Fundamentals of Physics II

PHYS 201 - Lecture 9 - Magnetism II

Chapter 1: Review of Magnetic Fields [00:00:00]

Professor Ramamurti Shankar: Okay, well I want to start by cleaning up a little bit of stuff done near the end of last time. So I told you the whole thing with magnetism is going to contain two parts, just like with electricity, namely how do things react to a magnetic field and how do things produce a magnetic field? Just like the electric charges, electric charges produce electric fields and electric charges react to electric fields. So the first part is very easy for magnetism. The force of magnetism on a charge q depends on its velocity, because if it doesn’t move, it doesn’t feel it. And it depends on the magnetic field, which I will denote by some symbol B. Then you can ask yourself, “How am I going to get a vector out of two vectors?” namely the velocity and B, and a natural candidate is the cross product. Now that doesn’t mean that’s the right answer.

For example, I can multiply this by the absolute value of velocity squared and that will still be a vector, because that’s just a scalar. Anyway, ignore that, because that’s not what happens. Nature is very kind and it’s just v x B. And you got to know how to do the cross products, okay? There are two rules you’re going to use all the time. One is the cross product, which is for me the screwdriver rule, that if you’ve got a vector v and you’ve got a vector B, you turn the screwdriver from v to B and the way it comes out is the direction of the cross product. So everyone’s doing that. You can also do it with your hand. Have you seen all the rap guys? They’re all computing cross products, because there turns out to be a right hand rule and a left hand rule. So you can do that, or you can do this.

Another one is whenever there’s a current or something, you don’t know which way the magnetic field is going, you’re supposed to wrap your hands around the current. Now don’t do this at home. Wrap it around the current, and the thumb points the direction of the field. So whenever you have doubts about which way something is going, if you see something looping around, wrap your four fingers around that loop, and thumb will point in the direction of the answer. I’ve been using it a lot without spelling it out for you, but you should know at least the technical definition of cross product. Another thing is, I didn’t tell you anything about units. So Bis measured in Tesla, and the meaning of a Tesla is that if you have a 1 coulomb charge moving at 1 meter per second, perpendicular to the magnetic field B of 1 tesla, it will experience a force of 1 Newton. That’s it.

So Tesla is cooked up so that in the mks units that we use, if you put in the numbers, you will get the force in Newtons. Then I said from this microscopic description of what happens to a single charge, you can go to a piece of wire carrying some current and take a segment dl, and the current is I going through the loop. The force, which I write as dF, meaning an infinitesimal force on an infinitesimal segment is dl x B. So if you stick a wire in a magnetic field, different parts of the wire have current going in different directions. I’m taking a tiny section, dl, and in that tiny section, I’m computing the cross product. That’s going to be the force. For example, if at this location the magnetic field points that way, then the force will be coming out of the board.

So this is the first half, and I tried to do a problem near the end. Let me repeat that, because I sort of switched coordinates in the middle. So I wanted to do this problem. There’s a wire. This is a semicircle. I’m not worried about this portion and this portion. I just want to know, what is the force on this segment here? Which way is it going to act? So B is like this. First of all, you can tell, the force is going to be in or out of the blackboard, because if B is in the plane of the blackboard, I mean, velocity of the current is in the plane of the blackboard, B is in the plane, cross product will be perpendicular to two of them, will come out. You’ve just got to figure out, is it coming out or is it going in? So take a segment here, if you like. That’s a segment dl, and here is the B. Turn a screwdriver from dl to B, it’s coming out of the blackboard. B is assumed to be uniform and in the plane of the blackboard.

So let’s now compute the contribution from this segment. The contribution from this segment is going to be — I’m going to ignore the vector nature, because it’s coming out of the board. I already know that. So there is the current, then there is the dl, then there is the B, and there’s the sine of the angle, which means these two guys. I claim that that is the same as this angle here. This is an ancient trick. The ancient trick says if the angle between two lines is θ, then the angle between the perpendicular to the two lines is also θ. So the two lines are this and this and the perpendiculars to them, this guy’s perpendicular to this one and the tangent vector is perpendicular to the radius. So that’s the same angle θ. So that sinθ is this one. So the total force will be IB — and what is dl? The segment length dl is just R times dθ. So you’ve got sinθ dθ integral from 0 to Π. The reason I repeated this is that normally we measure theta like that, but there’s nothing sacred. You can measure it any way you like. If you measure it this way, it’s increasing in this direction, you should go from 0 here to Π there. So integral of sine is -cosine.

You do all the limits, you get 2IBR. By the way, it’s very useful to know a quick rule, is that the force on a segment, where everybody is perpendicular to everybody, is BLl. Segment of length L carrying current I in a magnetic field B. If B and I are all perpendicular, then the force is just BIL. So if you look at it that way, it looks like B times I times 2R, 2R being this length. In other words, if the current, instead of going around like this, had gone straight from here to here, it would have a length 2R. B is this way, perpendicular. Take the cross product, you will get the same force. In other words, the force on the curved segment between two points is the same as the force on a straight segment joining the two points. This happens to be universally true in a uniform field. In other words, even if the wire did this, don’t bother to find the force. It’s just simply the force on that segment.

Now you will prove that in your homework. It’s not obvious, but I don’t want to tell you exactly why, but that’s something you can prove. So that was one line segment. But then I started taking a loop. So here is the loop, and let’s say the magnetic field is going up like this and the loop is oriented like this. It’s carrying a current I in the sense shown here. So what are the forces on the four sides. B is this way. Dl or I is this way, and B is up. If you turn a screwdriver from there to B, you will find this force here is like that, and this force is like that. Both are horizontal forces. There’s a force on this side and a force on that side, which are equal and opposite. Yes?

Student: What are the forces on _________ B inside and outside the force?

Professor Ramamurti Shankar: They are perpendicular. I will show you a side view where you will see exactly how they are pointing. At the moment, this is limited by my artistic skills, but I will show you in a moment a side view that will make it very clear. But it should be clear right now, the total force on this loop is 0, because it’s a uniform magnetic field, if there’s a current going this way, there’s the same current going the opposite way for the same length in the same field, so those forces will cancel. So the loop doesn’t feel a net force. But it feels a net torque, which I think you can imagine, is trying to straighten this loop out and make it horizontal. And if you want to see the torque, look at the loop from this end, from this edge. It looks like this. That’s the loop and that’s the B field. And that’s the normal to the loop or the area vector of the loop looks like that.

So the force looks like this here and looks like that in the top section. So let us say this length is l and this width is w. Then the force on this segment is BlI, I told you that, and this force is also BlI, and that is w and this angle and this angle are both theta, then the torque = BlIwsinθ. I hope you understand why it’s wsinθ, because wsinθ is that distance. By the way, when you compute torques, you can find the torque around this point or torque around that point, it doesn’t matter around which point, as long as the forces add up to 0, and finding it around this point. If you like, you can find the torque around the midpoint, then this will give a contribution, that will give an equal contribution. You will add them, but no matter what you do, you will get wsinθ times either force. So that’s the torque. But whenever we see a sine theta and a lot of vectors, we know we’re looking at a cross product. And the cross product, the vector formula for tau, is going to be written as μ x B, and μ is a quantity which is area vector times the current, because l times w is the area. Right?

The area of the loop which points perpendicular to that has a value equal to l times w. so l times w is area, so I times area is called the magnetic moment, and that times B times sinθ is the cross product. So what you notice is that the torque will turn it till μ x B is 0. That will occur when μ and B are parallel, or anti-parallel. There are the only two options. When they are the parallel, the area vector’s along B. When they’re anti-parallel, the area vector’s exactly opposite to B. And they are the two equilibrium positions. But one is the point of maximum energy; one is the point of minimum energy, just like with the electric dipole. Once you have a torque like this, you can find the potential energy and you can show the potential energy is just -μBcosθ, which is -μ⋅B. Right? The integral of sinθ is -cosθ, so basically the point I’m making is, this loop would like to have its area vector parallel to the magnetic field. That’s what you’ve got to understand. It will align itself so that the most amount of magnetic field lines go through it.

So one way is to be parallel to it, the other is to be anti-parallel, and they don’t have the same energy. The energy looks like this, - cosine theta means it looks like that as the function of theta. Theta = 0 is the best and theta = pi is the worst. If you keep it here, it’s a stable equilibrium. If the loop is perturbed a little, little bit, it will rattle back and forth. If you keep it there, it’s like being on the top of a hill. If you perturb it, it will come all the way down the other side; it will flip over. So this loop is really like a tiny compass needle. If this B field were due to the earth, then the normal to the area vector, if you can put a little — you could take a plane and you paint a little arrow on top of it, the arrow will line up with the magnetic field.

Chapter 2: Torque on Charge Moving in Magnetic Field [00:14:02]

All right, so you should remember then that this is called a magnetic dipole. Why? Because we studied the electric dipole in which if you got a plus charge and a minus charge, and you got an electric field E and the torque on it was p x E, where the magnitude of p is the charge times the distance between them. And a dipole would also align itself with the electric field, with the + charge along E and the - charge down here. The other configuration was exactly anti-parallel, but that’s the configuration of highest energy, and the energy again was -p⋅E. So a current loop is like an electric dipole, except it’s the magnetic dipole. It is to the magnetic field what the electric dipole is to the electric field, with one big difference. If you look at an electric dipole, it really has a + charge and a - charge at the two ends. A magnetic dipole, like a current loop here, doesn’t have a + and - magnetic charge there. It’s as if there is a + and - magnetic charge, but there are no magnetic charges.

There are no isolated magnetic charges that produce magnetic fields the way electric charges produce electric fields. If you had a magnetic monopole, it will produce a radially outgoing field, like 1/r², but there is no such thing. But there is something that looks like a magnetic dipole. In other words, if this was a + magnetic charge, that was a - magnetic charge, then they would feel a magnetic field and they would obviously line up with the magnetic field. The dipole will line up with the field.

Now it turns out you can get some money out of this. You can build a device. The device I’m going to build is an electric motor, because you can take a bar magnet or any other magnet, with some magnetic field coming here. This is the other end of the magnet. You put a current loop here. Here are some wires coming out of the current loop. If you drive a current through it, then let’s see, if I drive the current like this, the magnetic moment looks like that, and the magnetic field, B, looks like this. So it will flip till μaligns with B. Do you understand? It will flip so that the plane of the loop is perpendicular to the field lines, or parallel to this or that face of the two magnets. But to do that, of course, you have to drive some current. So I take this current, a source of current, and I connect it like that. Now, do you have any problem with this invention? Yes? What’s your problem?

Student: It will only go until new aligned with B, and then it will stop.

Professor Ramamurti Shankar: That’s right. It’s not going to run very long. It’s going to flip till it lines up and that’s the end. And if it’s already lined up, it won’t even do that. So this is not going to sell. So what do you have to do? You guys might know what you have to do.

Student: [inaudible] AC?

Professor Ramamurti Shankar: AC? Suppose you only have a DC voltage, what are you going to do? Suppose you’re very quick with your hands, what will you do? Pardon me?

Student: Switch the magnets.

Professor Ramamurti Shankar: Okay, two things. One is, switch the magnet. You can also switch the wires of the terminal, right? You can keep doing this and you can keep reversing the current. Every time this guy thinks it’s found happiness, you say no, and turn it around. And you go the other end, you do it again. But that means you can never leave this motor and go anywhere. It’s high maintenance. So here is a very clever device. If you haven’t thought about it or if you don’t know the answer, you’ll be very impressed the way — I was. Because that’s the gap between pure science and applied science. It makes all the difference.

Here’s what they did. These two wires feed into two semicircular pieces of metal. Imagine that comes there, that comes there. I’ve just blown it up for your view. Then the battery is connected like this, not with a hard wire, but with a spring loaded brush. There’s a brush here, there’s a spring holding it. So these brushes make contact with these two metallic hemispheres, if you like, semi cylinders, with them connected to this wire. Now you can see what happens. As the coil spins, as shown here, the + terminal is connected to the top guy, the - to the bottom guy, so the current goes like this and comes out. But half a revolution later, this guy will be touching this one, that guy will be touching this one, and the current will reverse. So the current reverses automatically because you have this split circle and you have the top half and bottom half split, and then the current enters this metallic part and goes like this, and then later on goes the other way. That will keep running.

Also, in real life what happens is, once it flips over, in a real system, made up of real coils, once it comes to equilibrium, it won’t stop there, even without the reversal. It will go oscillating, because it’s got some moment of inertia. You just cannot lose your potential energy and have nothing to show for it. So what will happen is, it will overshoot its goal a little bit, and then it will get carried away to the other side. Then it will keep going round and round.

Okay, so that is an application of the torque. That is for a DC motor and you’re quite right. If you have an AC motor, then you put an alternating current supply, but then you’ll have to make sure that the frequency of the current supply will be the frequency of this motor.

Okay, so let’s leave this board here, because this is the end of one section. You have to know where you stand. We have done the first part of magnetism, which is, what are the forces and what are the torques on either moving charges or electric currents, in some simple situations?

Chapter 3: Magnetic Effects Produced by Electric Currents [00:20:57]

Now we are going to come to the more interesting part, also the more difficult part, which is, how do currents produce magnetic fields? Now we’re going to the cause of the magnetic field. At the end, everything is produced by charges, microscopic charges, but the formula for that is quite difficult to calculate because if you look at all the charges, and they’re moving around producing magnetic field, the field at any one location, because of the delay of interaction, depends on what they were doing at various times in the past. Just like electric fields are also difficult to calculate if charges are moving because of the relativistic theory that says you cannot communicate instantly with another place. You can only do it at the speed of light, so the field here depends on what was going on earlier for the more remote charges. But in electrostatics, we beat the problem by saying, “Look, none of these charges ever moved. They’ve been there forever.” Then where they are now is where they were before and we can calculate the field. The analogous thing for magnetism is to say that if you have currents and the currents are steady, then it’s a case where the magnetic field is also going to be constant. Remember, a steady current doesn’t mean there’s no motion. It means the motion of these charges round and round the loop is constant. It’s not changing with time.

Do not confuse a steady current with a single particle moving at a steady velocity. That is not a steady current. You see the difference? Because if you sit at any one point in a wire in a closed loop, the current is always the same. If you have the instrument, called the ammeter, that reads the current, you can put it where you like, you’ll get the same answer. But if a single charge is carrying the current, there’s a current here, there’s nothing anywhere else. There’s a current only where there is a charge. So it’s like saying that when I go on the freeway at 40 miles an hour, I do not myself constitute steady traffic, because there is no traffic where I am not. It’s the traffic with me now. When I’m over there, there’s traffic there. What we mean by steady traffic is cars are flowing constantly so at any instant, every place looks the same. So a steady current is what we’re talking about, not a charge going at steady velocity. So the question is, what is the magnetic field produced by a tiny piece of current sitting somewhere? This is a piece of current, I times dl , do you understand?

You cannot have an isolated piece of wire. It’s part of a bigger loop, but I’m focusing on a tiny section. And let us say it is located at r’. Let’s draw a picture so you guys can see anything. I go to the point r’, and there I catch a piece of wire carrying a current. The current is I, the segment is called dl. I want the field at the point r. There’s nothing here. I just want the field here, and this vector separating them is r − r’. So every segment of wire will produce a little magnetic field, dB. And it’s going to depend on the orientation of the segment, the current it carries and the separation between where you are and where the current is. And this is called the law of Biot and Savart. It says the answer is μ₀ over 4Π times I times dl x e_rr’, divided by r − r’ squared. So let’s take a minute to understand this. μ₀/4Π, like 1/4Πε₀ is a number whose value happens to be 10^-7 in the units we are using. That number is cooked up so that if you’ve got 1 ampere here and 1 meter here and 10 meters here and so on, you crank out all the numbers, the field comes out in Tesla.

So whenever you pick a system of units, this has all the factors that control the problem, you have to put a number in front so that with all these measured in standard units, this comes out in standard units and that number is μ₀/4Π. That happens to be 10^-7. r −r’ is the separation vector between where the current is and where you are. I is the current flowing in that circuit. And e_rr’ is the unit vector from r’ to r. You can call it e_rr’or e_r’r, as long as you know it is from the cause to the effect. That’s the formula. That is a nasty formula, because unlike Coulomb’s law — see, why is it more nasty? Have you thought about it? Why is it so much harder now than for Coulomb? Remember, Coulomb’s law was just the unit vector divided by r². Why are we having all of these nasty cross products here? Yes?

Student: [inaudible]

Professor Ramamurti Shankar: What’s making it? What’s bringing in all the extra vectors?

Student: The magnetic field can never be worked [inaudible]

Professor Ramamurti Shankar: Right, but what’s the cause of the magnetic field compared to the cause of the electric field? Yes?

Student: Moving charges.

Professor Ramamurti Shankar: Right, but the electric field is a point charge. There is no vector associated with that point charge. It’s just sitting there. The only vector you’ve got is from the charge to the location. The current, on the other hand, has got its own direction, nothing to do with where it is. It’s the way the little wire is going at that point. It’s the presence of the extra vector and separation vector that gives you a chance to form yet another vector by combining the two of them. That’s where you get these cross products. But anyway, that’s how nature is.

That’s what happens, so we’ve just got to say what it is. So you’ve got to take this and learn how to use it, so that’s what we will do.

So the first application of this is going to be the field of a little circular loop of some radius r, and I’m going to find the field at a distance z on the z axis. Say the loop is in the xy plane. I want the field here. So what you will have to do — think about what you have to do now. You’ve got this formula, but what do you think you will do? The current here is I. You guys have a game plan, at least in principle? What will it take to calculate the B field? Anybody from this wing has an idea? Yes?

Student: Find the field due to a point charge on the circle and then you integrate —

Professor Ramamurti Shankar: He said find a field due to a point charge and then integrate. Except for one thing, you are right.

Student: [inaudible]

Professor Ramamurti Shankar: That’s right. Take a tiny line segment, dl, here. Join at that point, take the cross product. That’s a vector you’ve got to add to all the vectors coming from all the points in the circle. That’s your answer. So I’m going to show you. First of all, is it clear to you that by the fact that the problem is symmetric under rotations, with respect to this axis, I can pick any one point and whatever it does can be used to deduce what other points will do. They’re all in the same footing. I want you to take this point and this point for a minute. Here the current is coming out of the blackboard, and here the current is going into the blackboard. Are you with me now?

The loop is half behind the blackboard, half in front of the blackboard, and I slice it here. I just look at this guy and see what it does. There’s an element dl, if you want, coming out of the blackboard here. And the separation vector is this one. If you take the cross product of that vector with that vector, you’ve got to think about where it will be, and it’s got to be perpendicular to this vector, therefore in the plane of the blackboard. And it’s got to be perpendicular to this vector. And if you do your signs right, it will give you magnetic contribution dB that looks like that. But you know that the part of dB that’s going to survive is the component here, because this guy will produce a field like this. And by symmetry, they will have the same angle, so you only have to keep the z component, not the one in the xy plane. Because for every element I find here, I can find a compensating one on the other side of the circle.

So therefore, dB along the z direction is μ₀I/4Π times the segment coming out of the blackboard. What about the sinθ? Sine theta, here’s where you’ve got to be very careful. What do you think the sinθ is between the segment coming out of the blackboard and this guy? What’s the angle between a line that lies in the plane of the blackboard, and the line coming out of the blackboard?

Student: Ninety.

Professor Ramamurti Shankar: Ninety. Don’t be fooled by anything else. Anything coming out of the blackboard is orthogonal to anything in the plane of the blackboard. The wire is coming out of the blackboard, this vector’s in the plane. There’s no sine thetas to worry about. But you’ve got to divide by the square of the distance, and the square of the distance happens to be z² + R². That’s the square of the distance. That will give you this part, but I want that part. So that’s cosine of this angle α. Now let’s see where I can get the same cosine. The angle between two lines is the same as the angle between the perpendiculars, so that guy is perpendicular to this guy. This guy is perpendicular to that guy, so that will also be alpha. And cosine α will be R divided by z² + R², ½. So the sinθ didn’t come because of a sinθ in the cross product.

This factor, it came because you want the projection of this vector up the z axis. We are not done, because this is a contribution from a tiny link dl coming out of the blackboard, but notice the contribution is independent of where you are on the circle. Everything makes a contribution up the z axis. Each one makes a contribution dl, so if you add up all of these, there’s nothing to integrate here. That’s what I want you to notice. z and R are fixed. You’ve just got to integrate dl, which will be 2ΠR. So I get μ₀I/2ΠR squared, divided by 4Π, divided by z² + R² to the power 3/2. That is the magnetic field here. So along the axis, it will go like this. And what is it at the center of the loop? That’s a useful formula to remember. At the center, you put z = 0, you get μ₀I times 2ΠR², divided by 4ΠR³. That gives me μ₀I/2R Tesla. And it’s going to be pointing straight up. And if you go down here, you will find it looks like this.

Now if you want to find the field somewhere else, it’s a lot of work, because you don’t have the symmetries, but I will tell you what you get. You will get these lines that look exactly like — I’m not showing you the other half, but that looks the same. Now that is supposed to remind you again of a dipole. In fact, if you go to very, very long distances, when z goes to infinity, what does B look like? You go to the formula and let z go to infinity, you will get — maybe I’ll do it again here. z goes to infinity, B goes to μ₀IΠR², divided by 2Πz³. I’ll tell you why I don’t cancel the pi’s. I know you guys are dying to do that, but ΠR² times I is the dipole moment of the loop. Remember, it’s the area times the current. So ignoring all the factors, it looks like the dipole moment divided by the distance cubed, that’s exactly how the field went for an electric dipole.

So the magnetic dipole and electric dipole are, again, similar. So they’re similar in two ways. Just like the electric dipole aligns with the electric field, the magnetic dipole aligns with the magnetic field. That’s the first thing. The field of an electric dipole has got these little circles going out and the field of a magnetic dipole also looks exactly like that. The only difference is, if you go into an electric dipole, you’ll find two guys, + and -, producing the field. If you go into a magnetic dipole, you’ll find nothing at the center. You’ll just find a loop. So nature gives us magnetic dipoles, but not magnetic monopoles. That doesn’t mean they’re not there. People have been looking for them. They’ve been spotted once, but it’s not been spotted again. It could be because they’re very rare, and they’re very heavy and they’re very hard to measure, or maybe they’re not there. We don’t know. But we don’t see them the way we see electric charges to be sure they exist.

Okay, now suppose I take one coil and put under it a second coil and a third coil and I join them all up and wrap them around some little cardboard tube, you can imagine then these fields will all add up and you’ll get a field that looks like this. That looks exactly like a magnetic field around a bar magnet. Can you see that? You saw the field due to one. You can stack them up and then you can compare the field. As far as a compass needle is concerned, if you put it here or there, it behaves the same way.

So now we have a question. Here is a magnetic effect due to current carrying loop, or due to currents. Here is a magnetic effect due to no currents. Your magnet’s not connected to anything, so you have an option. Either you can say that’s a new kind of magnetism produced by god knows what, or you can say, “I believe that everything is coming from electric currents,” and the question is, where are the currents in this guy? Do you know where the currents are in a bar magnet? No? Okay, let me give you a hint. It comes from the atomic theory of materials. Don’t think of this as a lump of something. Yes?

Student: The electrons.

Professor Ramamurti Shankar: That’s right. You know the atom has got electrons going around, and every electron is a current. Now you can ask, let’s put one electron here going like that, then put another electron going like that, another electron going like that. What do you think they all do together? Suppose I’ve got nine electrons in the plane of the blackboard all going around their atoms. You can see that in the region between the atoms, they go in opposite directions. This guy’s going like that, this guy’s coming like this. They cancel. The only thing that doesn’t cancel is along the perimeter. Everybody — what are these guys doing? Here we go. Sorry. Yes, they do this. It’s a physical fact that when many orbits are put next to each other, they cancel on the interior and give you something on the boundary. So even though each atom is a very tiny loop, if you’ve got 10²³ atoms, they can look like a microscopic loop.

Now that is just a two dimensional loop, but you can take a solid, a magnetic solid. It’s go these little orbits here. In every layer it’s got orbits, and each one is like a current. So magnetic material is like having a sheet of current, and of course that will produce a magnetic field. There are a lot of subtleties, but this is roughly what happens. This is how you get magnetism. There are some questions. One of them is, why isn’t everything magnetic? Why not this guy? It’s got electrons, right? Why aren’t they lined up, and which way should they line up? Which way should the magnet point? So the answer is, many materials may contain equal number of electrons going in opposite directions, so that some are going one way, some are going the other way, and for that reason, each atom may cancel itself out.

Another possibility is, every atom may have one net uncancelled electrons, but the plane of the orbit for each atom is different. Some are rotating in this plane, some are rotating in that plane, so there are little magnetic moment vectors pointing in random directions and not lined up. And that happens because of thermal agitation. Things like to jiggle when you heat them. But when you cool them, they can all line up and you suddenly find the material is magnetic. So if you take a regular bar magnet in your fridge and you put it on a hotplate for a while, and you remove it, you’ll find it’s less magnetic. So it’s magnetic because you caused these guys to jiggle more. And beyond a temperature called curie temperature, which depends on the material, the jiggling will be so intense that magnetism is destroyed, not atom by atom, but because they’re no longer able to act together and produce a net magnetic moment.

Then you can also take materials in which every atom has an electron that’s willing to line up, but they don’t know which way to line up. They can line up in any way they like, but then if you put them in a strong magnetic field, then because every dipole likes to line up with a magnetic field, atomic moments line up with the magnetic field. And even if you take off the magnetic field, the dipoles remain aligned. Do you know why they remain aligned? Because they are producing their own magnetic field which keeps them in place. That’s called a self consistent solution. So the magnets, if they all agree, “Hey, let’s line up,” can actually make it a stable situation, because when they do line up, they produce their own magnetic field which encourages them to line up. These are all interesting things in magnetism, but the starting point always is to find out how to take these microscopic moments and make them line up.

So now I’m going to do — so this is at least an explanation of why bar magnets are magnetic, because they are really circulating currents, and it’s no different from this current. Okay, now I’m going to find the magnetic field for a very classic problem, and that’s the field of an infinite wire. So here is the infinite wire and we want to find the field it produces. You take a point here, and let’s take some segment of length dx. And this is the separation vector, and we want to find the field due to — dl is here. Now do the cross product. Turn screwdriver from dl to R, it comes out of the blackboard. So this little guy makes a contribution that comes out of the blackboard. In fact, you can check that, as you go along the line, everything acts in the same way and makes a contribution coming out of the blackboard. So what you can imagine is that I just happened to slice it in this plane, but the real picture is, the field lines will go around the wire.

Right now they’re coming out of the blackboard. Down here they’re going into the blackboard, and they are orbiting around the wire. If you look at the wire from the end, then the current is coming towards you. The field lines will look like this. And what we’re trying to calculate is how strong is that field? So you go back to the famous law, dB = μ₀I/4Π times dl, which is just dx, divided by x² + a². Let’s say the distance is a. Then you need the sine of the angle between the separation vector of the segment, and the sine of the angle, sinθ, is a divided by square root of x² + a². Then the integral of dB is the integral of all these guys. Now you can guess what the final answer is going to look like, just from dimensional analysis. dx is a length, a is a length. There are two lengths on the top. The bottom there are three powers of length, an x² or a² from this one, and the square root of a length squared, which is another length. You see that?

Two powers of length at the top, three in the bottom. The answer has to, in the end, have dimensions of 1 over length, and the only length in town is a. So you know it’s going to look like μ₀I/a. But then it could be half of a, or 10 times a. If you do that integral, you’ll find it’s over 2Πa. And I’m not doing the integral, because we have seen this integral before. dx over x² + a², to the 3/2. You make the tangent substitution and then get secant this and secant that. So I don’t want to do that integral again, but the answer is μ₀I/2a. so the magnetic field from an infinite wire falls from a distance like 1 over a. Very much like the electric problem. Even though single electric charge produces a field that goes like 1 over r squared, if you remember, infinite line produces a field that goes like 1 over r. Similarly, even though elemental wire has a contribution that goes like 1 over distance squared, when integrated over an infinite line, the final field goes like a, 1 over a, where a is the distance from the axis of the wire.

For a homework problem, I want you to notice that in this integral from - to + infinity, the segment for the + axis, the segment for the - axis make equal contributions. In other words, if you take this to be origin, that segment and that segment contribute equally to the field at any point, because it’s an even function of x. You will find it useful when you do the problem set. Okay, so here is an interesting result, μ₀I/2Πa. Now that’s going to explain a few things. So here is a wire and I told you long back when we started magnetism, I said, if you put a test charge here and it moves like this, it’s attracted to the wire. Now we can understand their attraction. We understand the attraction as follows: look at the thing from end on, the current is coming — what do you think the direction of the magnetic field here is?

Again, you put your thumbs around this one, it’s curling, therefore it’s going into the blackboard here and coming out of the blackboard here. That is the azimuthal magnetic field seen from the plane of the blackboard. You’ve got this guy going to velocity v here. Take the cross product of v with somebody going into the blackboard, you’ll find you get a force towards the wire. That’s why the charged particle going along the current is attracted to the current, if it’s positively charged. If it’s going in the opposite direction, if you reverse the velocity, you will get the opposite force, so it will go away from the wire.

Finally, I want to mention the experiment that people did before any of this which is to notice that if you’ve got two currents, l₁ and l₂, they attract each other if they are parallel and they repel if they’re anti-parallel. So if you went and asked yourself, “How am I going to find the force that’s attractive when they’re parallel or they’re anti-parallel?” you can think of all kinds of simple rules, maybe involving this current vector and that current vector. But the correct answer is a pretty complicated series of cross products on why these two wires attract. Let me tell you why they attract. I think you can see yourself. Take the force on this guy due to I₁. It produces a field that looks like this there and looks like that here. So just like the single charge here, this experiences a force towards the other wire, every part of it. So the force that current 2 feels due to current 1 is μ₀I₁/2Πa, where a is the distance between them. That is my B. I, and that’s my l. If l₂ is the length of this wire. I’m just using the formula BIl I told you about.

So the B is the B due to the first wire, I is the current of the second wire, l is the length of the second wire. So if you took an infinite second wire, you’re going to get nonsense, because you’ll get infinity as the force between two infinite wires. So it’s more common to take the force. Let me drop the arrow on this, because I’m looking at the magnitude per unit length. Then it is just this, μ₀I₁I₂ over 2Πa. And the current, you know, is measured in amperes, which means if you took 1 ampere in this wire and 1 ampere in that wire, and they were 1 meter apart, the force between them will be μ₀/2Π, which will be 2 times 10^-7 newtons per unit length. In other words, if you’re all holding this wire from running off to that wire, and you are given 1 meter of the wire to take care of, you will have to exert that force. So attraction between wires in the end has an explanation, but it’s very complicated. It does not simply involve the two current directions.

The first current, through a series of cross products, produces a magnetic field perpendicular to this wire, and another cross product of that perpendicular field and the current here gives you another vector which is the attraction vector. So the force of attraction, whereas force of attraction between two charges is simply q₁ q₂ over r squared, force of attraction between two wires, even though it has a simple formula, comes in the end from a whole bunch of cross products. That’s what you have to know.

Chapter 4: Ampere’s Law [00:51:27]

All right, so now I’m going to do a very basic result called Ampere’s law. Ampere’s law says the following: look at the current coming towards you. The field lines go like this. Let me call that angle Φ. Then the magnetic field is a unit vector in the Φ direction times μ₀I divided by 2Πr, where r is the distance. This is the wire, okay?

And I am looking at various circles. This is the circle of radius r. Here is what B looks like. So B is circulating around the wire, very different from E. E emanates from charges, but has no circulation. Its line integral on a closed loop is 0, electrostatically. The magnetic field is just the opposite. It’s always going around in circles and therefore has a line integral, and I will mention later on that it has no surface integral, because Bs don’t start and end anywhere. The lines of B, therefore if you took any surface, whatever line comes in has to go out, because it cannot stop. I’ll come to that in a minute, but let’s look at the line integral of B around a circle. B⋅dr around a circle. So on a circle, take a section of a circle. So here is my portion of a circle. dR = rdΦ times unit vector in the phi direction.

Do you agree with that? The tangent vector to the circle, a tiny little guy, has length r times dΦ. r is this one, and points in the direction of the unit vector in the phi direction. So E_Φ is unit vector like that. B is already this. Therefore B⋅e_rdr looks like μ₀I/2Π times integral… because the dot product of the two vectors, because they’re both azimuthal, is simply the product of this guy times that guy. I’m sorry. Please change this R to little r. In other words, I want that distance to be little r. You can also use big R. It’s going to cancel out, but I’m not taking a fixed length r. I’m taking r to be variable, so I want to use little r for it. So rdΦ is the length.

The point to notice is that the field goes like 1 over r, the arc length goes like r. So they cancel. Integral of dΦ is 2Π and I get μ₀I. So the line integral of the magnetic field are on a closed loop, penetrated by a current, = μ₀ times the current coming out of that surface. Independent of the radius of the circle. You can take a small circle, you can take a big circle, you always get the same answer. Because if you take a big circle, you’ve got a big circumference growing like r, but the field goes like 1 over r and the effects cancel. This remind you of something you’ve seen before? Which one?

Student: The plane.

Professor Ramamurti Shankar: No, where these things cancel. Pardon me?

Student: Gauss’s law _______________.

Professor Ramamurti Shankar: Like Gauss’s law, you mean?

Student: Yes.

Professor Ramamurti Shankar: Yes, Gauss’s law, where you took a single charge, the surface integral of Ewas independent of the size of the sphere, because the field went like 1 over r². Area of the sphere went like r², and surface integral just said it’s equal to the q that’s inside. Then of course we found it’s true, even if the shape is not spherical. Again, we find here, first you take a circle and you find the answer is independent of the radius of the circle. Then I’m going to show you that the answer is independent, even of the shape of the loop. So let the current be here. So take a loop like that. So take a portion of that loop, this portion here. It’s got a radial part and a tangential part. So you can see the dr of the loop is dr times e_r + rdΦ times E_Φ.

Let me blow this up for you if you like. So that is that segment there, dr, and that’s got an angular part and a radial part. The radial part is length dr, the angular part is length rdΦ. So integrating along that curve, which I can approximate by a straight line, I can break up that elemental vector dr into the tiny portion in the angular one, tiny portion in the radial one. But B = μ₀I/2Πr times E_Φ. So when you take B⋅dr, e_r⋅e_Φ is 0, because they are perpendicular vectors. All you get then is B⋅dr will be μ₀I over 2Π dΦ. So the line integral of Breceives the contribution equal to simply the angle of the two rays that bound this little segment. Then you can see if you go around the whole loop, you will just add up all the dΦ’s and you will get 2Π and you will get μ₀I.

So the answer is true, even if it is not a circle, as long as it is a closed loop, enclosing the wire, the answer is equal to current enclosed. What if you took a loop here and the current’s actually here? There’s no current going out of this guy, so we’d better get 0 for that and we want to know why. You’ll get 0 because if you do the same formula, it will give you integral of dΦ, but notice if you go round here and come back all the way here, the change in Φ is 0. Can you see that? In any loop bounded by some lines, the range of Φ as you follow it goes back to 0. Whereas if you encircle the origin, then Φ changes by 2Π. So if you take a point here and you go around in a loop, you come back to the same Φ. If you take a point here and you go around, phi goes up by 2Π. That’s why a loop not enclosing a current will get 0 contribution. A loop enclosing a current will get μ₀I.

And finally, by superposition, if you’ve got many, many wires coming out of this. Some are going in, some are coming out, some are going in, some are coming out, then integral of B⋅dr = μ₀ times sum of all the currents, I_j penetrates the surface bounded by the loop. That’s because B is made up of B due to this guy, that guy, that guy, but the integral of the total B is the individual integrals. Each one gives you its own current. So superposition is a great and powerful result. If you can do anything for one charge or one current, you’re able to add everything together. So this is called the law of Ampere.

Now I might as well tell you, the surface integral of B on any closed surface is actually 0. Because if you took electric charges, they’ve got lines coming out like this, right? And if you took a surface, you get some contribution from the dot product of the field vector with the area vector. Magnetic lines never end anywhere. They always close. So take whatever surface you want, any line that enters it will also leave it, because there’s nowhere to end. Had there been magnetic charges, magnetic monopoles, lines will either come out of the monopole or collapse into the monopole. Then the surface integral of B will count the magnetic charges. But we have not found any. So as of now, the surface integral of B is 0, the line integral of B is the current trapped by that loop. Do you understand? You take that loop and you draw a plane with that loop as a boundary. Then any current coming out of that is going to be equal to the line integral up to this factor mu. For the electric field, just for comparison, the line integral of the electric field is 0 and the surface integral of the electric field = the charge over ε₀. So here is this one integral. Let me write it here, B⋅dr = μ₀ times all the currents crossing it.

So it’s very useful for you to write these 4 mathematical equations. These are called the integral form of the Maxwell equations, namely, it’s the relation between electric and magnetic fields written in terms of the integrals of electric and magnetic fields. There’s something called the differential form of the magnetic field which is more useful, which requires more mathematics than we have done in this course. But this is all for the case of electrostatics and magnetostatics, very, very important. All the currents have to be constant; all the charges have to be fixed. Only then is this true. It turns out this is all you really need to determine electric and magnetic fields. Suppose someone comes to you and says, “I’m going to give you all the currents. I’m going to give you all the charges. Can you find E and B?” It’s a mathematical problem. The mathematical statement is you know the line integral of either field around any loop and you know the surface integral around any surface. It turns out that’s all you need. The line and surface integrals are known for every possible loop and every possible surface, then the solution is unique. That’s why this is all you really need to do any problem in magnetostatics. Because you might think, maybe this guy’s going to write two more equations. I’m saying mathematically, this is complete. On the other hand, if I stopped at one of them, except for the electric field, if all I told you was this one, then mathematically it’s not enough to solve the problem.

Okay, so I’m going to use Ampere’s law the way I used Gauss’s law to show you some simple applications. Suppose someone gives you an infinite wire and says, “Find the field around it.” Now you can do that nasty integral which I did on the top right corner, but now we don’t have to do it. Because you will argue by symmetry that the field here cannot be pointing to the left or pointing to the right, should lie in the plane, because there’s no reason to bias one way or the other. And in the plane perpendicular to it, the field should be invariant when you rotate the wire, because the wire looks the same when you rotate it.

So the only possible things I can think of are, the field can look like this, or the field can look like this. They both have the property that when you rotate the wire, they look the same. But you can rule this one out. One way to rule it out is to say that if all the magnetic field lines end on the wire, there must be magnetic charges and we don’t have them. Another way which is more subtle is to say, if this was the field configuration for this current, if I reverse the current, everything should reverse and the lines should go outwards now. On the other hand, if I took this infinite wire and I rotated it like that around this axis, the current will point the other way, but the field lines would still keep going radially in. That means if I took that current and I flipped it around, the field lines still go the center, but I has become - I, so that rules this one out. But you can easily satisfy yourself with this configuration of circling things will satisfy all the requirements.

So I know the field is azimuthal, goes around the wire. I know it depends only on the distance from the wire and not where I am in the circle, therefore I look at the wire end on and I know that there’s a field here in the angular direction. I also know the line integral of B⋅dr= μ₀I. This is like Gauss’s law. You cannot find from it B separately, except in the lucky case where B⋅dris a constant over the entire circle, which it is in this case, because it’s simply B times 2Πr. If B is azimuthal, which I’ve argued for, the line integral is B times 2Πr = μ₀I. That means B = μ₀I/2Πr. So what I’m telling you is just like Gauss’s law, will only give you one equation. If you told me to find the field due to an arbitrary distribution of currents, I can still write this true statement, but I cannot deduce from it the value of B. But if it’s due to this infinite wire and you’ve got enough symmetries and the only thing you don’t know is the magnitude of B as you vary r, then on a contour of fixed r, there’s only one unknown, which is, how big is B as a function of r in the azimuthal direction. Then the one number you get from this one equation.

Now here’s another variation of that problem. They all have analogs in electrostatics. Imagine that I have a solid cable carrying current I, coming out of the blackboard. And let’s say this cable has radius R. Then I want the magnetic field here. Now the real problem is much more difficult than before, because there you had a single line, you could do the integral. If you have a blob, a cylindrical blob, you do all the dl cross r’s for everything is going to be very, very difficult. But now we’re going to argue by symmetry that this one should also have necessarily fields that go around in circles, with a uniform strength on a circle of given radius. Then you conclude then that 2Πr times B = μ₀I. Therefore B = μ₀I divided by 2Πr, which is exactly the field you would get from a point wire carrying all the current.

Again, this should remind you of Gauss’s law, but if you have a ball of charge, the field due to it looks like the field of a point charge sitting at its center. Likewise the field of a uniform distribution, a tube of current coming towards you, outside this tube looks like it’s all coming from the center, because that’s what it will look like if the entire current were concentrated on the center. But you will see the difference now. If you go inside this and ask, what’s the magnetic field on a circle lying inside this, then again you can say 2Πr times B = μ₀ times the current that’s enclosed by that loop, not the entire current. And I think by now you guys should know, that will be this fraction of the total current. Or if you want to see the details, you can put a pi there and you can put a pi here. I/ΠR² is the current density and ΠR² is the area of this guy. But by now you guys should know that things scale like distance squared if you’re looking at areas. So what does that tell me now? It’s μ₀Ir² over big R². So for B, it looks like μ₀Ir divided by 2ΠR², with r less than or equal to R. And this is the formula for r bigger than or equal to R. And you can test the formula when little r = R. That’s correct. Little r = R, they agree. Therefore the magnetic field as a function of r grows inside the cylinder and then decays as 1/r. All this should be familiar to you if you still remember similar problems from Gauss’s law. If you’ve got a ball of charge, the electric charge grows linearly when you’re inside the ball, because as you grow, you’re enclosing more and more charge, which is growing like r³, but the sphere is growing like r². So it grows like r, but once you cross the sphere, you go to bigger and bigger spheres, you’re not enclosing more charge, but the sphere for Gauss’s law is getting bigger, so it falls like 1/r². Similarly here, it falls like 1/r when you’re outside the cylinder, and grows linearly with r when you’re inside the cylinder. Now there are a couple of other miscellaneous applications which I will do next time.

[end of transcript]

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