 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 in its 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. I'll call this a square loop, just to keep it simple. Square loop. So each side of this has a length L. So I didn't do a very good job of making this look square, but each side of this has a total length L. 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 going to be, they're going to 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 going to 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 going to do is we're going to 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 and be. 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 taking care of for us, okay? So L, I'm going to 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 Xs 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 going to 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 of those things, zero? Yeah. No force? No force. Any time 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 and anti-parallel lines are anti-parallel or parallel and anti-parallel? The vectors are parallel or anti-parallel. L? 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, sign of the angle between them and if theta equals zero or pi, so pi is 180 degrees, the sign of zero and the sign 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, sign 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 magnitudes 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. Similarly, on this loop of current, you could make one of these by taking copper wire and bending it very sharply, okay, and then hooking it together with a little battery at one end. And then you could actually make a little loop like this. So Lizzie, did you have a question, or I saw this, and I wasn't sure if this was like a wait stop. So yeah, and then Laura, do you have a question after Catherine? No, okay, all right, so Catherine. So are those the anti-parallel ones? The forces? Which directions with what? Oh, do the, oh, I see, do the wires, we haven't gotten there yet. Oh, we haven't gotten there yet. Yeah, yeah, yeah. There's a, that's the next step. So right now we're still investigating what does any magnetic field do to a wire. The next thing we're going to do is we're going to find out what exactly causes magnetic fields in the first place, and then we'll get into what happens when you have two wires that are, that are next to each other. So you guys are a little head in the reading, that's why I've slowed down a little bit. I'm going to lighten the reading load for next week, so that I'm going to skip Ampere's Law. So we'll catch up to where I want to be, but I'm trying to go carefully through this so you can see the steps, okay. So the prices, I have to slow down a little bit, 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, all right. 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. 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 like 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. And the early, like, airline industry illustrates this perfectly. The early commercial airline industry in Britain was building airplanes with materials that when received from factories were assumed to be perfect. And so when they manufactured their planes, the assumption was that they're perfect and that there's no tolerance for failure. But these planes started flying. And then after a while, they started ripping themselves apart in the sky and falling out of the sky, killing people on board. So there were two, I think there was at least two major airline accidents of planes tearing themselves apart in mid-flight before they stopped and tried to figure out exactly what was going on. And the flaw, essentially, was their reasoning. They reasoned that when the material shows up from the factory, it's flawless. And so they can assemble anything out of it and it will be flawless. But no material is flawless. They're always micro fractures, tiny cracks, imperfections, whatever. And so it's those imperfections that cause tears to begin. And what you have to do is engineer your aircraft so that if a tear begins, the material can absorb the spider webbing crack and stop it before the fuselage tears itself apart. So modern aircraft are designed with failure in mind. They have multiple redundant failsafe systems in the fuselage construction. And they use multiple kinds of materials so that if a crack develops, it stops. And so you have this kind of philosophy about designing now, which is much more accepting of faults and tolerant to faults. And interestingly, you can use electric currents and magnetic fields to scan the surface of a plane and detect microcracks. And this is in fact how airline safety professionals, when they're assessing the fuselage of a plane, they don't take a magnifying glass and look for little cracks, your eyes are not that good. But you can take a little coil of wire with a current running through it. And you can run it over the surface of the plane. And you can measure the current coming out of the loop. And from that, you can figure out whether or not there's a crack in the plane, even one you can't see with your visible eyes. You can take the plane out of production, and you can fix the crack. Okay, so fault is very important. Whenever we do physics problems, we're setting up perfect situations. But as you go into the world, the world is imperfect. And one has to be tolerant of that. So engineers have to, you know, mechanically fix this loop so that it doesn't wobble like this, because it does this little engine will tear itself apart. My electric lawnmower fails, because the rotor tilted out of alignment. And one day I was just mowing the lawn and he went, and that was it. That was the last sound it ever made. Because at that point, the wire, the loop of wire inside of it came in contact with the magnets and ripped the whole thing apart. So how would you know? That's us, you should take courses in the engineering school, they spend immense amounts of time on how do you build in tolerances, bridge failing fuselage failing, car failing. How do you assess the faults that are possible in materials as a whole branch of material science focuses on this whole thing, development and failure of materials. So that's a great question. And it is a zero time for it in an introductory physics course. But the engineering school focuses a lot on things like this. That goes into quality assurance and so forth. So yeah, Justin. Okay, so kind of back to what you're saying about the tour. Yeah. So you said something about it rotating one way, but if you flip something, it still rotates, right? So as let's say we have this loop, right? And we turn on the magnetic field or we start a current running through the loop. So now there's a force. And the loop starts to spin like this. Now when it gets up here, it doesn't stop going, it keeps going, right? So all you have to do now is when it gets up to the apex of the rotation, and it starts heading into the second half of the rotation, you flip the sign of the current, the force will flip direction, it'll just keep spinning like this faster and faster and faster. So as you cycle the current in different directions in the loop, you can make it spin faster or slower. It's all in the timing of the flip. So I'll show a motor demo later, because we're not quite yet ready to talk about motors. Flip the current? Yeah. Yeah, if you want to do two things you can do, you could flip the direction of the magnetic field, but it's really tough to do that if you're going to use big permanent magnets in your motor. It's cheaper to just to just have a circuit so that as the rotor comes around, it makes a different contact with the battery and changes the direction of current. So the battery might be hooked in in this mechanical assembly one way. And then as it spins half a cycle, the connections to the battery reverse. And so plus becomes minus and minus becomes plus and the current flows in the other direction. And this is mechanically how they design motors. I'll show you guys some some some videos of this, but we're not quite there yet. I just wanted to preview that. Okay, let's just do the Blase mundane thing of calculated torque. I know I really sold that right on a Thursday morning. God, it's hot in here. Why did you people complain about it being cold? I hate you all now. See, I told you you can't win with facilities. It's either too hot or too cold. There is no Goldilocks zone for this building or this room. It's like a cursed room. Happy Halloween. All right, torque. I'm going to stop trying to crack terrible jokes. There that one worked. 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, you know, let me relabel this. So we'll call this the right hand torque. And we'll call this that is not the left hand torque. So let me rotate the picture for you so that we can see the forces and we can picture the torques. So if to rotate the picture, all I'm going to do is I'm going to 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 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 towards us. So put a little circle of the dot there. And as the loop bends over here, the current goes into the board. So 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. Alright, so here goes into the board here 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 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 direction. Now I flipped my coordinate system around. I've got x going that way, I 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. Alright, so the forces here, 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. Okay, so again, I 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 this rotation. Okay, and the center of the rotation, if this is a nice uniformly distributed loop of wire board. Alright, 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. Alright, so let me just redraw this here. Here's the center of rotation. Okay, this is a length of one half L. That's a length of one half L. We have a force down here. We have a force up here. Okay, 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. We need to take the cross product of that with the force that's applied. It looks like it would be rotating the opposite way, rotating the opposite way. Well, what I did was I tilted this this way. So the bottom is now facing us. Okay, so that force points down that way and that force points up. So so this this loop is rotating out of the board like this. And in this picture is rotating up. Okay, so I did that consistently. I there's always a danger that I'll flip the sign here. So be vigilant. Okay, but I think that's okay. I think that I think those are both consistent pictures. Because all I've done is I've if I could take this board and tilt it like this so we could reveal the z direction. That's basically what I did is I ticked the picture so that the bottom is facing us this way. And that's what you see here. Okay, it's important to play around with this on your own. I'll give you torque problems. You'll have lots of fun with this. Don't worry. Okay. Alright, so we need to know the vector for the right hand side that points from the center of rotation out to here. Okay, so that's this vector. So this is our 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. Okay, and we have the force all ready. So let's write down some vectors. Alright, so keeping in mind that again, to the right is still the x direction, just like it was there. Our 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 our 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. Okay, now we need F. F, right? Well, we know that that is going to be I L cross B. Okay, so we need some vectors here, we need to write in terms of our coordinate system, these these vectors. Okay, so 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. Okay, so this is going to be I L. And the direction in which L points is the direction of positive current, or it's a current flow, positive charge flow. That's to the right, that's the positive x direction. So that's just I hat. Okay. And then we have to cross that with the B vector. Oh, hang on a second. Hang on. Hang on. Nearly went off the rails on that one. The force is acting on this length here. Okay, which we can't see anymore, because it's hidden behind the bottom of the 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, get 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. Okay, 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. Okay, so let's, let's do some cross product here. Let's, 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 got L, we got B, they don't participate in the cross product, they're just numbers. It's the length of the wire, whatever it is, one millimeter or two 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. Okay, 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. Okay. And as usual, I'm running out of board. So what I'm going to do is I'm going to just use this space here and finish off the cap relation. So continued from left board. Okay, so now we're going to do with R, which we have a vector for here, cross F, which we have a vector for. So the torque on the right is equal to one half L I hat cross negative I L B k hat. Okay, so let's get all the constants out in front, the negative I L B, the one half L, let's just get that out in front now. So we have negative one half L squared B. And then we have the cross product I hat cross k hat. That is an outstanding question doesn't current missing the current. Okay, negative one half I the current L squared B. Okay. And then we have unit vector in the x direction cross 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 one half I L squared B, negative j hat. And let's combine those minus signs, so that they cancel out. And we just want one half L squared B, j hat. Half the problem is now done. Yeah, we had an eye going and I had going. Yeah, this is where notation gets annoying. Would everyone like me to switch to capitalize for the currents? Yeah, I'm getting confused because that's fine. That's fine. So I big I big I big I anyway, I wanted to do this and then forgot to do it. So we'll just change. So we had a negative j hat and a negative one half here on so on the torque on the right. So you have negative I'll be and then you have negative, but then I pulled it. Oh, sorry, it's it got crammed in here. But there's a minus sign hiding right there. Yeah. Okay. Okay. Oh, and then yes. Thank you. And this is why it was a good idea for me to switch to capitalize because I keep dropping the little ice thinking that they are unit vectors. So big I, big I, big I think that's all of them. Big I, big I, there we go. So half the problem done. Ray, we need the other torque, the torque on the left side. Alright, so I'm going to slide things down a little bit here. Alright, come on. So the torque on the left is r vector left cross f vector left. Well, our 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 one half L, I have 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. Okay, 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 forces should 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 one half L i hat cross i L B k hat. So I can again pull all the constants out in front negative one 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 one half i L squared B j hat. And oh, look, torque on the right plus torque, sorry, torque on the yeah, 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 one half plus one half. Those equations are the same otherwise. So we just get i L squared B k hat j. Excellent. You're learning to question the teacher. My mission here is complete. I can quit and retire now. Yes, I doubt it's quick, but go ahead. On the T left on the right board. Yeah. Yeah. I just came from where did the left turn? Good question. So that is going to be our left vector. And our left vector is a vector that points from the center of rotation out to where the force is applied. So it's arrowhead points to the left. This is the x axis, positive x is that way. So this is a negative i hat direction. And it has a magnitude of one half the length of the loop. So that's where the negative one half l i hat comes from. And so that's that negative one half l i hat. So you saw that equation for F left first and then plug back in to get our love. Well, no, I plugged in. No, no, no, no, no, no, no, I wrote down our left that I figured out what F left was and I use that in the cross product to get torque. And then that's what I got. 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 and you'll see where I'm going with this. Okay. 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. Okay. 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. Okay, what I'm going to do is I'm going to put b back here in front of i hat. Okay, and I'm running out of board space again. So who gets sacrificed? You? What should I get rid of? In the box. Okay, does anyone know what needs this? Great. So we'll just reclaim this for spin. Okay, so we have work total equals i a b. And then I have k hat cross 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. Okay, 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 will 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 rotation rotations in a magnetic field due to currents. Okay, 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, 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 with writing r cross f and then I'm going to sum the torques 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, right? 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 that 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 fields. Does anybody remember that video I showed on the first day of the guy from the BBC? I'll show it again in a bit. He was talking, it was Humpty Dumpty sat on a wall, Humpty Dumpty had a great fall. It wasn't stuttering because he stutters normally. He was stuttering because they had a giant magnet next to the speech center of his brain. And all they did was they screwed with the currents in that section, and he couldn't talk anymore. Alright, so it's very important to have these concepts down because you can do all kinds of research with them mischievous or otherwise. Okay, so actually that research is very useful because what that research revealed, which I think have been known in other ways, but you can test it directly in the lab, was that singing and speaking are different things. I have a colleague who suffers from a very strong stutter when he talks. He's brilliant. He's a fantastic physicist. When he gives talks, though, you have to be patient with them and just let him talk because he'll catch on words and not be able to keep going. But he sings in a choir in Europe. And I asked him once, when you sing, do you stutter? And he said, no, and it's as well known that stutterers, extreme stutterers can sing just fine. And the demonstration that the video I showed you on the first day, which I will show again at some point, was the guy could sing Humpty Dumpty, I don't know. Well, Humpty Dumpty had a great fall and he could get through it, even though they were knocking out his speech center with the magnet. So this kind of tool is useful for revealing the behavior of the brain in a way that it's basically its functions are distributed. Okay, so neat little things you can do with magnets and currents and so forth. All right. So to solve force problems like this, though, you just need to know what the magnetic dipole moment anytime 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 that current is flowing, your thumb indicates the direction of the magnetic moment dipole moment. So you know it, you can just write it down. In this case, it's k hat. Okay, 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 two I 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 electric motor, electromagnetic motor. Okay. And again, I'll keep illustrating that we have other principles and 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 they 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, this compass needle, okay, 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, do protons have this to 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. Okay, 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 die and you cut them open. Alright, 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 image 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, okay, alright, so we did the loop. Okay, so this is just a slide. We'll put this I'll put this all up on the web, the webpage today. So 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 end points as a point out or does a point in you just curl your fingers in the direction of the current flow. That's it. Okay, so another right hand rule you'll have to absorb. Okay, 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.