 Hi, I'm Zor. Welcome to InDesert Education. We continue talking about Maxwell's equations for electromagnetic field. This is the third lecture about the third equation. Now, this one is about putting into differential form the Faraday's law of electromagnetic induction. So, whenever you have a changing magnetic field, it generates an electric field. Now, first of all, I would like you to strongly suggest to review material from the electromagnetism part of this course related to the law of induction. So, there is a topic, there are a few lectures there, familiarize it, because I will definitely use these results without going into many details. The only thing which I will do just to remind you, I will just remind you one particular experiment. Experiment is very, very simple. You have a viral loop and you have, let's say, a permanent magnet, which you are moving, let's say, down in this particular case. Now, as soon as you move down inside the ring, the electric current will be generated. So, that's basically the beginning of this concept of induction. Whenever the changing magnetic field induces the electric current in the wire. So, that's basically what we will be talking about today. Putting this into quantitative format after James Maxwell's third equation. Now, this lecture is part of the course called Physics for Teens, presented on Unisor.com. Every lecture on that side has textual explanation. In this particular case, textual explanation is very detailed. And during the lecture, I might actually just skip a couple of things, basically referring you to the text, when it's kind of tedious, long and obvious. Now, the site is completely free. There are no advertisements. You don't even have to sign in if you don't want to. This site has exams for many topics, not all of them, but for many topics. You can take them, again, free any number of times you want. So, I do suggest you to use the website. Rather than, for instance, you might find it on YouTube or somewhere else, this particular lecture, where it will be just a lecture without textual explanation, without previous lecture, which you know, have to find or something like this. From the website, there is a menu, so everything will be kind of obvious. You go sequential lecture after lecture. Okay. So, I have remind you about this thing. And now let me go into a little bit more specific things. First of all, why electric current circulates in this particular wire loop? Well, because there is some kind of electric field. It's the field which pushes electrons around in this particular case. Now, what if there is no wire loop at all? Will the electric field be there? Of course it will. I mean, wire loop is just a device which helps us to see what's going on. But basically, what's going on is going on regardless of whether wire loop is there or not. Same thing with this particular permanent magnet. What if I just say, okay, there is a magnetic field which is changing in time. Is it sufficient? Yes, it is sufficient. Basically, the point is, if there is a changing magnetic field somewhere, somehow, it generates the electric field as well. Now, we have agreed before that magnetic field and electric field have something which we call the field intensity. So, there is a magnetic field intensity and there is electric field intensity. What is intensity? Basically, it's the force which acts on some kind of a unit of whatever the unit is. If it's an electric field, there is something which we call a charge. So, a unit of charge. And the force which actually acts in that particular time and space location onto the unit charge would be exactly the electric field intensity. In case of magnetic field, we also measure magnetic field intensity. And there is also a unit of magnetic field amount, if you wish. Not magnetic field amount, amount of magnetism. It might be something like artificially created magnet, let's say, with one particular one ampere going through one meter of length of the wire or something like this. So, whatever we have assigned as a unit of magnetism and the force which is acting on that particular unit of magnetism would be a magnetic field intensity. So, let's talk about magnetic field intensity first. Since it's a force, it's a vector. So, B is a vector. Now, in three dimensional world, any vector can be represented as three components. Projections of this vector to three axes. Now, at the same time, now, these are vectors. At the same time, I can write it differently. If i, j, and k are unit vectors along the three axes, x, y, and z. Now, B, x, b, y, and b, z are magnitudes of this vector. So, I basically divided vector into magnitude and direction. And I can do that because I know the direction of this vector. We agree that this is the projection on the x-axis. So, it goes along the same as vector i, which is a unit vector along the x-axis. Same thing with y and z. So, we assume that there is such a vector. Now, it's kind of difficult to basically come up with some quantitative variation of all of these together in one shot. So, what I will do, I will just choose one particular axis and I will consider only vector b, z, which is b, z times k, if you wish. And I will find out what's the electric effect of this particular vector, magnetic vector, what kind of electric field it creates. Now, then I will do exactly the same with this one and this one. And now, I will use the principle of superposition. If they're together, well, the total effect should be a sum of effect of each one of these. So, right now, we are considering the vector field b, which is only directed along the z-axis. So, right now, its entire magnetic field is directed along the z-axis only. So, b, y and b, z are equal to zero. We are considering only these types of magnetic field. Now, obviously, b, z is a function of time and location. Now, we do know the direction of this vector. I can put the vector as well. We know the direction of this vector is along the z-axis. That's all we are kind of assuming right now. Okay, now, we know that from the electromagnetism part of this course, that if there is a change in magnetic field, it generates electric field. And if the change of magnetic field is always within the z-axis, then the electric field intensity would be perpendicular to it. So, let's do the same experiment as we did before. Instead of permanent magnet, I'm just assuming that there is some kind of a vector vertically directed. These are my three axes. And I will put my wire, well, in this case it would be not wire loop, it would be wire rectangle, perpendicularly to this guy, perpendicular to the direction of the magnetic field intensity. Now, if this is changing with time, still within the z-axis, but it can change in magnitude. So, whenever it's changing in magnitude, it's the same thing as permanent magnet moving back and forth vertically. So, there will be electric current generated in this particular wire rectangle. Now, what I will do is I will consider this wire rectangle to be very, very small. So, its dimensions would be delta x by delta y, and I assume they're small enough. Now, why I would like to do that? Well, because I don't want to be dependent on entire set of different values, I would like to be dependent only on the value of this vector bz at particular point x, y and z. So, I choose any particular point x, y and z, and I will center this particular rectangle in such a way that its middle, this middle of this would have a coordinate x, y and z. So, that would be this is, let's say, x, this is y and this is z. So, this will be x, y, and this would be z somewhere, okay? So, this is the point x, y and z, and I have a plane parallel to x, y, and in that plane I will put my wire rectangle. Now, from the view from the top would be like this. I will have x, y, and I will have this center would be x and y, and since the whole length is a, b, c, d, I don't know this anymore, so what's the coordinates of the points a, b and c and d? So, a will have coordinates x minus delta x divided by 2, right? y minus delta y divided by 2 and z. b would be at coordinates x plus delta x, y will be still minus delta 2 z, c would be at x plus delta x, y would be plus delta y half z, d is x minus delta x over 2, y would be plus delta y over 2 and z. So, this is my rectangle. Now, again, why do I need such a small rectangle? Because I would like to know the flux which goes through this particular wire loop, wire rectangle, and relate it to electromotive force according to the Faraday's law. Now, Faraday's law is very simple. And again, if you don't remember it, I refer you to electromagnetic path of this course. This is a flux which goes through this wire rectangle. Now, we are taking the first derivative by time, and this is electromotive force which is generated in the wire. Okay, now, again, let me remind you what the flux is. Well, if you have a vector b, and this is this rectangular wire loop, if b is constant everywhere exactly the same, then the flux is actually the b times area of the loop, or rectangle, whatever the area is. Now, in this case, we have a very small rectangle, so its area is this. Now, b, well, b in theory is different. All we know is it's vertically directed along z-axis, but it changes with time. However, we are getting this rectangle very, very small. So, approximately, in our case, we can have that this is phi approximately equal to b of... Well, it's bz, actually, because we are only talking about vectors directed along the z-axis, from x, y, and z times dx times dy. So, I took the value in the middle, x, y, well, and z, obviously. We took the value of the magnetic field intensity in the middle and multiplied by area, assuming that this particular wire rectangle is small enough, so the value of the vector b is not really changing much. And obviously, eventually, we will just say, okay, we consider this infinitesimally small wire rectangle, which basically makes this thing, in the limit case, absolute true. Okay, fine. So, here is the connection. This is something which depends on the magnetic field. This is something which is electric field characterization. So, this basically is the beginning of the future Maxwell's equations, which connect together basically two intensities, b, intensity of the magnetic field, and e, intensity of the electric field. But that's the beginning. So, we have to come up with this, basically, in terms of whatever we have, basically, in this particular case. All right. So, we know all this, and now let me just go back to this picture. If I have an electric field, and I have two points, and let's say electric field is constant, and let's say it's directed from one point to another point. Now, electromotive force, basically, it's basically a voltage between these two points, right? And the voltage is amount of work to move one unit of electricity from one point to another. Either the field does it or we do it against the field. But anyway, that's the amount of work. So, u is work. Now, if it's a field which is uniform, which means e, intensity of the field is exactly the same, and we're assuming it's from p to q, then this is equal to e times l, where l is the length of this particular segment from p to q. That's the simplest case, right? It's like we are saying the work is equal to force times distance, right? This is the force. This is the distance. This is the force, basically, on the unit charge, and this is also per unit charge. So force times distance. Now, if situation is more complex, let's say the force is at the angle. Then what we do? Well, we do projection of this force on this direction. Well, it's actually in vector form. This is our product, right? So you have a projection. And if the value of the intensity is changing, then basically we have to really divide it into small pieces and say that g, u, which is differential of infinitesimal incremental of the electromotive force is equal to e at that particular point, let's say, I don't know, coordinate, some kind of coordinate, times gx if this is x axis. Now, and then we should really integrate the whole thing from p to q to find out entire electromotive force. That's if e is changing. Okay, in our case, e is changing, but not too much. I mean, obviously, now if, for instance, our current is directed counterclockwise, so I will put these little arrows here. So the magnetic field is directed perpendicular to the white board and it's changing. This is the z axis. So it's changing along the z axis and it causes the electric current. Now, how to find the entire electromotive force? Well, again, considering it's a very small rectangle, I will assume that during this from a to b, there is no change in electric field and I can just take it in the middle here. Same thing from b to c I can take in the middle, from c to d and from d to a. So I'll take these middle points as the electric field intensity. And again, in the limit case, it's absolutely legitimate thing because whenever I will do it infinitesimally small, there is basically no difference between electric fields at point a and b. Now, the entire electromotive, the entire intensity of the electric field here is basically, well, first of all, again, it's function of, actually, time as well. It's function of time and coordinates. In general, if there is this field somewhere. Now, I can always represent it as three components. E x of t x y z plus E y of the same plus E z of the same. Okay? Y, well, it's actually very important. If I'm moving from a to b, I actually can take into consideration only E x component because E y is perpendicular. You remember, the work is equal to projection of the force on the direction of the movement, right? So projection of vector E y, if the movement is from a to b is zero, they're perpendicular to each other. So I can always take E x. Now, E z, which is the component, is always zero because all these movements are always in the plane which is perpendicular to the z axis. So when I'm talking about from a to b, I will take only the E x component. From b to c, I will take only y component. Then from c to d also x so let's just do it. And I will go in this cycle and I'll find out how much work I need to move from a to b and then from b to c, etc. And that will be my total electromotive force which on the other hand can be expressed in terms of magnetic field and that will be my equation. Okay? So, forget about this. We know this. And let's think about this. Okay, from a to b, I have to take basically the middle point of electrophilic field intensity which is E x and I will use, now t and z, time and z will always be the same so I will just skip them. But for x it would be, what's the coordinate of this? x and y coordinate would be y minus delta y over 2, right? And t and z I will just skip. So this would be intensity of electric field at middle point between a and b. It's x coordinate is x because this point is x and y coordinate would be y minus delta y over 2. And I assume this will be a constant electric field intensity along the way from a to b and I will have to multiply it by the length. Length is delta x. Okay? Now let's go from b to c. From b to c middle point is this one and I will use, instead of E x the horizontal component I will use E y. And so it will be E y. What's the coordinates of this point? By x it will be x plus delta x over 2 and by y it will be y and I will multiply it by the length of this bc segment plus. Let's move from c to d. Again, I'll take the midpoint which has coordinate of x and y plus delta. So it's again E x with coordinates x and in this case y plus delta y over 2. Now my length is delta x but now I'm moving to the negative direction so I have to put minus delta x. Now from d to a again I will use E y along this y direction. Now coordinates of this point would be x minus delta x over 2 and on y it would be y and minus delta y because I'm moving against the direction of the y axis. Now and this is my u. That's the total work which is needed to move once along this cycle. Now I don't need this anymore and I will use just this and what it's equal to let's collect delta x. Delta x would be E x of x y minus delta y over 2 times delta x minus E x x plus delta y over 2. I don't need this anymore and this would be my delta x plus. How about delta y? I will have delta y of x plus delta x over 2 comma y minus E y of x minus delta x over 2 y delta y. Another trick related to calculus. If you have a function and you have a difference between these two remember this formula? From calculus if you have a function from A to B then the difference between these two between this and this let's say this is middle point in theory right? Now what this theorem is saying that there is some kind of a midpoint C not necessarily exactly in between it all depends on the function but there is a point C derivative so it would be B minus A times the derivative which is tangent of this angle so basically the difference would be this piece. It's a very known theorem in calculus which I'm going to use and again if you don't remember you can always go back to unisor.com this course go to calculus it's presented there now from this actually if you remember we built the x plus delta x x that would be approximation x times delta x that's basically how derivative is defined. We take incremental of x and have the difference between these guys and define the tangent as this that's the definition of derivative so I'm using this basically so this and this I think I did something wrong here I missed something so that would be I forgot k plus it should be y plus x y plus delta y right? I think that's how it's supposed to be yeah so here is what we have here first of all we are dealing with function of two arguments but in this particular case I have only one argument changing from one position to another position so basically this is left end y minus delta y over 2 and this is the right end so again using my f of b minus f of a approximately equals to f of let's say a or b or c doesn't matter times b minus a so in this case x is fixed y is y minus delta would be this and y plus would be this alright so this difference is ex of x some middle point and we will have y so we will have a derivative I'm using the partial derivative because only y is changing x is not changing so it's derivative in some midpoint basically midpoint is y whenever I'm getting smaller and smaller the midpoint and both ends are actually going into one point so it's quite legitimate and I have to multiply it by the difference between b and a b is this, a is this so the difference would be minus delta y correct minus delta y by half and minus delta y by half would be minus delta y here plus it's x which is changing so I will use partial derivative of xy at point x by dx and difference between b and a would be delta x delta y that's it we have finished so this is the formula which we are looking for so it's dy of xy by dx times dx times well, I just wiped it out but there is a multiplier delta y so it's delta y and I will use in one expression here d ex of xy by dy times dx times dy so the first difference between ex and ex I replaced with derivative times dy and there was a delta x multiplier already that's why I have dx and dy and then this one would be exactly analogous because this is derivative by dx then I had to multiply by dx and then there was a multiplier dy, the delta y so that's why this would be my final formula and this is e this is this is my which is the total work which is needed etc now let's go here what is f in our case I can say that f is a function of obviously in the middle point now this vector is always perpendicular to my wire loop that's why I can use just straight the multiplication by the area right area doesn't depend on time so if I want derivative I have to have derivative of this right so and this is also what minus u right minus look at these two things they are very close to each other so what follows from it this part is equal to this part but with a minus sign so we have dy by dx now let me just put all the argument not only x and y minus dx of dx yz by dy is equal to minus db of dx yz with dt I'm using partial derivative here obviously because it's function of many arguments I should really put another d here ok this is a very important equation because this is basically almost everything I wanted to do all I have to do right now is expand it to other dimensions to x, y and z and I'll be done so this is all about vector called bz that's what I should really put here bz so if my b of magnetic field intensity is only along the z axis and equal to bz then this is true great for instance my b is only within y direction so instead of this formula which is directed along the z axis I should really direct it along the y axis well let me just write it down basically replacing what I will replace and how very easily z would be to y y to x and x to z I'll just turn the whole thing cyclical substitution so exactly the same formula so it would be d e instead of y x of all these three parameters divided partial by x goes to z minus d e x is z for d x equals minus d by y in this case for d t now I don't put all this t x y z because it's always the same so that's my analogous equation but in case my magnetic field intensity is along the y axis and now let's call what if it's along the x axis so I will have z would be going to x y goes to z and x goes to y if I'm not mistaken right z to x okay it's like x y and z so z to x x to y and y to z right that's correct okay so from here we will have y to z so it's d e z by d y right minus d e y by d z equals minus d b x by d t okay here are my three equations absolutely analogous this one is the result of this is electric field intensity based on vector of magnetic intensity stretched along the z axis this is my electric field intensity as b stretched only in the y direction and this is only if my magnetic field intensity is along the x direction now what if it's a combination like this I'll just add them up together basically and that's it so I will multiply this by by vector I will multiply this by j vector so it would be times i here times j here and here times k times k and if I add them up together on the right side I will have basically the combination of derivative by x direction derivative by b direction derivative by c direction which is a vector which can be which can be expressed as this what is the derivative of the vector well it's a vector which is sum of derivative of each component yj and k are not dependent on time so derivative of the entire thing is derivative of this and I would be a multiplier derivative of by j would be multiplier as a vector and derivative of bz so exactly like we have here that's on the right side with a minus sign okay what will be on the left side on the left side it will be this times i plus this times j plus this times k one square brackets another and the third one that will be on the left side looks like a very large formula right and now as the final accord of this particular lecture I will do a little trick now you know about the symbol nabla symbol nabla it's kind of a pseudo vector we used it when we were considering the first and the second Maxwell's formula now in the first and second we were using scalar dot product of this and electric field intensity in this case I will use a vector product now what is this basically again using again it's not a real vector it's a triplet of operators of differentiation but if I'm using a formal substitution again vectors in quotes i, j and k I just divided one vector into three components if I will multiply this vector by vector e which is e x times i plus e y j plus e z k as a vector product the final thing is you will get exactly the left part of this equation now if you don't believe me now you can go to a textual part of this lecture where I basically very very tediously multiply this by this it means like three of these three of those so it's nine different different terms here and then I have to basically multiply i by i which is zero vector product of vector by itself is zero i by j i by j vector product is actually k so if you are familiar with vector product I mean again if you don't go to mass routines where it's all explained so and that's some kind of a trick which brings you to a very nice formulas instead of all this instead of all this you will have just this formula but this would be expressed as nabla vector product with e equals minus db dT and this is the third equation of Maxwell now the pretanging can do everything 100% rigorously there are some shortcuts etc but anyway I just wanted you to feel basically this formula where it comes from and the very last trick from here to here I just suggest you to read the text of this lecture go to unisor.com it's part called waves field waves and that's where the equations number 3 is so it's all explained there but it's again it's a long and tedious multiplication of this by this and you will get exactly this well some smart people actually came up with this idea and well that's what it is so you've got this thing and again I suggest you to read the text complete text for this lecture it will basically put some more emphasis wherever it's necessary other than that thank you very much and good luck