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Hermholtz Coil differential coefficients

  1. Feb 16, 2013 #1
    1. The problem statement, all variables and given/known data

    I'm supposed to show that the first, second and third differential coefficients of the field midway is zero.

    The B-field at that location is given by:
    avqv4w.png

    which is field due to a ring x 2..


    3. The attempt at a solution

    I tried to differentiate it with respect to x but dont see how it equates to zero??
     
  2. jcsd
  3. Feb 16, 2013 #2

    marcusl

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    That's not the correct expression for B. If one coil is at x0 and the other at -x0, then the field is a sum of a term with (x-x0)^2 and one with (x+x0)^2. Do you see why? Do you know what to set x0 to?
     
  4. Feb 17, 2013 #3
    Sorry I don't get what you mean. The B-field due to a ring at distance x away is given by:
    6nwj2t.png

    according to Biot-savart law.

    Then the B-field due to both rings in between both rings is given by the above expression multiplied by 2:
    avqv4w.png

    I'm not sure what the differential coefficients here mean...Do they mean that a point Δx away from x the B-field remains approximately constant?

    i.e. ∂B/∂x = 0?
     
  5. Feb 17, 2013 #4

    marcusl

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    No, you have written the field a distance x away from two rings lying on top of each other, or equivalently from a single ring of twice the current. Think about the situation: moving away from the mid-point brings you farther from one ring and closer to the other so the B contribution from one must decrease and from the other increase. Obviously your expression doesn't do that.
     
  6. Feb 17, 2013 #5
    I see I see! So treating one ring as the origin, distance x away from the origin is distance h-x away from the other?
     
  7. Feb 17, 2013 #6

    marcusl

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    Yes, exactly.
     
  8. Feb 17, 2013 #7
    so i differentiate it with respect to x, (Where h is a constant) and show that it is equal to zero?
     
  9. Feb 17, 2013 #8

    marcusl

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    Correct. All odd x derivatives of the field at the mid-point should be zero due to symmetry, and the second derivative should be zero if you have chosen h properly (or, alternately, you can find h that makes the 2nd derivative zero).
     
  10. Feb 17, 2013 #9

    2qw1tso.png

    I have found the first derivative. Is it right to say that by setting h = 2x (midway) and thus the expression goes to zero?

    Do i do that for the second and third derivatives as well? (non-zero or zero)

    I've managed to show the same for the second derivative as well.

    Does this go on only for the first, second and third derivatives? or all odd derivatives other than the second? How do i show this relationship?
     
    Last edited: Feb 17, 2013
  11. Feb 17, 2013 #10

    marcusl

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    You can see that dB/dx=0 at the midway point x=h/2, regardless of h. That is because of the symmetry.
    The second derivative will be zero only for a specific value of h, which is the standard spacing of a Helmholtz coil pair.
     
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