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Fisika Teori Medan

Electromagnetic Wave Theory
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15: Momentum and its Conservation

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We're going to talk about a whole new concept, which is the concept of momentum.

We've all heard the expression, "We have a lot of momentum going." Well, in physics, momentum is a vector, and it is a product between the mass of a particle and its velocity.

And so the units would be kilograms times meters per second.

F = ma.

That equals m dv/dt.

That equals d(mv)/dt and that equals, therefore, dp/dt.

So what you see--

the force is dp/dt, and what that means is if a particle changes its momentum, a force must have acted upon it.

It also means if a force acts on a particle, it will change its momentum.

Let us now envision that we have a whole set of particles which are interacting with each other, and the interaction could be gravitational interaction, could be electrical interaction, but they're interacting with each other.

Zillions of them--

a whole star cluster.

I pick one here, which I call mi, and I pick another here, which I call mj.

And there is an external force on them, because they happen to be exposed to forces from the outside world, and so on this one is a force Fi external, and on this one is Fj external.

But they're interacting with each other, either attracting or repelling, and so in addition to these external force, there is a force that j feels due to the presence of i.

And let's suppose they are attracting each other.

This would be F that j experience in the presence of i.

Actions equals minus reactions, according to Newton's Third Law, so this force, Fij, will be exactly the same as Fji, except in the opposite direction.

We call these internal forces.

That's the interaction between the particles.

If these were the only two forces, then the net force on this object, on this particle i, would be... would be a force in this direction.

This would be F net.

Now, you can do the same here.

This would be F net on particle j, and this would be F net on particle i.

But since there are zillions of objects here, there are many of these interacting forces, and so I can't tell you what the net force will be.

The net force is ultimately the sum of the external force and all the internal forces.

What is the total momentum of these zillions of particles? Well, that's the sum of the individual momenta.

So that is p1 plus p2... pi...

and you have to add them all up.

I take the derivative of this--

dp total/dt.

That is p1/dt.

Well, that's the force on number one.

It's the total force on number one.

So it is F1, but it is the net force on F1.

And F... and dp/dt for this particle equals F2 net force, and the force on particle number i equals Fi net, and so on.

And so when we add up all these forces, obviously that is the total force on the entire system.

Now comes the miracle.

Ucapan Terima Kasih Kepada:

1. Para Dosen MIT di Departemen Fisika

a. Prof. Walter Lewin, Ph.D.

(http://web.mit.edu/physics/people/faculty/lewin_walter.html)

b. Prof. Bernd Surrow, Ph.D.

(http://web.mit.edu/physics/people/faculty/surrow_bernd.html)

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

Terima Kasih Semoga Bermanfaat dan mohon Maaf apabila ada kesalahan.

Physics For School

Physics For Junior High School

Specific Topics in Middle School Physics I:

Motion Forces Vector and Scalar Quantities Momentum Energy Circular Motion Center of Gravity Rotational Mechanics Universal Gravitation Gravitational Interactions Satellite Motion

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Kelistrikan dan Kemagnetan

Topics covered:

Transformers
Car Coils
RC Circuits

Instructor/speaker: Prof. Walter Lewin

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  » Download this transcript (PDF)

  Class average on the exam was 55.

  Three homework problems were on the exam.

  I'm a little puzzled by the fringe field problem.

  That was the one homework problem that was graded.

  It was graded on your homework, and I went back to your homework scores and they were 90%.

  You all did extremely well on that problem.

  But on the exam, it was only 40%.

  And so perhaps this is telling you something.

  And maybe it's telling me something too.

  If you turn in a correct solution for homework, it's not very useful if you do not understand what you wrote down.

  The understanding, of course, is what matters, not that you somehow, one way or another, get a correct solution.

  If I had to judge you only on the basis of exam one and two, forgetting the quizzes, forgetting the homework, then those with 80 or lower would fail the course, as of now.

  But of course those who have between 80 and 90 are by no means home free.

  They are still in the danger zone.

  And anyone who has 95 or even 100, I cannot guarantee you that you will pass the course.

  The depends, of course, on how you will do in the future, the third exam and on the final.

  Today I want to discuss with you RC circuits.

  We already discussed RL circuits.

  Now we get RC circuits.

  I have here a battery with an EMF V0.

  And I have here a switch connecting to a capacitor.

  And here a resistor.

  And I close the loop.

  When I have the switch in this position, the capacitor is going to charge up.

  I get a current I going in this direction.

  If I call this point A here and this point P and this point S to make sure that we are on the same wavelength, I will call the potential over the capacitor, I will call that V A minus V P.

  The potential over the resistor is of course always I times R, is then V of P minus V of S.

  The question now is what is the potential over the capacitor doing as a function of time and what is the current doing as a function of time as I charge up this capacitor.

  And your intuition will help you a great deal without any fancy differential equations.

  It is clear that at T equals 0, the capacitor is not charged.

  There is no charge on the capacitor.

  And it will take time to charge the capacitor.

  So at T=0, you expect that the potential over the capacitor is 0.

  If you make T a little larger than 0, you are going to charge up this capacitor, and so the potential over the capacitor will go up, and therefore the current will go down.

  And if you wait long enough -- we call that infinitely long -- then the capacitor will be fully charged.

  It will have the potential V0 of the battery, and then the current has become 0.

  No current is flowing anymore if the battery -- if the capacitor is fully charged.

  And so this capacitor is going to charge up, this becomes positive, and this becomes negative.

  And so you can construct a, a plot whereby you plot as a function of time here the potential over the capacitor -- we haven't used any differential equations yet.

  You know that if you wait long enough you will reach that value V0.

  And it's going to build up like this, asymptotically reach that value.

  And if C is very large, then the current will be more like this, and if C is very small, then it will go much faster, of course, small C.

  The current as a function of time.

  In the beginning, the current will be high, but ultimately the current will die down, when the capacitor is fully charged.

  So you expect something like this.

  So this you can do without any differential equations.

  Let's now do it the correct way.

  The closed loop integral of E dot dL happens to be 0, which will make Mr. Kirchhoff very happy.

  The closed loop integral of E dot dL in this circuit is 0.

  Mr. Faraday is happy and Mr. Kirchhoff is happy.

  There is no magnetic flux change here; we don't have self-inductances.

  The electric field inside the capacitor is in this direction, from plus to minus.

  The electric field in the resistor is in this direction, the current is flowing in this direction.

  And this battery, which has this as the positive side and this as the negative side, inside the battery the electric field is in this direction.

  So if I start at point A and I go around the circuit, then E and dL are in the same direction if I go from A to P, and so I get plus V of C, that's the E dot dL from A to P.

  Then I go through the resistor.

  Again, E and dL are in the same direction, so I get plus I times R.

  Then I come through the battery.

  Now the electric field is opposing the direction in which I go, so I get minus V0.

  And that is 0.

  So that's my differential equation.

  And I can write it differently, because I know that I, the current, is dQ/dT, Q being the charge on the capacitor.

  And it is only if that number changes, if the capacitor is either charging up or discharging, is there a current flowing.

  And in addition I know that V of C is Q divided by C.

  That's the definition of capacitance.

  And so I can write down for this V of C, I can write down Q divided by C.

  For I, I can write down dQ/dT.

  So I get R times dQ/dT minus V0 equals 0.

  And this is a differential equation in Q.

  We have seen an identical differential equation.

  It was not in Q, it was in I, but it had of course a completely similar solution.

  And the solution to this differential equation is actually quite simple.

  I will put it on this board.

  Q -- so this is going to be Q as a function of time -- is V0 C times one minus e to the minus T over R C.

  If I take the derivative of this Q then I have I, because I is dQ/dT.

  So I, which is dQ/dT then becomes V0 times C, I get a minus sign, I get another minus sign, then I get one over R C, and then I get e to the minus T / R C.

  So the two minus signs eat each other up and I lose one C here.

  And so I get that the current as a function of time is V0 divided by R times e to the minus T divided by R C.

  So that's the curve at the bottom.

  And the potential over the capacitor is now very simple, because the potential over the capacitor is Q / C.

  And I already have Q here, so I simply have to divide this C out.

  So I get V0 times one minus e to the minus T / R C.

  And so that is the upper curve.

  And so we can now make a small table and we can look at various values for T.

  Maybe I should do that here on the blackboard, because I don't want to erase anything yet.

  So we have T here, we have I, and we have V of C.

  And when T is 0, you go to your equation of I, so this is one.

  So you have a current V0 divided by R.

  And your V C -- V C of C is 0.

  You can see that.

  If T is 0, you get 1 - 1 -- oh sorry, I have to go here.

  You get 1 - 1, so you see indeed that the potential over the capacitance is still 0.

  If you wait long enough, then the current must go to 0.

  That exponential function goes to 0 if you wait long enough.

  And your potential over the capacitance then reaches V0, which is exactly consistent with our solution.

  If you wait a time RC, that's called the time constant of this circuit.

  In the case of the current, it's also called the decay time of the circuit.

  Then of course your current is one over e times V0 / R.

  And one over e is a roughly .37, 0.37.

  So your current is down to 37 percent of what it was at the beginning.

  And after a time R C, the potential over the capacitor is one minus one over E times V0, and that is then about 0.73 -- uh, .63, 0.63, 63 percent.

  In other words, if I go back here to my plot and if I draw a line at time R C, then this value here is 37 percent of the maximum and this value here is then 63 percent of the maximum.

  So it's in -- the solution is rather obvious, very intuitive.

  And the R C times can vary an enormous amount, as you can imagine, depending upon the values for R and C.

  If we have here an R and a C and we want to know what the R C time is, convince yourself that the product of R and C indeed has units of seconds.

  Remember we had L over R before, which had also units of seconds.

  R C also has units of seconds.

  If you have R equals 1 ohm and C is 1 microfarad, then the R C time is only a microsecond.

  But if you have R equals 100 megaohms and you have this 1 millifarad, then this is 10 to the 5 seconds, which is longer than a day.

  And that would mean that it would take you even three days to reach 95 percent of V 0.

  After three days, you would still have only 95 percent of the potential difference of the capacitor of the maximum value that you can get.

  Now what I want to do is I make a change here.

  I have here a conducting wire.

  I call this position 1 of the switch.

  And I'm going to put the switch in this position, position 2.

  And I can do that without any danger.

  Remember, I waited until this capacitor was fully charged.

  There was no current running.

  And so when no current was running, I can quietly take the switch and put it in this position.

  And what is going to happen now is of course this side is positively charged and this is negatively charged, so now you're going to get a current which is going to run counterclockwise, in opposite direction.

  And what is going to happen is the capacitor is discharging now.

  And the resistor will dissipate the energy that is in the capacitor.

  The one half C V squared energy stored in the capacitor is going to be dissipated in the resistor in terms of I squared R, in terms of heat.

  And if you wait long enough, the current will become 0.

  So it should be obvious what's going to happen.

  If I return to this curve here, if I redefine my time equals 0, and if this is the moment that I put the switch in position 2, then I expect that the capacitor will discharge.

  You get a curve like this.

  And I expect that the current, which now becomes negative -- it reverses direction and I call that negative.

  And so the current will come like this.

  And if you wait long enough, of course, the current will aga- again become 0.

  You have discharged the capacitor.

  So if you want the formal solution, you have to go back to the differential equation.

  And you take this term out, because it's not there.

  And now you have to solve this differential equation again, which is now utterly trivial.

  And I would like you to solve that differential equation.

  You couldn't have an easier one.

  I will give you the solution to I as a function of time, and you then will come up with this part.

  I as a function of time is exactly the same as this except with a minus sign, provided that I call this T equals 0.

  So I redefine the 0 time.

  And so I as a function of time is the equation that you have here, but now with a minus sign.

  So you get an exponential change again, but the current has flipped over.

  I can demonstrate this to you.

  I have a electronic switch, so I go between one and two -- every four milliseconds I throw the switch.

  And so what I have is, as a function of time, what we call a square wave.

  So this is my battery.

  And this time here, from here to here, is eight milliseconds.

  This is time.

Pengembangan Perkuliahan

1. Buatlah sebuah Esai mengenai materi perkuliahan ini

2. Buatlah sebuah kelompok berjumlah 5 orang untuk menganalisis materi perkuliahan ini

3. Lakukan Penelitian Sederhana dengan kelompok tersebut

4. Hasilkan sebuah produk yang dapat digunakan oleh masyarakat

5. Kembangkan produk tersebut dengan senantiasa meningkatkan kualitasnya

Ucapan Terima Kasih Kepada:

1. Para Dosen MIT di Departemen Fisika

a. Prof. Walter Lewin, Ph.D.

(http://web.mit.edu/physics/people/faculty/lewin_walter.html)

b. Prof. Bernd Surrow, Ph.D.

(http://web.mit.edu/physics/people/faculty/surrow_bernd.html)

Staff

Visualizations:
Prof. John Belcher

Instructors:
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

Acknowledgements

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.

Fisika untuk Universitas

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14: Orbits and Escape Velocity

» Download this transcript (PDF)

If you are standing somewhere on Earth... this is the Earth, the mass of the Earth, radius of the Earth, and you're here.

And let's assume for simplicity that there's no atmosphere that could interfere with us, and I want to give you one huge kick, an enormous speed, so that you never, ever come back to Earth, that you escape the gravitational attraction of the Earth.

What should that speed be? Well, when you're standing here and you have that speed, your mechanical energy--

which we often simply call E, the total energy--

is the sum of your kinetic energy--

this is your mass; this is your escape velocity squared--

plus the potential energy, and the potential energy equals minus m Mg divided by the radius of the Earth.

So this is your kinetic energy and this is your potential energy--

always negative, as we discussed before.

Mechanical energy is conserved, because gravity is a conservative force.

So no matter where you are on your way to infinity, if you are at some distance r, that mechanical energy is the same.

And so this should also be one-half m v at a particular location r squared minus m M earth G divided by that little r.

And so at infinity, when you get there--

little r is infinity, this is zero, potential energy at infinity is zero--

and if I get U at infinity with zero kinetic energy, then this term is also zero.

And that's the minimum amount of energy that I would require to get you to infinity and to have you escape the gravitational pull of the Earth.

If I give you a higher speed, well, then, you end up at infinity with a little bit net kinetic energy, so the most efficient way that I can do that is to make this also zero, so you reach infinity at zero speed.

So this is for r goes to infinity.

Ucapan Terima Kasih Kepada:

1. Para Dosen MIT di Departemen Fisika

a. Prof. Walter Lewin, Ph.D.

(http://web.mit.edu/physics/people/faculty/lewin_walter.html)

b. Prof. Bernd Surrow, Ph.D.

(http://web.mit.edu/physics/people/faculty/surrow_bernd.html)

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

Terima Kasih Semoga Bermanfaat dan mohon Maaf apabila ada kesalahan.

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Fisika Statistika

An Introduction to Stochastic Processes in Physics
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Statistical Physics
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A Modern Course in Statistical Physics
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Methods of Quantum Field Theory in Statistical Physics
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Renormalization in Statistical Physics
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Statistical Physics
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Statistical Physics and Spatial Statistics
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Stochastic Processes in Physics, Chemistry, and Biology
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Introduction to Mathematical Statistical Physics
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Fisika untuk Universitas

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

Kelistrikan dan Kemagnetan

Free Downloads

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» Download this transcript (PDF)

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 the blackboard.

I could use any surface, but I might as well use a flat surface.

And now I have to know what is the current going through that surface.

Right here, on this surface, the current is going into the blackboard, and right here, on this surface, the current is coming out of the blackboard.

The two magnitudes are the same, so the net current is 0.

And so the magnetic field outside the second cylinder is 0.

So let's now look at the area in between the two cylinders.

These are hollow cylinders, now, this is completely open here and this is completely open.

They are thin, thin material, thin shells, are both cylinders.

Now of course the closed loop is going to be one inside the opening between the two cylinders.

And again, I pick radius R.

And here I go.

I get B times 2 pi R equals mu 0, but now there is current going through this surface, and the current that is going through is I.

Not this one, but this one.

And so we get I.

And so now we get that B is mu 0 times I divided by 2 pi R.

One over R field.

The direction you will find from the right-hand corkscrew rule.

That's the way I normally do it.

You take a corkscrew and you turn it clockwise, it goes into the blackboard, so the magnetic field here is in this direction.

If you're not used to turning corkscrews in corks, think about when you try to tighten a screw with a screwdriver.

If you go clockwise, the screw goes in.

At least, that is the case with 99.99 percent of all screws that you find in this country.

They have a right-handed thread.

You could make one that has a left-handed threa- thread, but that's not done in general.

So now we can go to the area R less than R1.

So that's inside this thin shell cylinder, and again, you would take a surface, a closed loop, that is a circle with radius little R, and this is your surface attached, open surface attached to that closed loop.

And you will find for the same reason that you found here that B is 0, because there is no current going through that surface.

And so if you now make a plot of the magnetic field B as a function of R, here is R1, here is R2, then it is 0 here, it's 0 there, it has some value here, which is this value if you substitute for little R R1 and right here it is this value, if you substitute for little R R2.

And this is a curve that is proportional to 1/R.

And so this is the magnetic field, and inside, seen from where you were sitting, it is clockwise.

There is one case, and we did that in lectures, whereby we wanted to calculate the magnetic field inside a solenoid, where Ampere's Law works very well, even though we don't have then closed loops which are circles, we chose a rectangle.

I want you to revisit that.

It's undoubtedly in your book, and I covered it during my lectures.

We assumed that the magnetic field was uniform inside the solenoid and 0 outside the solenoid, which was not a bad assumption, and that allows you then, with Ampere's Law, even though you go rectangular and not circular, to get the magnetic field inside a solenoid.

Very classic.

Pengembangan Perkuliahan

1. Buatlah sebuah Esai mengenai materi perkuliahan ini

2. Buatlah sebuah kelompok berjumlah 5 orang untuk menganalisis materi perkuliahan ini

3. Lakukan Penelitian Sederhana dengan kelompok tersebut

4. Hasilkan sebuah produk yang dapat digunakan oleh masyarakat

5. Kembangkan produk tersebut dengan senantiasa meningkatkan kualitasnya

Ucapan Terima Kasih Kepada:

1. Para Dosen MIT di Departemen Fisika

a. Prof. Walter Lewin, Ph.D.

(http://web.mit.edu/physics/people/faculty/lewin_walter.html)

b. Prof. Bernd Surrow, Ph.D.

(http://web.mit.edu/physics/people/faculty/surrow_bernd.html)

Staff

Visualizations:
Prof. John Belcher

Instructors:
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

Acknowledgements

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.

Fisika untuk Universitas

Fisika untuk Universitas

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

Kelistrikan dan Kemagnetan

Topics covered:

Hysteresis
Electromagnets
Bohr Magneton
Maxwell's Equations
600 daffodils

Instructor/speaker: Prof. Walter Lewin

Free Downloads

Video

• iTunes U (MP4 - 112MB)
• Internet Archive (MP4 - 221MB)

  » Download this transcript (PDF)

  When I expose material to an external magnetic field, then we learned last time that the field inside that material is modified.

  And we expressed that in terms of an equation, that the field inside the material is kappa of M, which is called the relative permeability, times the external field, and I will refer to that all the time as the vacuum field.

  And when we have diamagnetic material, kappa M is just a hair smaller than one; with paramagnetic material, it is a hair larger than one; but when we have ferromagnetic material it can be huge.

  It can be thousands, 10 thousands, and even higher.

  Now in the case of para- and ferromagnetic material, the kappa of M is the result of the fact that the intrinsic dipoles of the atoms and the molecules are going to be aligned by the external field.

  And today I want to raise the question, how large can the magnetic dipole moment of a single atom be?

  And then comes the logical question, how strong can we actually, then, have a field inside ferromagnetic material?

  That means if we were able to align all the dipole moments of all the atoms, what is the maximum that we can achieve?

  To calculate the magnetic dipole moment of an atom, you have to do some quantum mechanics, and that's beyond the scope of this course.

  And so I will derive it in a classical way, and then at the very end I will add a little pepper and salt, which is quantum mechanics, just to make the result right.

  But it can be done in a classical way, and it can give you a very good, good idea.

  If I have a hydrogen atom, which has a proton at the center, has a charge plus e -- e is the charge of the electron but this is the plus charge.

  And let this have an orbit R, circular orbit.

  And the electron here e, I'll give it the minus sign to make sure that you know that it's negative.

  Say the electron goes around in this direction.

  This is the velocity of the electron.

  That means that of course the current around the proton would then be in this direction.

  If an electron goes like this, the current goes like that, that's just by convention.

  The mass of an electron -- you should know that by now -- is approximately 9.1 times 10 to the -31 kilograms.

  The charge of the electron, 1.6 times 10 to the -19 coulombs.

  And the radius of the orbit in a hydrogen atom -- it's often called the Bohr radius, by the way -- is approximately 5 times 10 to the -11 meters.

  We're going to need these numbers.

  That's why I write them down for you.

  If you look at this current running around the proton, it's really a current which the current, say, goes in this direction.

  And here is that proton -- trying to make you see three dimensionally.

  Then it creates a magnetic field in this direction, and so the magnetic dipole moment mu is up.

  And the magnitude of that magnetic dipole moment, as we learned last time, is simply this current I times the area A of this current loop.

  Now the area A is trivial to calculate.

  That's pi R square, R being the radius of the orbit.

  And so A, that's the easiest, is pi R squared.

  And if I use my 5 times 10 to the minus 11, then I find that this area is 8 times 10 to the -21 square meters.

  So that's easy.

  But now comes the question, what is I?

  What is the current?

  So now we have to do a little bit more work.

  And we have to combine our knowledge of, uh 8.02 with our knowledge of 8.01.

  If this electron goes around, the reason why it goes around is that the proton and the electron attract each other.

  And so there is a force in this direction.

  And we know that force, that's the Coulomb force.

  It's an electric force.

  That force is this charge times this charge, so that's e squared, divided by our famous 4 pi epsilon 0, and then we have to divide it by the radius squared.

  So that's Coulomb's Law.

  But from 8.01, from Newtonian mechanics, we know that this is what we call the centripetal force that holds it in orbit, so to speak, and that is M V squared, M being the mass of the electron, V being the speed of the electron, and V squared divided by R.

  And so this allows me to calculate as a first step, before we get into the current, what the velocity of this electron is.

  It's phenomenal.

  It's an incredible speed.

  So V then becomes -- I lose one R -- so I get the square root, I get an e squared upstairs here, my M goes downstairs, I have 4 pi epsilon 0, and I have here an R.

  And I know all these numbers.

  I know what e is, I know what capital R is, I know what 4 pi epsilon 0 is -- one over 4 pi epsilon 0 is the famous 9 to the power -- 9 times 10 to the power 9.

  And so I can calculate what V is.

  And if I stick in the numbers and if I did not make a mistake, then I find about 2.3 times 10 to the 6 meters per second.

  It's an immensely high speed, 5 million miles per hour.

  If this were a straight line, you would make it to the moon in three minutes.

  5 million miles per hour this electron goes around the proton.

  Now I have to go to the current.

  I have to find out what the current is.

  So the question that I'm going to ask now is how long does it take for this electron to go around.

  Well, that time, capital T, is of course the circumference of my circle divided by the speed of the electron.

  Trivial.

  Even the high school students in my audience will understand that one, I hope.

  And so I know what 2 pi R is, because I know R and I know V and so I can calculate that time, just by sticking in the numbers.

  And I find that it is about 1.14 times 10 to the -16 seconds.

  Just imagine how small that time is.

  You cannot even -- we cannot even imagine what it's like.

  It goes 10 to the 16 times per second around, because it has this huge speed.

  The 1.14 times 10 to the -16 really should have been 1.4 times 10 to the -16.

  Of course, it doesn't make much difference, but in case you substitute in the numbers, it is 1.4 times 10 to the -16.

  Now, we still haven't found the current, but we're almost there.

  Because when you look here, there is this electron going by, and every 1.14 10 to the -16 seconds, that electron goes by.

  So the current I, that's the definition of current, is the charge per unit time.

  And so every capital T seconds, the charge e goes by, and so this is per definition the current.

  And so this current, then, that you have, which is simply due to the electron going around the proton, is about 1.1 times 10 to the -3 amperes.

  And that is mind-boggling.

  A milliampere.

  One electron going around a proton represents a current of a milliampere.

  And now of course I have the magnetic moment mu, that is I times A.

  We already calculated A, and now we also have the current I, and so we now get that mu is approximately 9.3, if you put in all the decimals correctly, times 10 to the -24, and the unit is of course amperes square meters.

  This is area, and this is current.

  This A has nothing to do with that A, hey.

  This is amperes.

  Be careful.

  And this is square meters.

  But these are the units.

  And this has a name.

  This is called the Bohr magneton.

  Bohr magneton.

  What we cannot understand with our knowledge now, but you can if you ever take quantum mechanics, that the magnetic moment of all electrons in orbit can only be a multiple of this number, nothing in between.

  Quantum mechanics, the word says it is quantization.

  It's not in between.

  It's either or.

  It includes even 0, which is even harder to understand, that it can even be 0.

  In addition to a dipole moment due to the electron going around the proton, the electron itself is a charge which spins about its own axis, and that also means that a charge is going around on the spinning scale of the electron.

  And that magnetic dipole moment is always this value.

  And so the net magnetic dipole moment of an atom or a molecule is now the vectorial sum of all these dipole moments, all these electrons going around, means orbital dipole moments, and you have to add the spin dipoles.

  Some of these pair each other out.

  One electron would have its dipole moment in this direction and the other in this direction, and then the vectorial sum is 0.

  The net result is that most atoms and molecules have dipole moments which are either one Bohr magneton or 2 Bohr magnetons.

  That is very common.

Pengembangan Perkuliahan

1. Buatlah sebuah Esai mengenai materi perkuliahan ini

2. Buatlah sebuah kelompok berjumlah 5 orang untuk menganalisis materi perkuliahan ini

3. Lakukan Penelitian Sederhana dengan kelompok tersebut

4. Hasilkan sebuah produk yang dapat digunakan oleh masyarakat

5. Kembangkan produk tersebut dengan senantiasa meningkatkan kualitasnya

Ucapan Terima Kasih Kepada:

1. Para Dosen MIT di Departemen Fisika

a. Prof. Walter Lewin, Ph.D.

(http://web.mit.edu/physics/people/faculty/lewin_walter.html)

b. Prof. Bernd Surrow, Ph.D.

(http://web.mit.edu/physics/people/faculty/surrow_bernd.html)

Staff

Visualizations:
Prof. John Belcher

Instructors:
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

Acknowledgements

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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Fisika untuk Universitas

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

Kelistrikan dan Kemagnetan

Topics covered:

Magnetic Materials
Dia-, Para-, and Ferromagnetism
Prize Ceremony of Motor Contest

Instructor/speaker: Prof. Walter Lewin

Free Downloads

Video

• iTunes U (MP4 - 96MB)
• Internet Archive (MP4 - 190MB)

  » Download this transcript (PDF)

  Yesterday we had 225 motors, and six of those motors went faster than 2000 RPM, which is a reasonable accomplishment.

  And the elite is here.

  These are the elite, the six highest.

  The winner is, um, Jung Eun Lee, I talked to her on the phone last night.

  If all goes well, she is here.

  Are you here?

  Where are you?

  There you are.

  Why don't you come up so that I can con- congratulate you in person.

  I thought about the, the prize for a while, and I decided to give you something that is not particularly high tech, but come up here, give me a European kiss, and another one -- in Europe, we go three.

  OK.

  Um, the prize that I have for you is a thermometer which goes back to the days of Galileo Galilei -- come here.

  Uh, it was designed in the early part of the, um, seventeenth century.

  Uh, it doesn't, uh, require any knowledge of 8.02 to explain how it works.

  If anything, you need 8.01.

  It's not a digital thermometer.

  But it's accurate to about 1 degree centigrade, and if you come here, you can tell, you look at these floaters, and the highest floater indicates the temperature.

  It's now 72 degrees here.

  And I suggest that you brush up on your knowledge of 8.01 so that perhaps next week you can explain to me how it works.

  [laughter].

  And of course tell your grandchildren about it.

  You may want to leave it here.

  It's very fragile.

  Uh, there is also some package material here, so that you can take it home without breaking it.

  So congratulations once more [applause] and of course -- [applause].

  Terrific.

  And you will join us for dinner on the thirteenth of April with the other five winners.

  Thank you very much.

  There are two other people who are very special who I want to mention.

  And one is a person who is not enrolled in, uh, 8.02, but he did extremely well, and he was very generous.

  He was not competing.

  His name is Daniel Wendel.

  His motor went 4900 RPM.

  And then there was Tim Lo.

  Is Tim Lo in the audience?

  I hope he's going to be there at eleven o'clock.

  Tim made a motor -- when I looked at it, I said to myself, it'll never run, but it's so beautiful.

  It was so artistic that we introduced a new prize, a second prize, for the most artistic motor, and Tim Lo definitely is the one, by far the best, the most beautiful, the most terrific artistic design.

  And so for him I bought a book on modern art -- what else can it be for someone who built such a beautiful motor?

  It is here for those of you who want to see it later.

  It's very hard to display it on television because it's so delicate.

  It's like a birdcage that he built instead of having just -- looks like that it's a birdcage.

  It's very nice.

  The winning motor I have here, and I'm going to show you the winning motor, and I also want to teach you some, some physics by demonstrating the winning motor to you in a way that you may never have thought of.

  So this is the winning motor.

  And when we start this motor, the ohmic resistance of the current loop is extremely low.

  So the moment that you connect it with your power supply, a very high current will run.

  But the moment that the motor starts to rotate, you have a continuous magnetic flux change in these loops, and so now the system will fight itself, and it will immediately kill the current, which is another striking example of Faraday's Law.

  I will show you the current of this motor when I block the rotor so that it cannot rotate.

  It's about 1.6 amperes.

  And you will see the moment that I run the motor that that current plunges by a huge amount.

  Striking example of Faraday's Law.

  So I now have to first show you this current, so here you see the 1.5 volts, and on the right side you see the current.

  There is no current flowing now because the loop is hanging in such a way that the, that it makes no contact with the battery.

  And I'm going to try to make it -- there it is.

  Do you see the 1.6 amperes on the right?

  The current is so high that due to the internal resistance of the power supply, the voltage also plunges.

  But you saw the 1.6, right?

  Now I'm going to run the motor.

  See, the motor is running now, and now look at the current.

  Current now, forty milliamperes, thirty milliamperes, fifty milliamperes.

  It's forty times lower than when I blocked the rotor.

  And so this is one of the reasons why when you have a, a motor, whichever motor it is, it could be just a drill, you try not to block it all of a sudden, because an enormous current will run, and it can actually damage the motors.

  So you see here how the current goes down by a factor of forty between running and not running.

  All right.

  Electric fields can induce electric dipoles in materials, and in case that the, the molecules or the atoms themselves are permanent electric dipoles, an external electric field will make an attempt to align them.

  We've discussed that in great detail before when we discussed dielectrics.

  And the degree of success depends entirely on how strong the external electric field is and on the temperature.

  If the temperature is low, you have very little thermal agitation, then it is easier to align those dipoles.

  We have a similar situation with magnetic fields.

  If I have an external magnetic field, this can induce in material magnetic dipoles.

  And it, uh, induces magnetic dipoles at the atomic scale.

  Now in case that the atoms or the molecules themselves have a permanent magnetic dipole moment, then this external field will make an attempt to align these dipoles, and the degree of success depends on the strength of the external field, and again on the temperature.

  The lower the temperature, the easier it is to align them.

  So the material modifies the external field.

  This external field, today I will often call it the vacuum field.

  So when you bring material into a vacuum field, the field changes.

  The field inside is different from the external field, from the vacuum field.

  I first want to remind you of our definition of a magnetic dipole moment.

  It's actually very simple how it is defined.

  If I have a current -- a loop, could be a rectangle, it doesn't have to be a circle -- and if the current is running in this direction, seen from below clockwise, and if this area is A, then the magnetic dipole moment is simply the current times the area A.

  But we define A according to the, the vector A, according to the right-hand corkscrew rule.

  If I come from below clockwise, then the vector A is perpendicular to the surface and is then pointing upwards.

  And so the magnetic dipole moment, for which we normally write mu, is then also pointing upwards.

  And so this is a vector A, which is this normal according to the right-hand corkscrew.

  And if I have N of these loops, then the magnetic dipole moment will be N times larger.

  Then they will support each other if they're all in the same direction.

  I first want to discuss with you diamagnetism.

  Diamagnetism.

Pengembangan Perkuliahan

1. Buatlah sebuah Esai mengenai materi perkuliahan ini

2. Buatlah sebuah kelompok berjumlah 5 orang untuk menganalisis materi perkuliahan ini

3. Lakukan Penelitian Sederhana dengan kelompok tersebut

4. Hasilkan sebuah produk yang dapat digunakan oleh masyarakat

5. Kembangkan produk tersebut dengan senantiasa meningkatkan kualitasnya

Ucapan Terima Kasih Kepada:

1. Para Dosen MIT di Departemen Fisika

a. Prof. Walter Lewin, Ph.D.

(http://web.mit.edu/physics/people/faculty/lewin_walter.html)

b. Prof. Bernd Surrow, Ph.D.

(http://web.mit.edu/physics/people/faculty/surrow_bernd.html)

Staff

Visualizations:
Prof. John Belcher

Instructors:
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

Acknowledgements

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.