1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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?
  2. jcsd
  3. Jul 12, 2013 #2


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    $$(R^2+z^2)^{\frac 3 2} = \left(z^2( \frac{R^2}{z^2}+1)\right)^{\frac 3 2}
    \approx (z^2(0+1))^{\frac 3 2}=z^3$$
  4. Jul 12, 2013 #3
    Usually what is done, taking your example for clarity, is to write things in terms of the small quantity
    \frac{R}{z}\ll 1.
    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:
  5. Jul 16, 2013 #4


    User Avatar
    Science Advisor

    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


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Do you mean
  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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted