore [itex]\frac{\partial B_x}{\partial x}= - \frac{1}{2}\frac{\partial B_z}{\partial z}[/itex]
    This does not give you any magnetic field! It is just the derivative of the field at some specific point - probably along the central axis. Looking at the (x,y)-plane right in the middle of the coils, you have a field going radially inwards/outwards (depending on the orientation).
     
  4. Jun 23, 2012 #3
    Thanks, that was kind of you.
     
  5. Jun 23, 2012 #4

    marcusl

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    Science Advisor
    Gold Member

    The off-axis field from a current loop is written as an expansion in elliptic integrals (or sometimes other functions, depending on the coordinate system you choose). Here is a site that came up in a Google search for off-axis field from a loop:
    http://www.netdenizen.com/emagnettest/offaxis/?offaxisloop
    but there are many others. The field from a pair of coils is then a sum of the fields from the individual loops.

    More extensive derivations/explanations are found in advanced E&M texts like those by Jackson, Smythe, and Stratton.
     
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