# Frequent assumption made (z R)

1. Jul 12, 2013

### ShizukaSm

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:

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

But, according to my book, when $z \gg R$:

$B = \frac{\mu_0 i R^2}{2z^3}$

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 ($R^2$) in the example, surely there must be a procedure to conclude that precisely?

2. Jul 12, 2013

### LCKurtz

$$(R^2+z^2)^{\frac 3 2} = \left(z^2( \frac{R^2}{z^2}+1)\right)^{\frac 3 2} \approx (z^2(0+1))^{\frac 3 2}=z^3$$

3. Jul 12, 2013

### kevinferreira

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 $x$). In this way you have a meaningful expansion that you can control. In your example:
$$B=\frac{\mu_0iR^2}{2z^3(1+x^2)^{3/2}}=\frac{\mu_0iR^2}{2z^3}+O(x).$$

4. Jul 16, 2013

### HallsofIvy

Staff Emeritus
By the way, $z\gg R$, and similarly, $z/R\ll 1$, 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 $z\gg R$, z+ R is "indistinguishable" from z, just as if you have $100,000,000, you can "ignore"$1!

5. Jul 16, 2013

### haruspex

Do you mean
$=\frac{\mu_0iR^2}{2z^3}(1+O(x^2))$?

6. Jul 17, 2013

### kevinferreira

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 $O(x^2)$ is an $O(x)$ ).
Regarding the parenthesis, it makes no difference whatsoever, the notation $O(x^a)$ is sufficiently ambiguous to allow many different notations where you let constants get in or not. I'm using the standard Taylor expansion notation, $f(x)=f(0)+xf'(0)+...$.
So, what I wrote definitely holds, as what you wrote.