 Hi, I'm Zor. Welcome to a new Zor education. I would like to continue talking about electromagnetic induction by presenting a couple of problems, simple problems. But they do contain some mathematics and again the whole course of this particular course of physics is the course from the Looking Glass of Mathematician. So you will just see that math is really playing a very important role in physics and in this particular case you definitely need something in calculus. You have to understand and remember what differential is, what integral is, etc. So this lecture is part of the course called Physics for Teens, presented in Unizor.com and this is the second course, the first one is Maths for Teens. That's a prerequisite and that's exactly what I will be using today. I do suggest you to watch this lecture from the website from the Unizor.com. Every lecture on this side has textual notes or explanation, whatever, which is basically like a textbook. So for each lecture there is a corresponding text book. It's the same. I mean, I'm using maybe some different words, but it's basically the same material and sometimes text is more precise even than whatever I'm talking about here, like in calculations, for instance. Now the site is completely free. There are no advertisements, financial strings or anything. Also the site contains exams for those who would like a challenge and for people who prefer to be supervised in some way or another, if there is like a parent or a teacher who can tell, okay, this should be your next assignment, something like this. So there is a corresponding functionality in the website in Unizor.com because there are different categories of users. There are supervisors, parents, and there are students. There are also teachers who contribute the material, not those who are presenting it, but those who are contributing the material. And they might actually present it as well, like in case of myself. Okay, so the first problem. The first problem is analogous to whatever I was discussing during the corresponding lecture. It's about you have certain magnetic field uniform and these are magnetic field lines which are going that direction. From this part of the space in front of the board towards behind the board. Now in this space, I have the following construction. I have two rails, if you wish. What's convenient is to introduce system of coordinates. Now this is A and this is minus A. Along the y-axis and the rails are parallel to the x-axis and the magnetic lines are parallel to the z-axis which goes perpendicular to the board. Now these rails are connected here and there is some kind of a wire or another rail, whatever it is, which also here, but it's sliding this way. Now, so far it's exactly the same as whatever I was explaining during the lecture, but here is the difference. All these four different pieces of this rectangle are made of some material which has certain electrical resistance. Now let's assume that these are all of the same material in the same cross-section. So what I have is I have lowercase r, which is resistance per unit of lengths. So if you have a unit of lengths, let's say one meter or whatever, so lowercase r is resistance of this one meter of this material, of this wire, if you wish, or a rail. And obviously, whenever this rectangle is created, every side has resistance, which is equal to lowercase r times the lengths of the side. So lengths of this side, by the way, is 2a, right? It's a above and a below the x-axis. And same thing is this one. These are parallel. So this is a true rectangle, but this wire or rail, it moves back and forth. Now, here's my problem. Whenever I'm moving this particular wire, since I'm moving, crossing the magnetic field lines, there is a magnetic induction. So there is some kind of an electric current. There is electromotive force and on the ends of this wire. And that's why there is a circulation. There is an electrical current in this particular circuit. Now, if I am moving towards the right, towards increasing of the area of this rectangle. Now, we know that electromotive force, voltage is equal to the rate of changing of the magnetic flux. Where magnetic flux is basically the intensity of the field times the area of the rectangle. So the intensity of the field is given, b. And the area of the rectangle is this area of the wire frame. So as I'm moving my wire towards the right, I'm increasing the area of this frame. And that's why my phi, my magnetic flux is increasing. And that's why its first derivative is positive and then creating the positive voltage. So what's my problem? Here is the problem. As I'm increasing the voltage, I also increase the resistance of this wire, right? Because resistance is some of the four sides, resistances. And these two sides are increasing in lengths. These are fixed and these are increasing. So my problem is what should be my speed of moving my wire to the right? To make sure that the current is constant. So the electric current in this voltage in this circuit must be constant. Since I'm increasing the resistance, my voltage, my electromotive force should also increase. Which means I cannot just move with a fixed speed to the right. Because the fixed speed to the right means my area of the rectangle is growing linearly. Which means my derivative, b is constant. So the derivative of phi is b times the derivative of a stress. Now if s is linear, my derivative is constant. So this would be constant and constant voltage is not what I want. Because with constant voltage and increasing the resistance, according to the Ohm's law, my current would go down. I needed to be stable. I needed to be a stable current. Which means I have to increase the speed of my wire. If I increase the speed, my area would grow, not linearly, but faster than linearly. How it's a different question, but it should be growing faster than linearly. Maybe quadratically, maybe exponentially, I don't know yet. We have to make some calculations. So this is the subject of this problem. We have to find out what exactly is the dependency of the speed of time. Speed should increase with the time. So I need a formula. Alright, so it's actually not very difficult problem. All we have to do is to specify exactly what it is. Now first of all what is my resistance? Resistance is a function of t, right? Now let's say x is coordinate of this point. And this is function of t. Well, if x is coordinate, its first derivative is the speed. So we have to find back to this function and maybe it's derivative if you want to know the speed as well. So what is my resistance? Well, if I have this coordinate for simplicity, let's consider this to be at the beginning of coordinates. So it would be easier. So this is my beginning. So this is my beginning position of the wire. So my x of 0 is equal to 0. So in the beginning these two wires left and right are at the same x is equal to 0. And then I start moving. Okay, so that's probably enough initial conditions. Okay, so r of t is equal to lowercase r times the lengths. Now if this piece is x of t. So this is x of t. This is x of t. This is 2a and 2a, right? So my total resistance would be double x of t plus 4a right? This is aaa. That's my total resistance. Okay. Now what is my area? Area is equal to x of t times 2a. 2a times x of t. Okay. Now what is my magnetic flux? Well, magnetic flux also a function of t obviously is the combination of b times area. So it's 2a x of t. Finally, what is my electromotive force, my voltage? It's a derivative of this. So it's 2a b x derivative. Now we are using the Ohm's law. If you divide voltage by resistance, you will get the current. So let's say I want i0, which is a constant, as the constant current in this particular circuit. So u divided by r should be equal to i0 constant. Well, this is an equation. So let's just substitute whatever we have here. So my u is this 2a b x of t divided by r of t, which is this r times 2x of t plus 4a. And this is supposed to be constant. What is this? I know all the values. I know a b i0 and I don't know the function x of t. This is a differential equation. So differential equations were covered in the calculus part of mass for t and scores on unizor.com. This is a very simple equation. So let me just very simply solve it. Okay, so I will cancel 2 here. And so I will have a b x of t is equal to r x of t plus 2a. k is equal to i0. Or this thing is a constant. So I'll put it to the right. And instead of x prime, I will use d x by dt. And I also have this x of t plus 2a. And it's equal to i0 times r divided by a b equals I will use the letter c just to actually I don't want to use c. I'll use k for simplicity. Because these are all given numbers. This is the current which I would like to have. This is the resistance of the unit lengths of the wire. This is the coordinate of the horizontal rails a and minus a. And b is magnetic field intensity. All given. So k is given. So this is the only thing which I have. Now how to solve these differential equations? Very simple. You move dt to the right and integrate. So on the left I will have d x of t divided by x of t plus 2a. On the right I have k dt. Now differential of x of t is exactly the same as differential of x of t plus constant. So now I have the same here. And now I can integrate. Now the indefinite integral from this is logarithm of x of t plus 2a. Lager integral of this is equal to k t. Plus I have to put constant. Since we are integrating I can always add the constant. Now what is this constant equal to? Well it's very easy to find out. From here x of t is equal to, well let's use exponent on both sides e to the power of this. So I will have on the left x t plus 2a. On the right I will have e to the power of k t plus c. But again this is plus constant. I can put constant here. It's just any constant. So this constant is not the same as this one, but it's still a constant. Some constant. How to determine? Well we have initial condition. t is equal to 0. We have x of 0 is equal to 0. Remember my wire at the beginning of time is coinciding with the left wire. So if I put 0 here I will have c and this is 0. So c is equal to 2a. So I can safely put it to a e to the power of k t where k is this one. So my x of t, my x of t is equal to 2a times e to the power of k t which is a0 r divided by ab, capital B, ab t minus 1. Right? 2a, if it goes to the right, I will factor out and I will have this e to this power minus 1. This is dependency on t. As you see it's exponential dependency which means I have to move with my coordinate exponentially increasing. Now what is my speed? Well speed is the derivative of this, right? Derivative is equal to 2a times derivative of inner function, right? Now inner function obviously minus 1 is disappearing. So I have only derivative of this. Now derivative of e to some power, it's e to this power times derivative of the inner function. Inner function is just a multiply. So I will have i0 r ab times e to the power i0 r t divided by ab. So this is disappearing and what is my initial speed? Now my initial position is 0 but what's my initial speed? Well if you put t is equal to 0, this thing would be e to the power of 0 which is 1 so it would be 2 i0 r divided by b. That's my initial speed. Now you can actually check it out because you have a problem which is kind of investigated before. What if you have the wire in the magnetic field and you are moving this wire? What actually happens? Well if the wire has electric current in it it's a reverse. So if you put electric, you remember the Lorentz force? So if you have a current in the wire, this is your wire and you have current in this case it's i0. I know it's i0, right? Because that's how I specified all the parameters. Now the length of this wire is 2a, right? So I have a Lorentz force which is equal to remember what it is. It was the current times length which is 2a times b, right? So this is the force actually which moves to the right. You see we have very very close formulas here. Now if this is the force then you can check exactly what kind of a movement it is if you have the mass for instance. In this particular case what I would like to point out is that we are not really far. I mean the formulas are very very close to each other so whenever I'm starting this movement I have the formula which really kind of resembles the Lorentz force which actually moves the wire to the right. But in this case we know that the wire is supposed to be moving with the speed and I would like to point out that they kind of correspond. So basically that's what it is. This is my speed, this is my coordinate of the of the wire. So again it's moving exponentially faster and faster. See speed is exponentially growing. It's not a constant speed to assure the constant current in the circuit. Well that's it. So let's go to the second problem. Well as you see for relatively simple physical problem you do need to know mass. That was actually one of my main points. That mass is extremely important. You have to be very comfortable with all these tools. Differential integration, a very simple differential equation. Okay the second problem. So what do we have the second problem? Okay the second problem is the following. If you have a frame which is spinning around the z-axis. Now this is your magnetic field lines. Again they're resuming uniform. Okay actually in my text I have it parallel to the x-axis. That's the problem. This is parallel to the x-axis. And this is initial position of my wire frame. Now what happens for instance when you turn on the electric motor? Well first it's not moving. And then it's moving faster and faster and faster and during certain time it reaches its maximum right and moves with that angular speed. So the angular speed is changing. That's what I wanted to say. Now as angular speed is changing obviously the voltage generated in this magnetic field would be changing. So my question is if I give you a function of how my angular speed is changing with the time. Let's say in the beginning it's zero. And then as we are basically letting this frame to rotate well it actually generates a certain amount of electric motor force. So I would like to know what exactly is this electric motor force. So I need to know u of t. Now I have dimensions of the frame. It's a times b, a by b. So basically I have everything right. I have my magnetic field. I have frame a by b. And obviously I have my area. It's equal to a times b. This is area of the frame. But this is not the area which is supposed to be used when calculating magnetic flux. So to calculate the electric motor force, the voltage, we need to know the magnetic flux. U of t is equal to rate of change or first derivative of magnetic flux. And magnetic flux, in this particular case, it's not just b times s. It's times s of t, which is the function. You see whenever we have this direction of the magnetic field lines, my initially, my frame is parallel to magnetic field lines. So nothing is crossed. So whenever I'm trying to move it, whenever I'm rotating this, the area through which magnetic field lines are going through is increasing. Whenever I'm turning by 90 degrees, so the frame is parallel to the y-axis, the area through which magnetic field lines is the maximum. So from zero, it goes to a maximum. Then as it turns again, it goes to zero again. So my area through which magnetic field lines are going through, this is s of t. It's changing. Now what exactly is this area? Well, that's kind of obvious. If you will take a look at this from above, you will see, well, in any case, the area is a rectangle with one side equal to a. And another side, at its maximum, it's b whenever it's parallel to the y. And it's minimum, it's zero. So let's look from the top. From the top, you will see this is b, lowercase b. So if this is the direction of the x-axis and magnetic field lines are like this, then obviously the area of this rectangle is zero. But when I'm turning by angle phi of t, what is my area? It's basically this distance, right? It's this. Now I can multiply a by this. And what is this? Well, if this is, the whole thing is b. So this is b over 2. So this is phi. So this is phi. So this thing is sine. So we have b over 2 times sine of this angle. It's this one. b over 2 times sine of this angle would be this. And we have two of them. So it's b times sine phi of t. Okay? And if I will multiply it by a, I will have s of t. So this is my s of t. This is my magnetic flux. And that's why my electromotive force, which is derivative of the flux, is equal to b times derivative of s, which is a b derivative of sine is cosine times derivative of inner function. Inner function is phi of t. Phi of t is angle of turning. What is derivative? Derivative is angular speed. And we have it. It's omega of t. So we have all components here. Omega of t is given. B, A, B are given. So the only thing which is not really given is phi of t, the angle. But we have an angular speed. And we have the beginning. I said the beginning is parallel to the x-axis, which means the angle phi is equal to zero, right? So in the beginning, phi of zero is zero. So what is phi of t if I have omega of t? Well, it's integral of zero to t, omega of t dt. If t is equal to zero, from zero to zero obviously it's zero. Now it's integral because derivative of angular position is angular speed. Same thing as in linear derivative of the distance is speed, right? So distance in rotation is measured by the angle of turning, angular rotation. And angular speed is its derivative. So I'm just reversing this. If this is a derivative of this, then this is integral of that. All I need to do is integrate it with proper limits. And this gives me exactly what is the constant of the upper limit. Well, that's basically it. So this is my answer. That's my electromotive force. And it's expressed in terms of angular speed and angular parameters. Well, that's it. Now these problems are not really difficult from the physical standpoint. I would say it's a little bit more difficult mathematically. And maybe that's why I have chosen them. So I do suggest you to read the notes for this lecture. So you go to unizor.com, choose the physics 14th course. And there is an electromagnetism chapter in it. And this is where then you have to go to electromagnetic induction. That's where you will have these problems. Okay, thank you very much and good luck.