Deriving Magnetic Field from Biot-Savart Integral for Long Straight Conductor

[Pixter]

so trying to understand the derivation of magentic field due to a long straight conductor:

so biot-savat says in this case:


B= u0I/4pi * integrate over L and -L for: xdy/(x^2+y^2)^(3/2)

and i should turn up with the answear:

B= (u0I/4pi)*(2L/xsqrt(x^2+y^2))

but can't do the integration.. so well i understand biot-savat but yeah can't do the integration, would anybody maybe show me how to do this?
thanks!

 

[HallsofIvy]

Fundamentally, that's
[tex]\int_{-L}^L (a^2+ y^2)^{-\frac{3}{2}}dy[/itex]
and should make you think of a trig substitution (any time you have a square inside a square root and cannot simply substitute for the function in the square root itself, think of a trig substitutin). If I remember correctly, 1+ tan<sup>2= sec2 so try $y= x tan(\theta)$. Then $x^2+ y^2= x^2+ x^2 tan^2(\theta)= x^2(1+ tan^2(\theta))= x^2sec^2(\theta)$ and
$$(x^2+ y^2)^{-\frac{3}{2}}= x^{-3}sec^{-3}(\theta)=x^{-3}sin^3(\theta)$$
Also, $dy= x sec^2(\theta)$ so your integral becomes
$$\frac{1}{x}\int_{-tan^{-1}(\frac{L}{x})}^{tan^{-1}(\frac{L}{x})}sin(\theta)d\theta$$
which, except possibly for evaluating at the limits of integration, is easy.

Added in edit: I just noticed you say

"and i should turn up with the answear:
B= (u0I/4pi)*(2L/xsqrt(x^2+y^2))"

Well, that's obviously wrong, isn't it? If you are integrating, with respect to y, from -L to L, the result will not involve y.</sup>

 

[arunbg]

Not unless he means y for a

 

[Pixter]

yes ment L^2 not y^2 at the bottom..

thanks for pointing out that.

so nobdy who knows exatcly how one goes from the firs tpoint of the bio savart law to the later (stated in the first post)

yeah so let's sat that we can use the trig but then how do i get it out to "normal" form so i can evaluate?

how did the people who found this in the first place do it?

 

[HallsofIvy]

yes ment L^2 not y^2 at the bottom..

thanks for pointing out that.

so nobdy who knows exatcly how one goes from the firs tpoint of the bio savart law to the later (stated in the first post)

yeah so let's sat that we can use the trig but then how do i get it out to "normal" form so i can evaluate?

how did the people who found this in the first place do it?

Well, integral of sin(y) is -cos(y) so, evaluating at the limits of integration, we have $-2cos(tan^{-1}(\frac{L}{x})$.
To evaluate that, either use trig identities or imagine right triangle with "opposite side" of length L and "near side" of length x (i.e. the tangent of the angle is L/x). It's hypotenuse has length (Pythagorean theorem) $\sqrt{L^2+ x^2}$ and so the cosine of that angle is $\frac{x}{\sqrt{L^2+ x^2}}$. That should be simple enough.

 

[dextercioby]

Also a hyperbolic substitution ($y=x \sinh t$) might work.

Daniel.

P.S. And it's Jean Baptiste Biot and Felix Savart (sic!)

 

[arildno]

P.P.S:

Here's a character description of Biot:

A ...comment by Olinthus Gregory in 1821 is:-

With regard to M. Biot, I had an opportunity of pretty fully appreciating his character when we were together in the Zetland [= Shetland] Isles; and I do not hesitate to say that I never met so strange a compound of vanity, impetuosity, fickleness, and natural partiality, as is exhibited in his character.