## Rabu, 29 Desember 2010

### PUSTAKA FISIKA (PF)

1. Sebuah Visi Pengumpulan 100.000 Buah Buku yang terkait dg Fisika
2. Pengumpulan Data-Data Kefisikaan sebesar 1 Terra byte

Tempat Pengumpulan dan Pendataan Buku-buku fisika via Internet

Fisika Energi

• Hydrogen Energy and Fuel Cells
• Solar Hydrogen Generation
• Wind and Solar Power Systems
• Sustainability and Environmental Impact of Renewable Energy Sources

• Sumber:
FISIKA FOREVERMORE
Media Saling Berbagi Ilmu dan Informasi

## Minggu, 26 Desember 2010

### Fisika untuk Universitas

Fisika untuk Universitas

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

# 11: Work, Energy, and Universal Gravitation

Today, we will talk exclusively about work and energy.

First, let's do a one-dimensional case.

The work that a force is doing, when that force is moving from point A to point B--

one-dimensional, here's point A and here is point B--

and the force is along that direction or...

either in this direction or in this direction but it's completely one-dimensional, that work is the integral in going from A to B of that force dx, if I call that the x-axis.

The unit of work, you can see, is newton-meters.

So work is newton-meters, for which we...

we call that "joule.

If there's more than one force in this direction, you have to add these forces in this direction vectorially, and then this is the work that the forces do together.

Work is a scalar, so this can be larger than zero, it can be zero, or it can be smaller than zero.

If the force and the direction in which it moves are in opposite directions, then it is smaller than zero.

If they're in the same direction, either this way or that way, then the work is larger than zero.

F = ma, so therefore, I can also write with this m dv/dt.

And I can write down for dx, I can write down v dt.

I substitute that in there, so the work in going from A to B is the integral from A to B times the force, which is m dv/dt, dx which is v dt.

And look what I can do.

I can eliminate time, and I can now go to a integral over velocity--

velocity A to the velocity B, and I get m times v times dv.

That's a very easy integral.

That is 1/2 m v squared, which I have to evaluate between vA and vB, and that is 1/2 m vB squared, minus 1/2 m vA squared.

1/2 m v squared is what we call in physics "kinetic energy." Sometimes we write just a K for that.

It's the energy of motion.

And so the work that is done when a force moves from A to B is the kinetic energy in point B--

you see that here--

minus the kinetic energy in point A, and this is called the work-energy theorem.

If the work is positive, then the kinetic energy increases when you go from A to B.

If the work is smaller than zero, then the kinetic energy decreases.

If the work is zero, then there is no change in kinetic energy.

Let's do a simple example.

Applying this work-energy theorem, I have an object that I want to move from A to B.

I let gravity do that.

I give it a velocity.

Here's the velocity v of A, and let the separation be h, and this could be my increasing y direction.

The object has a mass m, and so there is a force, gravitational force which is mg, and if I want to give it a vector notation, it's mg y roof, because this is my increasing value of Y.

When it reaches point B, it comes to a halt, and I'm going to ask you now what is the value of h.

We've done that in the past in a different way.

Now we will do it purely based on the energy considerations.

So I can write down that the work that gravity is doing in going from A to B, that work is clearly negative.

The force is in this direction and the motion is in this direction, so the work that gravity is doing in going from A to B equals minus mgh.

That must be the kinetic energy at that point B, so that this kinetic energy at point B minus the kinetic energy at point A, this is zero, because it comes to a halt here, and so you find that 1/2 m vA squared equals mgh.

m cancels, and so you'll find that the height that you reach equals vA squared divided by 2g.

And this is something we've seen before.

It was easy for us to derive it in the past, but now we've done it purely based on energy considerations.

I'd like to do a second example.

I lift an object from A to B--

I, Walter Lewin.

I take it at A.

It has no speed here; vA is zero.

It has no speed there.

And I bring it from here to here.

There's a gravitational force mg in this direction, so the force by Walter Lewin must be in this direction, so the motion and my force are in the same direction, so the work that I'm doing is clearly plus mgh.

So the work that Walter Lewin is doing is plus mgh when the object goes from A to B.

The work that gravity was doing was minus mgh--

we just saw that.

So the net work that is done is zero, and you see there is indeed no change in kinetic energy.

There was no kinetic energy here to start with, and there was no kinetic energy there.

If I take my briefcase and I bring it up here, I've done positive work.

If I bring it down, I've done negative work.

If I bring it up, I do again positive work.

1. Para Dosen MIT di Departemen Fisika

a. Prof. Walter Lewin, Ph.D.

b. Prof. Bernd Surrow, Ph.D.

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

Terima Kasih Semoga Bermanfaat dan mohon Maaf apabila ada kesalahan.

## Rabu, 22 Desember 2010

### Fisika untuk Universitas

Fisika untuk Universitas

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

Kelistrikan dan Kemagnetan

Topics covered:

Electromagnetic Induction
Lenz Law
Complete Breakdown of Intuition
Non-Conservative Fields

Instructor/speaker: Prof. Walter Lewin

### Video

I'd like to thank you for your evaluations.

They were very useful to me.

I already sent e-mail to about 50 students and I had some interesting exchanges with some of you.

Many of you are very happy with their recitation instructors.

That's great.

Many are moderately happy.

Maybe that's OK.

But there are quite a few who are very unhappy with their recitation instructors.

If you are very unhappy with your recitation instructor, you are complete idiots if you stay in that recitation.

We have 13 recitation instructors, and I can assure you that it will be very easy to find one that agrees with you, and you can come and see me if that helps.

Some are better than others.

That's the way it goes in life.

Some students would like to see more cut-and-dried problem solving in my lectures.

I think that's really the domain of recitations.

Lectures and recitations are complementary.

In lectures, I prefer to go over the concepts and I always give numerical examples to support the concepts -- in a way that's problem solving -- and I show demonstrations to further support the concept, because seeing, obviously, is believing.

I try to make you see through the dumb equations and admittedly my methods are sometimes somewhat different from what you're used to here at MIT.

I try to inspire you and at times I try to make you wonder and think.

And I want to keep it that way.

I believe that hardcore probling- problem-solving is really the domain of the recitations.

Many of you found the exam too easy, and many of you found the exam too hard.

Some complained it was too hard because it was too easy.

[audience laughter] Quite ironic, isn't it?

They say we want more math, we want more standard problems.

Look, who wants more math?

I'm teaching physics.

I test you physics, I don't test you math abilities.

If you digest the homework, and that's very important that you make the homework part of your culture, that you study the solutions.

The solutions that we put on the web, today, 4:15, solutions to number four will go on the Web.

Believe me, they are truly excellent solutions, not cut and dried.

They give you a lot of background.

If you digest those solutions, then the concepts will sink in.

And now, at your fifty minute test, do you really want problems which are complicated maths?

Clearly, not.

I could try that, during next exam, but then I may have to buy myself a bullet-proof vest to be safe.

Concepts is what matters.

When I gave my exam review here, I highlighted the concept.

Each little problem I did here was extremely simple.

Conceptually, they were not so simple.

But from a math point of view, trivial.

Clearly, I can not cover all the subjects in a fifty minute exam.

I have to make a choice, so your preferred topic may not be there.

Some of you think that the pace of this course is too slow.

Some of you think it's too fast.

The score, the average score, was 3.8.

4.0 would have been ideal.

What do you want me to do?

I can't accommodate all of you.

Those who think it's too slow, and those who think it's too fast.

3.8 is close enough to ideal for me, 4.0.

And so I'll have to leave it the way it is.

Besides that, keep in mind you are now at MIT.

You're no longer in high school.

Now the good news.

There were quite a few students who said the homework is too long.

Not a single person said it was too short.

I can fix that.

I will reduce all future assignments by about 25%, effective tomorrow.

I have already taken off assignment number five, two problems.

You're down now to seven, and I will do that, all assignments that are coming up.

My pleasure.

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

1. Para Dosen MIT di Departemen Fisika

a. Prof. Walter Lewin, Ph.D.

b. Prof. Bernd Surrow, Ph.D.

#### 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

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.

## Selasa, 21 Desember 2010

### Fisika untuk Universitas

Fisika untuk Universitas

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

Kelistrikan dan Kemagnetan

Topics covered:

Ampere's Law
Solenoids
Revisit the Kelvin Water Dropper
Midterm Evaluation

Instructor/speaker: Prof. Walter Lewin

### Video

We've seen last time that using Biot and Savart's formula that if you have a current going straight into the blackboard perpendicular to the blackboard that we get a magnetic field at a distance R.

The magnetic field tangentially to the circle, B here, B here, and that the strength of that magnetic field equals mu 0 times I divided by 2 pi R.

If you walk around this circle, just walk around, and you carve up this circle in little elements dL, and you calculate the closed circle integral, so the closed circle of B dot dL, so everywhere locally you dot B with dL, the B and dL are in exactly the same direction everywhere, then you would find that this obviously is B times 2 pi R.

But B times 2 pi R equals also mu 0 times I.

This dL here has nothing to do with this dL here.

Don't confuse the two.

This dL is a small amount of length in the wire that goes into the blackboard which carries a current.

This dL is simply your dL when you walk around this current wire.

It doesn't matter at what distance you walk around.

You always get mu 0 times I.

You see it right in front of your eyes because B is inversely proportional to R.

And it was Ampere who first recognized that you don't have to walk around in a circle to get the answer mu 0 I, but that you could walk around in any crooked path as long as it is a closed path, something like this.

And now you have here the local B, which of course is perpendicular to this radius and here you have your local dL and if now you go around, closed circled any path, it doesn't have to be a circle, dot dL that now becomes mu 0 times I, which is known as Ampere's Law, and I then is often given an index enclosed.

It is the current which is enclosed by that path.

It is actually easy to prove this using Biot and Savart's formalism.

This is almost a third Maxwell's equation.

This is almost number three, not quite.

We're going to amend it in the future.

What is ill-defined a little bit in this equation is what we mean by enclosed, and I'm going to define that now so uniquely that there is never any misunderstanding.

If I have a very strange looking closed path that I have chosen, that's the path I walk, I have to attach to that closed loop a surface, an open surface.

That's mandatory.

You can make it flat.

That's fine.

You're free to choose it.

You can also make it sort of a plastic bag so it's open here.

You can put your hands in here, and here, like a hat.

Any surface is fine, but you must attach to that loop a surface, so here I have some path that you could be walking, and this would be perfectly fine open- open surface.

Could be flat, but it could also be open, so it's like a hat.

And now I can define uniquely what it means by- what it means by this I enclosed, because if now I have a current that goes through this surface and pokes out here, then I have a current penetrating the surface and that is uniquely defined, and if I have another one coming in through the surface, call it I2, this is penetrating that surface.

By convention, if you go clockwise around, we follow the same notation that we had before, in the right-hand corkscrew notation, the connection between magnetic field and current.

If you go around clockwise seen from this side, so you go clockwise, then I1 as I have it here, in your equation would have to be larger than 0.

I2 is then smaller than 0.

But if you decide to go counterclockwise, which is perfectly fine, Ampere's law doesn't at all dictate in which direction you have to march around, then I1 would be negative and then I2 would be positive.

So we follow the right-hand corkscrew notation.

And so if you want to amend now Ampere's Law to do me a favor, but you don't do books a favor because all the books use the word enclosed.

I would like to see this replaced by penetration.

It is the penetration of the surface of the current, that is uniquely defined.

But a current enclosed by a loop is ill-defined.

Because where possible, when we apply Ampere's Law, we will try to find easy passes around circles sometimes, sometimes rectangles, and since you are free to choose the surface that you attach to the loop if you can get away with it you use a flat surface, but you cannot always get away with a flat surface.

So the recipe is as follows.

Any loop is allowed.

It may not help you very much if you choose the wrong loop.

Any loop is allowed.

You then attach an open surface to that loop.

And I penetrate is now the current that penetrates through that surface, according to this convention.

And the direction of rotation is free to you.

How you go around the path is your choice, but that defines then the sign of the penetrating of the curve, of the- of the current, according to the right-hand corkscrew.

So now we can, for the first time, calculate the magnetic field inside a wire that draws a current using Ampere's Law.

I have here a wire that has a radius capital R and a current is coming to me, I, and let's assume that the current is uniformly throughout the wire, so it has a uniform current density.

And I would like to know what the magnetic field is everywhere.

Cylindrical symmetry, I want to know outside the wire and I want to know inside the wire.

Let's first look at radius which is larger than R, and so here we have the cross-section of that wire, radius R.

The current I is going through this surface.

I now have to choose a closed path.

Since we have cylindrical symmetry it is clear that we would choose a circle, with radius little r, so we can be sure that the magnetic field strength is the same everywhere because of reasons of symmetry.

Since the current is coming towards me and I am free to choose in which direction I'm going to march, I know that the magnetic field is in this direction, so I might as well also march in this direction so that my dL's are all in this direction.

I don't have to do that.

I could march the other way around, but if I march counterclockwise then both terms left and right of Ampere's Law will be positive.

I now have to attach an open surface to my path.

Well, this will be, the blackboard will be, that open surface.

And so now I apply Ampere's Law, so I get B times 2 pi little r, because dL and B are in the same direction so it's a trivial integral.

That now equals mu 0 times I, which now penetrates my surface.

Uniquely determined, all these current from this wire that comes to me penetrates my surface, so times I, and so B equals mu 0 times I divided by 2 pi R, and that's the same result that we found last time, when we applied Biot and Savart.

So that's no surprise that you see this.

But now we have a way of finding the magnetic field also inside the wire, so here we have the wire again, the cross-section, current coming out of the blackboard, and now I want a radius which is smaller than capital R and of course my closed path again for reasons for symmetry is going to be a circle with radius R.

And my surface that I attach is a flat surface, and so here I go, B times 2 pi little r equals mu 0 times- ah, now I have to be careful, because now not the full current I is now penetrating my surface, but it is only a fraction that penetrates the surface, and the fraction that penetrates the surface is now little r squared divided by capital R squared times I.

You see, because the total current comes through the radius capital R, but I only have now a circle with radius little r.

And so I lose one r here and so we get a very different result.

You get now that the magnetic field equals mu 0 times I is now linear in little r divided by 2 pi capital R squared.

And this grows linearly with r, whereas this falls off as 1 over r.

And if you substitute in this equation r equals capital R, which then would be the magnetic field right at the surface of the wire, you find exactly the same result here.

Little r becomes a capital R.

If little r becomes a capital R, you lose one capital R, you get the same result.

And so if you make a plot of the magnetic field as a function of little r, then it looks like- like so, so this is little r, this is capital R, and this is the magnetic field strength because we know that it is tangentially to the circles.

It would be straight line and then here it falls off as 1 over r, and the maximum value here is the value that you find there if you substitute little r equals capital R.

I will now show you that we can, using Ampere's Law, also come very close to calculating the magnetic field inside what we call solenoids.

Solenoids is like a slinky current that goes around in a spiral, one loop after another.

I want to remind you that if we had a loop, a nice current loop coming out of the blackboard here, and the current going into the blackboard so there's a circular wire but I only show you the cross-section.

I want to remind you that the magnetic field as we discussed last time, would be clockwise here, would be counterclockwise here.

In the middle, remember, it was like this.

And then in-between it was like so.

That was sort of the magnetic field configuration in the vicinity of a loop through which we have a current going.

But now imagine that you put another loop here, current again coming out of the blackboard, going into the blackboard, and another one, and so on, several.

What do you think is going to happen with these magnetic field lines which now diverge?

They're going to be sucked in here.

This loop also wants the field lines to come through its circle, so to speak, and this one too, and so you're beginning to get a near-constant magnetic field and the more tightly these loops are wound, the more accurately will your magnetic field be approximately constant, and I have some transparencies which will show that in more detail.

Here we have a figure from your book.

You see five windings, a spiral.

If you look from the left, the current is going in clockwise direction, and so the magnetic field is going from this side to that direction.

And when you look here you see that the magnetic field is approximately constant inside, and outside these current loops, outside the solenoids -- we call them solenoids -- magnetic field is extremely low.

And if you start winding these loops very tightly, then you get a configuration looks like this.

You get an almost perfect constant magnetic field inside the solenoids and the magnetic field outside the solenoids is extremely weak.

And now I would like to calculate with you using Ampere's Law what that magnetic field inside such a solenoid would be.

And we have to make a few assumptions.

Let this be my solenoid, and the length of the solenoid is capital L.

A current I is going through like so.

I, and I assume that if I look from the left side that the windings are wound clockwise, so I know that the magnetic field is then in this direction.

I make the assumption that the magnetic field outside the solenoid is approximately 0.

I will show you later with a demonstration that that's a pretty good approximation.

And so the question now is, what is the magnetic field there.

And I assume I have N loops, N windings, capital N.

So now I have to choose a path.

I have to apply Ampere's Law.

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

1. Para Dosen MIT di Departemen Fisika

a. Prof. Walter Lewin, Ph.D.

b. Prof. Bernd Surrow, Ph.D.

#### 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

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.

## Minggu, 19 Desember 2010

### PUSTAKA FISIKA (PF)

1. Sebuah Visi Pengumpulan 100.000 Buah Buku yang terkait dg Fisika
2. Pengumpulan Data-Data Kefisikaan sebesar 1 Terra byte

Tempat Pengumpulan dan Pendataan Buku-buku fisika via Internet

GeoFisika

• The Thermodynamic Universe
• Understanding The Universe

• Sumber:
FISIKA FOREVERMORE
Media Saling Berbagi Ilmu dan Informasi

## Sabtu, 18 Desember 2010

### Fisika untuk Universitas

Fisika untuk Universitas

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

# 10: Hooke's Law, Simple Harmonic Oscillator

LEWIN: You did very well on your first exam.

I was hoping for an average of about 75; the class average was 89.

So that leaves us with two possibilities: either you are very smart, this is an exceptional class, or the exam was too easy.

Now, this exam was taken by three instructors way before you took it.

None of them thought it was too easy, so I'd like to think that you are really an exceptional class and I'd like to congratulate you that you did so well.

Here is a histogram of the scores.

If we had to decide on this test alone--

the dividing line between pass and fail would be 65.

That means that five percent of the class would fail, which is unusually low.

Normally that is around 15%.

But time will tell whether you are indeed exceptionally smart or whether the exam was too easy.

The good news also is--

two pieces of good news--

that we promise that the books will arrive at the Coop today.

one of the key topics in 801.

If I have a spring...

and this is the relaxed length of the string... spring, I call that x equals zero.

And I extend the string...

the spring, with a "p," then there is a force that wants to drive this spring back to equilibrium.

And it is an experimental fact that many springs--

we call them ideal springs--

for many springs, this force is proportional to the displacement, x.

So if this is x, if you make x three times larger, that restoring force is three times larger.

This is a one-dimensional problem, so to avoid the vector notation, we can simply say that the force, therefore, is minus a certain constant, which we call the spring constant--

this is called the spring constant--

and the spring constant has units newtons per meter.

So the minus sign takes care of the direction.

When x is positive, then the force is in the negative direction; when F is negative, the force is in the positive direction.

It is a restoring force.

1. Para Dosen MIT di Departemen Fisika

a. Prof. Walter Lewin, Ph.D.

b. Prof. Bernd Surrow, Ph.D.

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

Terima Kasih Semoga Bermanfaat dan mohon Maaf apabila ada kesalahan.

## Rabu, 15 Desember 2010

### Fisika untuk Universitas

Fisika untuk Universitas

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

Kelistrikan dan Kemagnetan

Topics covered:

Biot-Savart Law
Gauss' Law for Magnetic Fields
Revisit the "Leyden Jar"
High-Voltage Power Lines

Instructor/speaker: Prof. Walter Lewin

### Video

Well, we have a current going through a wire, like so.

And we look at the magnetic field in the vicinity of this wire , then we know from experiment that if you put pieces of magnetite around the wire that they line up in a circle.

Put around like this.

If that circle has a radius R, then the magnetic fields, that's an experimental fact, is proportional with the current I and is inversely proportional with the radius of that circle.

By convention, the direction of the magnetic field is given by the right-hand corkscrew, rotate this way, the current goes up.

You've seen before, with electric charges, when you have a wire which is uniformly distributed say with positive charge, you've also seen that electric fields in the vicinity of that straight wire falls off as 1 over R, whereas the direction is different than the magnetic field but it also falls off as 1 over R, and the reason is that electric monopoles, individual charges, the electric field falls off as 1 over R squared.

And so when you integrate that out over a straight wire you get the 1 over R field.

So by analogy, it would be very plausible that if you took magnetic monopoles that the magnetic field would also fall off as 1 over R squared, but magnetic monopoles as far as we know don't exist.

In principle they could exist, but we've never seen one, and if any one of you ever find one, that would certainly be a Nobel Prize.

It's by no means impossible.

And so the simple fact that the magnetic field around a current wire falls off as 1 over R, sort of suggests that if you carve this wire up in little elements dL, that each one of those elements contributes to the magnetic field in an inverse R-squared law, and by integrating out over the whole wire you'd then get the 1 over R fields.

And this behind the idea of the formalism by Biot and Savart, who introduced the idea that if you have a little current element dL, and the current is in this direction, and you want to know what the magnetic field is, This is small contribution dB to that little current element, and the distance is R, and the unit vector from the element dL to the point where you want to know the magnetic field is R roof.

Then the idea is that dB, it's a little bit of curr- little bit of magnetic fields.

In this case it would be in the blackboard because of the right-hand corkscrew rule.

The current is in this direction, so these little elements would contribute to magnetic fields in this direction perpendicular to the blackboard.

Is some constant, proportional to the current no doubt, and then is proportional to the length of that little element dL, if it's longer then the magnetic field is larger, and in order to get the direction right perpendicular to the blackboard you take the cross product with the unit vector R.

The unit vector R has length 1 so you only do that in order to get the direction right.

And this, and that inversely proportional to R squared.

That's of course key.

And this is, the formalism by Biot-Savart and you can do experiments and measure the magnetic field in the vicinity of wires and this formalism works, so you then calculate the individual contributions of all these little elements dL and then you do an integration and this formalism works.

You can then also measure what C is, in SI units, C is 10 to the -7.

But we write for C something quite peculiar.

We write for C mu 0 divided by 4 pi, and we call this mu 0 the permeability of free space.

You've seen earlier with Coulomb's law that this constant 9 times 10 to the 9th, we call that 1 over 4 pi epsilon 0.

What is in the name?

And so here we call this mu 0 divided by 4 pi.

So now you can apply Biot-Savart's Law and you can go to a straight wire and you have a current I, and suppose you want to know what the magnetic field at that location P is at a distance capital R, and so what you now have to do, is you carve this up, in an infinite number of small elements dL, and this distance is R, and the unit vector is then like so, and you calculate the small amount of magnetic field due to this little element and you integrate this over the whole wire.

It's mathematics.

You've done it.

You've done it before, where we had uniformly electric charge on the wire.

So I'm not going to do this again for you.

It's a very straightforward piece of mathematics.

The magnetic field by the way, in this case, would come out of the blackboard.

Because of the right-hand corkscrew rule.

And what you find when you do this, we will find that B equals mu 0 times I, divided by 2 pi R, this being R, and so you indeed see that the inverse 1 over R comes out.

And so if you, for instance, take a radius of 0.1 meters, 10 centimeters, and you have a current through the wire of about 100 amperes, then you would end up with a B field.

You use this equation, 2 times 10 to the -4 tesla.

100 amperes.

10 centimeter distance is only 2 gauss.

The Earth's magnetic field is half a gauss.

So if you go 1 meter away from the wire, so we have a magnetic field which is 10 times lower, look, it goes with 1 over R, then the magnetic field of the Earth already dominates substantially.

So you need very high currents, actually, when you do these experiments.

It's nice to see that out of Biot-Savart's formalism the 1 over R pops out, but of course you must realize that Biot-Savart knew that the magnetic field falls off as 1 over R.

That was an experimental fact.

So the fact that it falls out is logical, because it was cooked into that formalism.

If you think about it, it all goes back to Newton.

Newton was the one who first suggested that the gravitational field falls off as 1 over R squared.

And then later a logical ex10sion was that the electric fields would fall off as 1 over R squared and out of that came the idea that the fields of magnetic monopole, if they only existed, would fall off as 1 over R squared, and that's all behind this and so the person who really deserves most of the credit for all this in my book is Newton.

Using Biot-Savart, we can calculate now quite easily the magnetic field at the center of a current loop.

Let this be a wire circle and let the current go in this direction, and I would ask you what is the magnetic field right at the center.

Well, the magnetic field right at the center of course is pointing upwards.

Each little element along the line here, dL, each little element will contribute a little bit magnetic field at that point right in this direction.

And if this radius is R, with Biot-Savart now, we can calculate quite easily the total field that you would get at this location, because that total field is then the integral of dB vectorially over the entire wire so the entire loop...

So if you go there, so you would get your mu 0, divided by 4 pi, you get your current and you get your 1 over R squared, and now you have to do an integral over that dL cross R.

Well, R is of course always perpendicular to dL.

Any element dL that you choose, the unit vector R is exactly perpendicular to the element dL, because that's characteristic of a circle.

And so the sine of the angle between dL and R is 1, and so all we have to do is do an integral over dL, which is the integral of the circle, which is the circumference of the circle, and that is 2 pi R.

And so now you find, you lose a pi, you lose an R, so you find mu 0 times I divided by 2R.

Just to show you an example, how in this case how easy it is to use Biot-Savart and calculate the magnetic field right at the center.

If you were asked what the magnetic field was here or there, that would be also relatively easy.

You've done that.

I've given you a problem earlier where we had point charges uniformly distributed on a wire and I asked you what the electric field was here.

So that can also be done now with magnetic fields.

If I ever asked you what the magnetic fields would be here, that of course is an impossibility to do that with Biot-Savart, practically an impossibility.

I wouldn't know how to do that.

But in principle it could be done and certainly with a computer you can do it.

So we can go to our same situation, we can take 100 amperes for I and you can take R 0.1 meters and then the B field, the strength of the B field right at the center of this loop that I found is then 6 times 10 to the -4 tesla.

And that would be 6 gauss.

It's clear that if you want to put in some field lines, magnetic field lines, as a result of this current going around in a circle, that the- through the center there would be a field line like so.

If you're very close to the wire here, which goes into the blackboard, I want you to see this three-dimensionally, then the magnetic field would go like this, clockwise.

Here the current comes to you, so it would be counterclockwise.

If the magnetic field line is here like so, and here it is curled up, then clearly I expect them to be here, sort of like so, and like so, and like so.

This is the kind of magnetic field line configuration that I would expect, then, in the vicinity of such a current loop.

And I want to show this to you in a little bit more detail.

I have here a transparency, and you see there on the right side, current goes into the paper and here it comes out of the paper.

That is a circular loop.

And you see here the field line configuration.

It's not too different from what I have on the blackboard there.

Very close to the wires, of course, you get circles because the 1 over R dominates there.

It's so close to the wire that the 1 over R relationship makes it come out like circles and here too, but then if you're farther away you get configurations like I have there.

When you're very far away from a current loop, the magnetic field configuration is very similar to that of an electric dipole.

I can show you that in the following way.

Let's first look at the electric dipole that you see up there.

This is a positive charge, this is a negative charge.

Don't look anywhere near the charges.

Don't look in between the charges.

Look far away.

Here you see electric field lines and you see them here.

Now look at your current loop here.

The current is going into the paper here, coming out of the paper.

There is a loop.

And look, you see the same configuration, field lines, field lines.

This goes like so.

This one goes like so.

Here, the electric field lines coming in, magnetic field lines are coming in.

Electric field lines are going out.

Magnetic field lines are going out.

They look very similar.

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

1. Para Dosen MIT di Departemen Fisika

a. Prof. Walter Lewin, Ph.D.

b. Prof. Bernd Surrow, Ph.D.

#### 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

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.

## Jumat, 10 Desember 2010

### Fisika untuk Universitas

Fisika untuk Universitas

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

Kelistrikan dan Kemagnetan

Topics covered:

Moving Charges in B-fields
Cyclotron
Synchrotron
Mass Spectrometer
Cloud Chamber

Instructor/speaker: Prof. Walter Lewin

### Video

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 that's clearly not possible.

The actual speed, if you make relativistic corrections, is 2.6 times 10 to the 8th meters per second.

And although I don't expect you to be able to make those relativistic corrections, I will make them today and you will see why I have to, and I want to show you that in fact this is not all that difficult even though I will not hold you responsible for these equations.

So what I have here is now kinetic energy, is again QV, that's not changing, but is no longer one-half MV squared but it is gamma minus 1 times MC squared, and gamma is defined there -- it's called the Lorentz Factor, and so if you know now for the electron that capital V is 500000, you can calculate what gamma is from the first equation and then you go to the second equation and you find what the speed is, and you will see then that you never find a speed larger than the speed of light.

And so we now have to make the correction also for the radii and those corrections become again relatively easy.

This now requires a factor gamma and you see that on the upper blackboard there, and this too now has to be replaced by this gamma plus 1 and then everything is OK.

So I don't expect you to know this, but I don't want you to think that all these relativistic corrections come out of the blue, nor do I want you think that it is very difficult.

It really isn't.

The equations are extremely straightforward.

So I want to show you now some of the results that we just discussed.

The 1 MeV proton and the 500 KeV electron, this is on the Web.

You can click on Lecture Supplements and you can make yourself a hard copy.

So here you see the kinetic energy, 1 MeV proton.

Notice the speed that we calculated there is non-relativistic, gamma is very close to 1.

You don't have to make a correction.

And in a 1 tesla field you get a radius of 15 centimeters, which we just calculated.

If you go to a 50 MeV proton, it's sort of in the borderline between relativistic and non-relativistic.

It's still non-relativistic enough, and if it is non-relativistic you can clearly see here that the radius goes with the square root of capital V.

And for 50 MeV, capital V is 50 million, and for 1 MeV, capital V is 1 million.

And since it goes with the square root of V, you expect roughly the radius to be the square root of 50 times larger, which is 7, and indeed, you see that.

So you see, from 15 centimeters the radius goes to about 1 meter.

Um, here is our 500 KeV electron, and notice that I did the calculation correctly.

This is relativistically corrected now.

You get your 2.6 times 10 to the 8th meters per second by applying the formalism that you see there.

I will leave this here throughout this lecture, because I will return to this several times.

I want to show you a cute demonstration.

I have an, er, electron gun here and the electron gun comes like so.

This is the velocity of the electrons.

I put a minus sign there to remind you that they are electrons.

If electrons go in this direction, the current goes in that direction.

And so if now I have a magnetic field which, let's assume the magnetic field is in the blackboard.

This is B.

Then I cross B is the direction of the force.

I is in this direction, B is in the blackboard.

So if I'm not mistaken, I think the force is in this direction and so you will see that it starts to bend in this direction.

If you change the direction of the magnetic field, the magnetic field is coming out of the blackboard, then the electron will go in this direction, and I can show you that here.

It is not too different from the distortion experiment I did when I had the television program there and I had the strong magnet and we distorted the image, but this of course is a little bit more controlled.

So, we're going to see the image there and we want to make it quite dark in the room.

Mmhm.

And turn on the electron gun.

So you see, the electron gun it strikes a fluorescent screen and that's how you can see it, and I have here a bar magnet and if I hold the bar magnet behind it then I can create more or less situations like this.

I can flip over the magnet and then the direction of bending should change, so here I come with the magnet, and you see, curve up the electrons.

I turn the magnet over and I come in again and they curve down.

Very straightforward, very simple.

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

1. Para Dosen MIT di Departemen Fisika

a. Prof. Walter Lewin, Ph.D.

b. Prof. Bernd Surrow, Ph.D.

#### 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

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.