8.03 - Lect 16 - Standing EM Waves, Reflection, Transmission Lines, Rad. Pressure
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- Опубликовано: 17 сен 2024
- Boundary Conditions at Perfect Conductors - Reflection - Standing EM Waves - Transmission Lines - Radiation Pressure - Comets
Assignments Lecture 15 and 16: freepdfhosting....
Solutions Lecture 15 and 16: freepdfhosting....
For those who are doing PCB or are going to do PCB designs, these are valuable lectures that we should watch over and over again!
Nowadays, ICs are switching faster and faster that we can hardly image! Rise/fall time is now in the range of pico-seconds. PCB designers can properly handle this high speed layout only if they starts to think about energy and EM waves and it all goes back to physics! Energy is IN the space (dielectrics of PCB layers), not on the copper trace. The good old times of circuity theory dealing voltages and currents no longer helps you a lot.
I found these lectures so precious that I watch them over and over again through these years. As a student I first watched them just for fun while these videos were still available in MIT OCW. Now, as a hardware engineer, I still watch them because they help me better understand signal integrity issues!
Professor Lewin, I want to thank you! Thank you so much for recording all these lectures! Even you are retired now, but these videos are still available through channel and keep reforming people everywhere around the world!!
yeah totally. thats why im in on this. also he sounds like the count from sesame street
I currently do couple of pcb design work for a semiconductor company, and I pretty much agree on this one. Lots of pcb designers from where I'm working on doesn't know the laws of physics that fundamentally explains the behaviour of the signals as it moves around the pcb traces, and how signals affect others trace to trace. Designing pcb for high speed applications requires going onto more fundental level thinking than the usual abstracted level of thinking.
Dear prof as an African studying communication engineering i found your lectures helpful and understandable. I wanna thank you sir for the good work 👏 and humanity thinking.
You are very welcome
These are timeless resources from theory to practical .. Excellent teaching
:)
Walter, you are really a GOAT! (Greatest Of All Time)
Your Transmission Line Demo is very impressive!
Mother of god. If I had EM lectures like these my life whold be different.
infinitely valuable ! 📐📚🎓 Thank You ! a blessing indeed ! 📊 🔬
What an amazing experiment of Transmission Line,today I saw what I have studied and taught.
At 50:04 that is the velocity factor, we use it a lot in applications such as amateur radio. Just about very cable manufacturer lists this number, it varies quite a bit between air core, foam, teflon, etc, dielectric medium.
:)
often called VoP
velocity of propagation
Is that a bagel in his front pocket?
Hello. I believe that the problem sets linked in the description of every video is offset by one. For example, I believe that the problem set linked on this video, which says lecture 15 and 16, is actually for lecture 14 and 15. I think this has occurred for the other videos as well, and I have checked on the OCW website as well to confirm that they are indeed offset.
Prof. Lewin, When you mentioned in your lecture (32:30) there is no electric field in the conductors (two parallel wires), do you mean there is no electric field inside the conductor along the wire and radially? However, based on your drawing, we can still have electric field normal to the conductor outside surface (due to surface charges) as shown in your blackboard drawing. Furthermore, based on the same drawing, can we also say that there is tangential electric field on the conductor outside surface (not inside), since there is a spatially modulated distribution of positive and negative charges on the surface as the voltage pulse travels on the wire? There are no words to describe my appreciation for your life long contribution to physics education, you have really set the bar extremely high! Thank You!!!!!
if an Electric current runs through a conducting wire (e.g., copper) there is an E-field in the direction of the current , thus in the direction of the wire, *it is not perpendicular to the wire*
How does DC energy propagate from source to load? We know it does, but the E and M fields are not moving.
In the AC wire you showed, you said no electric fields along the wire, however, consider the upper wire: one section is +ve, the next section is -ve. So why isn't there a field from the first section to the second section along the same wire ?
how many minutes into the lecture?
@@lecturesbywalterlewin.they9259 After 31:35: I have the same question. You put plusses and minusses ALONG the line. So why isn't there an electric field between these (horizontally placed) charges? Bedankt voor uw reactie!
Great videos, and deep concepts .What i don't undertstand is that for a di-pole antenna, if the normal component of the E-field is only present on the antenna, how is it that it that the propagating E-field is polarised along the antenna ? Meaning the tangential E-components of the antenna is propagating ? Tangential E-component is not present on the antenna as shown in the video.
I thought this was a particularly good lecture of interest to me. The radio amateur exam takes a lot of this in but without masses of maths. I could see we were heading towards Lecher lines and coax cable with it's velocity factor slowing down the signal. I wonder how much power those valves produce. I bought a radiometer a few years ago and always wanted one after seeing one in the school physics lab about 50 years ago.
sir, would you please see is my statement correct? 55.50 If the opening space of the comb is vertical, the EM wave is hardly penetrate through the teeth of the comb and the EM wave will be reflected back. I assume the EM wave is reflected because the charge inside the comb will is induced due to the EM wave and produce a opposite direction of EM wave ( if the Incoming EM wave reach the comb is crest, the reflected EM wave will be a valley, because of what you tought in lec14 the direction of the reflected E must be opposite in sign to the perpendicular vector of the acceleration ). And because of the skin effect and the skin depth, the charge (which will change with time in order to create the reflected EM wave(current) ) will only at the skin of the teeth of the comb because of the boundary condition derive from the Gauss's Law. If the assumption above are correct, then i have one question left, the current which will only flow at the skin of the teeth of the comb will act like an accelerating charge which will create EM wave propagate through all direction except the top and bottom of the accelerating charge. Which means it will create an EM wave propagate to the receiver too. So eventually the receiver will receive the EM wave. Would you please give some explaination? please
when E-field is parallel to the teeth AC current will flow in the teeth which will produce heat at the expense of the energy in the E-field (i.e., E-field will not get through) . That will not be the case if the E-field is perpendicular to the drection of the teeth.
@@lecturesbywalterlewin.they9259 Thanks for replying me sir, so my assumption is correct right? And the reflected EM wave is the result of the ac current at the skin of the teeth right?
Would the following be an acceptable answer to 1.3 us delay puzzle? The delay is caused by the presence of both the inductance and capacitance between the shield and the wire in the coaxial cable. Both play an important role, for example, if we take the same wire with the same dielectric and roll it into tighter loops (so as to change the overall inductance), the delay will be changed as well even if the dielectric material and its thickness remain the same.
How many minutes into the lecture?
This is at 49:36, the puzzle that the calculated delay is 0.84us while the observed delay is 1.3us. Thanks!
I gave the explanation in my lecture.
Dear professor i have a question...we say that there is no electric field inside a conductor. but how electrons move inside a cable when we connect it to a battery?? high school physics books say that there is an electric field that moves those free electrons along the wires...
there is NO E-field inside a conductor in STATIC situations. However, if a current is going through a conductor OFCOZ there is an E-field inside.
Explanation about polarizers is awesome
Glad to hear that!
if you can shape your electrical surface in a path desired then can that electric surface be implemented as an additional conductor of electricity from a second source and treat that initial electric surface as a superconductor or very close to it? This technique I believe has already been implemented in an EM Apparatus cluster.
question unclear
Sir,
can't we find tangential E field components using pillbox geometry and (div D) ?
why we're being forced to loop geometry and curl E ?
how many minutes into the lecture?
Lectures by Walter Lewin. They will make you ♥ Physics. thank you sir.
at 4:38 we used pillbox geometry and div E to find normal component En. at 12:33 we used loop and curl E to get tangential component Et.
what is the reason for shifting from pillbox geometry to open surface (loop)? is it accordingto maxwells equation or it is impossible to find Et using pillbox and DIV E??
To calculate B you MUST use Faraday's Law.
Sir atv 30:55 you told that no Potential Difference/E.Filed between 2 points on the conductor horizontally, but later you populated the conductor with different potential, which will cause E Filed between two conductors and also between 2 consecutive positions on same conductor...
Kindly clear my confusion if possible...
Thank you
when??????
@@lecturesbywalterlewin.they9259 sorry sir for incomplete question...
At 31:45
@@muhammadqaisarali The E field is NOT along the wires but across the wires. integral of E dot dL is the voltage across the wires.
@@lecturesbywalterlewin.they9259 Heartiest thanks sir....
Still confused but think again and again upon it....💟💟
@@lecturesbywalterlewin.they9259 you say " but then ofcourse they also going to be locations where this is -- and this + " so iam confused you putting a vloltage diffrence in a perefect conductor horizontally !!! Please proffesor ineed help in this
Professor you ar a great blessing to all of us!!! Is it possible to write to you directly?
Loads of fun to watch these.
Professor, at about 30:53, you mention 'no potential difference between any two arbitrary points on a single wire.' Whereas at about 31:50, you show charge accumulation of different polarities at different segments on the same wire. I am wondering how these two facts go in line with each other. Will charge accumulation of opposite polarities not give rise to an electric field along the wire?
I watched again. what I said is correct. there is NO current flowing in horizontal direction in either wire. Thus there is NO potential difference between any 2 points of each wire. *EVEN though there is a variable potential diff between the wires*
@@lecturesbywalterlewin.they9259 Thanks for your response.
Professor Lewin - incredible lecture. If I'm understanding what you're showing in the transmission line experiment, I take it that you could've used more than one light bulb. How about if you used 100 light bulbs at different anti-nodes or not at the nodes. I would surmise that you would have a good amount of power out than what is being put into the system....Am I correct in my assumption ?
I wonder if it is healthy standing right next to about 50W RF radiation like that.
Is there a transcript of this lecture available anywhere? I find it so much easier to search for things in a document than in a video.
no there is not
I can't understand why it is in this case that the EM wave is completely reflected ( 54:57 ). Any chance you could give me the answer or a clue or something? Thanks.
as I told my students :you have to think about this"
@@lecturesbywalterlewin.they9259 Ok. Could it be that when the comb is parallel to the E component of the wave, electrons have a chance to move up and down the comb and so it behaves a lot like a metal sheet and reflects the wave?
@@santiagoarce5672 yes that is correct
16:00
Prof, when an electromagnetic wave interact with a conductor at the boundary, the normal component of E field will suck electron onto the surface, the charge deposition at each instant will be proportional to the instantaneous value of E field, this modulated motion of charge carrier seems to create a surface current, I'm I right Sir?, Is this how reception of EM wave at receiving antenna take place ?is this how antenna reception take place.. or am I wrong somewhere?
the wave may reflect or some of it may enter and some may go through. This all depends on the size and the kind material of the conductor and of the wavelength of the EM radiation. Charge motion (currents) could occur but if the EM wave mostly ionizes the conductor (and gets absorbed) there are no currents.
@@lecturesbywalterlewin.they9259
Thank you prof for the answer 😍😍😍,you are a great teacher 😍😍
@@lecturesbywalterlewin.they9259
57:00
prof,
when we place the comb in the path of propagating EM wave, such that the axis of the comb is parallel with E field variating direction, it functions as a linear polarisor.that is total electric field is reflected back to obey boundary condition (tang E field component is zero at that point), but the normal component of E field will suck electron in the axis direction (when we consider the conductor the tangential E component to the curved surface, assume cylindrical shape, is effectively the normal component for the axis direction )then this in turn act as a source of EM wave, so total E field =incoming waves E field + that wave generated due to the surface charge buildups due to the normal component, together yield zero E value at the boundary, this inturn intuitively proves that the there is a connection between the maxwell's individual equations (consistency check 😊)
Is this concept is right, or completely wrong ?
At about 30:49 you connect the wires with an ac source but then say that the e field is 0 inside the wires. How can that be the case if a current flows thru them?
i have the same question.. do you have an answer?
@@sammyapsel1443 there is no current flowing, the only point of the power source is to setup the surface charges
Thank you very much .
Thank you
You're welcome
Nice trick question at the end😂
heat engine
If all particles must travel at speed of light what is happening to the wave at 0.6 Vc . Is it taking a non linear path ?
No particles (with rest mass) can travel with the speed of light
In Transmission lines, he mentions that there is no E inside the copper wire. However, if E is zero in the copper wire, how are the free charges moving inside the copper wire then?
how many minutes into the video did I say that?
30:44
make the wire out of suerconducting material. Then there can not be an E-field inside the wire. But there will still be variable E-fields between the wires. Copper has a very very low ohmic resistivity thus the E field along the copper wire will also be very close to zero and can be assumed to be zero for all practical purposes.
Would the following explanation be correct for the comb demo at 55:56? When the rods are vertical, the EM wave creates electric current in each rod, and this current in each rod produces the EM wave of its own. Depending on (1) how far from each other the rods are in the comb and (2) where exactly we position the comb w.r.t. source, these re-radiated waves may cancel out the original wave (and thus, kill the signal). Alternatively, they can actually amplify the signal (like in Yagi antennas). On the other hand, it is clear that when the rods are horizontal -- almost no electric current is induced in any of the rods so no EM wave is re-radiated and the original wave is undisturbed.
>>> (1) how far from each other the rods are in the comb and (2) where exactly we position the comb w.r.t. source, these re-radiated waves may cancel out the original wave (and thus, kill the signal). >>>
====>The geometry is not as critical as you think.
>>>>On the other hand, it is clear that when the rods are horizontal -- almost no electric current is induced in any of the rods ====> correct
Thanks but why is geometry not critical? As an example: if a comb is made of just 1 vertical rod then this becomes a Yagi antenna where the rod plays a role of a "director". If properly sized and positioned this rod (as I understand it) can be amplifying the received signal rather than killing it.
sorry but i can't understand the connection between wave and particle talking about the nature of light.
the question is:
a photon is a particle that have one specific direction and live in "ONE dimension" instead a wave is something that live in "TWO dimension". how could it is possible?
thank you so much for your attention.
@Lectures by Walter Lewin. They will make you ♥ Physics.
57:00
prof,
when we place the comb in the path of propagating EM wave, such that the axis of the comb is parallel with E field variating direction, it functions as a linear polarisor.that is total electric field is reflected back to obey boundary condition (tang E field component is zero at that point), but the normal component of E field will suck electron in the axis direction (when we consider the conductor the tangential E component to the curved surface, assume cylindrical shape, is effectively the normal component for the axis direction )then this in turn act as a source of EM wave, so total E field =incoming waves E field + that wave generated due to the surface charge buildups due to the normal component, together yield zero E value at the boundary, this inturn intuitively proves that the there is a connection between the maxwell's individual equations (consistency check 😊)
Is this concept is right, or completely wrong ?
when the Efield is in the same direction as the teeth a current will flow in the teeth which is converted to heat. Thus the wave will be absorbed. If the E-field is perpendicular to the direction of the teeth this will not happen and the wave can go through. This is analogous to the structure of linear light polarizers.
@@lecturesbywalterlewin.they9259
thanks a lot prof, 😍😍😍
Sir
if the parallel comb (vertical here) is a matched antenna, that is vertically polarised, the comb would absorb E field instead of reflect it. Is it right??
how many minutes into the lecture?
Lectures by Walter Lewin. They will make you ♥ Physics. thank you for your kind response Sir.
at 56:00 the comb killed(reflected) the wave. if the comb was a vertically polarised antenna, it would absorb total wave instead of reflecting it, right??
another question is
why the parallel rods are responding(killing) the wave? is it just because the E field and metal rods are parallel(both are vertically polarised )?
if the E-field is in the direction of the rods, the free electrons in the conducting rods will start to move and remove all energy of incoming E-field. HEAT!
Lectures by Walter Lewin. They will make you ♥ Physics.
thank you sir 🙏
@@ajathreya5005 According to google it seems only a small amount (7%) of EM radiation is absorbed as heat, and the rest of the radiation is reflected? (at normal incidence)
sir,
at 26.45 minutes in the lecture: according to boundary conditions EM wave must be reflected from conducting cu wires when they are parallel to E-filed, but bulb is glowing in the antenna when antenna is parallel to E- filed. how is it possible? can elaborate it?
currents are generated in the metal. The wave is absorbed by the comb (ohm heating)
not clear sir. when metal wire is parallel to E field, then in which
condition metal wire reflects? in which condition metal wire absorbs
(ohmic heat)? in which condition metal wire generates current?
use google if my lectures are not clear enough for you
The metal inside the black board is reflecting the E- filed but metal wire in the antenna absorbing the E- filed. can you give some hints to understand this?. reflecting because of boundary conditions. but absorbing because of ???
Reflection and absorption in conductors are frequency dependent and they depend also on the conductivity. In the case of the demo with the antenna with a light bulb in the middle, E fields parallel to the antenna caused currents in the antenna and the light dissipated the currents I^2*R. E-fields perpendicular to the antenna rod did not light the bulb. In the case of the comb I believe most (if not all) energy was also absorbed. Please remind me what the frequencies were in the case of the antenna and of the comb. In what lecture did I mention that EM radiation is reflected off the metal in the blackboard? Lect # please and how many minutes into the lecture? We know that radar (1 GHz - lambda few cm - tens of cm)) reflects off most metals
farside.ph.utexas.edu/teaching/315/Waveshtml/node65.html
Dear professor
Which book did u follow for this course ?
8.03
Vibrations and Waves by
Anthony French
CRC Press
ISBN 9780748744473
8.03
Electromagnetic Vibrations, Waves and Radiation
by Bekefi and Barrett.
The MIT Press
ISBN 0-262-52047-8
sir boundary condition from the amperes law comes as bt*db=mu0*I,help me how I get your result shown in the lecture that bt=J
question unclear
how many minutes into the lecture?
Lectures by Walter Lewin. They will make you ♥ Physics. 15:47
apply Maxwell's eq closed loop integral of B dot dl to the circuit on the black board.
Lectures by Walter Lewin. They will make you ♥ Physics. 55:10,why in one case wave will reflected and another case it will go straight throughout???I thought but can't came up with the answer.
help me
you figure that out.
An Atom can be see like a standing EM waves ?
use google
Thanks Professor, to answer my question, You are great teacher.
hi proff . can you tell us the explanation for the experiment at the end of the lecture?
you have to figure it out
is that because black surface absorbs most light and turns it into heat?
why would it rotate???? there is still one key thing that you nay be missing.
i can't reach the solution and also i can't sleep ;) .
you can find the answer on google
It is interesting thing that in America we had the most scientists but why we spent money on jet fighter air plane cost all children elementary lunch money?
Sir may be in last demonstration there is nothing to do with light. May be you had given Power supply to a motor who was rotating the plates.
This is not my syllabus
Hello professor,
I have some doubts regarding transmission lines. Could you please check them once?
1) If there is no electric field along the transmission line (no current along the transmission line), then how can there be a magnetic field around the wire?
2) In a simple circuit with a resistor and a battery, the electric field is along the wire which causes the current flow. How this situation is different from that of transmission lines in which the electric filed is zero along the line as no electric field can exist inside a conductor?
Thanks!!
1) how many minutes into the lecture?
At 30:46
standing wave between 2 copper wires - zero current.
@@lecturesbywalterlewin.they9259 If we terminate the line with a resistor then current flows along the line ( mostly through the surface) and ther by the flow of energy. Is this correct professor?
Hello sir
The answer to the tricky question is due to the high reflectivity of the white surface it will behave as the open end as in the transmission line example. So light will bounce back as a mountain which results in the rotation in the white direction. "standing waves :)"
@Lectures by Walter Lewin. They will make you ♥ Physics
Thank you
You're welcome
@@lecturesbywalterlewin.they9259 you are the greatest teacher to have ever lived I'm my lifetime that's for sure.