Senin, 14 Februari 2011

Fisika untuk Universitas

Fisika untuk Universitas

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

Kelistrikan dan Kemagnetan

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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.

b. Prof. Bernd Surrow, Ph.D.


Prof. John Belcher

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

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

Technical Instructors:
Andy Neely
Matthew Strafuss

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


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

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

Terima Kasih Semoga Bermanfaat dan mohon Maaf apabila ada kesalahan.

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