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Helmholtz coils

  1. Jan 26, 2008 #1
    Working in CGS units... hence i dont know if my answer is right or wrong anymore... It doesn't 'look' right
    1. The problem statement, all variables and given/known data
    Two coils are placed at l and -l on the Z axis as in the figure. They both carry a current I in the same direction (anticlockwise). Both have a radius of a. (Setup like Helmholtz coils.)

    a. What is the field on the z axis due to the two coils?
    b. Why are all the odd derivatives of B at the origin equal to zero?

    2. Relevant equations
    Due to the loop of current the field a hieght z above the center of the loop is given by
    [tex] B = \frac{2\pi Ia^2}{c(a^2 + z^2)^{3/2}} [/tex]

    3. The attempt at a solution

    a) The field is given by [tex] B = \frac{2\pi Ia^2}{c}\left( \frac{1}{(R^2 + (z + l)^2)^{3/2}} + \frac{1}{(R^2 + (z - l)^2)^{3/2}} \right)[/tex]

    b) I know that the first derivative of the B field is zero becuase there are no magnetic monopoles. But how does that explain why the octopole moment and so forth are zero? I need some kind of hint so if you could please advise... that would be awesome

    Thank you for your help
     
    Last edited: Jan 26, 2008
  2. jcsd
  3. Jan 26, 2008 #2

    pam

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    A lot of B fields have first derivatives.
    It is wrong to refer to "octupole moments" for the field between the loops.
    Odd derivatives at z=0 are zero, because B is an even function of z due to the symmetry of the coils.
     
    Last edited: Jan 26, 2008
  4. Jan 26, 2008 #3
    ok so even functions have odd first derivative and even second derivative odd third derivative and so on. But how does that explain that the odd derivative is zero at the origin? Aren't evne functions symmetric wrt reflection on the Y axis? Why aren't even functions zero at the origin?
     
  5. Jan 26, 2008 #4

    marcusl

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    Science Advisor
    Gold Member

    Mathematically, the derivative of z^(2n) is zero at z=0 for all positive integer n.
    Physically, we know the field at the origin is a minimum (it is strongest at +/-L), and is even about the origin. Therefore the tangent is zero.
     
  6. Jan 26, 2008 #5

    pam

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    An odd function must equal zero at the origin because f(0)=-f(0) for an odd function.
    An even function needn't be zero because f(0)=f(0) for an even function.
    For instance 2+z^2=2 at the origin.
     
  7. Jan 26, 2008 #6
    thanks for the help
     
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