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PHYS 201: Fundamentals of Physics II
Lecture 11
 Lenz's and Faraday's Laws
Overview
The electric effect of a changing magnetic field is described using Faraday’s Law. The direction of the current so generated is given by Lenz’s Law. The operation and energy accounting of the generator are described. The concept of inductance is introduced. The Betatron is described as an example of Faraday’s Law. Self and mutual inductance are introduced. The energy density in a magnetic field is derived.
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htmlFundamentals of Physics IIPHYS 201  Lecture 11  Lenz's and Faraday's LawsChapter 1: Review of Lenz and Faraday’s Law [00:00:00]Professor Ramamurti Shankar: All right, I’m going to begin by summarizing stuff done near the end. Usually stuff done near the end is maybe hurried, or also you may have forgotten where we were, so it’s helpful. The first of what I did last time is called Lenz and Faraday law, and it says the following: in any circuit made with a real conductor, if you integrate the electric and magnetic force on a charge, unit charge around a closed loop — this is called the electromotive force — that =  the rate of change of flux through that loop. And the flux is defined as follows. The flux is just the integral of the magnetic field over an area whose boundary is this loop. So this loop is a boundary of that, so if you have a loop, and there is some flux coming out of it, you integrate that, and you take the rate of change. That covers a variety of different phenomena. It’s what I was trying to explain last time. First of all, if you ask, “Why does this thing change? Why does the flux change?” it can change for two reasons. It can change because B itself is changing with time. Or even if B is not changing with time, but changing with space, if a loop happens to be moving, that can also change the flux through it. For whatever reason, it will change. So this total derivative means the total rate of change of flux for the two reasons I mentioned, or both. You can have a time dependent, space dependent field in which a loop is dragged along. Then at any instant, electromotive force, namely the total force on the charge, on the unit charge, is given by rate of change of flux. Then I said we can write this integral, this rate of change is made up of two parts. One was the rate of change due to explicit time dependence of the field on the old surface + another one that looks like v x B⋅dl. This was what I did the last time. And we got that result — I’ll remind you briefly, but I certainly don’t want to do it again, is that if you had a loop that was doing this initially, it was doing that a little later, the loop is moving, then if you want to find the change in flux between that one and this one, the natural thing to do is to integrate the flux on that loop later, subtract from it the original loop earlier. But calculationally, it’s easier to use the fact that if you want to find the flux through a surfacing bounded by this loop, you don’t have to simply take the easiest loop you see. You can take any loop with that as a boundary. And we cleverly chose the surface to be the one containing the original surface and the two walls, and the circular walls, if you like, that add onto it and produce a new surface. So if you take one flat sheet and glue the sides of the cylinder, you get a rim here, and that’s the final surface. The advantage of doing it is that when you find the flux at a later time using this joint surface, one contribution will come from exactly the old surface, and the difference there is only because of explicit time dependence. The other will come from the sides, and the sides, you can see, are spanned by little rectangles. This side of the rectangle is v times dt. This side is dl. And the dot product with the flux will be that term. And you rearrange the product you get this. Therefore now we can balance that with that, or if you take a problem where nothing is changing with time, then you can just forget this term, and this term will match that term. Then we are left with the following result: the line integral of the electric field on any loop = the surface integral of the rate of change of B over that fixed loop. So here is the fixed surface, and here is the boundary of the fixed surface. You have to understand the difference. We started out with a statement about a real physical loop. The physical loop was moving and there is an emf driving charges around the loop. It’s got two parts. One had to do with the motion of the loop. The other had to do with the changing of the field. This relation, you see, has no need of a real loop to be present. This one definitely needs a real loop to be present, a real conducting loop, because whose velocity is this? It’s a velocity of each portion of the loop that’s moving. But this one is about a fixed contour in space and has to do with the line integral of E around the fixed contour with this one. So it is true even if you remove that loop. It’s a basic statement about electric and magnetic fields that tell you that if you have a changing magnetic field, a time dependent magnetic field, it will generate an electric field which, unlike the electrostatic field, will have a line integral which is not 0. That’s the main point. It’s a non conservative electric field, whose origin is not electric charges, but really changing magnetic fields. So this is called Faraday’s law. Faraday’s law is a very profound statement about electric and magnetic fields and that’s what’s going to be more important. But the original formula is able to cover all situations where loops move in fields. This  sign is due to Lenz and the  sign is going to save your life. The  sign tells you, if you want to know when you drag a loop or when you change the flux through it, what is it going to do, the  sign tells you it’s going to fight the change in flux. Namely, if an emf generated, which way will it point? Will it move charges this way or that way? Answer is, it will move charges in such a way that they produce their own current, which will produce its own field. That field will fight the change that you’re trying to produce. In other words, if you take a loop here and you’ve got a bar magnet sending some lines of flux through it, you move the magnet closer to the loop. That’s an example of rate of change of flux through the loop. Which way will the current flow? The answer is very simple. You’re trying to increase the flux through the loop. It will fight it by decreasing the flux going upwards, so it will try to produce a flux going downwards. Therefore it will have a current that looks like this. If you have a current that looks like this, let me see, you do that and the thumb points down, and therefore the magnetic field it produces will oppose the field. So this rule is very important. That’s what distinguishes human being from primates. The primates cannot do this rule. In fact, there are lots of cave drawings of apes trying to invent solenoids, and they always kept saying this, and they thought the magnetic field was parallel to the coil. So one day, we figured this out. That’s the beginning of — it’s even better than fire, finding out that right hand rule. Okay, so this is the rule you should use. So whenever I do any calculation of emf, I’m not going to worry about the sign. In the end, we’ll fix the sign so it makes sense. That’s going to be our formula. All right, so now let’s continue with this one, and I remind you how we’d explained everything regarding the other experiment. Remember we had a magnetic field going into the board, only up to some point. To the right of this point, it’s everywhere uniform. Then you had a loop which we were dragging to the right. And I put a light bulb here, and I said if you drag it, the light bulb will glow. Now we can understand because there’s an emf, and the emf is coming because in this example, the term that really matters for the emf is the v x B term. There is no dB/dt. It’s fixed. Somebody’s holding a magnet, shoving flux into the blackboard. You’re carrying the loop. So it’s the v x B that gives you the emf. That’s fine. That’s the part we’re not very impressed with in this law. And let’s look at the  sign. The  sign will be telling you which way the current will flow. You are increasing the flux going into the blackboard, therefore the current that is generated will try to put flux coming out of the blackboard. And that means the current will flow that way. And that agrees with what you expect from fundamental principles, because if the wire is moving to the right, B is into the board, v x B will produce an emf this way. So we understand everything here. I also explained to you that the energy balance is that the work done on the wire, on the light bulb, is paid for by the person pulling this loop, because the minute you have a current, that current doesn’t like to be dragged across a field. There is some BlI force and when you overcome the force with your mechanical force, you convert mechanical energy to electrical energy. Now I come and say I don’t want to see it from the point of view of the laboratory. I want to ride with this loop. In that frame, the loop is not moving, but the magnetic field, whoever carries the field, is advancing to the left. So a little later, you’ll have magnetic field lines like that. And the question is, how do we explain the current now? You’ve got a static loop, but we’ve got a time dependent magnetic field, so this equation — it’s the same equation, but now we forget the v x B term, focus on this one. This tells you there’s a real electric field generated in this problem. So changing magnetic field generates electric field, to produce essentially a similar emf going this way. So I drew for you a picture of the electric field and what it does. The electric field will look like this. It will be constant and on 0 to the left of this interface, and at 0 to the right of the interface. Let’s see if that has the right line integral of E that we want. If you take this region here, there is no electric field so line integral is 0. There’s no changing magnetic field, because if you sit here, the field is not changing. If you go way over to the left, the field is not changing. It’s 0 everywhere. The line integral of E around this loop will vanish, because the two opposite sides will cancel. The only time you’ll get a nonzero line integral for Faraday’s law, is if you take a loop that’s partly in that region and partly in this region. If the loop straddles the region where there’s an E field and there is a B field, then you get a nonzero line integral, because there is no Ehere. There is E here, so the line integral will be nonzero. The very same loop also has the rate of change of flux, because in that loop, flux is moving in, so the equation will match. So this is a real electric field, meaning if you sit here, you put a test charge, it will move this way. So real electric field generated by magnetic field, that’s a very important result that you get from Faraday’s equation. (11:49:38:18)] There’s one subtlety that is usually overlooked, which is the following: if you go to this wire, here is the piece of wire. There are some charges in here the wire has moving at a velocity v. And there is a B somewhere, maybe like that. And we said v x B is a certain force. But if there’s a current flowing in the wire, it also has a speed u along the wire. This is something we saw even in the loop problem. So the real force of a charge on the wire is not simply v x B, but v + u x B⋅dl. Because it’s not just the motion of the wire that we take for the velocity, because charges are moving in the wire. The net velocity is v + u, to some number like that. But when you do the emf, you don’t have to worry about this part of the velocity along the wire. I’ll give you a second to figure out why. But just take that part. Why am I allowed to ignore it? Because u x B is perpendicular to u, and dl is parallel to u. u is the direction of motion in the wire and dl is along the wire. So if you take a vector, cross it with something else and take the dot product with another vector parallel to itself, you will get 0. That’s why you don’t worry about the extra term in computing the emf. This is a subtle point for those of you who wake up with a sweat in the middle of the night. I’m just trying to tell you, calm down, it’s okay. It’s really present as a force, but not as an emf. In fact, it’s going to push the charge perpendicular to the wire and not do any work going around the loop. So this is one topic. I’m going to illustrate the reality of this result by describing to you the operation of what’s called a betatron. You remember the cyclotron, what it does, right? You had these Ds and the charge was jumping around from one D to the other when you put a certain voltage, so this is positive and that’s negative. It bends in a magnetic field which is going in the plane into the blackboard. Then it gets another kick and when it comes here, you reverse the polarity, so that this becomes + and this becomes . It picks up another kick in velocity and it keeps doing that. And the remarkable thing that was noticed is if you write the equation mv^{2}/R, which is the force you need to bend into a circle, that’s going to be qvB. So if you cancel the velocity here, you find v/R to be qB/m. v/R is the angular frequency with which it rotates and that’s independent of the radius. Therefore you have to reverse the polarity with the same regularity, with the same frequency, even as the particle picks up more and more speed. It’s not that you have to keep changing the rate at which you flip the voltage. Then you don’t have any means of doing it. But if it’s flipping at a definite rate, then of course you can get a generator that generates voltage at that rate and you can make it work. The key to all of that was this remarkable fact that omega does not depend on the radius. But that has a weakness that as the particle picks up speed, eventually you will have to use the fact that the real momentum of a particle is not mv, but mv divided by this factor, which comes from relativistic effects. Then you can show the centrifugal force you need. The centripetal force you need is not given by mv/R, but rather it’s given by this momentum times omega, where momentum is given by this formula and not by this. Now that’s a homework problem where you’ll get enough time to think about it. That will show you that once the momentum formula deviates appreciably from p = mv to p equal to this, then condition for the cyclotron orbit, the frequency being independent of radius will fail. Essentially, as the particle gets more velocity, it’s harder to push it. Your ability to accelerate it changes and therefore this condition fails. So the cyclotron can accelerate particles only up to velocities where the relativistic corrections are unimportant. But the betatron, which I’m going to describe to you, is a device that actually manages to accelerate particles even as their motion becomes relativistic, even as the momentum is given by that formula. So I will tell you how that works. You’ve got these poles of some magnet, the side view, producing a magnetic field. Let’s look at the top view. In the top view, you’ve got some magnetic field. Let’s say it’s coming out of the blackboard. So if you put a particle here, some velocity v, v x B is to the right and it will bend. But here’s what we want to do. We’re going to have it go in a circle of definite radius and yet pick up speed. I’ll show you how that happens. So what you do is, this is not a fixed bar magnet. It’s an electromagnet, the current in which you’re changing, and therefore the field is in fact time dependent. Furthermore, the field will be strong near the middle and weak near the edges. Now if that field is changing, then if you draw any loop of radius r, the flux through that loop is changing. Therefore the electric field will obey this condition, 2Πr times E. The electric field will then also be going around in circles. It will encircle the magnetic field. If the magnetic field is coming from ceiling to the floor, electric field will be horizontal, horizontal circles. And by symmetry, they’ll be circles centered at the center of the magnet. That’s the direction of the electric field. This field coming from top to bottom, if it changes in strength, will generate a field in a horizontal plane that’s going round in circles. So if you put a charge there, you’ll push it and it will speed up. How big is the field? Well, the field is constant on the circle of radius r. It’s pointing azimuthally. So its line integral is just 2Πr times E at r. That’s going to be = (forget the  sign) d/dt of B⋅dA inside that circle. Now that B⋅dA I’m going to write as Πr^{2} ties some average B. That’s how we define the average. The total flux through the radius when the field is varying with distance can always be written as the area times the average B and the time derivative of that is, if you like, d/dt of B_{bar}. So I’m going to write now the equation I get. E of r, if you solve for it, is going to be r/2 times d/dt of the average B. Average is inside the circle of radius r. Now if that’s the electric field you multiply both sides by q, which is the charge of the object, that’s the force. That force is the rate of change of momentum. So rate of change of momentum is qr/2 times rate of change of the average field. Now if this is the rate of change and that’s equal to the rate of change of B, then p of t = qr/2 times the average Bat time t. Actually, I’m making one assumption here. Do you know what that is? I’m assuming the initial momentum was 0, because if you know only the rate of change, you can add a constant to it. So I’m assuming that t = 0, p was 0. So that’s the momentum of this guy after time t. So as the average B is growing, the momentum is growing, it has this value at this instant at that radius. Now what else do we need to make sure that solution makes sense. What do you need? Is there anything else to this story? Yes? Would you like to guess? If a particle picks up speed, why would it continue to be in that orbit that I’ve shown here. What does it take to keep it in that orbit. Pardon me? Student: The force of that entity. Professor Ramamurti Shankar: Which force, I’m sorry? Student: If it’s in an orbit of constant radius, then the velocity’s increasing the force, and that’s the increase. Professor Ramamurti Shankar: Right, but at a given velocity. Even at a given velocity, you need a force to bend something into a circle. You guys know that. Remember that force from last term? That force is mv^{2} over r. That comes from saying that the particle has a velocity momentum like that now, momentum like that a little later. There’s a change in momentum pointing towards the center and you can calculate that as mv^{2} over r. So who’s going to provide that force? In other words, things don’t move in a circle unless you pull them into the center. My question is, who’s going to do that for this guy? Who’s going to provide that force? Any idea? Don’t worry about the sign or anything. Is there anything that can push this guy towards the center? Want to make a guess? Student: The magnetic field? Professor Ramamurti Shankar: It’s the magnetic force, because that’s a charged particle, moving in a magnetic field, and the force of that is q times v times B. But this B is the B at the orbital radius. It’s not the average B. This is the B where the particle really is. So we’re going to keep the particle at a fixed radius. It’s the B at that radius that matters. Therefore you cancel the velocity, you find that mv = qrB_{0} and that’s the momentum. But we also saw the momentum at time t = this. They are both expressions for the momentum at time t. This just says the magnetic field hopefully is strong enough to bend it. That’s the field that you need, but I’m just going to equate the momentum computed two ways. That tells me this interesting result, that B_{0} is ½B_{average}. So let’s summarize what I’m saying. I want you to visualize this. You’ve got a magnetic field from ceiling to floor. It’s pointing down, let’s say. You put a little charge there, it won’t do anything. But if you change that field, because of this law, Faraday’s law, the change in field will produce a circulating electric field, that circles the changing magnetic flux and going around in circles, and it will accelerate the particle along a circle, because it is bending around a circle. But at the same time, you need the right magnetic field to keep it in that circle. But the particle is speeding up. Therefore the force needed is also increasing. So you’re in addition to increasing the average field over the loop over the circle, to produce the emf, you need to have the right field at the boundary to provide exactly the right force needed to keep it on an orbit of that radius at that time. Therefore the story is, as time passes, the field increases in strength, the particle’s tangential momentum increases. And the radial force to keep it in orbit at the tangential momentum also increases, and the demand when you look at it says that the field at the periphery, where it’s rotating, must be ½ the average field. So you’ll have to build a magnet very carefully. You can always build a magnet whose field varies as you go off center that gets weak when you go off the center. It should go off the center in such a way that by the time you come to this radius, the field strength there is half the average. Once you design a magnet that way you just crank up the current and let a particle go at the radius, it will pick up more and more and more speed. Even though it’s going faster and faster, the field at the orbit will be just right to push it towards the center with the right amount. One of the homework problems was to show something I made a big deal about, namely, here I’ve used relativistic kinematics. Momentum was mv, the force to the center is mv^{2}/r. But the homework problem shows you that even in the relativistic case, the momentum is mv divided by all of that, this is still true. So the betatron, in spite of the simple example I’ve given here, actually works, even if the particle is relativistic. In fact, the only reason the betatron eventually fails is that when particles start going in a circle at very, very high speed and they are charged, they start radiating energy. That’s another aspect that we have not discussed in our course yet, but accelerating charges radiate energy, and eventually, you cannot put in enough energy. It radiates more than you can give it. So then you need other things called synchrotrons. So that’s a constant struggle for people building accelerators. It’s easy to get more and more velocity, but if you bend them into a circle, they start radiating, emit gamma rays, emit light. And that energy eventually is so big that you cannot compensate it with your pushing force. Chapter 2: The Power Generator [00:25:39]So the next topic I want to discuss is in a totally different vein, and that has to do with more practical issues like this one. This is going to be a power generator. Remember, I told you one way to make electricity in your house is to take this coil and tell somebody to carry it and run. Then the light bulb will glow here. Or you can have somebody carry the magnet the other way and the light bulb will glow in your house. But there’s a more clever way to make the light bulb glow, which I think you guys have seen. By the way, I think these are all things you have seen in one form or another, so I’m not going to put too much energy into these things. I just want to tell you things you may have missed. So here’s some magnet. Here’s some field lines going from here to there, north to south. In that magnet you put a coil, which looks like this. The plane of the coil that’s pointing in that direction of the magnetic moment, or if you want, the area vector points that way, the B field is horizontal. Now if you spin that coil, I think you know what will happen. The flux through the coil is going to change and the rate of change of flux will give you an emf. So what is the emf? First you’ve got to find the flux through the coil. The flux through the coil is the area of the coil, the magnetic field, times cosine of the angle between them. This would be angle theta in the figure. But you have attached this to an axle and you’re going to just keep spinning it mechanically. And you agree to spin it at a uniform rate, omega. So theta is omega t. Then you can see that the emf, which is dΦ/dt = ABsinωt times another ω. That’s the emf. Now what about the  sign. Forget the  sign. We can figure out what the emf — which way the current will try to flow. Look at this coil here. Let’s decide how we want to turn this guy. If you turn it in the manner I’ve shown here, it’s going from some angle like that, eventually to an angle perpendicular to the magnetic field. What’s happening to the flux through it? Is it increasing or decreasing? Increasing. Increasing in this direction. So it will fight it by trying to produce a flux going in the opposite direction. Therefore it will try to produce a current that looks like this. That’s the direction of your emf. Emf, given a chance, will drive a current as shown here, because that one by right hand rule will have the flux going the opposite way. So that’s why I don’t care about the  sign. I know which way the current will flow if I let it flow. But now if this is an open circuit like this, the two wires sit here, what do you think will happen? What do you think will happen in that case? If you’re an electric charge in that wire, what will you do under the emf? You will follow the emf and you will run from this terminal to that terminal. So let me blow up that picture for you. If the current is trying to go that way, then charges will leave like this and pile up here. They cannot go very far because it’s an open circuit. At some point, these guys piled up here — so this is the induced electric field obeying that equation. In fact, it’s not induced electric field. This is just the v x B force on the magnet, on the charges. But then these + charges will produce and electrostatic force called the Coulomb field, which is the electric field due to charges, which will oppose it, till the two cancel inside the wire. In other words, this wire is assumed to be a perfect conductor. You all know that you cannot have an electric field in a perfect conductor. You can say, “Hey, why do you have an electric field now?” The real statement is, in a perfect conductor, you cannot have any net charge on the charged particles, because then they will pick up infinite speed. So if there’s a v x B force, that is actually canceled by an electrostatic force. The two of them cancel inside the wire. And the line integral of the v x B, which is the emf, will numerically equal the integral of the electric field from this terminal to that terminal. But now if you put the whole thing in a box and you don’t know anything and you come outside, what you will find is this will be positively charged, this will be negatively charged. There’ll be voltage between those two which will numerically equal your emf. This will be the polarity. So I’m giving you an option. If you don’t want to look under the hood, you simply say whenever you rotate a coil, you get an emf. That emf is like a voltage. Case closed. But if you really want to know what’s happening inside the wire that made up that coil, inside the wire the net force on the charges is actually 0, or an infinitesimal amount left to make them move this way. That’s because the v x B force is countered by an internal electrostatic force due to pile up of charges that opposes it so that inside it’s free. It’s just like the ski experiment I told you, where you have a ski lift. You come down the ski lift, you come here, and here is the lift that carries you up. But during the time gravity’s acting down and the force of the lift is exactly equal to mg and gravity. And once you’re outside the lift, it’s gravity that brings you down back here. Similarly, inside the thing the two guys are opposed to each other. Outside this region there is no v x B force, but the electrostatic force has a line integral that’s independent of the path. So integral that way the same as the integral that way. So if you put a circuit here, it will produce a voltage difference with this being positive, that being negative. So the bottom line is that if you have open circuit, and if you put a volt meter that measures voltage, you will get exactly that time dependent voltage. So this voltage is not constant. It will look like this, t versus V. So in the linear loop, when you drag it and run, you get a constant voltage. Here you get a time dependent one, and most of the supplies in the world are either 50 cycles or 60 cycles, and they come from rotating a coil in this magnetic field. Now how much is the work done by the person rotating this coil right now? I’m telling you the coil is made of massless, perfectly conducting wire. What work do you have to do to spin it? Any ideas? Student: [inaudible] Professor Ramamurti Shankar: Which one, dE? Student: Edq. Professor Ramamurti Shankar: Yeah, but I’m saying the total force — you don’t have to do any work because there is no current flowing in the coil yet. There’s no current. If there’s no current, it’s BlI force. So what we want to do then is do something more interesting where you bring it here, connect it to a resistor. Then the current will flow. Now we’re getting something. Till now, we were getting nothing from the generator. In other words, look, where’s a socket here? That’s a socket. There’s voltage there waiting for you to use. But you don’t pay, because you’re not drawing any current. You don’t pay just because someone gave you the voltage. You take a battery in your hand, you don’t run a current, you don’t pay anything. That’s what the situation is. But if I stick my fingers into that socket, then there is current. Then we’ll all pay. The loss is I^{2}R, right? That’s when you’ve got to explain to yourself, who is paying for this? Because we can see that the power in the resistor will be I^{2}R. I, if you want, is E^{2}/R, emf or voltage squared over R. That gives me ω^{2}A^{2}B^{2}sin^{2}ωt/R. That’s the rate at which this power is consumed by the resistor. Someone’s got to pay for that. The someone, I think you can imagine now, now that I’ve closed the circuit and a current is actually flowing, there is a torque on that loop in a magnetic field. The torque on any loop, you’ll remember, is μ x B. μ is the magnetic moment. That happens to be area of the loop, current in the loop, B times the sine of the angle between area and B. That is torque. So that’s the torque that you’ve got to fight. So you will have to apply mechanical force to turn that. If you’re turning the crank by hand, the minute you put a load, you’ll find it’s hard to turn it. And the power, just like power is force times velocity, for rotational motion it is torque times angle of velocity. That gives me ABωIsinθ. That’s the power, the mechanical power, P_{m}. But then that = ABωsinθ times I. What is I? The current in the loop is the emf divided by R. If you put the emf I got for you somewhere, then you’ll find it’s equal to A^{2}B^{2}ω^{2}/Rsin^{2}θ. So this is how you have to balance. We did a similar calculation here. This is also a generator and if you find out the minute there’s a current flowing here, you have I^{2}R or E^{2}/R energy loss here. But the minute the current flows through the loop, you will not be able to pull it to the right without paying the force. That force times the velocity will be the power and the two will balance. Similarly, here is a rotational problem, and the torque times angular velocity will exactly equal the power here. So the generators, you’ve got hydroelectric generators, or you’ve got steam, turbines that are spinning, initially, because the turbines in the real world have a real mass, it’s not easy to spin them. You spend some power, but if you ignore that, the minute you put a device in your house into the socket, your toaster, you start drawing current, that current’s got to flow right through the generator. It’s going to make it that much harder to rotate the generator. That’s when the steam turbines do their work. That’s when they pay for it. Chapter 3: Mutual and Self Inductance [00:37:47]Okay, so this is the end of the story about how to get power out of this and how it balances. I’m going to a fairly different notion called inductors and inductance. So here is the following phenomenon. I take some cardboard tube and around the cardboard tube I wrap some wire and I connect this to some alternating voltage. That means there is going to be some magnetic flux going through that coil. Then I bring a second wire, maybe I wrap it round a couple of times or a few times, and I leave it there. That’s it. The question is, what will happen if I look at the two ends of this wire? You can see what will happen. If this current changes, the flux through this solenoid changes. That means the flux through this little guy alsochanges. Every loop has flux going through it and that also changes. So there’s going to be an emf. For every turn of wire here, they’ll be an emf equal to the rate of change of flux. And therefore that emf will try to draw a current, just like before, and the charges will pile up maybe like this at some instant. Now if you are outside all of this, you will just think there is a voltage available to you. But I want you to understand the origin of that voltage. This is like the ski lift here. Charges, once they come into this region, are pushed up the wire and later on, if you’re connected to a load, they can drive a current. This is just like the generator I had there, except the flux is changing not due to any mechanical motion of a coil in a fixed field, but it’s fixed coils in a changing magnetic field. So you understand, when I change the current here, you will get a voltage here. This is the first thing Faraday noticed, is that he thought first that magnetic field may produce a current. It didn’t. Then he found a changing magnetic field is able to produce a current and this is why this happens. Once again, I want you to think about the following question: we are going to take such things and put them in electrical circuits and so on. We’ll draw all kinds of circuits later on. And what I will do in all those calculations, I will say I start here, I go around a loop and I come back and the change in voltage should be 0. I’m going to use that principle. But there is one flaw if you don’t think about it that says maybe I shouldn’t do that. You remember that you can define a voltage only for a conservative force, whereas you definitely have non conservative forces that work in this problem. That’s why they have a line integral not equal to 0. And yet we apply the laws of conservative forces, or a notion of a voltage in a circuit. I want to explain why that’s allowed. If it never bothered you, you can ignore this part. But it’s important to understand how we can have a notion of a voltage in the presence of time dependent magnetic fields. The line integral of E is not 0. Once again, what happens is, if you look at this coil, there may be an electric field that is induced now due to the changing flux of that. That will try to build up charges here and take them out of this end. After a while, these guys will say “Enough” and they’ll start fighting you, till they set up an electrostatic field, a Coulomb field, inside the wire that cancels this. You understand, the integral of this electric field from top to bottom will numerically equal the integral of the induced electric field from bottom to top, because at every point they’re equal in magnitude. Because you cannot have a nonzero field inside a wire. So the two fields have to be canceled. But this electrostatic field, built by these charges, is a conservative field, so if it does any work going this way, it will also do work going that way. Therefore if you don’t open the black box, you don’t know what’s inside, if you just took the two wires coming out of it, the static field coming from this will be able to do work going from + to  terminal. And if you put a resistor on its path, it will deliver some energy to you. The trick is, yes, there are changing magnetic fields, but they’re hidden inside the coil. In the region outside the coil, we don’t have to worry about the changing magnetic field, where an electric potential can be defined as the integral of the electric field. The electric field outside is entirely electrostatic. Electric field inside is a combination of electrostatic and induced one, which cancel. Once current begins to flow, you may worry that these + charges will go away. There’ll always be some electric charges making sure that the field inside the coil continues to be 0. Again, there’s one abuse of terminology here, because I’m saying electrostatic, and yet this is a problem where the electric field you need is actually changing with time, because the rate of change of flux is changing with time, induced electric field is changing with time. The compensating electrostatic field is also changing with time. So you really should not use the laws of Coulomb for this problem, but it turns out, even though Coulomb’s law is valid only strictly for fixed electric charges, as long as the velocities required is not too big, you can continue to use Coulomb’s notion of an electrostatic force. Basic question is, you’re telling the charges to constantly rearrange back and forth. First you want this terminal to be +. A short time later, you may want that to be +. How quickly can they rearrange? So that’s connected to something called plasma frequency of these materials. And unless the frequencies are like a trillion, you don’t have to worry about it. But any normal problem, like 50 cycles per second, the charges will be able to keep up with this changing field. In other words, can the charges go to the edge of a conductor and screen any internal field you’re trying to produce? If it’s a static field, they will kill it, because you’re giving them all the time in the world. But if you keep changing your mind, one minute the external field goes this way, so the charges in the metal go that way to kill it. Suddenly you reverse it,they’ve got to go this way. So how quickly can they do this dance? There’s a limit to how quickly they can do it. But that’s a very, very high frequency, so you don’t have to worry about that. Now we saw that the emf in the second coil = the flux in the second coil divided by d/dt. Now you’ve got to be a little careful about flux. Normally, the flux due to the magnetic field is defined as the integral of B⋅dA. But if you’ve got a coil made up of two loops, for example, and there’s a magnetic flux going through them, then the emf is not simply the rate of change of the flux through one of them, but you’ve got to double it, because it’s like two batteries in series. An emf is running round and round and round, so really, this is N_{2} times the rate of change of the magnetic flux to that. So this Φ is not simply the magnetic flux. It’s the magnetic flux times the number of turns the second coil. That’s what the real emf is. A real emf is really rate of change of the flux linking with your whole circuit, dt. That equals number of turns times the actual magnetic flux dt. Do you follow that? Emf around a single loop is the rate of change of flux, but if your coil loops around twice, the endtoend voltage will be double that. If I loops around three times, it will be triple that, assuming the same magnetic flux is going through all of them, which is the assumption I made here. So now you can see here that the flux in the second coil is due to the current in the first coil and the coefficient of proportionality is called the mutual inductance. So mutual inductance is how much flux you can get in the second coil per unit current in the first coil. Then you can write the emf in the second coil as −M_{12} dI_{1}/dt. M_{12} is called the mutual inductance of the first coil with respect to the second one. It’s true, but very hard to prove that M_{12} is the same as M_{21}. Not at all obvious that if you drove a current in the first coil, it’s going to have a magnetic field that’s threading the second coil and you can show that the flux per current in the first coil due to the current in the second, it’s given by the same number. M_{12} — yes? Student: How did the number of turns go away? Professor Ramamurti Shankar: Where did it go away? In this one you mean? That’s all included in the definition of M. I’m going to now calculate M so you will see the number of turns coming in. So let’s do the calculation of M for a simple problem. Here’s one solenoid. It’s got N_{1} turns. And there’s another guy. I’m just going to show you one turn of it, but it can have N_{2} turns. And the flux is going through this. Now you remember that the magnetic field due to any solenoid is μ_{0} times n, where n is the number of turns per unit length times I. Therefore the magnetic field in this solenoid = μ_{0}n_{2}I_{2} — I’m sorry about that one. This is the first wire of the first coil. It’s n_{1}I_{1}. And the flux of the magnetic field is μ_{0}n_{1}I_{1} times the cross sectional area of the coil. Now the flux linking with the second one = μ_{0}n_{1}I_{1} — sorry, n_{1}AN_{2} times I_{1}. You see that? So the first coil has some wrapping density of wires, n_{1} turns per unit length. I’ve shown you long back from Ampere’s law, the magnetic field that produced this has that flux, that B value. The integral of the field, which is the magnetic flux, is just area times that. But the linking with the second coil is that flux times the number of turns in the second coil. That by definition = the mutual inductance M_{21}I_{1} (I’m not going to call it 21 or 12. It’s independent of the order) = μ_{0}n_{1}N_{2}A. That’s the mutual inductance. So if I give you two closed loops and I say, “Find the mutual inductance,” here’s what you’re supposed to do. Drive a current in the first one and produce some flux lines, magnetic lines. Some of them will penetrate the second one. You count how many penetrate the second one, multiplied by the number of turns, if they exist in the second one, divide by the current producing it. That’s the mutual inductance. So inductance is measured in henries, another thing for you. And usually you may find millihenries or microhenries for common use. Notice a very interesting result. The flux is going through both coils, you understand that? The same flux is going through both coils. Maybe you’re happier if I drew the coil like this. Here’s a doughnut coil, right? You bring the wire here. I told you how you can wrap it around many times. Then the other coil, the secondary coil, is also wrapped around the same doughnut. It must be clear to you that the magnetic field is going through all these coils. The emf on the first one is proportional to the number of turns on the first one times the rate of change of the magnetic field. The emf on the second one is N_{2} times dΦ_{B}/dt. Therefore E_{1}/E_{2} = N_{1}/N_{2}. The same flux is going around the doughnut. One guy has N_{1} turns around it, other has N_{2} turns around it. Emf, you remember, is not simply rate of change of magnetic flux.. It’s that multiplied by the number of turns. They both have the same flux going through them, but this has N_{1} turns, this has N_{2} turns. You can see this ratio. That’s a very powerful result. You know what this gadget is called? It’s a transformer, because you put in one voltage and you get another voltage. You can have a step up transformer or a step down transformer. If you drive a current from here, and you pick it up here, that’s a step down transformer, because you’re going down in the number of turns. If you connect your power supply to this one and you pull it out of that one, it’s a step up transformer. So you can step up or step down, but you can only do it for AC. You cannot do it for DC, at least, not in any simple way. You need the changing thing to do a transformer. But I don’t have the time or techniques to convince you, in spite of the ratio of emfs, you don’t gain or lose energy by transformers. In other words, this is not a device for manufacturing energy. You will find out that if you connect a load here and it’s drawing some power, the same power has to be provided by the source. So it’s not a way to manufacture energy. It’s like a lever. You have a long thing and you’re trying to lift some object here, and you’re trying to life with — I got it backwards. So this is a huge object, a tiny object, you can balance them. By varying the distance, you can have a tiny guy lifting a big one in the inverse ratio of the distances. But you don’t get any free mileage out of this, because if you look at the work done by similar triangles, this will have to move a lot that will have to move a little. So the work done is the same, but doesn’t mean it’s useless, because this is the only way you can lift something very heavy. Likewise in a transformer, you may not have the ability to give 5,000 volts, starting with 200 volts, but you can if you use this transformer. Quite often you want to step it down. In all the gadgets you use in your house, you start with 110 volts, you want to step it down to some number, so you use a step down transformer. All right, so that’s the stuff on inductance. Now we are going to come — this part is really a curiosity. I’m not going to use it very much, the notion of mutual inductance. Mutual inductance is one coil trying to generate a voltage in a second coil, because they share a flux. And when it’s changing one of them, it’s also changing the other one. They don’t have to be really coaxial. You can put a second loop way over here, and maybe some other extra flux coming out is penetrating this. That’s the mutual inductance between this guy and that guy also. Any time the flux of one coil can go through another one, you have a mutual inductance, because if you change the current in the first coil, you’re going to generate a voltage in the second coil. That’s why you need to know that proportionality. All right, so now we are going to do the most important circuit element, which is an inductor, and it looks like this. That’s a coil of wire. Say some current is coming in like this. It’s wrapped around some solenoid, in the form of a solenoid. This wire is a perfect conductor, therefore it takes no voltage at all to drive a current through it. You put a battery there and it just burns immediately. But if you put a time dependent current you will need a voltage, because a time dependent current will produce a time dependent magnetic flux through this. So let us say the current was originally 0. You’re trying to increase it and produce a magnetic flux here. Then an emf would be generated that opposes it. And we can ask, how much is the emf? The emf is the rate of change of flux to that coil, and I’m going to assume that the flux through that coil = the current in the very same coil time a number called self inductance. The self inductance is how much flux do you produce by a current going through yourself? Not on another coil, on yourself. Every coil, when it carries current, will have some flux threading through itself. So that ratio is called the self inductance, also measured in henries. So this becomes −LdI/dt. You will calculate L in a moment, but I’m just telling you that as long as dI/dt is not 0, you will have to oppose that back emf with a voltage from some other supply. So in this example, the current is going like this. Let us say it’s trying to increase. If it is trying to increase, then the back emf will be set up, an electro emf will be set up to fight it. It will try to push charges that way, from this terminal to that terminal. It will pile up charges there. But if you don’t look under the hood and just went outside, you’ll have + charges and  charges and an electrostatic field here will be able to push them like this, maybe through a circuit. So it’s the same story again and again. There is no net field inside the coil. The electromotive force is canceled by a Coulomb force. But the Coulomb force, if it has a line integral here, will have the same line integral there, because it’s independent of the path. So if you don’t look inside, you will just find there is some field which has lines from here to here. That’s the voltage. But it will be a voltage drop, so that if you want another convention for the voltage, it’s like this. If the current is going this way and it’s increasing in this direction, this will be + and that will be  at the instant. Just like a resistor, it flows downhill, this will also flow downhill, meaning this is higher than this, provided the current is trying to increase in the direction of the current. So we can have the following very simple circuit. We have a battery here. You’ve got a resistor there and we have an inductor there. This is some voltage V, this is R, this is some L henries. So let us write a circuit equation and here is where I spend an inordinate amount of time justifying what I’m about to do. I say start here. There’s going to be a voltage defined everywhere except inside the black box. You understand? You cannot define a voltage inside the black box where there’s an inductor, because there is a nonconservative electric field inside. But we promise not to go there. Then from here to here, you go up by V_{0}. Then here, current flows downhill, so you drop RI and here, if this is the sense of the current and it’s increasing, your loop will go from there to there and it will jump this and come to this end. The drop from here to here is LdI/dt, and the whole thing should add up to 0. So here’s a one word summary for those of you who have heard enough: we learned the notion of voltage can be defined, or a potential, only in a conservative problem. But a changing magnetic field inductor definitely is producing a field which is not conservative. So if you go deep into the coil, you will have problems defining voltage. But if you come outside the coil, I’ve tried to show you over and over again, all it looks like is there is a voltage difference between the two plates, the two terminals, and that value is LdI/dt. So when you do your circuit equation, you go from there, all the way back here, you skip these funny elements, and around them you still have the notion of a voltage. So this is the equation to solve at any time for an LR circuit. Now prior to this, let me give you a very interesting result, which we will use a lot. If you take an inductor which had current I = 0 and you managed to drive a current through it and slowly built up the current, it’s got some other value I at the end, what was the work done to do that? I will now show you that when you drive a current through an inductor, you are doing some work, because when you start driving the current, it’s opposing you with a voltage LdI/dt, and you’re ramming it down that, up in spite of that opposition, so that the work you do, the power, is I times LdI/dt, which is, if you like the rate of change of energy. But I times LdI/dt is d/dt of ½ IL squared. I’m sorry, LI squared. You see that from the rules of calculus, the derivative of this guy is LI times dI/dt. Therefore the integral of the power is simply ½LI^{2}, assuming you started at 0 current, started at I = 0. So it takes some energy to build up a current in the inductor. That’s the point. Just like it takes some energy to charge up a capacitor. I showed you when you charge up a capacitor, here’s a capacitor. It’s got some charges, +, , you want to charge it even more, you’re going to ram more positive charges and more negative charges here. You fight it harder and the total work done is q^{2}/2C. Similarly, when you have an inductor and you’re trying to increase the current through it from 0 to some final value, this is the amount of work done by you. That energy is stored in the inductor. I want to look a little bit about inductors, but first let’s calculate L. So let’s calculate L for a simple solenoid. Here is my solenoid. Remember, L is defined as the flux linkage divided by the current. It’s going to be a one line calculation so it’s going to be very easy. The magnetic field = μ_{0} little nI. Little n you remember, always is the number of turns per unit length. The flux of the magnetic field is μ_{0} nI times the area of cross section. That’s the total magnetic flux. But the flux linkage is μ_{0} little n IA times big N, because every loop of the coil links with its own flux. But that is by definition LI. So you can see that L = μ_{0} little n big NA. That’s the self inductance of the coil. What does this mean? If this = 5 henries, it means that if you shove 1 ampere through this guy, a flux equal to 5 tesla squared meters will be linked to that circuit. Chapter 4: Energy Density of a Magnetic Field [01:04:11]So the thing I want to do now is to equate this energy, ½ LI^{2}, to the magnetic field inside the coil. So you’ve got 1 over 2. L is μ_{0} little n, big NAI^{2}. So let me rewrite that as ½ μ_{0}n^{2} times lA times I^{2}. In other words, I’ve written, using this formula to write little n as big N over l. times A, is the volume in which there’s a magnetic field. So this looks like 1 over 2μ_{0} times μ_{0}nI whole squared, times l times A. But who is μ_{0}nI. μ_{0}nI is the magnetic field. So this looks like B^{2}/2μ_{0} times the volume of the solenoid. From that we learn that when you have a magnetic field, there’s an energy of B^{2}/2μ_{0} per volume. So the energy density of the magnetic field = B^{2}/2μ_{0}. Let me remind you the electric field energy, energy density for electric field, is ε_{0}/2E^{2}. You might remember that formula. So they are very similar formulas, except μ_{0}, which is normally upstairs in every formula, comes downstairs here, and ε_{0} which is always downstairs in every formula, comes upstairs here. So let me summarize what you should remember from all of this. When you have a circuit element called an inductor, it’s just a coil of wire that’s wrapped around some solenoid. And when you change the current through the inductor, it’s going to fight it. It’s not like a resistor. A resistor fights any current. An inductor fights only a change in current, so that’s all summarized in this equation, the voltage = LdI/dt + RI. This is the circuit we’re going to look at. Maybe a switch is open like that. There’s R, that’s L, that’s a switch, that’s the voltage. When you close the switch, you’ve got to ask yourself, what’s the current going to be? What will be the current infinitesimally after the switch is closed? Yes? Suppose it was not 0, but .2 amps, what’s the problem? After all, you closed the circuit. Yes? Student: You have energy stored in the inductor without ___________. Professor Ramamurti Shankar: Yeah, first of all, if you had .2 amps, you’ll have ½LI^{2} and you can ask, “Who had the time to do that?” Nobody. More importantly, if LdI/dt is the voltage across this, it will become infinite if dI/dt is infinite. A current that jumps from 0 to something in no time, that’s got infinite derivative. So any quantity whose derivative is bounded cannot jump in its value just from calculus. So the current in the inductor will never jump. Likewise if you have a capacitor, with some charge on it, and you close the circuit, the charge on the capacitor initially was Q_{0}, will remain Q_{0} one femtosecond after you close it, because charge on it, the rate of change of the charge, is a current. The current is finite in any real problem. So capacitors cannot abruptly change the charge they have, and inductors cannot abruptly change the current they carry. If you want, they are connected to energy. The energy in the capacitor is Q^{2}/2C, therefore if q changes abruptly, the energy changes by a finite amount in infinitesimal time. Nobody can deliver that energy or take out energy at that rate. Similarly for the inductor. So I’ll tell you what’s in store on Wednesday. We’re going to come back and look at this LR circuits and look at LC circuits and look at LCR circuits. That’s the kind of stuff I think you’ve all done before in high school, but I think still have to do that, because that’s the kind of stuff that may be more useful than many of the other things I’m talking about. But it won’t be in the greatest depth. I just want to hit the high points. [end of transcript] Back to Top 
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