 Hi, I'm Zor. Welcome to Unizor Education. Last lecture was about magnetic field around a straight line electric current. And I would like to just solve a couple of problems related to this particular situation. Now, this lecture is part of the course called Physics 14 presented on Unizor.com. If you watch this lecture from the website, you will have an advantage of basically having the whole course in certain logical sequence. In addition, the same website contains Math 14's course, which I consider to be a prerequisite. You have to know Math to study Physics. Now every lecture has a very detailed written explanation notes, which is basically like a textbook. And for some people who like to be challenged, there are exams on this website. And the website is completely free. There are no advertising. You don't even have to look in if you don't want to. Okay, so, two problems. Well, first of all, let me just very, very briefly repeat what we were talking about last lecture. So, if you have a straight line with electric current, and the line is relatively long. I mean, really, from the physical perspective, it's long. From the mathematical perspective, it's infinitely long and thin. So, it's a thin wire. Now, there is a current I, which is running along this wire. Now, I was talking about magnetic field, which is caused by running electrons. And the field lines, the force lines, are circular around their concentrical circles at any point. Also, altogether, the lines of the same radius are making a cylindrical surface, with the wire being an axis of this cylinder. Now, the bigger cylinders, which have bigger radius, also have the magnetic field line, and the bigger one, and the bigger one. Basically, again, it goes to infinity, but getting weaker and weaker, because the field is spreading to a cylindrical surface of a bigger and bigger radius. And that's why per unit of area, the amount of energy is falling less and less. And it's proportional to number of electrons, which are going along the line. And it's inversely proportional to the circumference of the circle at the radius r. And there is a coefficient mu zero, which is called space permeability, which basically connects the units of measurements. This is Tesla, this is Ampers, this is Meters. So, that's how mu has certain value. Now, one more thing about the direction of the forces. So, if magnetic lines are circular, it means that, let's say, if you will put a compass here, it will direct the north towards some direction, which is tangential to the magnetic line anywhere. Here it will be tangential, here it will be tangential. So, what is the direction? The direction is determined by the rule of the right hand. So, if you will put, this is a direction from plus to minus. So, if you will put the right hand around this wire in such a way that your thumb points towards the direction from plus to minus, then your fingers will show you the direction of the magnetic lines. The direction, basically, of the force, because the magnetic line represents, basically, the force. And the force has a magnitude and direction, obviously. So, the direction is always tangential to the line, and the magnitude is this. So, in this particular case, as you see, force is always, since it's tangential to this circle, and circle is in the plane, which is perpendicular to the wire. So, all the magnetic forces are perpendicular to I. So, B as a vector is perpendicular to vector I as a vector from plus to minus. So, that's my kind of preamble, and I also told you about this rule of the right hand. By the way, there is another kind of rule, if you wish. And you know how the cork screw is working, right? You are turning it this way, and it goes this way. Or if you're turning this way, it goes down, right, into the cork. So, this is the same thing. If you consider this to be a cork screw, it's going this way along this, and it points towards the direction of the current. So, it's either the cork screw or the right hand rule helps you to determine the direction of the vector of intensity of magnetic field. Now, the problems. Problems are really simple. It's just, whatever I just said, that's the application of this. My first problem is the following. So, you have a Cartesian coordinate system. You have two wires, not one, two, two wires. One wire goes parallel to the y-axis, parallel to the y-axis. At the distance a, well, actually it's minus a. So, it's zero at x minus a, no, zero at y, and minus a at z. So, this is coordinate of this point. Zero at x, zero at y, and minus a. And you have the wire which is parallel to the y-axis. Now, here, on the opposite point, also a, you have the wire which is parallel to the x-axis. So, the coordinate of this point is zero, zero, a. So, these are two points on z-axis through which one wire goes parallel to the y, and another parallel to the x, to the x-axis. Now, the direction is towards the positive of the y, and this is towards the positive to the x. Now, let's consider you have the same current I. So, this is plus, minus, and this is plus, minus. My question is, what is the vector of intensity of magnetic field at the origin of coordinate? Now, you understand that this is actually sum of two vectors. You know, when you have two fields, you basically have superposition of forces. There is one force which is caused by this magnetic field force, and another force which is produced by this wire. So, these forces are two vectors, and they must be added together. That's the rule of superposition. If you have two forces acting at the same point, the resulting force is just a vector sum. So, you have to know how to sum the vectors, right? And if you don't know, you have to always go through the mass-frittin's course, which is presented on the same website. Okay, so, first we will have to establish what is the vector produced, a magnetic force intensity vector produced by one wire and then another one. Okay, so, we know the I, and we know the distance from the wire to the point which we are interested in. So, how the magnetic field around this wire will actually be? I mean, how exactly it will behave this magnetic field? Well, again, magnetic field lines, that's the very important thing. You always have to think about magnetic field lines. Magnetic field lines are lying in a plane perpendicular to the wire. So, you have a plane perpendicular to the wire and it's a circle with the center at the wire. Now, what is the plane which is perpendicular to this wire? Well, this wire is parallel to X. So, the plane which is perpendicular to this line should be parallel to YZ, right? And we also need such a circular line which is perpendicular to this and goes through this point, right? We are interested in magnetic field in this particular point. So, let's just think about it again. It's supposed to be in the plane which is parallel to YZ plane and it's supposed to go through this, it's supposed to be a circle with this line point at the center. So, it would be something like this. This is a circle which is in the plane YZ. Now, since YZ is perpendicular to the X and my wire is parallel to the X, so my wire will be perpendicular to this circle. So, you can put this as a dotted line, right? It's behind this circle, right? So, this part is in front and this dotted line is behind this circle. Now, let's talk about the force. Well, the force at this particular point should be tangential to this circle and the circle is basically lying in the YZ, so it's the plane of my board. So, everything, this is a real circle and it's supposed to be tangential and also supposed to be corresponding to right hand rule, right? So, I have to put my hand around. So, my current goes this way and it goes this way. I think, right? That's how it goes. Which means my vector of force is here. This is my B1. It goes along the Y axis and its magnitude is mu zero i divided by 2 pi and the distance is A, right? So, this is my magnitude and this is my direction along the Y axis. Okay, now let's talk about this wire. Now, this wire is parallel to Y, which means the plane which is perpendicular to this should be perpendicular to Y and it's basically XZ plane, right? This plane, XZ plane is perpendicular to Y. So, I have to find a plane which is perpendicular to the wire and goes through this point. So, what is this plane? Well, that's actually the Y-X-Y plane itself. And I have to find a point which... I have to find a circle, actually, which goes through this point and its center is this one. So, it's something like this. If you will just think about the direction, the direction is, again, it's this way. So, it looks like it's this way. So, the vector goes this way along the X axis and this is my B2. This is the direction, but since my current is exactly the same and distance is exactly the same, my value of magnitude will be exactly the same. So, again, my first magnetic field circle, which basically corresponds to the magnetic field line, is in the YZ plane. My second circle is, which represents the magnetic field line, which goes through this point. It's in XZ plane, the bottom part of it actually, negative part, negative towards Z. And so, I have basically these two. As you see, these are two vectors. They have the same magnitude and they are going one along X and another along Y axis. So, they are perpendicular to each other. So, what will be their result of their vector sum? Well, the vector sum will be along the... Well, it will be at 45 degrees, right? From both, because this is 90 degrees. These are perpendicular to each other, right? And the magnitude would be what? Let's just look from the top. It will be basically XY. This will be B1. This will be B2. And what's the hypotenuse? Well, the hypotenuse is the square root of this plus the square root of this. And they have the same value, so it would be mu0i over 2pii squared. Times 2, because you have to add them together and square root from it. So, it's muI2piA and then you have to multiply it by square root of 2. And that's the magnitude of this vector. So, this is the answer. The vector will be within XYZ starting from the origin going along the main diagonal at 45 degrees. And the magnitude is this one. So, this is my first problem. And the second problem will be very, very much like this one, just slightly different numbers, whatever. I'm basically trying to do these simple things. So, you will feel the right-hand rule or corkscrew rule and imagine the whole thing in three-dimensional space. So, you need a little bit of imagination of how it all looks. It's always good to have a little geometry in this. Speaking about geometry, it just emphasizes one more thing, one more time, that mass is very, very important for physics. You really have to know your mass. Calculus is a must for physics. And like in this and many other problems, especially mechanics, you need vector algebra. You need to know how to add, subtract, and multiply vectors by scalar or vector by vector, etc. So, my second problem is, again, you have the Cartesian system of coordinates, but we have different organization of wires. Now, they're parallel each other. One goes parallel to z, both wires. One crosses the x-axis at A. So, the coordinate of this point is A00. X coordinate is A, and y and z coordinate of this point are 0. Another is this B from the origin. So, the coordinate of this point is 0x B and 0 on z. Now, the direction is plus-minus. So, the direction is here, plus-minus. The direction is here. And I'm again interested in, yeah, this is I1 and this is I2. Now, we have different currents running. And again, it's ideal situation, which means that the wires are infinite, infinitesimally thin and infinitely long. So, there is a current. So, we need to calculate the magnetic field force at this particular line, at origin, magnetic field intensity force. Okay, so, again, we are, first of all, we have to think about magnetic lines. So, let's talk about this. Now, it's vertical. It goes parallel to z, which means that my magnetic lines are supposed to be perpendicular. So, if it's parallel to z, my magnetic lines are in the plane, which is perpendicular to z, which is xy plane. So, it's in this particular plane, and obviously in all parallel planes. I mean, they're all around. So, which plane, which is perpendicular to this wire, goes through this point. Well, that's actually xy plane, right? So, when we're talking about a circle, which is crossing this point, and at the same time, it's in the plane perpendicular to this, it's supposed to be centered at this, and the circle would be something like this. Okay, that's xy plane, and it's perpendicular, and since my direction is down, so it's here. This is the direction of the magnetic lines, the magnetic force, right? Which means that at this point, what is tangential to this circle, well, tangential to this circle, which has a center here, and the radius is exactly this, but this line is exactly perpendicular to the radius, so it would be here. So, this is my b1 vector. Okay, now let's talk about this. Again, this is parallel to z, which means perpendicular should be parallel to xy, and it should go through this point, and it's supposed to have a center, obviously, here, so that would be something like this, this circle. And my direction is right-hand, it's supposed to be this way, and at this particular point, perpendicular to this circle, which is completely in x, y, z, in xy, that's where xy plane should be perpendicular, and this is this one, so this is my b2. So, if you will look from the top, where you will see only xy, you will see one circle on x, that's a, goes this way, and the direction is this, so this is my b1, and another is y, this is the center, this is my circle, and the tangential would be here, this is my b2. So, these are two perpendicular vectors. Obviously, their magnitude is b1 is equal to mu i1 divided by 2 pi, distance is a, and b2 is mu 0 i2 divided by 2p b, because the b is distance, okay? So, this is b, this is a, and these are, this goes this way. So, these are two vectors. And again, this is the plane Pythagorean theorem. When you have to really find out what exactly is there, the magnitude of their resultant vector, it's just square root of 2, of sum of two squares, right? And the direction is obviously depending on a and b. So, if you have a rectangle of a and b, then this is what? This is arc tangent of b divided by a, right? b divided by a is tangent. So, the angle itself is arc tangent of this. So, we know this is our... So, this is an angle. So, we know the angle. And the magnitude is square root of b1 square plus b2 square. What will be that? Well, obviously, mu 0 divided by 2pi will be outside, and the square root will be of a1 divided by a square plus d2 divided by b square. So, this is the magnitude. This is the angle. So, what have we learned today? Well, number one, that the intensity of the magnetic field is always perpendicular to the wire itself. So, the vector of intensity is always... All vectors of intensity on the same magnetic line are in the same plane, and magnetic line itself is a circle in the plane which is perpendicular to the wire to the direction of the current, actually, right? So, that's what's very important. Important is the right hand rule or a corkscrew rule. So, whenever you are turning your hand around the wire pointing towards the direction of the current, your fingers show the direction of the intensity along the magnetic line which is a circle. And basically, everything else, again, I just applied the formula which we have derived in the previous lecture. And again, what's important is that whenever you have two sources of magnetism, then the vector of intensity is always a sum of vectors of intensity of components. This is just a plain kind of a vector algebra of things. And in this case, I was trying to do it in such a way that the vectors are perpendicular to each other, so you don't really have to go to these parallelograms. That's just a plain Pythagorean theorem. Okay, that's it. I suggest you to read the notes for this lecture. On Unisor.com, you have to go to, obviously, Physics 14's course. Then you choose electromagnetism subject. And in that subject, you have to find the magnetism and the current, electric current, where these particular problems, it's problems one you will find. So there are some nice pictures, much nicer than I did here, obviously. And textbook-style explanation, I think it's very helpful. All right, that's it for today. Thank you very much and good luck.