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Formula for Helmholtz Coil with a finite thickness?

  1. Feb 4, 2015 #1
    The formula for Helmholtz coil is given by mu*(0.8^1.5)*nI/R, where I is the current, n is te number of coil and R is the radius of the coil.

    Now assume the bunch of coils have a small 'thickness' w (so it looks like a hollow cylinder with a very small height), and the the two coils are separated by R' (R' is measured from the two inner most coils). How should I set-up the integral using Biot-Savart Law (or other method) to find the expression for B-field for this 'non-ideal' Helmholtz coil?
     
  2. jcsd
  3. Feb 4, 2015 #2
    You will need to know the current distribution through the coils. Is the current conducted on the surface of the coils or equally across the cross sectional area?
     
  4. Feb 4, 2015 #3
    I will assume the current flows evenly on the surface of that 'short, hollow cylinder'
     
  5. Feb 4, 2015 #4

    Philip Wood

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    If the coils have small axial length compared with their radii you won't make much error by placing their CENTRES one radius apart. If you still want a correction, then you'll find there's a simple equation out there for the field (at points on the axis) of a solenoid of finite length (which is what each of your coils is, if I've understood aright). It's a fairly standard formula, usually obtained by integrating up the fields due to 'flat' coils, into which the solenoid can be split. Anyway, you can use it to give you the field at the centre of your non-ideal Helmholtz pair. Come back if this isn't clear.
     
  6. Feb 5, 2015 #5
    It's probably easier to solve this numerically. It's a simple application of Biot-Savart but rather messy.
     
  7. Feb 6, 2015 #6

    Philip Wood

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    For what it's worth, I carried out the programme in post 4, expanding the cosines as Taylor series in terms of axial distance [itex]\Delta x[/itex] of the ends of the coils from their middles, and with the middles of the coils one radius (a) apart, found that B midway between the coils was
    [tex]B = \frac{\mu_0 I n}{a} \left[\frac{4}{5}\right]^\frac{3}{2} \left[1+\frac{3}{10} \frac{(\Delta x)^2}{a^2}\right].[/tex]
    The last set of square brackets contains the correction factor. As you can see, its very small, even if [itex]\Delta x = 0.1 x[/itex], that is even if the axial lengths of the coil is 20% of the coil radius.

    Of course, I may have made slips...
     
    Last edited: Feb 6, 2015
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