 So, now I'm going to do an example with Ampere's law. This is Ampere's law. It says that if you take the magnetic field and you integrate over some path, this is a closed path integral, not a closed surface area, integrate b dot d l along that path. It can be any path. Then that would be equal to this magnetic constant mu naught times the amount of current that passes through that surface area defined by the boundary of that path. So it's kind of like Gallus's law, but this is for path integral, so it has a path and a surface area instead of a surface area and a volume for Gallus's law. So let's use this to find the magnetic field due to a long straight wire of current, which we already know, so here's my current I going like that. So again, let me draw it coming out of the board, there's a current coming out of the board. We have to know something about the shape of the pattern or the field in order to use this effectively. So we already know that if I have current coming out, the magnetic field is going to make this circular type pattern like that. That's going to be b. So what kind of path can we choose to make that integral easy? Well, a circle seems like a likely choice. So if I have a circle like that of radius r, then I can do this integral. Let's do that. So in this case, the trick is that's b and if I integrate around this way, then this is going to be dL and at any location, b is in the same direction as dL. So it's going to be easy, right? If I take b.dL and they're always in the same direction and if I'm the same distance away from the wire and if I know b is constant at that distance, then this whole integral becomes just b because of b's constant integral of dL. So now if I go around the loop and integrate the length, I get the length of the loop. In this case it's a circle, so it's going to be b2 pi r. Now here is my, I want to do this side. Now here I have this area, right? How much current passes through that? Well, I only have one wire with the current I going through. So this is just going to be mu not I. Now I want to solve for the magnetic field b and I get b equals mu not I over 2 pi r. That's it, okay? And again you're like, man that's so easy, how come we just didn't do that before? Now before if you remember we said db equals mu not, let's see, we did I, delta L cross r hat over r squared. I mean that's really the more fundamental thing. That does tell you the direction, that's how we got those directions. So we had no idea what these magnetic fields looked like in order to do this without first doing that. What if, let me just, what if I had two wires but I had current going the other way right there, okay? Then how would this change? Well you may clearly see right away that the sum of the currents, in this case passing through that area, if I have I coming out and I going in, that's zero. Does that mean the magnetic field is zero? No, it doesn't. That means that the integral of b dot dl is zero. So in this case actually you would have in some places, if I heard about two wires, in some places I'd have the magnetic field going that way, in some places I may have the magnetic field going that way, okay? So I, and I'm, that could be wrong in those but as you integrate along the whole thing you get zero but individually they may not be zero. But this is, well that's fine, okay? So that's a, just a really basic example with ampere zone.