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 13 All right, you did well on the exam. Class average was 62. I always aim for 65, so I was very happy. 11 students scored 100. I believe that my exam review was extremely fair. According to some instructors, perhaps even too close for comfort. I did a problem with parallel resistors and a battery. I applied Gauss's Law for cylindrical symmetry. I spent quite a bit of time discussing where charge occurs and where charge cannot be located on conductors and I hit the idea of capacitors and dielectrics also quite hard. I prefer not to think about a rigid division between pass and fail, but I'd rather tell you that all of you who scored less than 47, in my book, are sort of in the danger zone. Now, that doesn't mean that you're going to fail the course, nor does it mean that you will pass the course if you scored 70. But those people are in the danger zone. I think you should talk to your instructor, and I would advise those people also to make frequent use of our tutors. Two exams to go, plus the final. Today I'm going to uncover a whole new world for you and you will see how 8.02 comes in there in a very natural way.The Lorentz force F is the charge times the cross product of the velocity of that charge and the B field that the charge experiences. If I have here a positive charge plus Q and it has a velocity V in this direction, and the magnetic field would be uniform and coming out of the blackboard, there's going to be a force on this charge according to this relationship and the force is then like so. Perpendicular to V, perpendicular to B. In this case the charged particle is going to go around in a circle. The Lorentz force cannot change the speed, cannot change the kinetic energy, because the force is always perpendicular to the velocity, but it can change the direction of the velocity. And so, what you're going to see is that the charged particle will go around into a perfect circle if the magnetic field is constant throughout. And the radius of this circle can very easily be calculated using some of our knowledge of 8.02. The force is QVB because I chose B also perpendicular to V, and so there is no sign, the sign of the angle between them is 1, and this now has to be the centripetal force that we encountered in 8.01, which is MV squared divided by R, M now being the mass of this particle. And so you'll find now that R equals MV divided by QB. And this, by the way, I want to remind you, is the momentum of that particle. If you look at this equation, it's sort of pleasing. If the charge is high then the Lorentz force is high so the radius is small. If the magnetic field is high then the Lorentz force is high so the radius is small. If the mass of the particle is high, there is a lot of inertia and so it is very difficult to make it go around, so to speak, so a very high mass, you expect a very high radius.And so that looks all intuitively quite pleasing. Let's do a numerical example. I take a proton, P stands for proton, and I take a 1 MeV proton. It's the same I took during my test review. 1 MeV means that the kinetic energy is 1 MeV, is the charge times the potential difference over which this proton was accelerated, in this case, delta V would be 1 million volts. And this now equals one-half times the mass of that proton times the velocity squared. In this case, if I have a 1 MeV, so it is a million volts, you will find that this is 1.6 times 10 to the -13 joules. I gave you there the charge of the proton, you multiplied it by a million, and this is the energy. And so now you can calculate the velocity because you know the mass of the proton. I gave you that too, there. And so you will find exactly what you found during my test review, 1.4 times 10 to the 7th meters per second, which is 5% of the speed of light, comfortably low so we don't have to make any relativistic corrections. If this proton now enters a magnetic field B, which is 1 tesla, then by using the equation I have up there, you know the mass of the proton, we just calculated the velocity. You know the charge of the proton and you know the B field. You will find that R is 0.15 meters, which is 15 centimeters, just a numerical example. It is more common, or at least often done, to eliminate out of that equation there the velocity and replace it by the potential difference, capital V, over which we accelerate these particles.And so, what you can do, you can replace this V by using the equation I have there, the one half MV squared, so we have that one-half MV squared equals Q times delta V, but I will write for that just a capital V, and I substitute this V now in here, and so I no longer see the velocity but I now see this potential difference. In the case of that proton, this V would be a million and you will find then that R is then the square root of 2M times that capital V divided by Q B squared. And so the two equations are of course the same physics, but it's different representation. If you put in for V now 10 to the 6th, mass of the proton, charge of the proton, and 1 tesla field, of course you find exactly the same 0.15 meters. Now this is all nice and dandy, but this works as long as the speed is much smaller than the speed of light. If that's no longer the case, then we have to apply special relativity and that is not part of this course but I would like to briefly touch upon that today. I can show you how things go sour because suppose we have a 500 kilo electric volt electron. So that means that in this equation here, the V is 500000, the Q is the charge of the electron, M is now the mass of the electron, and if I apply that equation I find that V is 4.2 times 10 to the 8th meters per second and that is larger than the speed of light, so …
View Full Document
Unlocking...