1 00:00:00,060 --> 00:00:07,110 So let us now come to our exercises, and if you had problems calculating these examples, please don't 2 00:00:07,110 --> 00:00:14,070 feel discouraged, because the truth is in theoretical physics, the exercises are most often quite 3 00:00:14,070 --> 00:00:20,280 difficult and you have to spend a lot of time solving them, and especially you have to spend the time 4 00:00:20,280 --> 00:00:21,750 to get the right idea. 5 00:00:23,040 --> 00:00:26,700 So now I will present the solutions for the three examples. 6 00:00:27,930 --> 00:00:33,740 So the first task was about the electric field and point charges, and we had three different tasks. 7 00:00:34,140 --> 00:00:39,240 First of all, we had to write down the electric field of the positive charge and we had to draw the 8 00:00:39,240 --> 00:00:45,480 electric field lines that after we had to do the same thing with a negative charge, and then we had 9 00:00:45,480 --> 00:00:51,270 to consider a combination of two charge that half a distance of the length A. 10 00:00:51,840 --> 00:00:57,750 So first of all, let us write down the electric field for A and B, and in both cases, the result 11 00:00:57,750 --> 00:00:59,830 is directly given by the Cullum's law. 12 00:01:00,600 --> 00:01:06,210 So we have a one over R-squared dependence and here we have a positive charge plus. 13 00:01:06,210 --> 00:01:10,470 Q And here we get the same results but with a negative charge minus Q. 14 00:01:11,480 --> 00:01:18,080 And in both cases, we also want to know about the orientation, so here the orientation is always along 15 00:01:18,080 --> 00:01:19,320 the radial direction. 16 00:01:19,910 --> 00:01:26,570 So from the center where our charge is located, the orientation E.R. goes along the radial direction 17 00:01:26,900 --> 00:01:29,990 and it also determines the direction of the electric field. 18 00:01:31,590 --> 00:01:39,360 Now, for see, the result is actually quite simple to write down, because we want to consider our 19 00:01:39,360 --> 00:01:42,150 first charge plus queue at the position of zero. 20 00:01:42,160 --> 00:01:46,800 So it is just the field of eight and then the other charge is negative. 21 00:01:46,800 --> 00:01:50,890 So it's similar results to be, but it's displaced by a. 22 00:01:51,510 --> 00:01:57,090 So we have to account for this by shifting our vector here so we get our minus a square. 23 00:01:57,510 --> 00:02:02,120 And also we need to be very careful because the orientation is now different. 24 00:02:02,130 --> 00:02:08,910 It is still a radial orientation, but not with respect to zero, but with respect to the position of 25 00:02:08,910 --> 00:02:11,400 the second charge, which is displaced by a. 26 00:02:12,740 --> 00:02:20,900 So while this is written down, quite simply, it's actually very hard to really write down the exact 27 00:02:20,900 --> 00:02:28,070 form because of this R minus a vector here, which is different from E r, so we have to be really careful. 28 00:02:28,700 --> 00:02:35,120 And you will see later in this course we will determine the electric field of such a dipole. 29 00:02:35,150 --> 00:02:36,320 This is what it is called. 30 00:02:37,490 --> 00:02:44,630 But we can already draw the field lines, which is quite easy because we know that for a the field lines 31 00:02:44,630 --> 00:02:45,350 look like this. 32 00:02:45,360 --> 00:02:47,080 So it's along the radial direction. 33 00:02:47,480 --> 00:02:51,410 So they pointier away from the center of this CU charge. 34 00:02:51,950 --> 00:02:53,960 And for B it's the same thing. 35 00:02:53,960 --> 00:02:56,390 But just with it, with the different sine. 36 00:02:57,700 --> 00:03:02,170 OK, so now let's really consider these two charges as pears. 37 00:03:02,200 --> 00:03:06,710 So this is the solution for see where these are displaced by a distance of a. 38 00:03:07,510 --> 00:03:11,580 So what we have to do is we have to add up all of these vectors here. 39 00:03:11,800 --> 00:03:15,010 So, of course, they will also be vectors in this wide space here. 40 00:03:15,940 --> 00:03:21,670 So if we do this, we get something like this, which is the electric field of the dipole. 41 00:03:24,000 --> 00:03:31,320 Now, the second task was about the induction law, so we had to consider a current loop with an area 42 00:03:31,320 --> 00:03:37,590 of a square and it is dragged from outside of the magnetic field to the inside of a magnetic field, 43 00:03:37,590 --> 00:03:43,370 which is constant and is pointing along the perpendicular direction compared to just plain here. 44 00:03:44,100 --> 00:03:46,660 So we could say it's pointing along the Z direction. 45 00:03:47,670 --> 00:03:52,140 And what we want to do is we want to bring this current look inside of the field. 46 00:03:52,380 --> 00:03:55,710 So as you have just seen, we've brought it inside and then out again. 47 00:03:55,710 --> 00:04:01,320 But here in this example, we just want to bring it inside and we want to calculate the voltage that 48 00:04:01,320 --> 00:04:01,950 emerges. 49 00:04:02,280 --> 00:04:09,000 And as we have learned, the induction voltage is given by the temporal change, this derivative of 50 00:04:09,000 --> 00:04:10,110 the magnetic flux. 51 00:04:11,670 --> 00:04:15,930 So basically, all we have to figure out is the magnetic flux of this example. 52 00:04:17,520 --> 00:04:24,240 So the magnetic flux is given by this integral here where we have to integrate the magnetic fields in 53 00:04:24,510 --> 00:04:28,860 the area of this current loop in this area. 54 00:04:29,610 --> 00:04:32,550 And this this here is a vector that points along the Z direction. 55 00:04:32,560 --> 00:04:33,750 So it's parallel to be. 56 00:04:34,140 --> 00:04:37,610 So this means we can just calculate B times day. 57 00:04:38,700 --> 00:04:47,670 And since our yeah, our area does not change, we can also write it down as B zero times the area of 58 00:04:47,670 --> 00:04:50,250 aid that is inside of the field region. 59 00:04:50,760 --> 00:04:58,080 So now we just have to think about how does the field rate region change over time when we move this 60 00:04:58,460 --> 00:05:00,960 this loop here with the velocity of V? 61 00:05:02,190 --> 00:05:07,290 So of course the area that is included is given by the length A, which was this one here. 62 00:05:07,560 --> 00:05:13,800 And then we have another length which I called A in, which is the length of this side here that is 63 00:05:13,800 --> 00:05:14,970 inside of the field. 64 00:05:15,360 --> 00:05:16,880 So right now it is zero. 65 00:05:17,250 --> 00:05:25,200 But if you bring this loop completely into the field, then it is a and this of course, given by the 66 00:05:25,200 --> 00:05:31,350 velocity times the time because we considered a motion with a constant velocity. 67 00:05:31,370 --> 00:05:38,310 V So we have our flux which is given by that one, and we can easily calculate the partial derivative 68 00:05:38,310 --> 00:05:39,280 versus back to T. 69 00:05:39,870 --> 00:05:41,760 So it's just minus B zero times. 70 00:05:41,760 --> 00:05:42,750 Eight times V. 71 00:05:44,850 --> 00:05:50,280 Now, of course, you also have to consider that once this whole loop is inside of the field and you 72 00:05:50,280 --> 00:05:54,640 keep on moving it inside of the field, then the voltage will drop to zero. 73 00:05:55,380 --> 00:05:57,990 This is because then the flux will be constant. 74 00:05:57,990 --> 00:06:03,000 It will just be be zero times a day and then the time derivative will be zero. 75 00:06:05,470 --> 00:06:08,870 Now, the third task is a bit more difficult. 76 00:06:08,980 --> 00:06:15,490 So here we have to consider a straight wire that goes along the direction and we have the magnetic field 77 00:06:15,820 --> 00:06:20,830 that has a profile that is given by this by this polar direction. 78 00:06:20,870 --> 00:06:22,840 There's a certain cylindrical coordinates. 79 00:06:23,270 --> 00:06:26,490 So these we have discussed in the mathematical tutorial section. 80 00:06:27,280 --> 00:06:32,950 But even if you're not familiar with the with cylindrical coordinates, you can still solve this example 81 00:06:32,950 --> 00:06:38,320 if you just use this representation of this vector here in Cartesian coordinates. 82 00:06:39,130 --> 00:06:45,370 So it's it's helpful to know about cylindrical coordinates in this example, but it is not necessary. 83 00:06:46,930 --> 00:06:50,320 So what we have to do here is we have to take amperes law. 84 00:06:50,890 --> 00:06:55,930 And I just want you to check that it is correct so that this can be a correct solution. 85 00:06:56,410 --> 00:07:04,190 So we have to verify that the Kerl or the rotation of B is given by zero times to the current density. 86 00:07:05,920 --> 00:07:07,810 And here I have already given you a hint. 87 00:07:08,350 --> 00:07:14,950 It's very important to realize that our current density is zero everywhere, except for the position 88 00:07:14,950 --> 00:07:18,930 of the wire, which is for real equal to zero. 89 00:07:20,020 --> 00:07:27,040 So this means we have to verify that this one here is zero for every point except for row zero. 90 00:07:27,940 --> 00:07:28,620 So let's do it. 91 00:07:29,770 --> 00:07:31,860 And here you can see the whole solution. 92 00:07:31,870 --> 00:07:33,580 I mean, I already show you that result. 93 00:07:33,580 --> 00:07:36,750 But let me briefly go through this calculation. 94 00:07:37,210 --> 00:07:42,310 So what I have done here is just I have just written down the Knobler vector and then here we have the 95 00:07:42,310 --> 00:07:43,170 magnetic fields. 96 00:07:43,480 --> 00:07:49,600 And here I have used this F.I. vector, which is this one here, and I have used that row is given by 97 00:07:49,600 --> 00:07:52,010 the square root of X squared plus Y square. 98 00:07:53,380 --> 00:07:54,760 So this is sincere. 99 00:07:54,770 --> 00:08:00,940 We have row and here we have one overall, we get one overall square, which is why we have here X squared 100 00:08:00,940 --> 00:08:01,630 plus Y square. 101 00:08:03,580 --> 00:08:09,100 Now the next step I have calculated this vector product here and if you look at it carefully, you will 102 00:08:09,100 --> 00:08:12,410 see that this expression here only has a Z component. 103 00:08:12,460 --> 00:08:20,170 This is where you calculate the X derivative of the Y component of this one here, minus the Y derivative 104 00:08:20,170 --> 00:08:22,160 of the X component, which is this one here. 105 00:08:23,260 --> 00:08:31,300 So we get these two terms and in both cases we have to use the product rule because we have a product 106 00:08:31,300 --> 00:08:34,660 of X times one over X squared plus Y square. 107 00:08:34,990 --> 00:08:37,970 And both of these functions are X dependent here and here. 108 00:08:38,680 --> 00:08:42,610 So this means for both of these terms we get two additional terms. 109 00:08:43,840 --> 00:08:45,470 So, of course, the first one is very easy. 110 00:08:45,490 --> 00:08:51,730 We just have to differentiate X, which gives us one times the other function, which is one of X squared 111 00:08:51,730 --> 00:08:52,480 plus Y square. 112 00:08:53,260 --> 00:08:59,470 And then for the other derivative, we have to take X times the derivative of one over X squared plus 113 00:08:59,470 --> 00:09:08,200 Y square and one of X equals Y square is the same as saying in brackets X squared plus Y square to depart 114 00:09:08,200 --> 00:09:09,010 from minus one. 115 00:09:10,200 --> 00:09:16,560 So we have to multiply by minus one, increase the Powerball, but I just say decrease the power, which 116 00:09:16,560 --> 00:09:19,440 gives us power of minus two, which is this one here. 117 00:09:19,920 --> 00:09:24,450 And then we have to account for the inner derivative, which is 2x X, which is this one. 118 00:09:25,890 --> 00:09:29,100 And then we do the same thing for the other component. 119 00:09:29,480 --> 00:09:31,010 We have to I have to apologize. 120 00:09:31,020 --> 00:09:31,880 Here's a mistake. 121 00:09:31,890 --> 00:09:35,190 This is not X, this is, of course, Y and this is X. 122 00:09:35,250 --> 00:09:37,950 There's a I've made a typo when I wrote down the equation. 123 00:09:38,370 --> 00:09:45,120 So this one must be y times to Y divided by X squared plus Y square squared. 124 00:09:46,320 --> 00:09:53,520 OK, so if we have done this, we have this term here twice, so we have to over X squared plus Y square 125 00:09:54,450 --> 00:10:00,270 and then we have here two Times Square over X squared plus by square. 126 00:10:00,600 --> 00:10:09,000 And then the other term where we have this typo here we have two Times Square over X squared plus Y 127 00:10:09,010 --> 00:10:09,990 square squared. 128 00:10:11,070 --> 00:10:15,630 And then you can see that you can here get rid of this power too and get rid of this one. 129 00:10:15,790 --> 00:10:21,460 So we have to over X squared plus Y square, minus two of X squared plus Y square, which is zero. 130 00:10:22,770 --> 00:10:26,970 So this means we have verified that this is indeed zero. 131 00:10:27,660 --> 00:10:28,740 But please be careful. 132 00:10:28,920 --> 00:10:35,010 We have divided here by row, which means our whole calculation is only true for row different from 133 00:10:35,010 --> 00:10:38,190 zero, which in this case is all we wanted to know. 134 00:10:39,290 --> 00:10:45,000 We will learn later, underscores how we can really determine this relation here. 135 00:10:45,000 --> 00:10:46,590 So we will really derive it. 136 00:10:46,590 --> 00:10:48,120 We will go the other way around. 137 00:10:48,510 --> 00:10:53,580 We will start from the umpire's law and then we will derive that the field looks like this. 138 00:10:54,330 --> 00:11:01,710 And at the moment we cannot really do this because this calculation here is not valid for the position 139 00:11:01,710 --> 00:11:03,000 are equal to zero. 140 00:11:03,840 --> 00:11:09,570 So please stay tuned and continue watching the course and then you will see that we can also solve the 141 00:11:09,570 --> 00:11:10,830 whole problem later on.