Sabtu, 22 Januari 2011

Fisika untuk Universitas

Fisika untuk Universitas

Ditujukan untuk meningkatkan kualitas proses dan hasil perkuliahan Fisika di tingkat Universitas

Kelistrikan dan Kemagnetan

Topics covered:

RL Circuits
Magnetic Field Energy

Instructor/speaker: Prof. Walter Lewin

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    Today, I will quantify the ability of a circuit to fight a magnetic flux that is produced by the circuits themselves.

    If you have a circuit and you run a current through the circuit, then you create some magnetic fields, and if the currents are changing then the magnetic fields are changing.

    And so there will be an induced EMF in that circuit that fights the change, and we express that in terms of a self-inductance: L, self-inductance, and the word self speaks for itself.

    It's doing it to itself.

    Magnetic flux that is produced by a circuit is always proportional to the current.

    You double the current, the magnetic flux doubles.

    And so it is the proportionality constant that we call L, that is the self-inductance, and so therefore the induced EMF equals minus d phi/dt, that is Faraday's Law.

    And so that becomes minus L dI/dt.

    L is only a matter of geometry.

    L is not a function of the current itself.

    I will calculate for you a very simple case of the self-inductance of a solenoid.

    Let this be a solenoid and this is a closed circuit, and we run a current I through the solenoid, and the radius of these windings is little r.

    Let's say there're N windings and the length of the solenoid is little l.

    Perhaps you'll remember that we earlier derived, using Ampere's Law, that the magnetic field inside the solenoid is mu 0 times I times capital N divided by l.

    This is the number of windings per meter.

    If we attach an open surface to this closed loop, very difficult to imagine what that open surface looks like -- we discussed it many times -- inside this solenoid you have sort of a staircase-like of surface.

    That magnetic field penetrates that surface N times because you have N loops.

    And so the magnetic flux, phi of B, is simply the area by little r squared, which is the surface area of one loop, because I assume that the magnetic field is constant inside the solenoid, and I assume that it is 0 outside, which is a very good approximation.

    So we get pi little r squared surface area of one loop, but we have N loops and then we have to multiply that by that constant magnetic field.

    So we get an N squared because we have an N here, mu 0 I divided by L.

    And this we call L times I.

    That's our definition for self-inductance.

    And so the self-inductance L is purely geometry.

    It's pi little R squared, capital N squared divided by L times mu 0.

    Let me check this.

    Pi little r squared, I have a capital N squared, mu 0, that's correct, divided by little l.

    And so we can calculate, for instance, what this self-inductance is for a solenoid that we have used in class several times.

    We had one whereby we had 2800 windings.

    R I think was something like 5 centimeters -- you have to work SI of course, be careful -- and we had a length, was 0.6 meters.

    We had it several times out here, and if you substitute those numbers in there, you will find that the self-inductance of that solenoid is 0.1 in SI units and we call the SI units Henry, capital H.

    It would be the same as volt-seconds per ampere, but no one would ever use that.

    We call that Henry.

    Every circuit has a finite value for the self-inductance, however small that may be.

    Sometimes it's so small that we ignore it, but if you take a simple loop, a simple current, just one wire that goes around -- whether it is a rectangle or whether it is a circle it doesn't make any difference -- it always produces a magnetic field.

    It always produces a magnetic flux through the surface, and so it always has a finite self-inductance.

    Maybe only 9- nano Henrys, maybe only micro Henrys, but it's never 0.

    And so now what I want to do is to show you the remarkable consequences of the presence of a self-inductor in a circuit, and I start very simple.

    I have here a battery which has EMF V.

    I have here a switch, and here are the self-inductor.

    We always draw a self-inductor in a circuit with these coils, and we also have in series a resistor, which we always indicate with this, these teeth.

    And I close this switch when there is no current running.

    In other words, at time T equals 0 when I close the switch, there is no current.

    When I close this switch the current wants to increase, but the self-inductance says uh-huh, uh-huh, take it easy, Lenz law, I don't like the change of such a current.

    So the self-inductance is fighting the current that wants to go through it.

    There comes a time that the self-inductance loses the fight, if you wait long enough, and then of course the current has reached a maximum volume, which you can find with Ohm's Law, because the self-inductance itself has no resistor.

    Think of the self-inductance as made of super-conducting material.

    There's no resistance.

    And so without knowing much about physics, you can make a plot about the current that is going to flow as a function of time.

    You start out with 0 and then ultimately, if you wait long enough, you reach a maximum current which is given by Ohm's Law, which is simply V divided by R.

    And you slowly approach that value.

    And how slowly depends on the value of the self-inductance.

    If the self-inductance is very high, it might climb up like this, so this is a high value for L.

    If the self-inductance is very low, that is a low value for L.

    If the self-inductance were 0, it would come up instantaneously, but I just convinced you that there I no such thing as 0 self-inductance.

    There's always something finite, no matter how small.

    And so this is qualitatively what you would expect if you use your stomach and if you don't use your brains yet.

    There's nothing wrong with using your stomach occasionally, but now I want to do this in a move civilized way, and I want to use my brains, and when I use my brains I have to set up an equation for this circuit.

    And if you read your book, you will find that Mr. Giancoli tells you to use Kirchhoff's Loop Rule.

    But Mr. Giancoli doesn't understand Faraday's Law, and he's not the only one.

    Almost every college book that you read on physics do this wrong.

    They advise you to use Kirchhoff's Loop Rule, which says that the closed loop integral around the circuit is 0.

    That, of course, is utter nonsense.

    How can it be 0?

    Because there is a change in magnetic flux, and so it can only be minus d phi/dt.

    I advise you to go to the 8.02 website and download a lecture supplement that you will find in which I address this issue and hit it very hard.

    So the closed loop integral of E dot dL, if you go around the circuit, is not 0, is minus d phi/dt -- Faraday's Law -- so it's minus L dI/dt.

    So we have to go around to circuit and we have to apply Faraday's law and not Kirchhoff's Loop Rule, which doesn't apply here.

    This is the plus side of the battery and this is the minus side, so the electric field in the battery is in this direction.

    The electric field in the self-inductance is 0 because the self-inductance has no resistance, it's super-conducting material, and so the electric field in the resistor -- if the current is in this direction, which it will be -- then the electric field in the resistor will be in this direction.

    So now I am equipped to write down the closed loop integral of E dot dL.

    I start here and I always go in the direction of the current, and I advise you to do the same.

    I don't care that you guess the wrong direction for the current.

    That's fine.

    Later, minus signs will correct that, they will tell you that you really guessed the wrong direction, but always go around the loop in the direction of the current, because then the EMF is always minus L dI/dt.

    If you go in the direction opposed to the current, then it is plus L dI/dt and that could become confusing.

    So I always go in the direction of the current, and so I first go through the self-inductance.

    There is no electric field in the self-inductance, so the integral E dot dL -- in going from here to here -- is 0.

    This is where the books are wrong.

    It is 0.

    Now, I go through the resistor, and so now I get plus IR, E and dL are in the same direction.

    Ohm's Law tells me it's IR.

    In the battery, I go against the electric field, and so I get minus V.

    That now equals minus L dI/dt, and this is the only thing and the only correct way to apply Faraday's Law in this circuit.

    You can write it a little differently, which may give you some insight.

    For instance, you could write that I can bring V and the L, and L, to one side -- so I can write down that V minus L dI/dt equals IR.

    It's the same equation when you look, and the nice thing about writing it this way is that since dI/dt is positive here -- it's growing in time -- the induced EMF, which is this value -- notice it's always opposing the voltage of my battery -- and that's what Lenz's Law is all about.

    It's not until dI/dt has become 0 that V equals IR, and that happens, of course, if you wait long enough.

    And so we have to solve that differential equation, and what is often done that you bring all the terms to the left side and that you get an, a 0 on the right side.

    And so what you often see is that L dI/dt plus IR minus V equals 0.

    And because we have a 0 here, some physicist thinks that this is an application of Kirchhoff's Rule.

    This is nonsense.

    You can always make it 0 here by bringing all the terms to this side.

    The closed loop integral of E dot dL is not 0, the closed loop integral of E dot dL is minus L dI/dt, but when I shift minus L dI/dt to this side I get 0 here.

    And of course the people who write these books know that this is the right answer and so they manipulate it so they get this equation and they call that Kirchhoff's Rule.

    Sad, and also embarrassing.

    So, this is the equation that you have to solve.

    Some of you may have solved this equation in 8.01 already.

    Surely you didn't have an I here.

    You may have had an X here for the position, but you probably solved it when you had friction.

    Maybe you didn't.

    I will give you the solution to that differential equation.

    It's a very easy solution.

    The current as a function of time is a maximum value times 1-e to the minus R divided by L times T, and I max -- that is the maximum current -- is V divided by R.

    And let's look at this in a little bit more detail.

    First, notice that when t equals 0 that indeed you find I equals 0.

    Substitute in here t equals 0, you get 1-1.

    So you find, indeed, that I equals 0.

    Substituting there t goes to infinity, then you find that I indeed becomes V divided by R, which is exactly what you expect.

    If t becomes infinity, then clearly the self-inductance has lost all its power, so to speak, and the current is simply V divided by R, the maximum current that you can have.

    And so that's a must, that's a requirement.

    If you wait L over R seconds -- and believe it or not, if you have some time, convince yourself that L over R indeed, as a unit, is seconds -- then the current I is about 63% of I max, because if t is L over R, then you get 1-1 divided by E, and that is about 0.63.

    And if you wait double this time, then you have about 86% of the maximum current.

    In other words, right here -- if I wait L over R seconds -- this value here is about 0.63 times the maximum value possible, and it's very, it's climbing up and asymptotically approaches then, ultimately, the maximum current which is V divided by R.

    Make sure you download that lecture supplement that you'll find on the Web.

    Now, what I'm going to do is all of a sudden I'm going to make this voltage 0.

    The way I could do that is by simply shorting it out.

    Of course on the blackboard, I can simply remove it.

    So it's not there.

    The current is still running and all of a sudden at a new time -- t 0, I define the time t 0 again -- the voltage is 0.

    And now comes the question what is now going to happen?

    Well, the self-inductance doesn't like the fact that the current is going down, so it's going to fight that change, and so you expect that the current is not going to die off right away, but you expect that the current is going to go down sort of like so.

    And you want, when you wait long enough, you want that, at t goes to infinity, where t equals 0 the current is still maximum, so it's still V over R, but when you go to infinity -- if you wait long enough -- then, of course, the current has to become 0.

    And as the current dies out, heat is being produced in that resistor at a rate of I square R joules per second, and then there comes a time that the current becomes almost 0 and then the whole show is over.

    And so we can also calculate the exact time behavior by going back to our differential equation and make V 0.

    Where is that differential equation?

    Is it hiding?

    Oh, there it is.

    So I solved this differential equation, but this is now 0.

    And the solution to that differential equation is that I as a function of time is I max times e to the minus R over L times t, and that exactly has all the quantities that you want it to have, because notice that at t equals 0 -- when you put in t equals 0 -- the current is indeed maximum, and that's what you require.

    That's the moment that you make that capital V 0, the current was still running.

    But notice that when t goes to infinity -- if you wait long enough -- that indeed the current goes to 0.

    And if you wait L over R seconds, then you are down to about 37% of your maximum current.

    So if you now go to I, I re-define t equals 0 here, so if now I wait L over R seconds then this value here is about 37% of that value.

    So you've lost 63%.

    And so you see, this is the consequence of the fact that the circuit is capable of fighting its own magnetic flux that it is creating.

    When the current was running happily here, with the battery in place, the current was, let's say, all the time I max -- at least very close to that value -- V over R.

    And so all the time, there was heat produced in the resistor.

    I square R joules per second.

    Who was providing that, uh, energy?

    Well, of course, the battery.

    But now, when I take the battery out, there is still current running, and that means while the current is dying there is still heat produced in that resistor, and that heat slowly comes out until the current ultimately becomes 0.

    Now where does that energy come from?

    Well that energy must come from the magnetic field that is present in the solenoid, and this idea -- that we have energy that comes out in the form of heat, which really was there ear-, earlier in the form of a magnetic field -- allows us to evaluate what we call the magnetic energy field density.

    Let me first calculate how much heat is produced as the current goes from a maximum value down to 0.

    Hmmmm, I'll have to erase something.

    I'll erase this part here.

    So at any moment in time, the current is producing heat in the resistor, and so if I, if my voltage becomes 0 at time t equals 0, then this is the amount of heat, uh-uh, no square here.

    I square R dt, integrated from 0 to infinity, is the total heat that is produced as the current dies out, but I know what this current was -- I just erased it, if I still remember it -- so I can bring I max outside and I can bring the resistance outside and then I get the integral from 0 to infinity of e to the minus R over L times t dt.

    And this is a trivial integral.

    This integral is, uhm, L divided by 2 R.

    Oh, by the way, it is I squared, so I have a 2 here.

    It's very important.

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Ucapan Terima Kasih Kepada:

1. Para Dosen MIT di Departemen Fisika

a. Prof. Walter Lewin, Ph.D.

b. Prof. Bernd Surrow, Ph.D.


Prof. John Belcher

Dr. Peter Dourmashkin
Prof. Bruce Knuteson
Prof. Gunther Roland
Prof. Bolek Wyslouch
Dr. Brian Wecht
Prof. Eric Katsavounidis
Prof. Robert Simcoe
Prof. Joseph Formaggio

Course Co-Administrators:
Dr. Peter Dourmashkin
Prof. Robert Redwine

Technical Instructors:
Andy Neely
Matthew Strafuss

Course Material:
Dr. Peter Dourmashkin
Prof. Eric Hudson
Dr. Sen-Ben Liao


The TEAL project is supported by The Alex and Brit d'Arbeloff Fund for Excellence in MIT Education, MIT iCampus, the Davis Educational Foundation, the National Science Foundation, the Class of 1960 Endowment for Innovation in Education, the Class of 1951 Fund for Excellence in Education, the Class of 1955 Fund for Excellence in Teaching, and the Helena Foundation. Many people have contributed to the development of the course materials. (PDF)

2. Para Dosen Pendidikan Fisika, FPMIPA, Universitas Pendidikan Indonesia.

Terima Kasih Semoga Bermanfaat dan mohon Maaf apabila ada kesalahan.

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