Let's finish up our integral here. We had said that the B field, we're going to integrate these dBs. What did we have? We had Mu_naught I/4Pi and we had an a that can go with constants like that. We had dx over x squared plus a squared to the 3/2. We get all this by combining the sine Theta and the 1/r squared. The last thing we need are limits on the integral. We're describing a physical system. This isn't calculus where the limits are meaningless. This is a real system, so we have to have real limits. In this case, we're doing a really long wire, so we're going to put it from minus infinity to infinity. This same integral can be used to find a finite wire. If you put this at the origin and just made it for minus over two to over two. You could plug those in and get the answer just fine. But let's do the infinite case. B equals, let's keep these constants, Mu_naught Ia/4Pi. Then how to do that integral. Now it's not an easy one. That's one of those that you look up in a table or you put in your computer, or you go through and integrate it by parts or some horrible thing like that. But that integral in the end, you get x over a squared times x squared plus a squared to the minus 1/2, if you put in the top, over that to the 1/2. Then you evaluate that from minus infinity to infinity. Now, if you don't believe me, you can take the derivative of this and you'll get that. I promise I've done it. It's a lot of fun, but it does work. A whole bunch of terms cancel and it works. Now, we have to plug in our limits to get B. Let's see, B Mu_naught I and we can see one of those a's goes away, basically, this a is now in the bottom, over 4Pi a, 4Pi this a stays and that one gets canceled with that one. We're just going to keep this on the outside and plug in infinity and minus infinity. Now, plugging in infinity and minus infinity, plugging in infinity. What we're going to do is just plug it in and think about what do we think we should get? If you have infinity over the square root of infinity squared, which doesn't seem like it should be a real thing plus a squared, what do you get? Well, if you do infinity, things are going to go wrong. Let's just say big. Let's do big and little. Infinity is really big compared to a, this measly little separation. It's big and a is small. Let's just think, what if we plugged in BIG over the square root, no I'll just write it the same way. BIG square, which is really big, plus what? Small squared to the 1/2. What would that be? Basically, in physics and math, this comes up all the time. You have something BIG plus something small and if the thing that is big is a lot bigger than the thing that's small, you just ignore the thing that's small. Who cares? In the limit if something is actually infinity, then it's actually perfectly true. Often we do this for an approximation, but here it's really true. If we just ignore the small because the BIG is so BIG, then the square root of BIG squared which is BIG, and BIG/BIG is one. That's how you plug in infinity. This is fast and loose. It can get you in trouble, but it's good enough for right now. We plug in one and now minus what? Now we're going to plug in minus infinity. We got to do this limit and you might think it's going to give us 1 minus 1, but it won't. Let's see what's actually going to happen. If we plug in minus infinity, it's minus BIG and this would actually be minus BIG squared plus small. If you do that, again, this is minus BIG squared, it's something really BIG and this is small so the small goes away and this remains positive, but this one was negative. You have minus BIG/BIG that's minus 1. It's 1 minus minus 1, which of course is 1 plus 1. Let me get this out of the way. Then you get a B, I'll indicate head direction. You get B, this is going to be a two and it's going to knock that four down to a two, is Mu_naught I over 2Pi a and the k head direction. That's telling us once we add up all these B fields, that's the B field we're going to get. Now, I've seen something that's a little bit ugly here is this vector I drew in the wrong direction because I was thinking of an electric field. I don't want to reshoot this whole thing, so what I'm going to say is that's an Easter egg for you to notice during the whole lecture. Of course, dB is like that. That's a little dB, now we've added up all the dBs and we have the total B field. When I circle it, I'm going to leave room to write something here because there's one thing we got to think about. That's the direction of the B field in this plane. In the plane of the board. That can't be the B field everywhere. If I just write the B field everywhere it goes in the k direction, then it's going to be this thing that varies, and even out here it would be in the k direction. That's not the case. We know the field does something else. We're going to talk about it in a minute. We've really only solved for it here and where we go, the reason it has to be is when we figured out this cross-product. We cross ds along the wire with r-hat going this way. They were in this plane that gave us the direction of the B field. If you start coming out into space, out of the board, r-hat changes, and then the B field direction would change. We'll talk about that in a minute. But just to be clear, this shows that the B field, it goes down is one over the radius, if a got bigger, the B field gets smaller. You can think of that as we've solved it as a function of radius away. But it's only in the plane of the board. Now the second here, we'll draw it and see what it looks like.