 It is extremely common in nature for charges to come in large numbers, and so it's not enough to ask what will a magnetic field do to a moving electric charge, rather one needs to consider the details of what happens to currents of electric charge, large flows of electric charge in the presence of external magnetic fields. Such large collections of charge moving in the presence of strong magnetic fields created by external phenomena is extremely common in the universe. This is a phenomenon that we will begin to explore in a bit more detail now. I want to build on this equation. I want to go one step further and I want to introduce another concept that will be helpful going forward in the discussion of electric currents, magnetism, and so forth. So let's start from the force equation on this current carrying wire. So the force is given by the magnitude of the current, i, times a vector, l, which is units of meters, it's a length vector, and it represents the length of the wire that's exposed to the magnetic field, and it's direction, it's where it points is the direction that current points. So for instance, if we were to look at this loop of current that I've drawn here, the bottom section points to the right. So here, down in this section, that's where l vector for the bottom points to the right. For this section over here, l vector for the right-hand side points up. And then on the top, it points to the left, and on the left, it points down. Okay, so l vector is just following the current direction in each section of this square loop that I've drawn here. So, well, square or rectangular. Don't call this a square loop, just to keep it simple. Square loop, all right? So each side of this has a length l, all right? So I didn't do a very good job of making this look square, but each side of this has a total length l, all right? And it's immersed in a magnetic field, b, which is uniform in strength, and it points always in the same direction, and that's to the right. So b vector in this picture are these sort of fainter lines with arrows that I've drawn here pointing to the right. And so we can start thinking about what the force is going to look like on each piece of this square loop. And then the total force is just the sum of the forces, okay? So the forces are gonna be, they're gonna add just like vectors because they are vectors. So the total force is just the sum of the individual forces on each of the four sides of the loop, okay? So I equals side of loop drawn over here, okay? So let's start looking at the forces qualitatively. I'm not gonna calculate anything quite yet, but we can look qualitatively at the forces. Let's start with the right-hand side of the loop. So to figure out the direction that the force due to the magnetic field is pointing, what we can do is we can apply one of the two versions of the right-hand rule that I've been talking about. We can take the fingers on our right hand, flatten out our hand, and make sure that that points in the direction that current is flowing, because that's the direction L vector points. And now what we're gonna do is we're gonna reorient our hand until my palm basically faces the direction that the magnetic field is pointing. That is, that's the direction I would have to curl my fingers to point in B. And wherever my thumb is pointing, if I do thumbs up at this point, that's the direction the force would then point on this current, okay? And since current is defined as the direction that positive charge flows, we already are dealing with positive charges here. So we don't have to worry about a weird sign change like you do with Qv cross B where it could be positive or negative. I is always defined in the direction that positive charge is moving. So that sign is already taken care of for us, okay? So L, we're gonna have to tip my palm to the right to point in the direction of B. And then my thumb indicates the direction of the force on this wire. So the force over here, and I'll represent that as these little arrows with X's in them, so the tail feathers of arrows fired into the board. Those are the directions that F right points. So the force on the right-hand side of the loop points into the board. Let's look at the top, okay? So the top of the loop, you now have a current going to the left, all right? So I take my fingers and I point in the direction of the current. And now I'm gonna orient my palm in the direction of the magnetic field. Well, the good news is I don't really have to sweat this one too much because what's true about the magnetic field and L vector in that particular case, they're parallel. And so the cross product is anyway, zero, right? So up here, you have no force in this particular case. Because L vector, which is the direction that current points is, in this case, it's anti-parallel, so it points in the opposite direction of B. But nonetheless, it means that L and B lie on the same line. And there is no cross product. The cross product of two parallel or anti-parallel lines is always zero. So the top is easy. And then let's just do the bottom while we're at it. So the bottom L vector points to the right, the B field points to the right. So the force is zero, okay? So there is no force down here, either, okay? So let's then focus on the left, yeah? Sorry to clarify, is it the cross product with those things, zero? Yeah, and there's no force? No force. Anytime if L is ever parallel or anti-parallel to B, the force is zero. Because the force is the result of taking the cross product. If the cross product is zero, the force must be zero. So parallel, A parallel, do you mean the line that is parallel or anti-parallel? The vectors are parallel or anti-parallel to B. Yes, yeah, well, these two are to each other. These are the only two vectors involved in the cross product. So if L is parallel or anti-parallel to B, then it's zero, okay? So let me just make this a bit more concrete by reminding you that the magnitude of the force is I, L, B, sine of the angle between them. And if theta equals zero or pi, so pi is 180 degrees, the sine of zero and the sine of pi are zero. So that's the other way you can remember this is if you remember, okay? The magnitude of the force is I, L, B, sine theta, okay? Then if theta is zero or pi, force magnitude is zero, there's no force, okay? So that's another way you can do this. All right, so for the left, so rather than playing this game where I have to kind of get my fingers to the point, what I'm going to do is I'm going to use the other right hand rule, which is this one. You make sort of coordinate axes out of your index, middle and finger and thumb, okay? This is x, this is y, this is z, that's the way I remember it at least. So you're going to take your finger and you're going to point your index finger in the direction that L vector points. You want to point your middle finger in the direction that B vector points and your thumb will indicate the direction of the resulting force. So it's not very comfortable, but it's not too bad. So here I'd have to do this. So my index finger points in the direction of current, which is down. My middle finger points to the right, which is the direction of the magnetic field. My thumb indicates the direction of the force, which is out of the board over here, okay? So draw these as circles with dots in them, arrows flying at your face. So this is f vector left. So to do the total force, we just need to add f vector left plus f vector right. So let's just look at the magnitude of those for a second. We know that their directions are opposite. We know that f left points out of the board and f right points into the board. So they're opposing each other. What are their magnitudes? What are the magnitudes of f left and f right? The same. So these vectors oppose each other, and there's no net linear force. This is net linear force on this loop of current. So that's not so bad. But you've got the forces applied at different locations on the loop. So if I take an object, and I apply, so I don't have a good example here, but I can do it with this. If I have an object like this chalk, which is very rolly, okay? And I apply opposing forces to the ends of the chalk. What happens to it? It tilts, yeah? So I can apply equal but opposite forces. I can let gravity do the force on the left, and I can push up on the right. It tilts, it rotates. This is causing rotational motion to put opposing equal magnitude forces on the ends of the chalk, okay? So let's just focus on the forces. So we've got one force going in over here, pushing on that side of the loop. We've got one force coming out here, pulling on the left side of the loop. The loop is going to begin to rotate. So we have a torque. We have a force that's displaced from the center of rotation, and this causes a rotation. So just like you can put your force on the end of the door, and the hinges are fixed, so you cause a torque that opens the door, this magnetic field is causing two torques. One that tries to rotate the right end down into the board, and one that tries to rotate the left side of the board up and out, okay? So what I want to do now is I want to look at the sum of those torques and get a sense of what the total torque is on this loop, right? Once we do that, we can do all kinds of things like figure out how fast this thing's going to spin in a circle, and so forth. And this comes in handy. Let me preview why this is useful. If you can get a rigid mechanical loop exposed to a magnetic field, and I hinted at this in the video lecture that I gave you guys on, just magnetic phenomena, if you can expose a loop like this to an external magnetic field, it will begin to rotate. And if you time your device just right as the loop rotates so that it's now perpendicular to the magnetic field, but spinning, and you flip the sign of the magnetic field, it will then continue to rotate around. And then you flip the magnetic field again, and now you have a device that's spinning. And if you were to hook into that rigid mechanical loop, something like a drive shaft, you could move a car. You could run an electric generator. And in fact, this is the exact principle, with a few extra pieces we'll get to in a lecture or two. This is the exact principle behind electric motors. So you use a current and a magnetic field. You can make the magnetic field from permanent magnets, big strong permanent magnets, and all you have to do is get something to change sign whenever the loop rotates so that there's no more force due to the magnetic field. In the case of an electric motor, you change the direction of the current. If you flip the direction of the current, now the force points in the other direction and it keeps spinning the loop. And if you time this just right, you can make a motor. And so mechanically, electric motors are set up to make that timing work perfectly. Okay, they do this with various tricks, but that's how you get a motor, basically, okay? So basically the significance of not having a force is that it doesn't translate, it just rotates? Exactly, so this loop isn't like doing this, okay? It's just spinning in place. Okay, now, of course, there's no perfect loop. There's no perfectly uniform magnetic field. So engineers have to engineer those motors with some mechanical assembly that keeps the loop from tilting because that can cause the whole thing to just rip itself apart. So there's, of course, good engineering that goes into this to take account of the fact that there is no such thing as a perfect system. If you design systems and assume that they're perfect, you will fail. So we have two torques caused by two forces in this problem. So we need to figure out what each of the torques is. Let's focus on, we'll call one, let me re-label this. So we'll call this the right hand torque, and we'll call this, that is not how you spell that, the left hand torque, okay? So let me rotate the picture for you so that we can see the forces and we can picture the torques. So to rotate the picture, all I'm gonna do is I'm gonna draw the loop. So this is now the side. So what I've done is I've taken this loop and I've just tipped it like this so that we're looking at the end here, all right, where there's no force. And so we have the current on this bottom side is now facing us. And it's moving to the right, and it's coming out of the left side. The current is coming toward us. So I'll put a little circle of a dot there. And as the loop bends over here, the current goes into the board. So we have I out of board over here. And I into board over on the right. So, and then in this section, it is going to the right, all right? So here it goes into the board, here it comes out of the board, and this is the bottom of the loop. The magnetic field still points this way, okay? But now we've revealed the z direction. So this, if I label that up here, but the B field is pointing along the x direction. Here the current was going in the positive y direction. Here the current was going in the negative y direction. And the force here is going into the negative z direction. It's going into the board, so it's going down. The positive z direction is indicated here with this dot coming out of the board. And the force here is coming out of the board in the positive z direction. So negative z direction there, positive z direction there. So out of the board is positive z directional. Now I've flipped my coordinate system around. I've got x going that way. I've got z going that way. And I have positive y coming out of the board. So again, I've just tilted the picture so that we can see the forces now. So the forces here, I'll draw this over here, I into board. So the force here points down. This is f right. And the force here points up. This is f left. So again, I've just tilted the picture. So now we can see the force arrows. The current, the L vectors, are coming out of the board here and into the board here. So let's think about what happens. We have equal forces on either end of this loop. And so we get this rotation. And the center of the rotation, if this is a nice uniformly distributed loop of wire board. So the distance from where the force is applied to the center of rotation is half the length of either side of this loop. So the whole side is of length L. That's half L and half L. So let me just redraw this here. Here's the center of rotation. This is a length of 1 half L. That's a length of 1 half L. We have a force down here. We have a force up here. Now for torques, let's do the torque on the right-hand side first. This is equal to r cross f. So we need to know the radius vector that points from the center of rotation to where the force is applied. And we need to take the cross product of that with the force that's applied. So we need to know the vector for the right-hand side that points from the center of rotation out to here. So that's this vector. So this is r right, points from the center of rotation out to where the force is applied. The force is applied on the end of this arm here of the loop. And we have the force already. So let's write down some vectors. So keeping in mind that, again, to the right is still the x direction, just like it was there. r right is a magnitude times a direction. Well, the magnitude of that r is just the length of the conductor between the center of rotation and the end. That's a half L. We need a direction. And for that, we just need a unit vector that points in the direction that r vector points. That's in the positive x direction. So that's going to be i hat. That's it. We're done with r. Now we need f, f right. Well, we know that that is going to be i L cross b. So we need some vectors here. We need to write in terms of our coordinate system these vectors. Let's see. So for this, what I'm going to do is, what we know when L is already, it's the total length of the wire that's exposed to the magnetic field. Force is acting on this length here, which we can't see anymore because it's hidden behind the bottom of a loop that's now facing us. So this is the direction of the current flow that's experiencing the force. And it points in the positive y direction. Positive y direction. Positive y direction. That's where current is flowing. So we have to consider the length that's exposed to the magnetic field that's getting the force put on it. So that's L. That doesn't change. But now we're in the positive y direction, which is j hat. And then finally, we have to get a vector in for magnetic field. Well, the magnitude of that vector is just b. And it points in the positive x direction. That's where I drew b originally, positive x. So this is i hat. OK. So let's do some cross product here. Let's get that cross product sorted out. And then we'll do the torque cross product. So let me pull the constants out of here. We've got L. We've got b. They don't participate in the cross product. They're just numbers. It's the length of the wire, whatever it is, 1 millimeter, or 2 millimeters, or 10 centimeters. And then the magnetic field string, b, whatever it is, a tesla, a milli tesla, we don't care. It's just a number. So we pull it out in front, i L, b. And then we just have this cross product, j cross i. OK. So let's dig back a lecture. i cross j is k hat. So i hat cross j hat is k hat. So what is j hat cross i hat going to be? Negative k hat. Great. Thank you. So we have i L, b. So negative i L, b k hat. So that is the exact expression for the force acting on the right-hand side of the loop. And that's the force that enters the torque equation here. So the torque on the right is equal to 1 half L i hat cross negative i L, b k hat. OK. So let's get all the constants out in front. The negative i L, b, the 1 half L. Let's just get that out in front now. So we have negative 1 half L squared, b. And then we have the cross product, i hat cross k hat. Put it to that high. That is an outstanding question, doesn't it? Current, missing recurrent. Yeah, negative 1 half i, the current, L squared, b. And then we have unit vector in the x direction across unit vector in the k direction. So again, if we dig back to last time, we had i hat cross j hat equals k hat, k hat cross i hat equals j hat, and j hat cross k hat equals i hat. Flip any of those two, and you put a minus sign in front of the right-hand side of the equation. So we have i cross k. Here we have k cross i is j hat. So this is negative j hat. Excellent. Negative j hat. So we're done with the torque on the right. The torque on the right is negative 1 half i L squared b negative j hat. And let's combine those minus signs so that they cancel out. And we just want to line up with 1 half L squared b j hat. So how's the problem done? Hooray. We need the other torque, the torque on the left side. All right? So I'm going to slide things down a little bit here. All right, and we'll come back to those in a second. So the torque on the left is r vector left cross f vector left. OK? Well, r vector to the left will point out to where the force is applied. So it goes from the center of rotation out to where the force is applied. And the force is applied right on the end of the loop here. So this now points in the negative x direction. So our vector for r left is negative 1 half L i hat. The force on the left is, again, i L vector left cross b vector. Now, L vector left is this. This is the side that actually gets the force that causes the torque that has a length of L. And it points in the negative y direction. So negative j hat. It points down. So this is just equal to i negative L j hat cross b still points in the same direction, b positive i hat. So we can group these terms together. And we get negative i L b j hat cross i hat. OK? J hat cross i hat is negative k hat. So we wind up with just i L b k hat. So how did I screw that up, if at all? I did not. These torques, yeah, this is fine. So the force is your point in the opposite direction. This one points in the negative k hat direction. This one points in the positive k hat direction. We figure that out anyway from the right hand rule. The equation reflects that fine. So we're good. So now we just need the torque. And that is going to be the cross product of this vector, negative 1 half L i hat cross i L b k hat. So I can, again, pull all the constants out in front, negative 1 half i L squared b. And then I just have i hat cross k hat, which is negative j hat. So I just wind up with 1 half i L squared b j hat. And oh, look, torque on the right plus torque on the left. They add up. You get a rotation. There's a net torque. That's good, because we kind of assume that that was going to happen in the math bears that out. We get a net torque. The total torque is the sum of the two. And it is just 1 half plus 1 half. Those equations are the same otherwise. So we just get i L squared b. I want to unpack this for a moment, because in order to introduce the key concept here that I'm using this calculation to motivate, I need to kind of step back for a moment and put a cross product back into this equation. And I know that's going to seem a little odd, but let me go ahead and do it. You'll see where I'm going with this. One thing that I want to point out, what is L squared equal to for a square loop? You have a square, and you calculate L squared. What's that? The area. Yeah, it's the area of the loop. So let me just make a quick substitution here and write this as i times the area. This is the area times b j hat. And then let me go one step further, and let me unpack j hat into a cross product. So what cross what gives me j? See if we can do this backward. k cross i, yep. So this is i a b k hat cross i hat. Let me unpack this one step further. What I'm going to do is I'm going to put b back here in front of i hat. So I'm going to put the b back with the i hat. No reason I can't do this. Totally legit. What is b i hat equal to? What was it originally? The b vector. Yeah, so let me go ahead and put that in. OK, we're nearly there. The new concept is nearly upon us. What I will now do is I will define this thing i a current times area gets a new name, mu. And it stands for magnetic dipole moment. And it is analogous just like an electric dipole. In an electric field, we'll execute rotation. And it has a moment. That moment is equal to its length times the magnitude of the charge on either end. A current loop behaves just like an electric dipole, but in a magnetic field. Remember I mentioned that the simplest fields in nature we've ever seen are dipole fields. So it's very convenient to simply define the equivalent for the magnetic field in force that we had for the dipole and the electric field in force. There's a dipole moment for electric charge. There's a magnetic dipole moment for magnetic rotations in a magnetic field due to currents. OK, and what is it? It looks very similar in construction to the electric dipole moment. Electric dipole moment was q times d charge times the length of the separation. This is i moving charge current coulombs per second times the area of the loop. So it is very similar in its construction. And this is no accident. As you'll see in a bit, I'll demonstrate it in a moment. There is no accident that there's a weird symmetry here between current loops in magnetic fields and two charges bound together in an electric field. There's a reason for this. And it's very convenient to define this quantity as mu. And here's why. If I want to know the total torque on a loop, I just have to know mu vector cross b vector. And mu vector is just i a, in this case, k hat. How do you figure out the direction of mu in an arbitrary problem involving a current loop? Well, it's actually not as hard as it seems. Where does k hat point in this picture? Out of the board or into the board? Out of the board. K hat points in the positive z direction, and positive z comes out this way. So if I wanted to just draw mu, that would be mu. And its magnitude would have been the current times the area of the loop. That's it. You don't have to do this nonsense garbage was writing r cross f, and then I'm going to sum the ports, and then, nope, it's easier than that. You need to know the area of the loop. You need to know the current in the loop. And you need to know one direction, the direction of a vector perpendicular to the area, which is what k is. The area is in the plane of the board. Mu vector points out of the board. And the way you figure it out, take your fingers, curl them in the direction the current is flowing. Current is flowing counterclockwise in this loop. Your thumb indicates the direction of mu vector. That's it. Whole lot easier. Whole lot easier to figure these things out. With this trick, you can figure out, for instance, how a microscopic current loop that might be present, for instance, in the electrochemical processes in a cell would respond to the cell being exposed to a magnetic field. With this information, you might be able to figure out how currents in the brain, current loops in the brain, would be affected by magnetic field. So to solve force problems like this, though, you just need to know what the magnetic dipole moment. Any time you have a loop, and that loop has an area, and there's a current in that loop, all you do is take your fingers, curl them in the direction the current is flowing. Your thumb indicates the direction of the magnetic dipole moment. So you know it. You can just write it down. In this case, it's k hat, because it points perpendicular to this area. And you just put i a k hat. If I increase the current by adding more loops, let's say I add a second loop in here carrying the same current with basically the same area. I've doubled the current. I have 2i instead of i, but the same a. So I can double the magnetic moment by doubling the current. I can have the magnetic moment by having the current. I can control the behavior of a loop of wire by altering the current flow. And that is how you control an electromagnetic motor. And again, I'll keep illustrating that. We have other principles in magnetism and electricity we need to illuminate. But this is a really fundamental principle. And in fact, what we know now is that even subatomic particles like electrons behave like little current loops and that have an irreducible little magnetic moment inside of them, which is related to something called spin. And so the actual origin of magnetic fields in a terrestrial magnet, like those craft magnets I showed you were like in this compass needle, OK? The spins of the electrons are all pointing, let's say, 1% or 10% of them are all pointing in one direction. And those little current loops are all oriented. And you'll see why that matters in a moment, because I'm going to demonstrate the phenomenon. But this makes magnets. So each electron is a little magnet, and it's a little dipole magnet. It has an north end and a south end. And that's really useful because we can then, for instance, protons have this too. Neutrons have this as well. So if you can get those magnetic moments to flip in response to an external magnetic field, you can do things like image the chemicals in the body in different slices of the human body. That's what a magnetic resonance imaging scan is. You take the spins of the atoms and you flip them in resonance with an external magnetic field or an electric field. And then by doing that, you can figure out what chemicals are present in different parts of the body and build up these gorgeous 3D images without ever cutting a single human body part, which is really important, right? I mean, when you think about the way that we used to learn about human beings, you wait till they died and you cut them open. But you can't learn about a human being who's breathing when they're dead and you're cutting them open. You can learn a lot of things, but there are some things like the functioning of the brain or the way that the organs behave or how a cancer grows in a human body. You can't learn that unless you can cut open a human noninvasively. And that's the technology that we as a species have been developing now for decades, noninvasive human imaging. And MRI and spin and magnetic fields are essential to that. So let me demonstrate the next important principle here. All right, any questions on this before I move on? This will all tie me together at some point, so yeah. All right, so we did the loop. OK, so this is just a slide where I'll put this all up on the web page today. Here's the example of how you calculate the direction of the magnetic moment. That little n vector is the vector that's so-called normal to the area. Normal means that are right at a 90-degree angle. So the only way if you've got your area in a plane like this, the only way to get a normal vector is to point perpendicular to the plane. And to figure out the direction that n points as a point out or as a point in, you just curl your fingers in the direction of the current flow. That's it. So another right-hand rule, you'll have to absorb. And then this is just pointing out what I said here. You can align loops of current along or against a magnetic field, just like you can align dipoles, electric dipoles, with or against the field. And this is the aligned is the low energy configuration and anti-aligned against the field is the high energy configuration, just like electric dipoles.