[reachmanxx]

In this question here:
http://planetphysics.org/encyclopedi...SavartLaw.html
How did they deduce that dl x r = dl sin90?

And, in general, how to you break the dl x r part down to be able to do the integral?

[bestformaldress23]

vector cross product by the looks of it, A x B = |A||B|Sin (Theta)

here you have dl crossed with the r unit vector, so the cross product is just the magnitude of dl x the magnitude of r unit vector which is just 1, and then from the diagram you can see that dl and r are at right angles to each other.

[reachmanxx]

So whenever I have dl x r, r is always 1? And theta's the angle between dl and r? Also, does dl = 2*pi*dr, since l = 2*pi*r?

[Sopsneork]

So whenever I have dl x r, r is always 1? And theta's the angle between dl and r? Also, does dl = 2*pi*dr, since l = 2*pi*r?

Yes, r is always one because it's a unit vector meaning it has a length of 1.

Like Chocobo said, dl x r is the vector cross product. This is the way you see it in a text book or, in your case, the website. Instead of looking at it like that, read is as: (dl)(1)(sinθ) or (Δl)(1)(θ).

Generally, you set θ=90°

As for the integration, there are two types you use depending on the magnetic field.

If B is everywhere perpendicular to a line then:

∫B*dl=0

If B is everywhere tangent to a line then:

∫B*dl=Bl

Basically, the way you integrate B is by using

B = (μ_0*I*dl)/(4πr^2)

To sub in for B, obviously, you then integrate from whatever your limits are. So it looks like this

= ∫ (μ_0*I*dl)/(4πr^2)
= (μ_0*I)/(4πr^2) ∫ dl

If you integrate with limits of 0 to 2π you will get the equation for the center of a coil, which is what the website you posted gave. (even though the factor of π did not cancel for some reason)


Was that it?

[reachmanxx]

So cos l relies on θ, but not on r, I should use l = rθ -> dl/dθ = r instead of l = 2πr -> dl/dr = 2π?

Also, this bit here:

If B is everywhere perpendicular to a line then:

∫B*dl=0

If B is everywhere tangent to a line then:

∫B*dl=Bl The integrals are normal integrals, right? Cos the one involving the closed integral is ∫B.dl = (μ_0)(I_enclosed), yes?

[Sopsneork]

So cos l relies on θ, but not on r, I should use l = rθ -> dl/dθ = r instead of l = 2πr -> dl/dr = 2π?

Also, this bit here:



The integrals are normal integrals, right? Cos the one involving the closed integral is ∫B.dl = (μ_0)(I_enclosed), yes?

Yeah, since you're doing a circle you're using a differential equation (sort of) with respect to θ, because whenever you take the derivative/integral you take it to what variable you're changing. The radius remains the same so the only thing that is changing is θ.

They are normal integrals, but they are general equations and not for a closed surface. That's the only difference between and normal integral and a closed loop integral (∮). You treat both of them the same way, it's just that when you have the ∮ you're integrating in a closed surface.

[reachmanxx]

Righto then, I think I get it now. Thanks a bunch, everyone. ^.^