Let's use the Biot-Savart law to calculate the magnetic field at the center of a loop of current. We've talked about how a loop is very fundamental in terms of magnetism, a loop of current. We will be talking more about loops of current, but for now let's just get B in the center of a current loop. For our loop, we'll assume a current I and radius a, just at the center. Let's go ahead and draw a current loop like this and the current we'll say is going around in that direction. Where is it coming from? I don't know. There's no battery. It's just a current loop of a superconducting wire, I say. No resistance, so the current will go on and on forever. We know we start with Biot-Savart, that the little differential element of magnetic field dB, that is due to a little piece of the current, ds, is equal to Mu naught I over 4 Pi. Permeability of free space, current 4 Pi times, let say it's ds crossed with r-hat over r squared. That's Biot-Savart. If we want to draw these vectors in here, I is already there, ds we've already drawn, r is from the ds to the place where we're measuring or we're calculating the magnetic field. We're doing it in the center of the circle, r would look like this, like that. For that ds, there's the r. Therefore r-hat is the little unit vector in that direction. It changes as you go around the circle, and r with nothing on it, we'll say is just the magnitude, that's just the value of the radius. All we got to do then to get big B is to do the integral, integrate around the circle. It always a bother just to have our differential involved in a cross product inside of our integral, that all basically. Let's think about what will always be true for ds and r. As you go around the circle, ds will always be perpendicular to r-hat. When you're down here, there's ds, there's r, there's r-hat. You can imagine as you go around, they'll always be perpendicular. Another thing is that the direction, ds cross r-hat is always in what direction? It's always going to stick out. If we have ds cross r, it sticks out, so the B field will stick out here or that component of the B field to stick out, ds crossed with r out. Db for that piece will stick out. Db for every piece will stick out. Let's make a coordinate system because I know it bothers you when I just write out with a hat on it. Here's x, here's y, x cross y is z, so z sticks out. It will always stick out in z or as we like to call it a k-hat direction. That's what we know. If they are perpendicular, then the angle between them is 90 degrees and sine of 90 is one, so it's really down to just the magnitudes. Let's see what that does. Let's write then B is the integral, add up all the dB's. For the path, let's just write a circle. We want to go around the circle. This a closed loop. It doesn't have to be, you can do Biot-Savart on an open path, but for this purpose, we're going around a closed loop. That's equal to the integral of this thing. Let's go ahead pull out all the constants. Anything that's constant doesn't need to be in the integral. The permeability of free space is constant I hope. I, the current is constant over 4Pi, 4Pi is constant. The ds has to stay in the integral that'll just be the magnitude of ds because we took care of the cross-product, r-hat, it's magnitude is just one, so don't worry about that. Little r, this is the radius, but it's actually a constant. The magnitude of the r vector is always a constant. The magnitude of the r vector, which is what we just called r always equals a, the radius of the circle. We can actually pull that r squared out and just go ahead and call it a square like that. We have Mu naught I over 4Pi a squared and all that's left is the integral around the loop of ds. Everything else is gone. The sine of 90 is 1, r-hat magnitude is 1 that came out. It's just the integral of ds around the circle. If you're familiar at all with calculus and maybe a little bit of vector calculus, you know that if you integrate this little ds around the circle, that's a length, and it's just the circumference of the circle. That's all it is. When you see that, you can just call it 2Pi times a. But if that step bothers you, I'll do that for you. Let's see. I'm just going to go super explicit Mu naught I over 4Pi a squared. If we wanted to convert this from ds to something we can actually integrate, well, you can integrate in terms of Theta. Could you? You can call it Theta. We could say this is the integral from 0 to 2Pi going around a circle. Then you've got to think about what is ds, ds is this vector. You could also describe it as the radius a times d Theta. If you're going from looking at polar coordinates, the way you define a differential around the circle is the radius times the d Theta. You could, if you wanted to write this as a d Theta. Integrate with respect to Theta and the a is just a constant, it comes out, integral of d Theta is Theta evaluated at 2Pi and 0 and you just have 2Pi. What do you get? You get 2Pi times the radius, which is the circumference. Often you'll see in books and when people lecture, you'll see that you'll just write the circumference, not really doing an integral, which is also why I did a circle on it. It's like we're just going around a circle. The integral around a circle, ds 2Pi times radius. Or you can do this. Either way you do it a 2Pi a ends up in the top and its direction of creating right of here is k-hat. Either way you do it, you get a B, that's Mu naught I, 2Pi times a, that a is going to cancel. There's no a in the top and the 2 from there cancels that 4 into a 2, the Pi's cancel. This Pi cancels that Pi, and you're left with that. You get that the B field at the center of the current loop is Mu naught times I over 2 times the radius of the loop. That's only in the center, and it sticks out in the k direction. To get the direction, if you don't want to do all the ds cross r vectors, use the right-hand rule. Everything's a right-hand rule. In this case, something goes around, goes with your fingers, your fingers go along the current and your thumb will stick out always in the direction of the B field. We'll do another complicated case in a minute.
