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MIT OpenCourseWare http://ocw.mit.edu 8.02 Electricity and Magnetism, Spring 2002 Please use the following citation format: Lewin, Walter, 8.02 Electricity and Magnetism, Spring 2002 (Massachusetts Institute of Technology: MIT OpenCourseWare). http://ocw.mit.edu (accessed MM DD, YYYY). License: Creative Commons Attribution-Noncommercial-Share Alike. Note: Please use the actual date you accessed this material in your citation. For more information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/termsMIT OpenCourseWare http://ocw.mit.edu 8.02 Electricity and Magnetism, Spring 2002 Transcript – Lecture 23 Here are the topics the way I see them. They're on the web. Look on the lecture supplements of today and you can download them. I want to point out that, uh, the discount is very reasonable because these magnets are broken, and when you break a magnet, you end up with two monopoles, so you get 50 percent off. You should know better by looking at the key equation there, that magnetic monopoles don't exist, but that's a detail. I want you to appreciate that I cannot possibly cover today these topics in any depth, nor can I cover all of these during a 50 minute exam. So please understand that this review is highly incomplete, and what is not covered today can and will be on the exam. I'm interested in concepts. I'm not interested in math. There will be seven problems. Five of the seven problems have only one question. Two problems have two questions. I don't think that time, that the length of the exam is going to be an issue. This exam was taken by several instructors. It took them 15 to 18 minutes, and that's normally my objective.Courtesy of Professor Belcher, we have old tests on the website. I don't have the solutions. There's only so much I can do. I'm very grateful to Professor Belcher that he made these exams available. If you can't do some of the problems, I would suggest you see your tutors or you see your instructor. I will also be available all day afternoon, not tomorrow, because tomorrow afternoon I'll be in 26-100 all afternoon to work on demonstrations for the next week. Many of these problems are straightforward. You may also want to consult your study guides. All right. Let's first start with Biot-Savart. There are not too many problems that one can do with Biot-Savart. dB equals mu 0 divided by 4 pi times the current dL cross R / R squared. That's the formalism. A classic problem that you probably have done. We have point P here at a distance D from a wire, and the current through the wire is I. If you want to know what the magnetic field at P is, you can use Biot- Savart. It would be a stupid thing to do, but you can do it. You take then a small element dL here of the wire. This distance is R.This is the unit vector R, which you have in this equation. And you can calculate, then, what the contribution to the magnetic field right here is -- it comes out of the blackboard -- due to this section dL. This angle is theta, and the sine of theta is D divided by R. And then you have to do an integral over the whole wire, theta 0 to pi. And then you get the magnetic field here. Not very smart thing to do. A waste of time. Because clearly the way to do this is to use Ampere's Law, which is the one at the bottom there. In which case, you would construct a closed loop with radius D. This loop is perpendicular to the blackboard. I'll try to make you see it three dimensionally. You have to attach an open surface to that closed loop. Any open surface will do. Let's make the open surface flat. And then we apply Ampere's Law, which is the one at the bottom there. We don't have the second portion because there is no changing electric flux. We don't deal with kappa M at all. So we simply have that B times 2 pi D, that is going around in this circle, because I know that B is tangentially to that circle.I also even know the direction according to the right-hand corkscrew rule, coming out of the blackboard here. So as I go around the circle, B and the dL that you see there at the bottom are in the same direction. So I get 2 pi B times D equals mu 0 times the current that penetrates that surface, and that is I. And so the answer is very simple, mu 0 I divided by 2 pi D. That's the way you would do this problem, and you would stay away from Biot-Savart. There is one particular problem whereby Ampere's Law will fail. Of course, Ampere's Law in general works when we have cylindrical symmetry. This is cylindrical symmetry. There is one problem where Ampere's Law bitterly fails and where Biot-Savart is highly superior. I have here a conducting loop. It's a circle. And you're being asked, it runs a certain current I, and you're being asked, what is the magnetic field right at the center? It only works for the center. You could not find what the magnetic field is here. We did that in class. You probably also did that for your homework. Biot-Savart will immediately give you the answer. And I will leave you with that. And Ampere's Law won't work.So let's now turn to Ampere's Law and do a few problems with Ampere's Law. We need cylindrical symmetry, with very few exceptions. I have a hollow cylinder here, radius R1. Concentric another cylinder with radius R2. These are very long cylinders. And assume that there is a current flowing, I, in this direction on the surface of the inner cylinder, and the current I is returning on the surface of the outer cylinder, and the two currents are the same in magnitude. And I want to know what is the magnetic field everywhere in space. I will make a, a drawing whereby we only see the cross-section. So this is R1 and this is R2. And let's first calculate the magnetic field for R being larger than R2. It's immediately obvious that your closed loop that you choose itself is going to be a circle. That is a must. Radius R. We use a symmetry argument. Whatever the magnetic field is at that distance R, little R, it must be the same everywhere. It cannot be any different here in terms of magnitude than there because of the symmetry of the problem. We have cylindrical symmetry. We also know that if there is any magnetic field, that it is going to be tangential, either in this direction or in that direction.And so we go around. We make the closed loop integral, use the equation that we have there at the bottom, and so you get B times 2 pi little R equals mu 0. And now I have to attach an open surface to this loop. I will use the surface in


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