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Frequent assumption made (z R)

  1. Jul 12, 2013 #1
    Frequently, when a length is bigger than another (say, an axial distance is bigger than the radius of something related) a certain assumption is made and the smaller length is "removed" from a formula. It's hard to describe, so let me show an example:

    The magnetic field from a bobbin at an axial distance Z is:

    [itex]B = \frac{\mu_0 i R^2}{2(R^2+z^2)^{3/2}}[/itex]

    But, according to my book, when [itex] z \gg R [/itex]:

    [itex]B = \frac{\mu_0 i R^2}{2z^3}[/itex]

    Even though this is just an example I've seen similar assumptions being made multiple times. How do I arrive at such conclusion? I hardly think that I must simply "delete" the smaller length ([itex]R^2[/itex]) in the example, surely there must be a procedure to conclude that precisely?
     
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  3. Jul 12, 2013 #2

    LCKurtz

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    $$(R^2+z^2)^{\frac 3 2} = \left(z^2( \frac{R^2}{z^2}+1)\right)^{\frac 3 2}
    \approx (z^2(0+1))^{\frac 3 2}=z^3$$
     
  4. Jul 12, 2013 #3
    Usually what is done, taking your example for clarity, is to write things in terms of the small quantity
    [tex]
    \frac{R}{z}\ll 1.
    [/tex]
    Then you do a Taylor expansion in terms of this small parameter (call it [itex]x[/itex]). In this way you have a meaningful expansion that you can control. In your example:
    [tex]
    B=\frac{\mu_0iR^2}{2z^3(1+x^2)^{3/2}}=\frac{\mu_0iR^2}{2z^3}+O(x).
    [/tex]
     
  5. Jul 16, 2013 #4

    HallsofIvy

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    By the way, [itex]z\gg R[/itex], and similarly, [itex]z/R\ll 1[/itex], do NOT just mean "z is greater than R" and "z/R is less than 1". They mean "z is much greater than R" and "z/R is much less than 1". "much greater" and "much less" is generally interpreted as "one can be ignored, compared to the other". That is if [itex]z\gg R[/itex], z+ R is "indistinguishable" from z, just as if you have $100,000,000, you can "ignore" $1!
     
  6. Jul 16, 2013 #5

    haruspex

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    Do you mean
    [itex]
    =\frac{\mu_0iR^2}{2z^3}(1+O(x^2))
    [/itex]?
     
  7. Jul 17, 2013 #6
    It doesn't really make a difference.
    Yes, indeed, the first derivative vanishes at 0, but it doesn't make a difference because we're interested only in the zero order, so what I wrote holds anyway (an [itex]O(x^2)[/itex] is an [itex]O(x)[/itex] ).
    Regarding the parenthesis, it makes no difference whatsoever, the notation [itex]O(x^a)[/itex] is sufficiently ambiguous to allow many different notations where you let constants get in or not. I'm using the standard Taylor expansion notation, [itex]f(x)=f(0)+xf'(0)+...[/itex].
    So, what I wrote definitely holds, as what you wrote.
     
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