1 00:00:00,540 --> 00:00:02,380 That was all very theoretical. 2 00:00:02,400 --> 00:00:11,520 So in this lecture, we will consider a real example, we will calculate the magnetic field for a current 3 00:00:11,640 --> 00:00:12,420 in a wire. 4 00:00:14,070 --> 00:00:16,930 So this is the set up that we are considering here. 5 00:00:17,460 --> 00:00:26,040 So we have such a straight wire and it has the length L and here this is the equation that we have established 6 00:00:26,040 --> 00:00:28,170 previously for the vector potential. 7 00:00:28,500 --> 00:00:33,160 And you can see that it includes an integral of the current density. 8 00:00:33,480 --> 00:00:38,370 So our current flows in this wire and at the position are prime. 9 00:00:38,400 --> 00:00:41,250 And then we have here two different position vectors. 10 00:00:41,910 --> 00:00:45,270 And so these are the different position vectors. 11 00:00:45,610 --> 00:00:55,110 So this one here includes all the positions that are basically inside of the wire and the other one 12 00:00:55,260 --> 00:00:56,970 includes the whole space. 13 00:00:57,120 --> 00:01:02,610 So because we want to know the vector potential at every possible point in space. 14 00:01:03,180 --> 00:01:06,240 So let us now go ahead and calculate this integral. 15 00:01:07,470 --> 00:01:11,220 So the first thing I do is I apply an approximation. 16 00:01:11,880 --> 00:01:15,300 I want to consider here an infinitely thin wire. 17 00:01:15,890 --> 00:01:22,920 So this means that the square root of this area, so basically you could say the radius or the diameter 18 00:01:23,430 --> 00:01:24,870 is very, very small. 19 00:01:25,020 --> 00:01:26,400 It's infinitely thin. 20 00:01:27,330 --> 00:01:32,140 So what this means is that at every position outside of the wire. 21 00:01:32,640 --> 00:01:38,970 So when we are when this R is outside of this wire, we can say that this distance to the wire is much 22 00:01:38,970 --> 00:01:40,660 larger to the area. 23 00:01:41,370 --> 00:01:48,210 Of course, this is not really true if we are very, very close to the wire, but we are here only interested 24 00:01:48,570 --> 00:01:53,340 in the vector potential and in the magnetic field far away from the wire. 25 00:01:54,600 --> 00:01:58,260 Or we could also say we consider an infinitely thin wire. 26 00:01:59,700 --> 00:02:07,130 So what this does is it allows us to separate disintegration of the volume into an integration over 27 00:02:07,140 --> 00:02:12,900 the cross section of an area and an integration over the Z component. 28 00:02:12,930 --> 00:02:14,210 So along the wire. 29 00:02:15,180 --> 00:02:21,660 So this means we have here is integration of all of the whole length and here this integration of this 30 00:02:22,020 --> 00:02:24,120 cross section, surface area. 31 00:02:25,470 --> 00:02:34,110 And now we can really apply or we can carry out this integration of a D over this area because we know 32 00:02:34,440 --> 00:02:38,040 that this term here is much larger than the surface. 33 00:02:39,210 --> 00:02:47,220 So this means we can calculate integral over F then this G of R and D. 34 00:02:47,220 --> 00:02:52,430 S, and this just gives us the current that is flowing through the wire. 35 00:02:53,460 --> 00:02:59,070 And this current is not dependent on the position because our wires straight and has the same cross 36 00:02:59,070 --> 00:03:04,170 section area along all the length so we can pull it out of the integral. 37 00:03:04,980 --> 00:03:06,540 So this is what we are left with. 38 00:03:07,470 --> 00:03:12,690 Please keep in mind this is an approximation where we say that the current does not change. 39 00:03:13,170 --> 00:03:15,340 Yeah, 30 the area. 40 00:03:15,360 --> 00:03:18,240 So we can just separate these two integration's. 41 00:03:20,150 --> 00:03:26,270 Now, what this means is we have simplified our expression for eight, so that means we can pull out 42 00:03:26,300 --> 00:03:26,550 here. 43 00:03:26,550 --> 00:03:32,280 It is current as well, and we are left only with this integration over the length of the wire. 44 00:03:32,870 --> 00:03:39,710 So what we have done essentially is we have simplified our problem to a quasi one dimensional problem, 45 00:03:39,890 --> 00:03:42,920 which is because our wire is infinitely thin. 46 00:03:44,420 --> 00:03:48,500 So what we have to do is we have to integrate here from minus L to L. 47 00:03:48,650 --> 00:03:55,150 So let's say here zero and here is minus Alah divided by two and here is plus Al divided by two. 48 00:03:55,160 --> 00:03:57,680 And these are the boundaries for our integration. 49 00:04:00,040 --> 00:04:08,560 Now, this distance in these two position vectors is actually given by this distance here, and you 50 00:04:08,560 --> 00:04:16,450 can apply now this Pythagoras theorem where you have here the square root of L Square and square, and 51 00:04:16,450 --> 00:04:22,390 then you integrate over the L and we know that this wire is always along the Z direction. 52 00:04:22,570 --> 00:04:28,660 So this D-R vector here, I have expressed in terms of the L, it was just one dimensional scalar, 53 00:04:28,660 --> 00:04:31,930 you could say, and then some vector easy. 54 00:04:34,300 --> 00:04:41,170 Now we carry out this integration here and the integral of this one of a square root thing here can 55 00:04:41,170 --> 00:04:44,100 be expressed in terms of this logarithm. 56 00:04:45,130 --> 00:04:51,520 So we have as a solution this logarithmic function with the boundaries that we still have to put in 57 00:04:51,520 --> 00:04:52,060 for our. 58 00:04:54,280 --> 00:05:01,990 Now, I apply another approximation, which is that we consider an infinitely long wire, so this means 59 00:05:03,850 --> 00:05:07,380 that it doesn't really matter at which position we are looking at. 60 00:05:07,390 --> 00:05:13,050 It doesn't matter if we are right in the middle or bit above because our wire is infinitely long. 61 00:05:13,060 --> 00:05:17,240 So we are basically always sitting in the middle, roughly speaking. 62 00:05:17,650 --> 00:05:24,910 So this means we always have an infinitely long wire above us and below us, which allows us to simplify 63 00:05:24,940 --> 00:05:25,690 this expression. 64 00:05:26,820 --> 00:05:34,140 Because now we can see that this vector potential is given by the same prefecture, but the multiplied 65 00:05:34,140 --> 00:05:41,280 by two, then we have our current again and is natural logarithm simplifies just to al the length of 66 00:05:41,280 --> 00:05:44,360 our wire divided by S, which is this distance here. 67 00:05:44,970 --> 00:05:47,750 And then we have the orientation, which is a long this direction. 68 00:05:48,870 --> 00:05:55,500 And just so you know what it is, this s is of course the square root of X squared plus Y square, because 69 00:05:55,620 --> 00:06:00,570 remember, even though this looks two dimensional, here are real problems, of course, three dimensional. 70 00:06:02,520 --> 00:06:06,480 So now we know what our vector potential looks like is the solution. 71 00:06:07,410 --> 00:06:13,500 And we have applied here to the approximation of this infinitely long wire and also the approximation 72 00:06:13,500 --> 00:06:15,060 of an infinitely thin wire. 73 00:06:16,020 --> 00:06:23,730 Now, what we do is we calculate the magnetic fields, and for this we only have to calculate the rotation 74 00:06:23,730 --> 00:06:25,200 of this expression here. 75 00:06:26,730 --> 00:06:33,930 So, of course, you could now apply polar coordinates and do some fancy tricks, which I think would 76 00:06:33,930 --> 00:06:34,890 be very elegant. 77 00:06:35,310 --> 00:06:41,460 But sometimes I think you understand more if you just do it straightforwardly and don't use any fancy 78 00:06:41,460 --> 00:06:47,610 tricks and just write down all of the individual components of your vector and just carry out these 79 00:06:47,610 --> 00:06:48,560 differentiations. 80 00:06:49,110 --> 00:06:51,660 So here I have written down. 81 00:06:52,710 --> 00:06:53,100 Yeah. 82 00:06:53,100 --> 00:06:57,000 The derivatives of the sector potential in terms of the three components. 83 00:06:57,630 --> 00:07:01,290 So here can see our vector potential only has a Z component. 84 00:07:01,560 --> 00:07:07,210 So this means we only get two entries here for the calculation of this vector product. 85 00:07:07,800 --> 00:07:15,120 So here I have calculated the derivative of the Z component with respect to the Y coordinate and here 86 00:07:15,450 --> 00:07:20,820 the derivative of the Z component with respect to the X component with a negative sign. 87 00:07:21,810 --> 00:07:25,380 And yeah, I can I can go through it for one of these components. 88 00:07:25,860 --> 00:07:32,940 So d that revetted of the natural logarithm is one divided by this argument here. 89 00:07:32,940 --> 00:07:37,770 So we get, you know, basically the reciprocal value of this one here, which is written down here. 90 00:07:38,880 --> 00:07:44,660 Next, we have to calculate it in a derivative, which is the derivative of this term here. 91 00:07:45,240 --> 00:07:51,260 So we have basically this X squared plus Y square in brackets to the power of minus one half. 92 00:07:51,750 --> 00:07:57,910 So we get the factor of minus one half and then we have to decrease the exponent by one. 93 00:07:58,170 --> 00:08:02,160 So we have now an exponent of minus three 1/2. 94 00:08:02,490 --> 00:08:03,870 So this is here, this one. 95 00:08:03,890 --> 00:08:08,040 So the square root, which is exponent of one half to the power of three. 96 00:08:08,700 --> 00:08:14,490 And then we also have to consider it in a derivative of this thing here, which is if we derive for 97 00:08:14,520 --> 00:08:17,550 the Y component, which is just to Y. 98 00:08:18,520 --> 00:08:21,770 So we have these three different parts that we have to multiply. 99 00:08:22,420 --> 00:08:26,320 And for the X component, you can see the first two parts are exactly the same. 100 00:08:26,320 --> 00:08:29,740 And then for the third part, we just get to X instead of two Y. 101 00:08:31,880 --> 00:08:38,870 So now we can see that most of this expression of yours is the same for these two components, so we 102 00:08:38,870 --> 00:08:44,720 can pull this whole thing out of the of the vector and write it in front of it. 103 00:08:45,650 --> 00:08:50,590 And also, you can see here we have the square root of X squared was twice grand. 104 00:08:50,600 --> 00:08:53,450 Here we have the same thing, but to the power of three. 105 00:08:54,050 --> 00:09:00,290 So we are left with the power of two, which is just one, which just compensates the square root. 106 00:09:00,470 --> 00:09:02,380 So this is why we get this expression here. 107 00:09:02,900 --> 00:09:06,180 And for the vector we just have here, minus Y, X and zero. 108 00:09:07,940 --> 00:09:14,750 Now, the next thing or we could say, OK, this is our solution, but we could also make it a bit more 109 00:09:14,840 --> 00:09:15,560 useful. 110 00:09:15,830 --> 00:09:24,050 So you can see here this is pointing along the Y directions depending on X, and here we have basically 111 00:09:24,050 --> 00:09:32,030 one over the square root of X squared plus Y square squared and the square root of X squared plus Y 112 00:09:32,030 --> 00:09:37,110 square is basically the, you know, the length in the X Y plate. 113 00:09:37,930 --> 00:09:46,100 So this means we can re express this in terms of X Square and we can take one of these S's to this vector 114 00:09:46,100 --> 00:09:53,670 here and this will give us the unit vector in terms of our indirect and pull our direction. 115 00:09:53,720 --> 00:09:56,420 So just to pull our unit vector. 116 00:09:57,350 --> 00:09:58,700 So this is our solution. 117 00:09:59,000 --> 00:10:07,460 We have here a magnetic field that is determined by the current and by the distance, and it is oriented 118 00:10:07,460 --> 00:10:09,550 always along the polar direction. 119 00:10:10,160 --> 00:10:16,640 And this is, of course, something that you already know from high school, where you just use this 120 00:10:16,640 --> 00:10:20,600 right hand rule, where you have your wire and the current flowing along this direction. 121 00:10:20,840 --> 00:10:28,100 You take your thumb and then you take the fingers of your right hand to see when which directions the 122 00:10:28,100 --> 00:10:29,300 magnetic field lines go. 123 00:10:29,840 --> 00:10:32,980 And this is something that we have derived from scratch. 124 00:10:33,650 --> 00:10:38,030 We have not relied on any phenomenological explanations. 125 00:10:38,030 --> 00:10:45,800 We have really considered the Maxwell equations we have derived abusable are also the law for calculating 126 00:10:46,100 --> 00:10:48,230 the vector potential based on currents. 127 00:10:48,770 --> 00:10:56,120 And then we have gone through all of these mathematical derivations and have ended up with this simple 128 00:10:56,120 --> 00:10:57,310 looking expression here. 129 00:10:57,590 --> 00:11:01,870 Now, as a final comment, I want to discuss these two approximations that we have used. 130 00:11:02,570 --> 00:11:09,420 So these approximations led to the fact that this expression looks so simple. 131 00:11:10,010 --> 00:11:15,620 So, for example, if we would not have considered an infinitely long wire, then we would, of course, 132 00:11:15,620 --> 00:11:21,050 get some XY dependence because then it would make a difference where we cut the wire, where we look 133 00:11:21,050 --> 00:11:26,450 at the magnetic field, for example, if the wire would really end here, then of course our magnetic 134 00:11:26,450 --> 00:11:31,520 field would look very different if we would look here at the top or the bottom compared to the middle. 135 00:11:32,300 --> 00:11:37,170 But since our wire is infinitely long, we are, so to speak, always in the middle. 136 00:11:37,370 --> 00:11:41,930 So every cut through the wire looks the same in terms of the magnetic field profile. 137 00:11:43,430 --> 00:11:49,220 And then for the other approximation here, we have considered that our wire is infinitely thin, which 138 00:11:49,220 --> 00:11:54,530 means that the wire is not three dimensional anymore, but basically just one dimensional. 139 00:11:55,400 --> 00:12:01,240 And this means that when we are very close to the wire, we would see different effects. 140 00:12:01,250 --> 00:12:06,500 We will get we would get additional terms that look more complicated than this one here, because then 141 00:12:06,890 --> 00:12:12,770 there would be really a different distance from this position to the right end of the wire and to the 142 00:12:12,770 --> 00:12:13,930 left end of the wire. 143 00:12:14,600 --> 00:12:20,690 But due to our approximation, all the points in the cross section of the wire have the same distance 144 00:12:20,690 --> 00:12:21,900 to our reference point. 145 00:12:22,550 --> 00:12:24,470 So this is why this looks so simple. 146 00:12:25,040 --> 00:12:30,620 But it turns out it's really a good approximation, because also this is the equation that you learn 147 00:12:30,620 --> 00:12:31,160 in school. 148 00:12:31,400 --> 00:12:36,230 Maybe not this if I hear about this magnitude at least, and also this right hand rule. 149 00:12:36,620 --> 00:12:38,330 So this really turns out to be correct. 150 00:12:38,600 --> 00:12:44,720 And these approximations may seem a bit strange at first, but they are really useful and really sensible 151 00:12:44,720 --> 00:12:45,140 also.