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PHYS 201: Fundamentals of Physics II
Lecture 8
 Circuits and Magnetism I
Overview
After a description of more complicated electric circuits, the basic ideas underlying magnetism are discussed and the relationship between electrical charges and magnetic fields is explored. Magnetism is caused and experienced only by moving charges. The Lorentz force on a charge is described and used to deduce the force on a current carrying wire. The cyclotron and velocity selector are described.
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htmlFundamentals of Physics IIPHYS 201  Lecture 8  Circuits and Magnetism IChapter 1: Review of Electric Circuits [00:00:00]Professor Ramamurti Shankar: Okay, let’s start. I’m going to do some leftover stuff from circuits, because I didn’t finish it last time. So I want you to remember the following things about a circuit. I’m always thinking of different ways to explain it. So the things we normally learn is, if you’ve got some battery here, some number of volts, let’s give it the symbol E, and you have a resistor, we just say there’s a 1.5 volt difference. So you can say this is 0 this is 1.5. Current goes around and comes back. And we say the work is done, because the battery is somehow making the current go around. But we look under the hood and found it’s somewhat more complex. It is not done entirely with the electrostatic force. The electrostatic force is due to these charges here, sitting on the plate, but they cannot do work for a charge that goes round and round. Because if you go round in a full loop in an electrostatic field, the net work done by the field is 0. That’s because the net work done is the integral of the electrostatic field and because it’s conservative, you’ll get 0. Yet when charges go round and round the circuit, they are doing some work, because the resistor is heating up, delivering heat. And the hidden secret of course is not the electrostatic force, but another force inside the battery that’s doing the job. So I gave an analogy which I think is worth repeating. So here is a ski slope and you start here and you come down. Let’s imagine on the way down you bump into a lot of trees, that you pretty much come down at 0 speed. Then you stagger back to the foot of this ski lift. Now that’s the end of your skiing. There’s nothing else, because gravity is not going to help you anymore, because gravity is not going to help you go from here to there. So it’s not obviously run by just gravity. That’s when you have the ski lift, which I’m going to portray as a vertical thing, where you’ve got a lot of chairs, so you can sit on one of them and you can ride to the top. So that ski lift exerts a force. Gravity, which is acting down here, is also acting down here, and the ski lift, F_{ski} and F_{gravity}, ski lift cancels gravity here and takes you to the top. So if you define the analog of an emf, there is no “electro” about this one. Let’s just call it emf. As the integral of the forces acting on this one skier, as he or she goes around the loop, you can write that as the force due to gravity ⋅ dr and the force due to the lift — or what did I call it? Hey guys, you know what this was called? Student: Ski. Professor Ramamurti Shankar: Ski. Okay, ski dot dr. this is no good. Gravity does something to you when you come down and takes it all back when you go up. This guy, the force inside the ski, has a nonzero line integral, because it’s a nonzero force only in here. The line integral’s basically the ski force times the distance. The ski force is just mg, the magnitude. That times the distance is mgh. That’s why you can say this point is a potential mgh that lifts up to that. But inside this region, gravity is pointing down and the machine takes you against gravity and puts you on the top. And that’s the nonconservative force that does the work. So the main point was, electrostatics is conservative, and yet when a charge goes around and around the loop, the line integral of the total force is not 0, so it’s not entirely driven by electrostatics, but electrostatic forces are important. Just like gravity’s important here, because gravity’s what helps you come down here. The ski lift is all hidden inside, and then it takes you back up. So I’m going to draw the circuit in a way that really indicates the gravitational analogy. So I’m going to draw it not this way but this way. Here’s the top plate and here’s the bottom plate and you may directly visualize the height here to be the potential. Then there is a wire that has no resistance. It’s going to be flat. Potential doesn’t have to be up or down for current to flow down this. Then here is a resistor. Then there’s another little wire, which is pretty much flat and joins at the bottom. There are charges piled up here, there are charged piled up here, and they will produce an electrostatic field that will do something there, but inside, they’re bad. Electrostatic field goes down here. So if this is the whole story, you cannot do anything, because maybe if you charged it once, it will go round and come back here and it’s stuck, because it cannot go up. That’s when the tiny little hands of the chemical forces will come and drag the charge, against this field. That’s the F_{chemical force}. And it’s the line integral of the chemical force around a closed loop that’s called emf. If you write the chemical force per unit charge, integrate over a closed loop, that’s what we call the emf. But that’s very easily computed. That’s what I didn’t realize, a much easier way to do the computation. This thing is just a product of that force times the distance, but in magnitude, that force precisely cancels the downward electric force, just like the lift cancels the force of gravity. So the line integral in magnitude is just the electric field times the distance between the two plates, and that is just the voltage. Therefore the voltage difference between this point and this point will be the emf. Now if you don’t want to know the whole details, you can go back to drawing pictures this way, and if you have a circuit like this, and maybe there’s a switch here, you can be assured that between this point and this point, if you go around and measure the voltage difference, that will be the emf of the battery. That’s all you really need to know. The way it comes out is that this battery is like a pump that lifts you up by an electrical height equal to 1.5 volts. The emf is literally the voltage difference. Yes? Student:What’s the symbol under F_{c} in the first? Professor Ramamurti Shankar: Charge. Because electric field is the force per unit charge. We should take whatever force there is and write it as a force per unit charge. All right, so I wanted to do some simple problems. This stuff you’ve done already in high school, but I want to go over it in view of everything we have learned. So here’s the simplest circuit in the world. You have an emf. There is a resistor and it comes right back. You want to know what current is flowing in the circuit. So the logic, there are two equations you write now. One equation is, if you start at any point on the circuit and you go around a path and you come back to the same point, the change in voltage should be zero, because voltage is like height. You can go anywhere you want and you can come back, then the change in height should add up to zero. If you want, sum of all the change in voltage should be zero around a closed loop. Second thing is, current is conserved. These are the only equations you need. Current doesn’t disappear. It’s uniform here. So current I comes here, drops through the resistor and comes out of the other side. So I’m going to write the following equation. I’m going to start at this point and keep track of all the changes in voltage. First when I go around this battery, I told you many times, the change in voltage will be just E. When I come here, current flows downhill. I know the voltage drop here is RI, but if I follow it, I’m dropping in potential. It’s a skier; it’s going downhill. It’s a ski lift where it’s going up. And then you come back here to where you started, the total change must be 0. Now this is pretty trivial. You probably wrote this without any further thought, but we then just say we get from that E = IR. Then we’re going to do a more complicated problem, one resistor, second resistor, R_{1}, R_{2}. Again, if you apply an emf here, that current is going to go through this guy and go through that guy, and it’s the same current because it has nowhere to go. So the equation now will be E, which is increased when you go up here. Then you drop R_{1} times I + R_{2} times I and that equals 0, so current is E/(R_{1} + R_{2}). That means if you put two resistors in series, it’s equal to a single resistor equal to their sum. In other words, if these two were enclosed in a black box, you didn’t know what’s inside, you’ve got two leads coming out and someone says, “Hey, what’s inside the black box? What’s the resistor inside?” you will apply voltage, you will measure the current and that ratio you will call R. E/I = R, but here, you can see E/I = R_{1} + R_{2}. So that’s the value you will ascribe to the two guys in series. Now if you put them in parallel — I think again I’m repeating it, but maybe it’s harmless right now — so here is one resistor, here’s the second resistor. This is R_{1}, this is R_{2}. This is some E. If you put this inside a black box and someone says, “Hey, tell me what’s inside,” you don’t know. It may not be this, but there’ll be some effective resistance. And we compute that by saying let a current I leave this battery here, do whatever it wants and come out of the other side. Inside the box, this current will split into an I_{1}. I can call it I_{2}, but I’m just going to call it I  I_{1}, because that’s what I_{2} should be. And it comes out of this side. So let me say the total current = I_{1} — I changed my mind, guys. Sorry. Let me call this I_{2}. The total current is I_{1} + I_{2}. But I_{1} is V/R_{1} because the voltage V is acting across these two ends. The voltage V is also acting across those two ends, so it will be V/R_{2}, and that’s going to be equal to V divided by whatever the resistance inside the box is, by definition. If you compare the two, you find 1/R is 1/R_{1} + 1/R_{2}. I mention this just to tell you that it comes from current conservation, where I used I, is I_{1} + I_{2}. We might as well deal with the two others, one other circuit element, which is the capacitor. So suppose someone takes two capacitors. This guy is C_{1}, this guy is C_{2}, puts them in a black box and says, “Tell me what’s inside.” So you will take your battery, apply voltage E, then see how much charge leaves the battery. Then the capacitance will be the charge that leaves the battery divided by the emf or the voltage. If a charge comes from here, a charge Q_{1} will go from there and Q_{1} will go there. Q_{2} will go there and Q_{2} will go there. So the Q that’s going in will be Q_{1} + Q_{2}… over E. But Q_{1} = C_{1 }times E. Q_{2} = C_{2} times E. So divided by E, you’ll get C_{1 }+ C_{2}. This will tell you that the two capacitors in parallel will act like a single capacitor whose value is the sum of the two. That’s actually very easy to understand because here’s one capacitor. Here’s another guy. Let me just glue those plates together. It doesn’t matter because they are the same voltage. And you don’t need these two wires. You can just dump that and dump that. Current can just come in this way. You’ve got a single capacitor with the area equal to the sum of the two areas. Capacitance was ε_{0}A/d. They have the same d, but the areas add up so capacitances add up. And the final thing which I don’t want to do, because I don’t want to move onto more interesting things, is when you put two capacitors in series, what happens is, you can take it as a homework problem, that if you put a charge Q on this guy, and a Q comes out of the other terminal, these two guys have no choice but to take Q and +Q, because these two plates are isolated from the world. The charge they can have has to be equal and opposite, but Q has to balance this, and +Q has to balance that. Now if you use the fact that the voltage drop on this one + the voltage drop on that one is the total emf, you will find 1/C = 1/C_{1}+ 1/C_{2}. So capacitors and resistors combine in exactly the opposite way. This is just in series, they’re additive; capacitors in parallel are additive. Whereas for resistance in parallel, you add their inverse to get the inverse of the total, and capacitors in series, you add the inverses to get the inverse of the total. So now I want to do one slightly more interesting problem. It looks like this. You have a voltage E here. Then you have a capacitor C and a resistor R, and there’s a little switch which we will close at some point. So we want to know what will happen here. Initially you are told the capacitor’s empty, there’s no charge on it. Once you close the switch, you should sort of imagine what will happen. This plate is full of positive charges. They don’t want to be there. And this is full of negative charges. They don’t want to be there. So positive charges would like to go there, negative would like to come around, but the switch did not allow that. But now that it’s closed, some charges will leave this and some negative charges will be formed on the other side and they’ll come back here. It’s very interesting to note that current doesn’t really flow through the capacitor. Instead what happens is, positive charges come here, and positive charges leave that plate, leaving behind negative charge. And that current will come here, that’s the current I, and goes back there. But once the capacitor begins to charge, it bites the hand that feeds it, because it’s then trying to drive its own current. If you ask this guy, “What do you want to do?” he’ll say, “I want to drive the current that way.” The capacitor will start opposing the external voltage, and as it charges up more and more and more, eventually, its voltage will equal the applied voltage and then the current will stop. And when that’s happened, the charge on this capacitor at the end, I’m going to call it Q_{infinity}, you’ll see why, will be C times E. Then the current would stop. So let’s understand in detail what happens to current from the time you close the switch. So you write down the basic equation which says start here, go around and keep track of the change in voltage and equate it to 0. So I get a +E when I go through this battery. When I come from here to here, I drop by an amount Q/C. Then I drop by an amount RI. Then I’m back to where I started, which was my 0. That’s the equation. So now you can write this equation as — now what’s the relation between I and Q here? You should think about it. When we drain the capacitor, remember, I did a problem earlier on where I only had a capacitor and a resistor. There the current was −dQ/dt because as the current flows, the capacitor’s getting drained. Here actually the current is +dQ/dt, so if you’re not careful, you’ll get in trouble. It’s +dQ/dt, because any charge that piles up on the capacitor leaves the other plate and goes through this. So the current flow is the charging of the capacitor so I is in fact +dQ/dt. Otherwise if I is positive and the current is going this way, charge is building up on the capacitor. So if you remember that, you get the equation E = Q/C + R dQ/dt. So that’s the equation you want to solve. So how do we think of this equation? We tell ourselves if we wait a long, long time, until the capacitor is completely loaded, so that it doesn’t allow any more current to flow, its voltage will equal this one and at that point, dQ/dt will vanish. And let me call that final charge Q_{infinity}, because you will find out it happens only after infinite time. That Q_{infinity}/C = emf. So the capacitor is empty to begin with. After infinite time, it’s fully charged, meaning this battery cannot charge it to any higher voltage, because its voltage is precisely balancing that of this. So let me solve — our goal is still to find Q as the function of t, starting with this equation. So we are going to write Q as a function of t, as Q_{infinity}, which is the asymptotic value + some Q twiddle. You can always write your answer as 96 + something else, and something else will adjust itself to make this true. But let’s find the equation for Q twiddle. So take this Q, put that into the equation and see what you get. Left hand side is E. Right hand side is Q/C, + Q_{infinity}/C + Q twiddle/C. And when you take dQ/dt, this is a constant. So this is just dQ twiddle/dt. But Q_{infinity}/C is exactly equal to E, so these terms cancel, and the equation I get is this combination = 0. Right? This says dQ/dt, dQ twiddle/dt = RC — I’m sorry, Q/RC. Q twiddle/RC. And we know how to integrate this. We’ve done it many times. With Q twiddle = Q twiddle at time 0 times e to the −t/RC. Now Q twiddle at time 0 has to be chosen to satisfy the following condition: what is Q at time 0? You guys know what charge is on the capacitor at time 0? Do you know? Student: Sorry, my mind is elsewhere. Professor Ramamurti Shankar: Okay. Welcome back here. So what is the charge on the capacitor when I started everything? Student: Charge on the capacitor. Professor Ramamurti Shankar: At the beginning of the experiment. Student: The charge on the capacitor should equal the 0. Professor Ramamurti Shankar: No, where I just started. Anybody know? Yes? Student: 0? Professor Ramamurti Shankar: 0, because I told you the capacitor started out with nothing on it. That’s not a mathematical result. It’s a historical fact. You can’t begin an experiment with charge on the capacitor. I told you today that the charge on the capacitor is 0. So I’ve got to get 0 at t = 0 and this guy is E/C + Q twiddle, that’s Q twiddle of 0 over C. I’m sorry. Q_{infinity} is EC. Is that right? Yes, Q_{infinity} = EC + Q twiddle of 0 and that’s got to be 0. Therefore Q twiddle of 0 = EC. That means Q(t) = Q_{infinity} + Q twiddle. That becomes EC times 1  e^{t/RC}. So if you draw the graph of this, it will satisfy all the expectations we have. If you draw Q as a function of time, at t = 0, e to the 0 is 1. 1 cancels the 1, you start with 0 charge. At Q = infinity, this guy is gone. It is EC, which is just that value. So the charge builds up to that value, but it never quite reaches the value EC. In other words, the voltage in the capacitor is never quite equal to that of the battery. There’s always some left over. Yes? Student: Where does the negative come from? Professor Ramamurti Shankar: Here? Student: Yes. Professor Ramamurti Shankar: Okay. I said Q at time 0 is this guy at times Q_{infinity}, which is EC + Q twiddle at time 0. But that had to be equal to 0. Student: Oh, okay, yeah. Professor Ramamurti Shankar: So that’s how I got that. This is a trick to solve the equation. See, we all sort of know how to solve this equation. You may not know how to solve an equation with an extra term here, so this is a trick for solving that. So this is the part that’s interesting, in the sense that once you write the basic rules of physics, you get some equations. You’ve got to solve the equations and it’s no longer up to you to see what happens. The mathematics rules after that point. And whatever it tells you, you rush out to the lab and see if it’s true. And with the capacitor, whose capacitance you have measured carefully, and a resistor which you have measured carefully, you put them together, here’s a nice prediction on what will happen as it charges up. For example, you may want your capacitor to hold 80 percent of its maximum charge and you may like to know, “How long should I wait?” If you want it to hold 100 percent of the maximum charge, it’s never going to get done. So pick some number. They’ll tell you, if you want 75 percent, that’s how many seconds you wait. So this is how you’re supposed to have interplay between mathematics and physics, because what happened was, you got into a situation where you have to solve an equation. So this is a differential equation, but differential equations are just questions which are opposite of derivatives. You’re trying to guess a function about whose derivative you know something, namely this one. And it’s all guess work. You keep on guessing and keep on guessing and you make a table of integrals where people tell you what they guessed, and that big fat table of integrals, you can turn to. But in easy cases, you can solve it yourself. Okay, now there’s a homework problem I’m going to give you guys, but I’ll tell you what’s in store for you. We want to ask ourselves, what happens to the energetics in this problem. When I started, how much energy did I have, and where was it? Anybody know? Was there any charge in the capacitor? Nothing. There’s nothing to begin with. Yes. Student: It’s all in your battery. Professor Ramamurti Shankar: Yes, the battery had some internal energy. That’s correct. But now, during this experiment, once I close the switch, I think of the battery as outside my universe so it gives me some energy. How much energy did the battery give me during the whole process? Yes? Student: Equal to the potential energy stored in the capacitor + the energy dissipated by the resistor. Professor Ramamurti Shankar: Right. So let me repeat what he said. He said it’s got to be = to the energy stored in the capacitor + the energy dissipated in the resistor. And that’s got to be equal to the work done by the battery. It’s like saying, how much work did the ski lift do? Well, the work done by the ski lift is the work it takes to carry each person from the bottom to the top, multiplied by the number of people. That’s the work done. In electrical language, that just means the charge transported from the bottom to the top times the voltage. That means the total work done by the battery = the emf times I(t) dt from 0 to infinity. I(t)dt is a charge that flows in a small time dt, and I is not a constant. In fact, we can find I(t) by the formula of Q(t). Remember, Q(t) = EC times 1  E to the −t/RC. Then I, which is dQ/dt, you can calculate is EC divided by RC times e^{−t/RC}. It becomes E/R times e^{−t/RC}. So the current falls exponentially in this form. At t = 0 the current is as if the capacitor were not even there, because the capacitor’s inert at t = 0. It has got no charge. It’s not opposing you. And eventually, the current goes to 0. So take that formula for the current, put it here and do some integral. You get some answer. That’s the energy given by the battery to us. And what have you done with it? You charge the capacitor to ½Q^{2}/2C, where Q in the end should be Q_{infinity}. The resistor burns power at the rate VI, which is I^{2}R. So you take this I^{2}, take this I, square it, do the integral from 0 to infinity, multiply by R. That is the work done in the resistor, or work dissipated in the resistor. You should make sure that this energy delivered by the battery = what is stored by the capacitor + what is dissipated in the resistor. That’s your homework. That just means doing these integrals and making sure everything works out. Yes? Student: Shouldn’t it be negative, negative t/RC? Professor Ramamurti Shankar: Here? Student: Yes. Professor Ramamurti Shankar: Okay, finally I have a chance. This negative sign is going to cancel that. Student: Oh, okay. Professor Ramamurti Shankar: Look, don’t give up, okay, because just like you thought, I make mistakes too. So I never mind it when you guys do that. In this case, I’ll tell you why I sort of knew what the answer was. I know the answer’s positive because the current’s going to be positive. So if I’d kept the extra  sign, it will mean the current’s going against the battery and I would know that doesn’t work. Quite often, we’re allowed to use the sign of the answer, we’re allowed to guess the sign. For example, what’s the height of a person? You said the answer was 6 feet, but I don’t know if it’s + or . Well, you should know it’s +. You’re talking about the height of a person. What’s the depth to which he sank? Well, there it can be negative. So you should know by context whether the answer is positive or negative. That’s a very useful check as far as signs go. Now as far as overall formulas go, you can always take extreme limits of every answer and see if it’s correct. In the beginning, just when you close the switch, there’s the battery, there’s the resistor, the capacitor doesn’t oppose you, so you get e/R. At the end of the day, the capacitor’s fully charged, it’s neutralizing the battery by driving the current in the opposite way with equal voltage, so the current should vanish. These are some tests. All right, so let’s continue now to the last thing in circuits, which I’m not going to do any more, but I’ll tell you the kind of stuff people throw at you. So here is some guy, some resistor, branches into two other branches, some other voltage, join it here. This is E, this is R_{1}, R_{2}, R_{3}. This is E_{2}, let’s say. You’ve done stuff like this before, but let me remind you the trick. The trick is very ancient. We all know what the trick is. First is, you’ve got to know how many unknowns there are. If you can say the current here is I_{1}. I don’t know what I_{1} is; that’s an unknown. The current flowing through this guy is I_{2}. The current flowing through this guy is I_{3}, but I know that I_{3} = I_{1}  I_{2}. Do you agree? I_{1} comes here. If I_{2} goes there, I_{1}  I_{2} goes there. So I’m actually solving one of the equations, which is that I_{1} = I_{2} + I_{3}. Let’s put that into the equation. Then you have to write, so how many unknowns do I have? I have two unknowns, I_{1} and I_{2}. So I have to get two equations for two unknowns, and one equation will be — you can do many things. You can start here, go around like this and say the change in voltage is 0. That would tell me E_{1}  R_{1} I_{1}  R_{2} I_{2} = 0. Are you guys with me now? You drop here, you drop here, then you come back. Then you can take another loop that goes like this. Then you will say E_{1}  R_{1} I_{1}  R_{3} times I_{1}  I_{2}. Now let’s look here. You keep going the same way, but here you drop by an amount E_{2}. Then you’ve come over here and you come to the other end and you get 0. Now somebody can say, “Hey, why don’t I start here, go around that loop and say that voltage difference is 0?” I’m getting the third equation, but there are only two unknowns. So you mathminded people should sort of know what will happen if I write a third equation. What do you think will happen? Student: Nothing. Professor Ramamurti Shankar: What do you mean, nothing will happen? Student: It’s not going to help you get any more. Professor Ramamurti Shankar: In what manner will it prove to be useless? Student: It will be a combination of the other two. Professor Ramamurti Shankar: Right. You can deduce the third equation by fiddling with the other two. Maybe 9 times one equation  6 times the second equation will be the third equation. Therefore it’s not an independent equation. It will always turn out that if you’ve got two unknowns, you need two independent equations. If you can get three and four and five, you will find out they’re not telling you anything new, so that’s when you stop. If you don’t know that, you may start writing all kinds of equations. You may think, here’s another loop. Here’s another guy who wants to do this. You can do all that, but you’ll keep getting the same stuff. Chapter 2: Introduction to Magnetism [00:34:33]Okay, now we will really start new topic: magnetism. Well, you start this because every time you think you’re done with physics, somebody does some experiment and it doesn’t fit what you know, so you’ve got to make up new stuff. So magnetism you know was discovered in Ancient Greece, when parents noticed kids are sticking stuff on the refrigerator using some little black things. So more experiments. I obviously don’t know all the details. But it was discovered in many ways. One was in little compass needles which told you which way was north. So I’m going to give you a string of things that happened that tell you there is something going on that is not covered by anything I’ve written down so far in this course, new phenomena that don’t make sense. Here’s the simplest one. There’s a wire carrying some current. There’s a little charge sitting here. Nothing happens. Because the wire is electrically neutral, the charge Q doesn’t do anything. Now the charge begins to move at velocity v. Then you suddenly find the charge attracted to the wire. It starts bending in. That is not the electrical force, because the electrical force is 0. The wire is electrically neutral. It doesn’t care if the charge is moving or not, so it is a new force. And if the charge goes this way, the force is repulsive. That’s just one phenomenon you notice. Another thing you notice is there are little things called bar magnets. They seem to have a north and a south. And if you bring it next to another bar magnet, which has got a north and south, you find they repel and south and north attract. Then you take this compass needle that people use to find the north direction for the earth, and you put it somewhere here. You find that it swings, if it’s free to pivot, and points in a certain direction. And if you use that as a direction of a certain field, you can draw these pictures and you sort of know what it’s going to look like. They look like this. So this means if you put a compass needle here, the north will point like that. That’s what I mean. That’s another phenomenon. Then once you found out about electric currents, you also found that if you took a coil of wire, then a compass needle somewhere here began to respond to a magnetic field, which if you plot, seemed to look like this. It seemed to look just like a magnet with a north pole here and a south pole here. There are no magnets at all. You just had a coil of wire carrying the current in a certain direction. It goes like this. Then the magnetic field lines, which you plot by moving the compass needle, they do this. So all I’m trying to say is there are several phenomena going on, and the clue that you get is, why didn’t I need this before, and why do I need it now? What is new in this problem compared to the problems I’ve solved? Can you see what’s making this problem outside the realm of what we studied? What’s the one feature you notice from electrostatics? Yes? Student: The charges are moving. Professor Ramamurti Shankar: Everything is moving. The wire carries charges which are moving and this little guy who got attracted or repelled, he’s also moving. So magnetism is caused by moving charges and it’s felt by moving charges. They’ve got to move to play this game. So magnetism is caused by moving charges and it’s felt by moving charges, whereas in electrostatics, we didn’t have that motion. Yes? Student: What about reference frames? Professor Ramamurti Shankar: Yes, her question was, how about reference frames, namely moving according to whom, right? So I will come to that near the end when we do relativity, when I remind you of some ideas of relativity and see what that has to say about electricity and magnetism. But you are free, even in the relativistic theory, to take the view that you are not moving. As long you’re an inertial frame, that means a frame in which Newton’s laws are valid, you can apply all the laws of physics as if you were not moving. I’m just saying for that observer, who’s inertial, it’s found that when charges are moving according to him or her, they produce currents which produce a field, and the charges in that field will also respond. Now you can say, “What’s so special about you? I will go to a new frame of reference,right?” I’ll come back to that later, but I cannot resist telling you at least the answer to some of the questions. For example, if this charge is moving at a speed v and I’m completely stymied by this velocity, because I don’t know how to deal with it. There’s one way to deal with it, which is to get on a train that goes at the same velocity as this charge. Then this charge is at rest and if it still bends towards the wire, which it will — you agree that if you go on a moving train, a charge attracted to the wire will continue to be attracted to the wire. You’ve got to say, “How come? How do you explain that?” Neutral wire attracting a charge. Either you can say it’s all happening because you’re in a moving train, but relativity tells you, people in a moving train are entitled to the same laws of physics as people in a nonmoving train. Now it’s true for Amtrak. As long as you can get the train to move, you can make the same statement. So how does a person in a moving train explain it? What do you think happens? Yes? Student: The wire looks like it’s moving back in that direction. Professor Ramamurti Shankar: The wire may look as though it’s moving backwards, but still, a neutral thing moving forwards or backwards shouldn’t matter, right? A neutral rod, let it move. Why does it matter? So what do you think happens? First of all, do you think it will be attracted to the wire, even in the moving train? Yes, no? It will be attracted, because if I see it moving towards the wire, you can go on a train, you can go on a plane, you will also see it’s moving towards it. Maybe the rate will be different and so on, but the fact that it’s getting closer to the wire is undeniable. And there’s going to be no new physics. You’re just going to rely on good old electromagnetism, because electromagnetism is supposed to work for everybody. In fact, that’s how the relativity — even though electromagnetism was discovered before relativity, it obeys all the principles demanded by relativistic theory. So everybody should have the same laws of motion. So have you found a way out now? Yes? Student: Would the wire look like a bar magnet? Professor Ramamurti Shankar: No, because bar magnet means you’re going to magnetism, right? So let’s say I don’t know any magnetism, because the charge is at rest. A charge at rest doesn’t care if it’s near a bar magnet. Yes? Student: ________ lines. Professor Ramamurti Shankar: So what should be happening to the wire, you guys? Tell me? Yes? Student: If you’re moving and then the charges in the wire are moving, say you’re moving at the same speed as the charges on the wire, it would seem stationary. So it’s just like the point charge being attracted to a bunch of — Professor Ramamurti Shankar: Right. Why is the point charge attracted to a neutral wire? That is my question. Yes? Student: The wire isn’t neutral, because you’re moving along with the negative charges, so they stay there, but then the positive charges keep — Professor Ramamurti Shankar: That’s correct. So let me explain what he said. In fact, the answer is, the wire will not be neutral. So how did that happen? You might say, “Oh, there’s a Lorentz contraction,” meaning the lengths get contracted, but if the wire is neutral, the positives and negatives all get compressed, it still should look neutral, but that’s not how it is. Remember, in a real conductor you’ve got positive charges, which are at rest and negative charges which are moving in the opposite direction, right? So I do that by taking a rod, which is positively charged, and I take another rod, which is negatively charged, and I simply move this rod. That will produce a current. Now if the wire is neutral, I think you guys can understand that the density on this, in its own rest frame, will be different from the density of charges here, because lengths get contracted. It is a contracted wire that should have the same density as the positive wire, therefore in reality, the contracted wire, length contracted with increased density, should match the density of this. But if I go to the moving frame, what will happen is this guy will freeze, and that guy will move the opposite way. This guy will stop moving; the other will move the opposite way. But then you can show — I will do that calculation for you later — that the densities that previously canceled will no longer cancel, and the wire will have a net positive charge. The point is, there’s a lack of symmetry between the positive and negative charges, because one of them is moving. In fact, if you took the simple example where the charges in the wire are moving at the same speed as this guy — let’s imagine positive charges can move — and if they’re moving at the same speed as this guy, if you stop this, you will also stop those, and they will go the opposite way. But if they go the opposite way, they will contract, and when they contract, they will no longer be balanced by this, and the positive charge will be attracted to the negative wire, and not only attracted. It will be the same force that you will get when you learn about magnetism. So I’ll do this in detail, but I’m just telling you that you don’t have to worry about which frame of reference you’re in. The laws of physics are guaranteed to work for all frames of reference in uniform relative motion. So when I say velocity is v, I mean according to any one observer. That could be you. Chapter 3: Fundamental Equations of Magnetostatics [00:45:31]So you’ve got all this phenomena. Now I’m going to give you the fundamental equations of magnetostatics, that will explain to you everything I’ve described so far. So it will have two parts. The first part will be, what is the force felt by a charge that’s in motion in a magnetic field? The next thing is, how do you produce a magnetic field? Well, the answer is electric currents. Then you can ask, how does a current here produce a magnetic field there? What’s the analog of Coulomb’s law? Coulomb’s law is charges here producing the electric field there. Magnetism is a current here producing a magnetic field there, so you will have the second part. So I’m saying that this formula, q times E is the electric force. Then there’s the fact that the E here due to this guy is 1/r^{2} times the q here, etc. You need both parts. I’m going to give the two parts. The second part will come later, but first is, if you have a charged particle in a magnetic field, what’s the force on it? So here’s the answer. The total force of a charge particle q is this guy that we already know, + the cross product of the velocity with the magnetic field. The direction of the field is determined by the compass needle. That’s how the B is directed. But if you want, this force is called the Lorentz force. It is not invented by Lorentz, but he did so many things in this field, it’s named after him. You can take this to be a postulate, the summary of years of experiment, and you can say, “I want to begin here,” you can begin here. You’re never going to derive this, but this is an experimental fact. So if I go to a part of the room and say, “Find the electric field here,” I think you all know what to do. We’ve done it many times. Take a coulomb and put it there. Find the force on it, and that’s actually the electric field. If you put 5 coulombs, you divide the force by 5 and that’s the electric field. If I tell you, what’s a magnetic field, you can take a coulomb there, but you still have to find out — by the way, the electric field problem’s very easy, because the direction of the electric field is simply the motion of the charge. In the magnetic problem, there are lots of directions involved. There’s the velocity of the charge. There’s a magnetic field, there’s a magnetic force, and they form a cross product. v x B is the force. So if someone says, “Which way is B pointing?” you’ll have to shoot a few particles and find out how they bend. If you shot it exactly parallel to B it won’t bend at all. That’s the direction of B. If you sent it perpendicular to B, it will bend the most, because the cross product will be the biggest. Then you can slowly determine by some experiments what the value of B is. So let’s do a couple of simple problems where we just use this part, v x B. But there’s a very important aspect of v x B. Wherever a force acts on a body, you know force dot velocity is the power or the rate at which work is done. If you take the electromagnetic force dot velocity, you get q times v ⋅ E + q times v ⋅ v x B and that is 0. So you know why v ⋅ v x B is 0? It’s again a purely mathematical result. Yes? Student: Because v cross Bis going to be perpendicular to — Professor Ramamurti Shankar: v x B is orthogonal to both v and to B, so it’s coming out — in this case, if B and v are in the blackboard, v x B is outside the blackboard. Its dot product with anything in the blackboard is 0. So whenever you have a vector dotting itself cross something else, the answer is 0. That means the magnetic field is always perpendicular to the motion of the particle. That means it doesn’t do any work, never. Not just weekends. The magnetic field doesn’t do any work. So you can say, “Who cares about such a thing?” Electric fields do a lot of work. They speed up particles, they slow down. The kinetic energy of a particle will never change due to the magnetic field. And yet you will see, even though it’s not able to do anything by itself, it’s extremely useful in getting a lot of things done, like generators and so on. They rely on the magnetic field. So we’ll see that, but at the moment, it’s fact. The force is always perpendicular to the velocity for a magnetic field. So now we are going to do one or two simple problems. First problem, let me take the easiest one. I want to find a way to select from a beam of particles, which are all going from left to right, those which have a certain velocity. I want a velocity filter, and I cannot seem them microscopically; they are little guys. But they’re moving, they’re all moving like this. Here is how we can make a velocity filter. You take two parallel plates, charge them up so the electric field looks like this. This particle q will then bend like this in the electric field. Now put a magnetic field into the paper. Let me make sure I got this right. Yes. Put a magnetic field into the board that’s shown by this symbol. Magnetic field coming out towards you is shown by dots, and coming away from you — coming towards you is the dot, and going away from you is the cross. That comes from the arrow. If you have an arrow, if you look at the arrow from behind, it looks like this, and if it looks like this, you run like hell. These are the two things. So I want to show you that B is going into the blackboard. Now what’s the force due to B? Take the cross product — that’s what I was trying to do in my head, make sure I got it right — v x B is turning a screwdriver from v to B. Let’s see, B is into the board, v x B will be like this. Therefore the total force on a charge will be q times E, which is downstairs — I mean, pointing down, and vx B which is pointing up. You can see then for a given value of E and B, for any randomly chosen velocity, these forces will not cancel. So some guys may bend like this, some guys may bend like this, or some guys will go straight through. They are the ones for whom E = v x B or the magnitude of the velocity is E divided by B. So only those particles with that velocity will make it. In other words, imagine this thing is very, very long. Some things will go hit it there, others may hit it here. If you are just right, you will make it to the other side, so it’s a velocity filter. Now you should be able to answer the following question without doing long calculations. The guys hitting the top, are they faster than the desired speed or slower? Student: Faster. Professor Ramamurti Shankar: Because what? You understand that? If they’re going up, the magnetic force is winning. The magnetic force is the only one who cares about your velocity, so the velocity is too big. If they’re falling down, electric force is winning. Electric force is a constant, magnetic force is velocity dependent. So the fast guys go here, slow and the right one comes here. Here’s another exercise you can do. There is a magnetic field coming out of the blackboard and it’s uniform in density. And I send a particle here with the velocity like this. What will it do? So B is coming towards me. B is here. vx Bis a force in this direction. It will bend; it will go there. So it’s constantly being applied a force perpendicular to its velocity, so it’s like planetary motion. It will go in a circle. Never speeding up. It’s not speeding up, because the force is always perpendicular to velocity. You don’t change the kinetic energy, but you change the actual direction of motion. So this is the way to trap a particle. If you want to trap particles, you put them in a magnetic field, you don’t have to touch them. They will not go anywhere. They’ll form an orbit. So let’s find out what we can say about this orbit. If that distance is R, the dynamical equation is mv^{2}/R, is the force you need to bend something into a circle of radius R, if it’s going at speed v. And we can take V to be constant, because v is not going to change. And that’s going to be equal to the magnetic force, which is v times B times q. I didn’t write the cross product, because everything is perpendicular to everything. The magnetic field and v are perpendicular, so the cross product and magnitude is just v times B. The direction of the force is of course towards the center. If you balance these two equations, you find v/R is qB/m. So what is v/R? Let’s look at what’s v over R. If the particle has a velocity v, the time period due in orbit will be 2ΠR/v. The distance divided by speed is the time period. So v/R = 2Π/T or 2Πf. That’s called the omega or the angular speed. It goes around and around the circle with an angular speed, ω. So v/R = ω. So this fact was known for some time. It’s not very hard to find. But do you know who made a living out of this? Notice something very interesting. The angular frequency does not depend on the speed or on the radius. It just depends on the charge to mass ratio of the particle in a given magnetic field. So that circle, that circle, they all go around at the same rate. As long as the particle has a given charge, a given mass, it’s in a given magnetic field, you can launch them in orbits of various velocities, they will all be in synch. Now do you know who used this? Does it look like anything, these guys? That person was actually associated with Yale. That tell you anything? Okay, his name was Lawrence. Do you know what Lawrence invented? No? Pardon me? He invented the cyclotron. I’ll tell you what — in fact, he had the ideas when he was a young faculty member here, but he didn’t build it here. He went off to Berkeley and he built the cyclotron there. Then he built bigger and bigger cyclotrons. And I want to tell you what the idea is behind a cyclotron. Here’s what you want to do. You want to accelerate particles so you can smash them against other things. So one thing you can do is you can take a little battery here, connect it to two plates. This is positive, this is negative. If you release a proton here, for example, it will fall down the potential. So it will lose a potential energy, q times V. That will turn into ½Mv^{2}. And just when it hits the bottom plate, if you have a hole there, it’ll go through the hole and you have a particle accelerator whose output velocity will be v, given by this equation. But now if you want to accelerate it to a higher and higher energies, you’ve got to get bigger and bigger voltage. So how are we going to do that? Here’s what Lawrence did. So take a magnetic field. In fact, there are two metallic halves called Ds, because of the way they look. You send a particle here and at the instant you release it, imagine that this plate is positive and this plate is negative, and the particle is positively charged. That means the whole plate has one potential, and the other plate has a negative potential. So this particle will speed up when it comes here. Now the whole thing is immersed in a magnetic field, which I’m not showing you. So this guy will bend and come out of this side. Now what will happen to it? What will it do now? If you don’t do anything, what will happen to the particle now? Student: It will slow down. Professor Ramamurti Shankar: It will slow down, so that’s not the accelerator. You’ve got one gain and immediately lost it. But suppose when it’s here, you very cleverly change the polarity of this? Just when it’s leaving this hole and going to the other D, you change the polarity, so it’s falling again. So it picks up a lot of speed and goes on a bigger spiral. Here is doesn’t experience any field, goes around and comes here. When it comes there, you quietly change it one more time, so it’s always falling downhill. So every time it’s ready to cross the gap, you flip the voltage. How do you do that? You’re not going to stand there with the battery and keep switching the wires, right? So what do you think you do? Student: Alternating current? Professor Ramamurti Shankar: Put an alternating voltage. Take the power supply from your house, 60 cycle a second, denoted by this symbol, AC voltage, will reverse its polarity automatically. And the beauty of this process is the following: once it’s got a certain frequency, it will produce a cyclotron orbit whose omega matches that frequency. Even though the particle is speeding up — here’s the whole point — even though the particle is speeding up and going in bigger circles, the ratio of the speed to the radius is such that it takes the same time to go around these semicircles, no matter how big they are, no matter how fast the particle’s moving. It’s the fact that v over R does not depend on R, R on v. So that’s the omega that you will take for this guy. Then it will keep on picking up speed and at the very end, it will do a few more such things, then it will come shooting out of here. So you don’t need — in the end, you can get 1 million volts, but you don’t have a 1 million volt battery. You have a 1 volt battery. It gives it a kick a million times. A million times you cross the midpoint. It’s a very clever way to make the particle accelerate. The Stanford linear accelerator is somewhat different. There you have a lot of cylinders and charged particles go from one to the other. But once again, the polarity is reversed, so every time it goes there, it’s constantly falling downhill. It’s just this thing taken out and made into a straight line, that’s about two miles long, I think, in Palo Alto. And that was the machine that discovered a lot of great things, including what’s called the J particle. So the linear accelerator’s one, but this is the cyclotron. So the next question you want to ask is — these are all examples of forces on a single charge — what’s the force on a wire? Chapter 4: Force on a Current Carrying Wire [01:02:44]Sometimes you do microscopic experiments with tiny particles. Other times you do macroscopic experiments with wires carrying current. So I want to look at that problem. So let’s look at the wire carrying current. So here’s a piece of wire, and I want to find just a little segment, which I write as vector dl. There’s some current going here and the whole thing is bathed in a magnetic field B, which is constant over the tiny segment. We want to know what’s the force on the little segment. There’s going to be a force because there are guys moving in the wire. That’s it. Each one feels a v x B force. You’ve got to add it up. So if this has got length dl, how many charges am I talking about? The number of charges is the density of carriers times the charge of each one. That’s the number of charges, per unit volume. If the cross section of the wire is A, the length is dl, that’s how many charges are involved here. Do you understand? This cross section is A. A times dl is the volume of the cylinder. That times the number per volume times the charge for each guy is the total charge. And each one of them experiences the force v x B. Now I’m going to do a little switch here. I’m going to write it as Anev dl x B. So you see if you follow that. The velocity of the carriers is along the wire. dl is also along the wire. Therefore either you can take the magnitude of this vector times the vector v, or the magnitude of vector v times the dl. It’s the same thing. They’re both the vector parallel to either one. But if you write it this way, if you go back to what we did earlier, this is just the current in the wire. So we have a nice formula that says the force on a segment of wire carrying current I is I times dl x B. So here’s an example. Suppose you’ve got a magnetic field like this. And I’ve got a wire. I don’t know where it’s coming from, I don’t know where it’s going. Let’s say it’s just carrying a current. And imagine this is in the real vertical plane, so that there are some weights hanging from it. It’s trying to fall down. You can balance that by applying a force of magnetism to that wire, because the magnetic force will be dl x B. Let’s see, so dl is this way, B is out of the board, so let me set the current like this. Then dl is this way, cross Bwill be a force upwards. There’ll be an upward force of B times the length of the wire times the current. And if that’s equal to the mass of the wire times g, then the wire won’t fall up or down; it will just stay there. So you can hold up a piece of wire by driving a current through it and putting in a magnetic field. But here’s another simple homework problem, or some problem you can find in a textbook. This is a semicircular wire. Now this wire cannot just start and end here because the current has to come from somewhere and go from somewhere, so maybe it’s doing this. But I’m just focusing on this section. And imagine there is a magnetic field going like this. I want to find the force on the semicircle. This is supposed to be a semicircle. What’s the force on it? So take a portion of the wire here. This is my dl. This is my B. Take the cross product of dlwith B. Can you see which way it will go? Do the old screwdriver, your favorite thing. Do this. It’s coming out of the blackboard. And what’s the force that’s coming out of the blackboard due to the little section = I times dl times B times the angle between these two. And that will turn out to be a sine theta. If you want, you can measure theta from here and you can see that dl is like that, B is like that, and the angle between the horizontal and the radial is the same as the angle between the perpendicular to the radial and the perpendicular to the horizontal. So this theta is the same as that theta. Then you integrate the force, you get I. dl will be just 2ΠR, times B. Sorry, dl will be ΠR time B, times integral of sinθ dθ, which is cosθ from 0 to Π, that will give you a 2. So that will be the force coming out of the blackboard on this section. I should say theta from 0 to Π. I think there’s something wrong with this formula. Hold on for a second, guys. It should not be done this way. dl = Rdθ. There’s I times B and Rdθ sinθ from 0 to Π, and that will give you 2IBR. Do you understand that the length of the wire, this segment dl, is R times dθ. Theta is measured in radians. Anyway that is a force and this section tells you how to calculate that force. All right, the last thing which I will start but which I will not finish today, and you can try to look at some good pictures before you come to class next time. There is just no way I can get this right. So here is a current loop. And the current goes like this. And it’s in a magnetic field like this. The question is, what happens to it? So you’ve got to go to the four sides of the loop, in this side the v x B, your current is going this way. B is going straight up. v x B will act like this. There the v x B will act like this. Here it will act like this and here it will act like this. And this is the normal to the plane, it’s the area vector. So let me draw you a side view, because that’s the only thing I can even try to do. In the side view, it looks like this. These two edges are getting pulled. This edge that you can see here is getting pulled out of the board. The other one’s getting pulled away from the board. They just try to distort the loop, but they cancel each other and they don’t do anything. But these two are equal as forces, but as a torque, they are not equal and opposite, but they’re additive. So they together produce a torque and let’s find out what the torque is. The force on this segment is B times l times I, where l is this. Let the other dimension be w, which stands for width. This is w here. Then the torque will be that times wsinθ. That is the torque. But l times w is the area of the loop. B times A times sinθ. If I represent the area as a vector, pointing on this direction, then the torque is simply given by μ x B, where μ is a vector whose magnitude is equal to the current in the loop times area. I forgot the I. And this is called the magnetic moment of the loop. You call it a magnetic moment, because it’s just like a dipole moment. You might remember, if I take a dipole, put it in a field, the torque on it is p x E and the energy of it is p⋅E. So magnetic loop looks like a dipole in a magnetic field. In other words, if there were magnetic charges in nature, which there aren’t, then this guy behaves exactly like a + magnetic charge and a  on the other side of the loop. So if magnetic charges existed, this would be simply a magnetic dipole, aligning itself with the magnetic field, but in reality, there are no magnetic poles. But there is just the loop that behaves like a dipole. And the dipole moment of the loop, which is the analog of distance times the charge is the current times the area. So we’ll take it from here next time. [end of transcript] Back to Top 
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