8.02x - Lect 20 - Inductance, RL Circuits, Magnetic Field Energy
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- Опубликовано: 13 фев 2015
- Inductance, RL Circuits, Magnetic Field Energy,Nice Demos
Lecture Notes, Faraday's Law - Most Physics College Books have it WRONG!: freepdfhosting.com/0813df09f5.pdf
Lecture Notes, Driven L-R Circuits: freepdfhosting.com/d83e186b92.pdf
Assignments Lecture 20, 21 and 22: freepdfhosting.com/b7a29b6b50.pdf
Solutions Lecture 20, 21 and 22: freepdfhosting.com/c27e7839e3.pdf 
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This website contains all my 94 course lectures (8.01, 8.02 and 8.03) with improved resolution. They also include all my homework problem sets, my exams and the solutions. Also included are lecture notes and 143 short videos in which I discuss basic problems.
ENJOY!
Great lectures. On a finer note, the E-field doesn't have to be zero inside an ideal conductor! As Feynman points out in his famous "Lectures on Physics" book Ch-22 , if the conductor is moving then you can have E-fields inside. He gives the example of a rotating coil in a magnetic field.The only requirement is that the total force is zero inside an ideal conductor. And the force is a sum of E-field and B-field components.
Prof. Lewin First and foremost, thanks for the physics education!!
I would like to comment on your lecture on "RL CIRCUITS". I noticed that in the light-bulb experiment where it was shown that the light-bulb had great delay coming to full brightness when switched on through the huge inductor...that it was NOT shown by demo, the similar DELAY of the light-bulb going off.
I just wanted to let you know that I have come up with a VERY SIMPLE modification of the circuit where the delays in BOTH directions are illustrated!
ALSO--as mentioned in that lecture--I am VERY curious as to the construction of such a large inductor with this very low resistance.
>>>NOT shown by demo, the similar DELAY of the light-bulb going off.>>>
NO and that is by choice. Look at the circuit!!!!! You may then understand why.
thank you walt, you've gone down in the annals of history in education as one of the great educators of one of the strangest fields in human existence.
Which i book follow for this cource?
Mr. Lewin, glad to witness your presence, thank you for existing. I am really enjoying your lessons I will probably learn mathematics again because of you and these lessons.
Thanks a lot Dear Sir!!
It's really remarkable to see how we young students in India can learn from a maestro like you.
You've helped us in infinitely many ways
Respect and Love from the heart!!
- Ayushman
Really sir you are awesome
Amazed to see such a beautiful explaination that how nature maintains electric field by charges at junctions of different resistors in the document given in description.
For anyone wondering what music Professor Lewin plays for the demonstration, it's J.D.Hienichen -Siebel 215. The exact point of time in the piece that Professor plays is the starting of the 4th movement(Allegro).
This professor is the magician of physics 😄
The best thing about it is that these lectures were filmed before I was born, and now I can benefit from them in a wonderful way. Thank you
same for me !!
Thanks for the comprehensive explanation of Faraday's law. I appreciate it.
Hello Dr Lewin! After failing to understand how inductance in circuits work after so long, I finally stumbled on your video...Thank you very much for your insight on the KVL! Your lecture videos are a treasure (I wish I had gone through them before picking up an Electrodynamics book 2 years ago, but better late then never haha)
Greetings from Germany :)
Electromagnetism is a huge subject for me! Thank you for presenting it with such clarity!
My pleasure!
As Mech E major, I used to hate physics. Professor Lewin has really sparked my interest in this subject.
Mine too 🤩🤩
And hence this lecture has demystified why impedances are complex for me. It all comes from the fact that inductors can produce lags in the current, which is rooted in Lenz's law! That rewrite of Kirchoff's voltage law into integral of Edl = -dphi/dt is an eye-opener. Amazing, thanks professor.
Professor Lewin,
Can we conclude that as a result of the induced B field inside the solenoid (and it being zero outside) and the current passing through it, an induced Lorentz force acts in the outwards direction, everywhere on the solenoid wire?
Thank you so so much for these lectures. They have been a true inspiration!
Why when in the 30:13 min you turned off the power energy of the bulbs both died at the same time, instead of having the one with the inductor a bit more time ON (as it is fighting against the change in the current from I_max to zero)? Thank you Sir :)
Prof. Lewin, I thought that the induced emf in a self inductor is always 90 degrees out of phase with the current through it. If so, how is it the case that the phase angle is a function of omega and L (not constant 90 degrees).
Thank you :)
+Yotu Watch this lecture ruclips.net/video/FWMhk6x785Q/видео.html
In principle you're absolutely right in pointing out that the KVL not holding for varying current. But don't we have to add the 'parasitic' self inductance of the circuit loop(L'), which we would have if we replace the coil with a straight (superconducting) wire, to the self inductance of the coil(L)? And that is, I think, even after moving the term involving L to the other side of the equation, makes the deviation from the conventional KVL argument.
You do have the self-inductance of the loop as well, but it is most likely either negligible, or assumed embedded in the inductor.
The conventional KVL method of solving these problems is mathematically correct, but physically inaccurate. It is similar to using pseudo-forces as shortcuts to solve mechanics problems. You can get the right answer, but with a method that doesn't reflect what is really happening.
Somewhere I have seen that you shared a list of recommended books but where is it now I am not able to find it out. Please share here.
If the magnitude of the self-induced emf of an inductor is equal to L*di/dt, then how can there be no "voltage drop" through the inductor and how can there be zero E field? I read the link in the description. All I could conclude from the article was that Faraday's law cannot tell us at what points the E field is zero or nonzero. The article simply states that E = 0 in an inductor. In the explanation of why that is so, I didn't see it mention E = 0 anywhere. It talked about nonuniform E fields and the production of E fields via charges (only when the resistance across the one-loop inductor was appreciably large and nonuniform). My book (University Physics, 12th edition), explains that there indeed is a "genuine potential difference between the terminals of the inductor, associated with conservative, electrostatic forces, despite the fact that the electric field associated with the magnetic induction effect is nonconservative" (bottom of page 1035). The question of whether there is an E field between the terminals of an inductor has made me interminably confused! I would greatly appreciate any help you could give! Thanks, Andrew
+Andrew Guo No E-field is allowed in any superconducting material as it would lead to an infinitely high current. Since E is zero everywhere in an "ideal" self inductor the integral through the self inductor of E.dl is zero (people used to call this integral the potential difference). If you attach a voltmeter between the 2 ends of a self inductor you will not measure zero, but you will measure L*dI/dt. The reason is that now you create a loop with the L and the Voltmeter. According to Faraday's law the closed loop integral of E.dl in that loop is -dphi/dt. The conclusion is that in case where Faraday's law must be used, the integral of E.dl depends on the path. I suggest you watch my 8.02x lecture #16. You will see in action that the integral of E.dl depends on the path. Two voltmeters connected to THE SAME points in my circuit will read very different values - even the polarity is different. It's not very intuitive, because Faraday's Law is not intuitive. If you have problems with this, you are not alone. Keep in mind though that Faraday's law runs our entire economy. You may also want to watch (but you will only see the demo in my 8.02x lecture #16). ruclips.net/video/LzT_YZ0xCFY/видео.html
Dear Walter,
confirm please "Two voltmeters connected to THE SAME points"? as it looks that really voltmeters connected to resistors ends (NOT THE SAME points) in that experiment but not same points - and there present gap of conductor line, if so this gaps of conductor line between resistors connections can be replaced with equivalent battery sign as there inducted emf from external magnetic field - and after redrawing this schema to reflect real processes this experiment became clearly explained : first resistor (. here one voltmeter connected) -eq. battery+ (. here second voltmeter connected) second resistor (. here second voltmeter connected again ) -eq. battery+ (.connection to first resistor, here first voltmeter connected )
Sir what will happen after the connection.
If I connect a finite current carrying inductor(say 5A) across a inductor having no current initially. (We assume small series resistance of inductor.)
Will the current just after connection be 5A or 0 A. As we say that inductor resist sudden current change.
Professor Lewin: THANKS FOR BRINGING THE WORLD'S GREATEST PHYSICS LECTURES INTO MY HOME!
One tragic thing about the 30H inductance delay demo at timestamp 20:30 ...
The experiment only demonstrates HALF of the lecture. This is sad, especially since it is SO VERY trivial to modify the circuit and demonstrate the RL delay SLOWING the light bulb going OFF when the power source is switched off. All you need is a (low current) SPST switch and a small fuse...that's it!!! PLEASE let me know if you are interested. It truly completes the demo!
The power of my demo is that students have to explain on their own why there is NO delay when I open the switch.
@@lecturesbywalterlewin.they9259 OK, agreed and I understand. I was just trying to make the demo more complete, but if this is your reason, I am sure you are right. Thanks so much for the reply. I wish I had you for a physics teacher. If I had made it to MIT in the 80's when I majored in physics, I always knew that I would have met you and had you as my physics professor. I regret missing it. I don't know what I would do if it wasn't for MIT's physics OCW, even now.
I so much want to recreate that demo, but have yet to find how to make such a low resistance high "L" inductor.
I am an Electrical Engineer and I have read and applied all that stuff. But I watch these lectures only those parts where the Professor Lewin is explaining the equations "intuitively" and relating it with our common sense.
I agree Professor, Physics is not Mathematics, it is when you understand the meanings of mathematical equations.
thats amazing, i like how you teach and your lecture.
The solution of the differential equation for the case of the AC power supply has also an exponential term c1*exp((-R/L)t) added to the one that you showed (where c1 is a constant). Why did you omit that term? Did you assume that it is negligible?
I have heard that a battery always maintains the potential difference across its terminals even in the presence of external electric fields. So shouldn't the battery's voltage dynamically increase in response to the induced electric fields' opposition to the voltage ?
I think at low frequency or for dc currents kirchhoff loop law is a good approximations as the change in flux in the ckt loop can be neglected or is approximately equal to zero only at very high frequency ac currents where the change in flux can noticed kirchhoff laws would be violated hence we can say for the transient condition ie when the current is rising from zero to its steady state value kirchhoff loop is not applicable but for the steady state condition where I is a constant or reached it max value kirchhoff loop law is applicable anyone please correct me if I am wrong
in DC current there is no dphi/dt. Thus according to Faraday's Law, the closed loop integral of E dot dL = zero. Some people call this Kirchhoff's Loop rule without realizing that it REALY is Faraday's Law which is one of Maxwell's 4 eqs. The statement that dphi/dt can be ignored for low frequencies is not meaningful as it all depends on the magnitude of the inductance in the system. After all -dphi/dt=-L*dI/dt
Dir professor, I have problem solving the differential equation at 34:35, can anyone help me out with this? By the way, this video is really helpful, thanks a lot.
The way I'd solve it, is to assume the solution is a linear combination of sine and cosine functions at the same frequency as the source, with arbitrary coefficients A & B. Then apply this prototype solution and its derivative to the original equation, and set up a pair of linear equations to solve for A & B. The sine terms have to add up to zero, and the amplitudes of cosine terms have to add up to V0.
The solution:
I(t) = A*cos(ω*t) + B*sin(ω*t)
A = V0*R/((ω*L)^2 + R^2)
B = V0*L*ω/((ω*L)^2 + R^2)
Correcting the texts awesome put a line through kischoffs law. I think Lewins pointing out why its misapplied : because it's not a lump in the circuit where the currents induced ie maybe it's a person who will act against the induced magnetic field .
Proffessor Lewin,
During Switch On ( t=0+)
If inductor oppose the current changes ??
Then how the current rises through the inductor??
Why induced voltage drop across the inductor dies out ??
solve the differential equation of an RLC circuit - that will answer your question.
ruclips.net/video/FWMhk6x785Q/видео.html
Good Morning Walter,
I hate how in school we were told that the phase lag of the current when entering an inductor is always 90deg... This is just lazy.
At 43:23 I like how we see that there is always a 90deg phase lag between the CURRENTS in the two bodies having mutual EM induction. That obviously comes from the fact that dcos(wt)/dt = -sin(wt) in dI/dt in Faraday's law. That is a solid assumption. Otherwise the two bodies would attract 100% of the time, but that is obviously not the case. I thank you for pointing that out. And the fact that the self-inductance in the aluminum plate is non-zero is the only reason why we get levitation is now very clear.
Now, I would like to know what happens if we apply this exercise to self-induction in an inductor. Let's say we excite an LR circuit with an AC current I_s*cos(wt). Then, there will be an opposing EMF self-induced (Faraday-Lenz) that will generate an AC current -I_i*sin(wt), that has a 90deg phase lag with the driving current, as for the woman levitation demonstration. Now do we have to add a phase lag tan(phi) = wL/R on the induced current? Now how do these two currents (source and induced) collide? They are definitely not in phase! At least 90deg difference or maybe even 180deg...! Is it how we can derive tan(phi)= wL/R ?
I checked google and I could not find a fulfilling answer. Maybe I can try to simulate that by modelizing the inductor with a AC current source that has a phase lag and goes against a driving AC current...!
Thanks!
Xavier
Thank you Professor Lewin, your teaching style is awesome and inspirational! I was also wondering if it’s wrong to say there’s a voltage drop in an inductor, because that would be as if it behaved by KVL, which of course it doesn’t.
the E field inside an ideal inductor is zero. Thus if you measure the integral of E dot dL through the superconducting wire of the coil (from one end to the other), you measure ZERO!
Lectures by Walter Lewin. They will make you ♥ Physics. Ohh, now I see why. Thank you a lot for answering me and making this content available and free, it’s truly a gift that all the world should thank you for. Take care Professor and be well!
@@lecturesbywalterlewin.they9259 what if we measure voltage across inductor using a voltmeter, what value will it show?
@@manjeetyadav9287 closed loop integral of E dot dL = -dphi/dt. since the intergal through the inductor is zero, the integral of E dot dL across the voltmeter is + or-dphi/dt depending on the polarity of the voltmeter in the circuit.
@@lecturesbywalterlewin.they9259 so, is there any way to measure this E.dl through inductor? (Other than the one in the lecture in which we saw different voltages measured between two same points through resistors).
Great video! Got a feel for the topic! Thanks professor.
one piece fan huh??
Someone knows how can there be a induced emf in a inductor if it is made somehow of superconducting material?
Sir what is effect on inductance on inserting golden rod in inductor??????????
@11:00 kvl can be used since emf induced also a voltage so sum of voltage =zero IR-Vbat+emdinduced=0?
AKRAM A. A. AL-KHAZZAR kvl has no terms that deal with changing magnetic fields, so kvl can not be used here
@@robinhaas6895 *you are dead wrong*
*I appreciate the fact that this Physics may not be familiar to you. Don't feel embarrassed about it; you are not alone. However, it's never too late to get educated. The choice is yours. Maybe this will help you.*
*I demonstrated at the end of my lecture #16 of my 8.02 E&M course at MIT that two identical voltmeters attached to the same 2 points in a circuit can show very different values.* ruclips.net/video/nGQbA2jwkWI/видео.html *The reason is that in the case of an induced EMF (Faraday's Law) potential differences are no longer determined; they depend on the path. This also applies to the secondary windings of transformers as the EMF in the closed loop of secondary windings is induced.* ruclips.net/video/b7i2uMx7gHo/видео.html
*Of course, in cases where Kirchhoff's loop rule (KVL) applies, 2 voltmeters attached to the same 2 points in a circuit will always show the same value.*
*My demo was first suggested and published by Romer in December 1982 in the American Journal of Physics.* *This demo has now become a classic; it's done all over the world at many colleges and universities.*
*Kirchhoff's original text can be found in the following link, pages 497-514:* books.google.de/books?id=Ig8t8yIz20UC. *Clearly he was fully aware of the prerequisite for his "loop rule". KVL is a special case of Faraday's Law. That's why Faraday's Law is one of Maxwell's equations and KVL is not.*
*By teaching students that KVL always works without telling them when it does not work, makes many believe that the closed loop integral of E dot dL is always zero. ElectroBOOM and Dirk Van Meirvenne therefore believe that 2 voltmeters attached to the same 2 points in a circuit must always show the same value which is not true as demonstrated in my lectures.* *They each posted a video (see @ footnote below) on their channel in which they claim to have proof for their wrong ideas which violate Maxwell's equations.* *Apparently they do not know, or do not understand, that in the case of an induced EMF potential differences are no longer determined; they depend on the path.* *MIT students who took my 8.02 course (Electricity and Magnetism) would not make this mistake!* *I therefore believe that to introduce a "modern version" of KVL and then teach students that KVL always holds is not advisable as you may set them up for making the same embarrassing mistake that both Dirk Van Meirvenne and ElectroBOOM made.*
*Also read Professor John Belcher's thoughts on Faraday's Law and on KVL. Professor Belcher is my former colleague at MIT.* freepdfhosting.com/0813df09f5.pdf
*(@) ElectroBOOM insulted me in his video by mentioning that the reason why my 2 voltmeters read different values was due to "bad probing". He stated that if I had done the demo correctly I would have found that both voltmeters would have read the same value. He could not have been more wrong*
*This is my only msg on this issue. This is too basic. I do not argue either with people who believe that the Earth is flat.*
This video may also help some of you to digest why 2 Voltmeters attached to the same 2 points in a circuit can give very different readings. ruclips.net/video/T6S5VS03xZc/видео.html
my understanding is that if we connect a voltage battery to a wire, initially there will be resistance and also self inductance that opposes current flow. am i right?
yes you're right. the opposing of inductance always exists. but it's strength is decreasing.
Professor, regarding the comment you made about inductors not having a E field or a Voltage drop across it; do you mean that whenever a EMF is induced (Faraday) there needs to be a resistance? But dont you induce eddy currents in your superconducting (cooled) ring?
Any induced emf (Faraday's Law) will show up in the ohmic resistance portion of a circuit.
Feynman:
In a ''perfect conductor'' there is no resistance whatever to the current. So if currents are generated in it,they can keep going forever .In fact,the slightest emf would generate an arbitrarily large current -which really means that there can be no emf at all.Any attempt to make a magnetic flux go through such sheet generates currents that create opposite B fields -all with infinitesimal emf's,so with no flux entering. If we have a sheet of a perfect conductor and put an electromagnet next to it, when we turn on the current in the magnet, currents called eddy currents appear in the sheet, so that no magnetic flux enters.
At 11:40 why did you add IR( potential across resistor) it should've been subtracted right as we are travelling in the same direction of the current
inside the resistor E and my dl are in the same direction
Thank you so much! Your lectures are truly amazing.
Although I am already an engineer, I cant believe curiousity to watch some of you lectures will lead me to still learn something new by givingme some clarifications to the foundations of my profession. If only I can turn back time I would love to be a physicist
:)
As an a freshman in electrical engineering I just passed my finals in circuits but only now do I begin to understand what a terrible mistake I did in my exam but unfortunately it will NOT be considered as such because the professor deals with RL circuits the same-wrong way
I am really sorry I didn't have a chance to ask him this question.
Thank you professor for clearing this up. I am going to watch all your lectures about induced emfs and solve your problems in search for the truth and not comfortability.
:)
In 35:00 it surprised me that he didn't mention the antiintuitive fact that a large enough R vanishes the change in phase between I and V
He mentioned after, when he used the ring with the slot jeje. Greetings!
Professer Lewin,
In the experiment at
30:00
Why does the light turn off immediately? Isn't it supposed to turn off gradually?
Thank you
+Lectures by Walter Lewin. They will make you ♥ Physics. That's a good question. When I open the switch the current will now continue through the loop that contains both bulbs. Thus R total is now 20 ohm. The current will decay as e^(-Rt/L). For t=1 s, this is 0.5, thus i^2 is 0.26*i^2_max, Thus after 1 sec the heat production is only 26% of the maximum. After 1,5 sec the heat production is down by a factor 7.
Simple answer: L is low enough to be not noticable.
Professor Lewin, isn't the resistance of the parallel circuit be 1/R_total=1/R1=1/R2?
Will the self inductance affects the current in the loop that without self inductance, in a way that it slows down the turns off speed of the bulb?
Thank you.
@@ongchinyuan2901 no because the original power source is no cut off the circuit so you only need to analyze the top loop
@@lecturesbywalterlewin.they9259 the link for the lecture notes is not working sir.could you please help me with that sir.
Hello professor Lewin,
About halfway through your lecture on inductance, you brought out a monster inductor 30H @ only 4 Ohms). At about timestamp 28:30 you say that later on in the course you will tell how such an inductor is made. Is that information in one of your videos somewhere? I would love to know the construction logic.
Thanks so much for all of your wonderful videos. I almost got to MIT to major in physics--and the timing was such that I would have known you--I regret having to go to a more local university.
In a later lecture I discuss how solenoids are made and how you can calculate the B field for given dimensions of the solenoid.
Sir, how exactly does a wattless device work? Does the source deliver power? Where is it going if the inductor doesn't use it?
any source that delivers power uses energy, thus the electric company will charge you for kWh used.
Hello, Hello, Hello Dr. Lewin❗️
Needless to say, I’m a huge fan...
@30:14...Why didn’t the bulb on the right (on the inductor leg), dim out slowly (just as it did glowing slowly)...with the inductor fighting the change to zero volts/current ?
draw the circuit in the 2 cases you will then understand why.
sir pls give details about the experiments
that is no of turns,gauge of wire, amount of current in circuit...to demo the experiment in classes..
According to the equation given in 14:08 ,will current be zero when the resistance in the circuit equal zero (an ideal case)?. If so will self induction of the coil be responsible for that ?
If R=0, Imax goes to infinity. 0 x infinity can be ANYTHING!
I was down today. But as I grasped why inductor's current lags voltage by 90 degress reinvigorated me. Thanks Professor!
Referring to 32:50, I see that in many books, they mention that the current thru the inductor lags behind the voltage across the inductor. You mention that the current lags behind the driving voltage. By voltage across the inductor, do they also mean driving voltage? If yes, then is it true that the driving voltage and the back emf induced in the inductor are completely out of phase(180 degrees) (ideally where there is zero resistance)?
I cover in my lectures voltages and currents as a function of time and phase angles. I cannot add to that here. Perhaps you should watch more lectures (e.g., LRC circuits).
Sir. What is your opinion on text book - concepts of physics by Dr. Hc verma? I am just curious sir. 😊
it's fine & perfect for India
@@lecturesbywalterlewin.they9259 outside India ??
Professor Lewin,
I have a query that the magnetic filed the is produced by the closed circuit in the RL circuit does it have any effect on the magnetic field produced by the solenoid/inductor, because the current flowing in the circuit will itself generate a magnetic flux which is going in the loop and would be perpendicular to flux generated by the inductor (as per my intuition), could you please elaborate on it though. Thanks for the amazing lectures
Make a calculation! You will then conclude that the effect is so small that it can in general be ignored
@@lecturesbywalterlewin.they9259 Thanks for responding Prof Lewin, will do it 😊
Anyone know if there is a way to get the problems from 8.01, 8.02 and 8.03 online without buying the books? I have my own books I use I just need the problems and the problem sets direct me to the books.
are available in video description
Here at time 45:17, what is phase difference between green, red and blue lines. What should be the condition of "always repel"?
45:17 the figure is very very clear. red and blue 180 degrees out of phase, green 90 degrees ahead of blue.
Prof Lewin, the magnetic field outside the inductor is approximately zero and therefore the electric field everywhere outside the inductor is conservative. Thus, when applying Faraday's law, you can choose the closed loop that bypasses the inductor, i.e. you do NOT go through the inductor - you start at one end of inductor, go outside and enter at the other end. That way, the closed loop integral will be ZERO and therefore you CAN use kirchhoff law. You just need to carefully specify the curve.
that is 100% correct. I mention this in one of my videos. It is important for students to realize that. Becoz when you attach a voltmeter over a superconducting coil it creates a loop with the selfinductor. The integral E dot dL through the volrmeter is +or -Ldi/dt but zero through the selfinductor. You can then also draw a closed loop (bypass the selfinductor) in which KVL applies. However, if you want to know what the voltmeter indicates you *must* include the loop voltmeter-selfinductor (see above).
@@lecturesbywalterlewin.they9259 thank you for the explanation, professor
Thank you very much, for the gorgeous lectures, but I have a doubt, If we have a battery, switch(open), inductor and resistor in series, when I close the switch( at t=0), then at t=0+(a moment after t=0) the current is zero, because of self inductance of the inductor.
But my question is, self induction is possible only when there's a change in flux for which to happen there has to be a current flow, atleast a current with very small amplitude but steep slope, so back emf becomes quite large stopping current flow momentarily. How can it be zero?
*it's NOT the current that matters, it's di/dt that matters!!* and di/dt can be huge when i is zero. Brush up on your math and on Faraday's Law.
@@lecturesbywalterlewin.they9259 Thank you very much for responding sir. I understand that change in current is what matters. But initially when we switch on the supply, if there's no actual flow of current (that is, current *changes* from 0 to some value) which is supposed to change the flux across the inductor and produce induced emf which inturn opposes that change in current flow.
"di/dt can be large when i=0" I agree this is possible, when we have been running the circuit for sometime, for example with a sinusoidal voltage, the change is maximum when i=0 like in a pendulum, but when we switch on the supply, the pendulum is at rest ... Isn't it?
So my question is; is there a change in current (from 0 to some very small value, with steep slope) the moment we switch on the circuit?
Kindly bare with me if you had already addressed this in your answer.
you still don't get it. when you throw the switchat t=o i=0 but di/dt is not zero. this is my last msg.
@@lecturesbywalterlewin.they9259 Thank you sir. So a steep *change* in current occurs when I throw the switch which in turn generates the emf.
Sir,please add lectures on electronic devices,semiconductors,bias,transistor etc.I am in a trouble with these.
Does anyone know the name of the song he's playing in his experiment at ~min 38??
Heinichen: Concerto in G major Seibel 215 - 4. Allegro - 2/4
please provide me lecture supplement for this lecture because my textbook also used kirchoff's rule. they dont know change in magnetic field
I posted Lecture Notes as pdf files below the video thumb nails. Not all lecture have lecture notes.
Professor, you've changed my concepts of physics. Electromagnetism WAS a mystery to me. You made them so easy to understand. I love watching your lectures. They made me realize physics is something that is an art. You're an artist of physics. Thank you professor for your beautiful lectures.
:)
Hi, professor Lewin, how come so many books are wrong (in this issue of Kirchoff's loop rule vs Faraday's law) and so many smart people go along with it?
Professor lewin, 9:30. How can integral of E.dl be equal to d(phi)/dt, where phi is the magnetic flux thru the solenoid. When we are integrating on the left hand side over the entire circuit, is it right to consider the magnetic flux change only thru the soap film surface attached to the solenoid(which is the rhs of the equation), the surface considered here doesn't even touch the whole current loop (referring to lecture 17).
your should try to understand Faraday's Law. If you take a copper ring and you quickly move a bar magnet into or out of the ring, a current will flow through the ring. I do this demo in 8.02 where the ring is a copper coil which hundreds hundreds of windings but it's only 1 wire. There is a light bulb in the wire and when you move the magnet quickly in or out the light bulb lights up. This is the way we generate electricity. Faraday's Law runs our entire economy. The closed loop integral of E dot dl -dphi/dt. Any dphi/dt, no matter how or by whom a dphi/dt is produced, will produce an induced EMF in the closed loop.
Watch this lecture in which I introduce Faraday's Law + demos ruclips.net/video/nGQbA2jwkWI/видео.html
Is it fair to say that the self inductance of any circuit is the magnetic flux it gives per unit of current running through it? L = phi-sub-B / I?
I discuss self inductance in my 8.02 lectures. You will find your answer there.
I live the way Physics is being taught in an interesting way and I am using this to generate interest in studies. Could anyone please recommend a similar chemistry channel 🙏 whereI can enjoy Chemistry like this?
hello sir! i have a question and this might be a stupid question but at 31:34 you said v=vcoswt for ac supply. shouldnt it be v=vmaxwsint? in other words i dont understand why is this expressed here as a cosine function when ac is all sinusoidal? thanks
cos or sin will work. The 2 are IDENTICAL except for a phase angle of 90 degrees.
WL, is there any link between the equation at 22:15 1/2 LI^2 and the equation for Ek = 1/2mv^2 ?
No, actually. There's no analogy here.
Sure. Think of L as analogous to inertia, and current as analogous to speed.
You'll see an analogy between circuits with capacitance, resistance, and inductance, and vibration setups with a spring, a damper and a mass. The same differential equation governs both kinds of systems.
Capacitance is analogous to the flexibility of the spring (inverse of the Hooke's law constant).
Resistance is analogous to the viscosity rate of a linear damper.
Inductance is analogous to the mass, as it resists a change in current, just like mass resists a change in speed.
As omega increases what happens to the phase relationship between the direction of the current flowing in the inductor and the direction of the magnetic field created by the inductor .
phase angles have only meaning if we deal with AC. I suggest you watch my AC lectures on RLC circuits.
It is at 32.30 in this lecture that you discus AC , induced EMF , current and phase angles in Self Inductors , what is happening to the phase relationship between the direction of current flowing in the inductor and the direction of the magnetic field created by the inductor as omega increases.
after 32:30 I introduce AC Power. I give you the solutions also the tan of PHI. I cannot add anything to the clarity of my lecture. My equations allow you to increase omega.
Later in the lecture the magnetic field is discussed, in the discussion on magnetic levitation, and I see that the magnetic field and current direction alternate together.
11:36 I find this easier.
Add emfs together, and it's the sum
of the components:
V-LdI/dt=IR+0
Energy sources = Energy losses.
The self inductor is a 'negative source'
wrt the input voltage V, because it reflects back energy put into it rather than let it out as heat or some other loss.
The sound of chalk is the most beautiful
Sir,
In the exiperiment at 30:0 , if we use ac supply instead of dc supply, ( let bulb withstand Ac voltage).
Den how is the illumination of bulb which connected series with inductor ?
Is it glow ? >> induced current in the inductor is 180degree phase shift with respect to source current ??
ac current in the light bulb
Is it bright or dim ??
Today i did an experiment with small transformer 230v/12v ,2A.
I connected 230v,40w incandescent lamp serial with primary of the transformer to 230v Ac supply..
But the bulb didnt glow..
.
Professor, what is the music that it played in the demonstration for the self-inductance cancelling out higher frequencies?
Beethoven's 5th
@@lecturesbywalterlewin.they9259 I've replayed the entirety of Bethoven's 5th Symphony, professor, but I don't seem to find it. I really am addicted to that, I find myself humming this in random times of the day!😅😁
In I- t grap, During t = 0+ (transient state)Concave down graph reprsentation.And in switch off condition Concave up graph reprsentation..
Is it any wrong if concave up graph reprsentation in Switch on ( t=0+)Also concave down reprsentation in switch off condition.. ??
What abt V- t graph during these interval.?
I cannot improve on my lecture
In the experiment with the 2 bulbs, how come the bulb in series with inductor turns off immediately.
good question. The resistance of the light bulbs increases with temperature. When both lights are fully lit, each light bulb probably has a resistance of 50 ohms *or even more* (I discuss this in a different lecture + demo) . When we open the switch the current is now going around in the loop with L, 4 + >50 + >50 + 4 ohms. Thus L/R is now 30/>108=
At about timestamp 26:26 professor Lewin stated that: there will be an explanation later on in the course on how such a large 30H (4 ohm) inductor is made!!
I HAVE BEEN SEARCHING FOR ...YEARS... FOR THIS!!!!
Can someone tell me where I can find it????
THANKS MUCH!!!!!!!!!!!!!!!!
Sir, you are the best teacher I have ever seen.. I enjoy your lectures so much😍😍😍
Glad to hear that
Proffesor how can there be phase difference between the emf induced in the loop and the current produced i am afraid this going to give me some sleepless nights if you would be kind enough to explain it hopefully i can get some sleep
37:40 passive filters???? Wow !
Sir, there is a college physics book where it discusses a very beautiful and nonintuitive example of the consequences of nonconservatives E-fields. That book is physics for scientists and engineers 6th edition by Serway. Look at the example page 973. That example is even more bizare than two different voltemeters reading different values. I guess that example will be the best to choose as one of your weekly problems.
Hello, At about timestamp 26:27 it is stated:
"LATER IN THE COURSE, I WILL SHOW YOU HOW WE MAKE SUCH A HIGH SELF-INDUCTANCE."
Can you please direct me to this explanation?
watch my lectures
Awesome lecture! But pdf links seem broken does anybody know where to get it? Is it possible for you to reupload?
they are *not* borken but you have to know how to get them. ruclips.net/video/CGDZr7e-4JE/видео.html
hello professor ! referring 20:35. earlier you said that magnetic fields doesnt do any work so how are they providing energy? is it because these fields are dynamic?
the work is done by electric charges which create the B-fields. When all the magnetism comes from classical currents, the magnetic field does no work directly on the currents. However, by steering the electrons in new directions it can cause them to bounce off things and do work. Once the current carrying loops start moving, they create electric fields that do work on the currents.
@@lecturesbywalterlewin.they9259 got it sir thankyou and love you
In school we learnt about a concept called "back EMF" and it seems very similar to self inductance... is there a difference between back EMF and Self conductance? Or are they the same?
Lenz Law opposes the change in magnetic flux - that's Faraday's Law. The induced EMF is sometimes called "back EMF".
Hello professor , in RL circuit the E-field is zero inside the wires of the inductor although there is a changing B-field in the region where the inductor exists so there must be an induced E-field in that region but after reading your lecture supplement i think that the induced E-field will be zero due to the build up of static charge ?
>>>i think that the induced E-field will be zero due to the build up of static charge ?>>> yes that is correct. Read this file all the way to the end. freepdfhosting.com/0813df09f5.pdf
@@lecturesbywalterlewin.they9259 actually this is the paper i read and me conclude my conclusion .
nice music,which Mr Lewin plays in the demo , shazamed it but couldn't find anything , does anyone know wtge name of the composition?
+InventTwig If I played on a wooden trombone, it was a famous christmas song "Jingle Bell"
+Lectures by Walter Lewin. They will make you ♥ Physics. They will make you ♥ Physics. Mr Lewin you played a wooden trombone ? was it in 8.02or 8.03 I would like to see that :D ,I'm talking about the music played through the speaker in the inductance demo at 39:06
+InventTwig That was probably a radio station.
Love you sir♥♥♥sir in experiment why does both bulb become off at same time, bulb with self inductance should produce a current when battery is tutnedoff.
look more closely at the new circuit. There is no longer a LR in series.
Professor Lewin,
At 47:41, You calculated the self-inductance of the ring in order to find the phase delay of the current. But as far as I believe, we should have taken mutual-inductance. Because the magnetic flux is generated by the coil and hence the inductance should depend upon coil geometry too. Please correct me if I am wrong.
P.S. - Thanks a lot for the amazing lectures. I love them and wish to see you teach every subject in physics :D
Nevermind, I see my mistake now.
near the end of the lecture, if the aluminum ring was placed inside the electromagnet, would it oscillate or just get repelled in one direction?
behavior would be nearly identical
to be clear I drew what I described in this link: drive.google.com/file/d/11efMmfX1-KuzYLaLmXfhTB8hcu-fHY9q/view . I was thinking that if the ring is completely above the electromagnet it will be repelled upward and if it is completely below the electromagnet it will be repelled downward. if it is in the middle it may still go in one direction or the other or it may oscillate.
In the self inductance demo, why does the 2nd bulb go off almost immediately like the first bulb? Shouldnt the current in it drop to 37% of max value by 3 seconds and so it should go off slowly?
good question. the answer is that you have changed the circuit substantially when you turn the power off. Make a circuit diagram and you can see that for yourself
@@lecturesbywalterlewin.they9259
Thank you!
Sir if we wait long enough then it makes sense to me that Electric field should be zero In the inductor but initially before system reaches a steady state. .. There's a change in B flux around the inductor which should produce a induced (non conservative) E field!...so there will be a E filed initially in the inductor .....please tell where I am going wrong!
you are mistaken - do your homeweork!
These lectures are simply amazing and deserves the highest praise.
at 7:20 you said that think of inductor as made up of superconducting material so if thats the case then i think that induced current will always gonna run and it will always fight the change and as the current increase, induced current also increases so how can an inductor loses the fight?
solve the diff eq. you will find that the current will no longer change after some time. Then dI/dt=0 Thus L*dI/dt=0 thus the induced EMF= 0'.
got it sir thankyou so much ! i love you and your lectures ❤
sir when you switch of the light at time 30:11 I thought the self-Inductance would delay the bulb " I " to go off. But both the bulb turned off at the same time.
look closely at the 2 circuits - before the light is turned on and when the light is trurned off.
Thanks for the reply sir
Excellent lectures. Reminds me why I became an engineer. Also reminds me how much I have forgotten over the years.
:)
my dear friend, thanks for the subtitles!
all my 8.01 and 8.02 videos have subtitles. The help session modules don't have subtitles. It's very time consuming. I may start doing a few a week.
That would be great (maybe with helpers). Because the whole world needs your whole ultra bright mind.
to subscribe an 8 minute video takes me about an hour.
I understand you. It's hard (and perhaps boring) to do that (if without helpers). Then I will just start and ask straight questions.
Gor
I just noticed that all my 144 help sessions have closed captions. You should click on "cc". The captions are not very accurate (I don't know who did them) but they do help. Try them again and tell me whether they work for you.
Thank you sir for the lectures .. I have a question though.. You said that the inductance will also fight the change when the voltage source is immediately taken out.. then shouldn't the light bulb take equal amount of time to get to 0 light as it took to get full light .. why does the light is off immediately as you turned off the switch?? the experiment is at 30.00 minutes in the lecture
That's a good question. When I open the switch both lights go out very quickly. The reason is that when the filaments are cold the resistance is 10 Ohm but when they are hot the resistance is much larger then 10 Ohm. In addition when I open the switch the current can now only go in the upper loop which contains both lights + twice 4 ohms. The net result is that L/R of both "effects" is now much smaller than 3 sec.
Thank you sir for replying .. I understand your first logic but i am having trouble to understand the 2nd one (the one you tell by saying in addition)
When the switch is closed current flows separately from the battery through the L + lamp and through R + lamp. That is no longer possible when I open the switch. When I open the switch, the ONLY way that current can flow is "circling" around in the loop which contains BOTH lamps and BOTH 4 Ohm resistance. Thus now the same current going through L MUST also go through the other lamp and resistor (and that was NOT the case when the switch was closed). That in creases R, thus L/R goes down.
Thank you sir .. I understand this time .. but it was non-intuitive for me that after closing the switch the current will circle through the loop .
You mean "after opening the switch" (not closing).
Sir....can you explain that electric field energy or magnetic field energy will be greater if they are measured equivalently....thank you sir
watch my 8.02 lectures
So professor, If you connect a voltmeter across an inductor, it will *always* read 0?
*The answer is NO*
if you place a voltmeter over an ideal conductor, made of super-conducting material you are creating a LOOP. The closed loop integral of E dot dL in that loop is -L*dI/dt. Since the integral of E dot dl through the ideal conductor is ZERO, the integral of E dot dL through the voltmeter is --L*dI/dt and that is what the voltmeter will show.
Professor Lewin,
At 11:36,
Are both the loop's induced emf and inductor's induced emf taken into account in a single term of Ldi/dt?
Thank you!
Maxwell's eqs!!! I cannot add to the clarity of this lecture
@@lecturesbywalterlewin.they9259
I still haven't studied the equations yet but after studying it I'll come back to the lecture.
Thank you Professor!
@@lecturesbywalterlewin.they9259 I hope you are having a great health!
sir please tell how did you write V=V0(cos omega*t) , love from india !!
Thank you so much this is too important but sir i have a question how is the current passing through the hole circuit if it isn't passing through the inductors wire is it passing through the air between the coil of the soilinoid and if that the cause why don't you consider the air as a resistors? and isn't that means that the potential diffrence across the inductor is zero?
Thank you
>>>>the current passing through the hole circuit if it isn't passing through the inductors wire is it passing through the air between the coil>>> *ofcoz the current is going through the wire of the coil of the inductor.* Please watch my lectures.
@@lecturesbywalterlewin.they9259 sir i have seen this lecture and aslo lecture 16 and many other of your lecture but I'm confused because if there is no electric field in the inductor isn't that means that there is no current in it?
@@zahraa-dm5cy ofcoz current can go through any wire that has zero resitance in which there is NO E=field. V=IR V=0 R=0 I can be anyhting.
Thank you so much sir
@@lecturesbywalterlewin.they9259 yes indeed sir you are right and sir you are great teacher i can't sleep because of curiosity your lecture created in me again after my childhood
Engineers who design and layout circuits need to be aware of how magnetic flux in inductors can couple into other parts of a circuit. Circuit Theory textbook authors want to describe everything in terms of voltages and currents. Parameters that depend on the physical layout of a circuit are a complicating detail authors either don't know about (sadly), or are unwilling to take the time, effort, pages to explain properly. Some circuit theory books will use a separate dependent voltage source to represent the emf from a coupled inductor (mutual inductance), so that's a step in the right direction I suppose. I teach an introductory circuit theory class for electronics technicians at a local community college, and watching this lecture has definitely given me some things to consider when teaching students about inductors and magnetic induction.
I'll admit, I am watching your lectures not so much for the physics (although it's a good review), but more for observing your teaching methods, as I am relatively new to teaching. Thank you Dr. Lewin!
Sir if integral E dot dL is zero across inductor...what reading will i get if i connect a voltmeter across inductor..?
if you connect a V meter over the self ind it will read -Ldi/dt.
You have to apply FAraday's Law in the closed loop V meter - inductor. The intergral of E dot dL through the wire of an ideal indcutor is zero, but the closed loop integral is -Ldi/dt.