[{"Name":"Introduction to Time Dependent Fields","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Explanation","Duration":"2m 59s","ChapterTopicVideoID":21358,"CourseChapterTopicPlaylistID":99494,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/21358.jpeg","UploadDate":"2020-04-06T21:45:13.8970000","DurationForVideoObject":"PT2M59S","Description":null,"MetaTitle":"Explanation: Video + Workbook | Proprep","MetaDescription":"Time Dependent Fields - Introduction to Time Dependent Fields. Watch the video made by an expert in the field. Download the workbook and maximize your learning.","Canonical":"https://www.proprep.uk/general-modules/all/physics-2-electricity-and-magnetism/time-dependent-fields/introduction-to-time-dependent-fields/vid21438","VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.815","Text":"Hello. In this lesson,"},{"Start":"00:01.815 ","End":"00:05.205","Text":"we\u0027re going to be speaking about time-dependent fields."},{"Start":"00:05.205 ","End":"00:10.740","Text":"Here we have Maxwell\u0027s third and fourth equations."},{"Start":"00:10.740 ","End":"00:13.365","Text":"What we can see is, over here,"},{"Start":"00:13.365 ","End":"00:17.130","Text":"if we have a magnetic field which is dependent on time,"},{"Start":"00:17.130 ","End":"00:19.995","Text":"as in, T is one of its variables,"},{"Start":"00:19.995 ","End":"00:23.688","Text":"then this derivative will not be equal to 0,"},{"Start":"00:23.688 ","End":"00:24.915","Text":"and in that case,"},{"Start":"00:24.915 ","End":"00:28.620","Text":"if we have a magnetic field which is time-dependent,"},{"Start":"00:28.620 ","End":"00:32.352","Text":"then therefore we must have an electric field."},{"Start":"00:32.352 ","End":"00:38.040","Text":"Similarly over here, if we have an electric field which is time-dependent,"},{"Start":"00:38.040 ","End":"00:40.920","Text":"that means that this derivative of the electric field with"},{"Start":"00:40.920 ","End":"00:44.038","Text":"respect to time will also be non-zero,"},{"Start":"00:44.038 ","End":"00:50.060","Text":"and then that will mean that we must have therefore also a magnetic field."},{"Start":"00:50.060 ","End":"00:54.980","Text":"Wherever you are given an electric or a magnetic field,"},{"Start":"00:54.980 ","End":"00:56.959","Text":"which has time as a variable,"},{"Start":"00:56.959 ","End":"01:01.886","Text":"you know that that field is going to form the other one."},{"Start":"01:01.886 ","End":"01:03.605","Text":"If you have a magnetic field,"},{"Start":"01:03.605 ","End":"01:05.370","Text":"then you\u0027re going to get an electric field,"},{"Start":"01:05.370 ","End":"01:07.640","Text":"and if you have an electric field dependent on time,"},{"Start":"01:07.640 ","End":"01:09.560","Text":"you\u0027ll get a magnetic field."},{"Start":"01:09.560 ","End":"01:13.069","Text":"Now let\u0027s see how they form one another."},{"Start":"01:13.069 ","End":"01:16.775","Text":"They form one another,"},{"Start":"01:16.775 ","End":"01:21.286","Text":"just like current can cause a magnetic field."},{"Start":"01:21.286 ","End":"01:26.120","Text":"Let\u0027s imagine that I have"},{"Start":"01:26.120 ","End":"01:31.619","Text":"a magnetic field going into the page like so,"},{"Start":"01:31.619 ","End":"01:33.358","Text":"this is a B field,"},{"Start":"01:33.358 ","End":"01:39.380","Text":"and let\u0027s imagine that there\u0027s some cylindrical symmetry like so,"},{"Start":"01:39.380 ","End":"01:48.520","Text":"that will mean that I will have an electric field in the Theta direction."},{"Start":"01:49.290 ","End":"01:53.905","Text":"This is exactly the same as if I had current"},{"Start":"01:53.905 ","End":"01:58.066","Text":"going into the page in this direction of the magnetic field."},{"Start":"01:58.066 ","End":"02:00.430","Text":"If I have current going into the page,"},{"Start":"02:00.430 ","End":"02:05.380","Text":"then I would have a magnetic field in the Theta direction."},{"Start":"02:05.380 ","End":"02:08.305","Text":"Now notice with the electric field,"},{"Start":"02:08.305 ","End":"02:12.044","Text":"the direction of the magnetic field could be clockwise or anticlockwise."},{"Start":"02:12.044 ","End":"02:14.320","Text":"We\u0027ll speak about that later,"},{"Start":"02:14.320 ","End":"02:19.299","Text":"but what we do know is that the electric field will be in the Theta direction,"},{"Start":"02:19.299 ","End":"02:22.420","Text":"be it the positive or negative Theta direction,"},{"Start":"02:22.420 ","End":"02:25.675","Text":"and that we will speak about later on in this lesson."},{"Start":"02:25.675 ","End":"02:29.730","Text":"How do we know which equation to use?"},{"Start":"02:29.730 ","End":"02:35.227","Text":"Whatever we\u0027re trying to find is on the left side of the equal sign."},{"Start":"02:35.227 ","End":"02:38.390","Text":"Let\u0027s say I\u0027m trying to find the electric field,"},{"Start":"02:38.390 ","End":"02:42.350","Text":"then I\u0027ll usually have a magnetic field which is dependent on time,"},{"Start":"02:42.350 ","End":"02:44.338","Text":"which will give me the electric field,"},{"Start":"02:44.338 ","End":"02:47.060","Text":"and let\u0027s say I have an electric field dependent on"},{"Start":"02:47.060 ","End":"02:52.145","Text":"time and I\u0027m trying to find the magnetic field, that\u0027s what I\u0027ll do."},{"Start":"02:52.145 ","End":"03:00.660","Text":"In the next lesson we\u0027re going to be solving a question to see exactly how this works."}],"ID":21438},{"Watched":false,"Name":"Exercise 1","Duration":"20m 54s","ChapterTopicVideoID":21359,"CourseChapterTopicPlaylistID":99494,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.710","Text":"Hello. In this lesson,"},{"Start":"00:01.710 ","End":"00:05.550","Text":"we\u0027re going to be answering the following question,"},{"Start":"00:05.550 ","End":"00:09.105","Text":"which is going to use Maxwell\u0027s third equation."},{"Start":"00:09.105 ","End":"00:19.155","Text":"An infinite solid cylinder of radius R is spinning at an angular velocity of Omega,"},{"Start":"00:19.155 ","End":"00:24.540","Text":"where Omega is equal to Alpha t. So it\u0027s dependent on time"},{"Start":"00:24.540 ","End":"00:30.995","Text":"and the cylinder has a charge density per unit volume of Rho."},{"Start":"00:30.995 ","End":"00:35.820","Text":"Question Number 1 is to calculate the magnetic field."},{"Start":"00:36.200 ","End":"00:42.775","Text":"Of course, the magnetic field is going to be caused by these charges moving."},{"Start":"00:42.775 ","End":"00:45.590","Text":"These charges are spinning around,"},{"Start":"00:45.590 ","End":"00:46.670","Text":"so they have motion,"},{"Start":"00:46.670 ","End":"00:49.175","Text":"which means that we have a current and of course,"},{"Start":"00:49.175 ","End":"00:51.347","Text":"current causes a magnetic field."},{"Start":"00:51.347 ","End":"00:54.515","Text":"We\u0027re going to be using Ampere\u0027s law."},{"Start":"00:54.515 ","End":"00:56.570","Text":"If you can\u0027t remember this,"},{"Start":"00:56.570 ","End":"01:00.020","Text":"please go back to the chapter where we speak about Ampere\u0027s law."},{"Start":"01:00.020 ","End":"01:05.330","Text":"The first thing that we\u0027re going to do is we\u0027re going to calculate the current density,"},{"Start":"01:05.330 ","End":"01:10.860","Text":"which is equal to Rho multiplied by v. Over here,"},{"Start":"01:10.860 ","End":"01:12.555","Text":"our Rho is this,"},{"Start":"01:12.555 ","End":"01:16.235","Text":"our v is equal to Omega multiplied by the radius."},{"Start":"01:16.235 ","End":"01:20.762","Text":"Notice that the radius is changing because it\u0027s a solid cylinder."},{"Start":"01:20.762 ","End":"01:26.330","Text":"Of course, it is moving in the Theta direction,"},{"Start":"01:26.330 ","End":"01:31.080","Text":"so going around in circles."},{"Start":"01:31.550 ","End":"01:37.685","Text":"Now I\u0027m going to use Ampere\u0027s law to calculate the magnetic field."},{"Start":"01:37.685 ","End":"01:43.775","Text":"To remind you, Ampere\u0027s law is the integral on B.dl,"},{"Start":"01:43.775 ","End":"01:49.955","Text":"and that is equal to Mu_naught multiplied by in."},{"Start":"01:49.955 ","End":"01:54.920","Text":"First of all, if I can see that the current is flowing in the Theta direction,"},{"Start":"01:54.920 ","End":"02:04.760","Text":"then that must mean that my magnetic field is in the z-direction, like so."},{"Start":"02:04.760 ","End":"02:06.905","Text":"What I\u0027m going to do is I\u0027m going to draw"},{"Start":"02:06.905 ","End":"02:12.560","Text":"an Ampere\u0027s loop like so at this point over here."},{"Start":"02:13.880 ","End":"02:19.400","Text":"This edge over here is a distance of"},{"Start":"02:19.400 ","End":"02:28.370","Text":"r away from the center of the cylinder and the loop carries on to infinity."},{"Start":"02:28.370 ","End":"02:30.140","Text":"If it carries on to infinity,"},{"Start":"02:30.140 ","End":"02:36.170","Text":"I know the magnetic field at infinity is equal to 0 so I don\u0027t have to calculate it."},{"Start":"02:36.170 ","End":"02:43.985","Text":"Then I\u0027m going to say that the length of this side of the Ampere loop is equal to"},{"Start":"02:43.985 ","End":"02:52.640","Text":"l. Then I can assume that the magnetic field along this edge of the loop is constant."},{"Start":"02:52.640 ","End":"02:57.040","Text":"That means that this integral is just equal to Bl."},{"Start":"02:57.040 ","End":"03:02.210","Text":"B multiplied by this length l and that\u0027s because this length is parallel to"},{"Start":"03:02.210 ","End":"03:08.270","Text":"the magnetic field and these sides are perpendicular to the magnetic fields,"},{"Start":"03:08.270 ","End":"03:11.730","Text":"so it will be equal to 0 over here."},{"Start":"03:11.730 ","End":"03:13.280","Text":"Of course at infinity,"},{"Start":"03:13.280 ","End":"03:15.575","Text":"the magnetic field is also equal to 0."},{"Start":"03:15.575 ","End":"03:19.120","Text":"This integral becomes B.l,"},{"Start":"03:19.120 ","End":"03:21.820","Text":"or B multiplied by l,"},{"Start":"03:21.820 ","End":"03:26.300","Text":"and this is equal to Mu_naught multiplied by I_in."},{"Start":"03:26.300 ","End":"03:28.720","Text":"What is I_in?"},{"Start":"03:28.720 ","End":"03:36.450","Text":"I_in is equal to the integral on our J.ds."},{"Start":"03:36.820 ","End":"03:40.570","Text":"What area is our ds?"},{"Start":"03:40.570 ","End":"03:47.435","Text":"In principle, our ds is all of the area inside our Ampere\u0027s loop."},{"Start":"03:47.435 ","End":"03:54.080","Text":"However, we can see that this section over here is empty. What does that mean?"},{"Start":"03:54.080 ","End":"03:56.855","Text":"The cylinder doesn\u0027t fill out this area,"},{"Start":"03:56.855 ","End":"04:00.620","Text":"which means that there\u0027s no charges in this area moving,"},{"Start":"04:00.620 ","End":"04:03.785","Text":"meaning that there is no current over here."},{"Start":"04:03.785 ","End":"04:10.045","Text":"The only area which is relevant for us to sum up to find J.ds,"},{"Start":"04:10.045 ","End":"04:13.340","Text":"to find out current is this area over here where we can"},{"Start":"04:13.340 ","End":"04:16.550","Text":"have current because we have charges."},{"Start":"04:16.550 ","End":"04:21.690","Text":"Because it\u0027s inside the cylinder."},{"Start":"04:22.190 ","End":"04:31.325","Text":"That means that this integral becomes so ds means that we have a double integral of J,"},{"Start":"04:31.325 ","End":"04:35.870","Text":"which is Rho Omega r and then ds."},{"Start":"04:35.870 ","End":"04:37.880","Text":"We\u0027re integrating along this side,"},{"Start":"04:37.880 ","End":"04:43.885","Text":"which is dr and up over here along this axis."},{"Start":"04:43.885 ","End":"04:46.145","Text":"So that\u0027s in the z-direction dz."},{"Start":"04:46.145 ","End":"04:49.490","Text":"Notice if we were also integrating along d Theta,"},{"Start":"04:49.490 ","End":"04:52.050","Text":"then we would have to include the Jacobian."},{"Start":"04:52.050 ","End":"04:55.210","Text":"We would have to multiply all of this also by r,"},{"Start":"04:55.210 ","End":"04:57.020","Text":"if we were also integrating along"},{"Start":"04:57.020 ","End":"05:00.820","Text":"d Theta but because we\u0027re not integrating along d theta,"},{"Start":"05:00.820 ","End":"05:04.260","Text":"we don\u0027t have to add in the Jacobian,"},{"Start":"05:04.260 ","End":"05:08.870","Text":"or rather we don\u0027t have to multiply by the Jacobian."},{"Start":"05:08.870 ","End":"05:14.855","Text":"Whoever is having trouble understanding why the ds is equal to drdz,"},{"Start":"05:14.855 ","End":"05:19.955","Text":"if the direction of J is in the Theta direction,"},{"Start":"05:19.955 ","End":"05:26.990","Text":"which is as so ds is always the plane perpendicular to the Theta direction."},{"Start":"05:26.990 ","End":"05:30.230","Text":"Of course, the axis which are perpendicular to"},{"Start":"05:30.230 ","End":"05:35.165","Text":"the theta direction is the radial axis and the z-axis."},{"Start":"05:35.165 ","End":"05:42.855","Text":"In other words, ds vector is equal to drdz,"},{"Start":"05:42.855 ","End":"05:46.930","Text":"and then in the Theta direction."},{"Start":"05:47.720 ","End":"05:50.895","Text":"Now let\u0027s sub in our bounds."},{"Start":"05:50.895 ","End":"05:54.144","Text":"The bounds for r are from this radius,"},{"Start":"05:54.144 ","End":"05:57.280","Text":"r until the total radius,"},{"Start":"05:57.280 ","End":"06:02.460","Text":"which is R and the bounds on z are from z=0,"},{"Start":"06:02.460 ","End":"06:04.120","Text":"so at the bottom over here,"},{"Start":"06:04.120 ","End":"06:06.490","Text":"until the full length of this loop,"},{"Start":"06:06.490 ","End":"06:10.810","Text":"which we said is equal to l. Now we can see that"},{"Start":"06:10.810 ","End":"06:17.740","Text":"B.l is equal to Mu_naught multiplied by I_in."},{"Start":"06:17.740 ","End":"06:20.165","Text":"So what is I_in?"},{"Start":"06:20.165 ","End":"06:25.615","Text":"I_in is Rho Omega and then from the integral of z,"},{"Start":"06:25.615 ","End":"06:32.130","Text":"we\u0027re just going to multiply this by l. Then the integral of rdr,"},{"Start":"06:32.130 ","End":"06:37.055","Text":"what we have is r^2 divided by 2 and then when we substitute in our bounds,"},{"Start":"06:37.055 ","End":"06:43.600","Text":"so we have R^2 minus r^2 divided by 2."},{"Start":"06:43.600 ","End":"06:46.385","Text":"Then the ls in both sides cancel out."},{"Start":"06:46.385 ","End":"06:50.480","Text":"Then what we get is that the magnetic fields,"},{"Start":"06:50.480 ","End":"06:52.190","Text":"soon we\u0027ll look at the direction,"},{"Start":"06:52.190 ","End":"06:54.470","Text":"is equal to Mu_naught,"},{"Start":"06:54.470 ","End":"07:00.840","Text":"Rho Omega R^2 minus r^2 divided by 2."},{"Start":"07:00.840 ","End":"07:04.279","Text":"Of course, we already said that this was in the z-direction."},{"Start":"07:04.279 ","End":"07:07.505","Text":"We can also see that from the right hand rule,"},{"Start":"07:07.505 ","End":"07:11.720","Text":"if the current is flowing in this direction from the right-hand rule,"},{"Start":"07:11.720 ","End":"07:14.435","Text":"your thumb will point upwards in the z-direction,"},{"Start":"07:14.435 ","End":"07:18.970","Text":"symbolizing the direction of the magnetic field."},{"Start":"07:18.970 ","End":"07:24.694","Text":"Of course, this is the magnetic field when the radius"},{"Start":"07:24.694 ","End":"07:29.870","Text":"is smaller than the total radius of the cylinder."},{"Start":"07:29.870 ","End":"07:33.935","Text":"So long as we\u0027re located inside the cylinder."},{"Start":"07:33.935 ","End":"07:37.610","Text":"We can see that if we\u0027re located outside of the cylinder,"},{"Start":"07:37.610 ","End":"07:39.785","Text":"the magnetic field will be equal to 0."},{"Start":"07:39.785 ","End":"07:46.070","Text":"We can show that by adding an Ampere\u0027s loop outside of the cylinder where"},{"Start":"07:46.070 ","End":"07:52.415","Text":"this radius is of course still r but here,"},{"Start":"07:52.415 ","End":"07:56.885","Text":"r is greater than the radius of the cylinder."},{"Start":"07:56.885 ","End":"08:02.570","Text":"Of course, because there\u0027s no charged particles inside this Ampere\u0027s loop,"},{"Start":"08:02.570 ","End":"08:07.970","Text":"that means that there is no current in Ampere\u0027s loop."},{"Start":"08:07.970 ","End":"08:09.530","Text":"If there is no current,"},{"Start":"08:09.530 ","End":"08:10.630","Text":"there\u0027s no magnetic field."},{"Start":"08:10.630 ","End":"08:12.800","Text":"We can also see that over here,"},{"Start":"08:12.800 ","End":"08:14.885","Text":"if there\u0027s no current,"},{"Start":"08:14.885 ","End":"08:18.050","Text":"I is equal to 0 and then this whole equation is equal to 0,"},{"Start":"08:18.050 ","End":"08:21.450","Text":"leaving that B is equal to 0."},{"Start":"08:22.760 ","End":"08:26.190","Text":"That\u0027s the answer to question Number 1."},{"Start":"08:26.190 ","End":"08:29.910","Text":"Now let\u0027s look at Number 2."},{"Start":"08:29.910 ","End":"08:32.555","Text":"Number 2 is to calculate the electric field."},{"Start":"08:32.555 ","End":"08:35.540","Text":"Now, why will we have an electric field?"},{"Start":"08:35.540 ","End":"08:41.150","Text":"We already saw that if our magnetic field is dependent on time,"},{"Start":"08:41.150 ","End":"08:44.448","Text":"as in this integral is non-zero."},{"Start":"08:44.448 ","End":"08:48.055","Text":"Then that will mean that we have an electric field."},{"Start":"08:48.055 ","End":"08:50.870","Text":"If we look at our equation for B,"},{"Start":"08:50.870 ","End":"08:56.990","Text":"we see that B has Omega as one of its variables and"},{"Start":"08:56.990 ","End":"09:03.185","Text":"we see that Omega also from the question or from the diagram,"},{"Start":"09:03.185 ","End":"09:06.220","Text":"is equal to Alpha, some constant,"},{"Start":"09:06.220 ","End":"09:09.290","Text":"multiplied by t. In other words,"},{"Start":"09:09.290 ","End":"09:15.315","Text":"the magnetic field is dependent on t over here."},{"Start":"09:15.315 ","End":"09:17.960","Text":"If our magnetic field is time-dependent,"},{"Start":"09:17.960 ","End":"09:21.010","Text":"that means that we have an electric field."},{"Start":"09:21.010 ","End":"09:27.920","Text":"Now, the direction of the electric field is,"},{"Start":"09:27.920 ","End":"09:30.420","Text":"we already saw, dependent on the magnetic field."},{"Start":"09:30.420 ","End":"09:34.175","Text":"If our magnetic field is in the z-direction,"},{"Start":"09:34.175 ","End":"09:39.170","Text":"that means that our electric field will be somewhere in the Theta direction,"},{"Start":"09:39.170 ","End":"09:41.195","Text":"be it the positive or negative direction,"},{"Start":"09:41.195 ","End":"09:45.150","Text":"but along the Theta axis."},{"Start":"09:45.380 ","End":"09:49.410","Text":"Let\u0027s take a look over here."},{"Start":"09:49.410 ","End":"09:53.555","Text":"What we\u0027re going to do is we assume that"},{"Start":"09:53.555 ","End":"09:59.225","Text":"the electric field will be traveling in this type of direction."},{"Start":"09:59.225 ","End":"10:04.730","Text":"We have here a radius of r, like so."},{"Start":"10:04.730 ","End":"10:09.095","Text":"Over here, we\u0027re going to have the electric field,"},{"Start":"10:09.095 ","End":"10:14.840","Text":"the integral of the electric field multiplied by dl,"},{"Start":"10:14.840 ","End":"10:17.690","Text":"the length of this root over here."},{"Start":"10:17.690 ","End":"10:19.460","Text":"Because our electric field,"},{"Start":"10:19.460 ","End":"10:22.445","Text":"we have no reason to think that it isn\u0027t constant."},{"Start":"10:22.445 ","End":"10:29.975","Text":"We can just say that this is equal to E multiplied by the length of the circle,"},{"Start":"10:29.975 ","End":"10:34.250","Text":"which is of course equal to the circumference of the circle."},{"Start":"10:34.250 ","End":"10:38.090","Text":"That is equal to 2Pi multiplied by this radius,"},{"Start":"10:38.090 ","End":"10:43.195","Text":"r. Now let\u0027s take a look at what we have over here."},{"Start":"10:43.195 ","End":"10:46.210","Text":"First of all, we have dB by dt."},{"Start":"10:46.210 ","End":"10:52.080","Text":"In other words, what we have is our vector B dots."},{"Start":"10:52.080 ","End":"10:54.050","Text":"The time derivative of our vector B,"},{"Start":"10:54.050 ","End":"10:58.450","Text":"remember this is another symbol for a dB by dt."},{"Start":"10:58.450 ","End":"11:05.310","Text":"This is simply equal to this over here with the time derivative."},{"Start":"11:05.310 ","End":"11:09.200","Text":"We remember that Omega is equal to Alpha"},{"Start":"11:09.200 ","End":"11:13.654","Text":"t. If we substitute an Alpha t and we take the time derivative,"},{"Start":"11:13.654 ","End":"11:19.725","Text":"we just get rid of the t. So we have Mu_naught Rho Alpha,"},{"Start":"11:19.725 ","End":"11:24.585","Text":"the t is gone and then R^2 minus r^2"},{"Start":"11:24.585 ","End":"11:30.422","Text":"divided by 2 and in the z-direction."},{"Start":"11:30.422 ","End":"11:34.420","Text":"Now we want to plug this into our integral."},{"Start":"11:34.420 ","End":"11:37.600","Text":"We have on this side of the equals sign,"},{"Start":"11:37.600 ","End":"11:43.915","Text":"so we have the negative double integral of B.ds."},{"Start":"11:43.915 ","End":"11:53.530","Text":"B. is mu naught Rho Alpha R^2 minus r^2 divided by 2."},{"Start":"11:53.530 ","End":"11:58.990","Text":"Then our ds is the area of this circle."},{"Start":"11:58.990 ","End":"12:00.985","Text":"Let\u0027s show it in green."},{"Start":"12:00.985 ","End":"12:11.395","Text":"The area of this circle is simply equal 2 if we\u0027re using cylindrical coordinates."},{"Start":"12:11.395 ","End":"12:16.690","Text":"It\u0027s r dr d Theta,"},{"Start":"12:16.690 ","End":"12:19.480","Text":"because we\u0027re going in the Theta direction,"},{"Start":"12:19.480 ","End":"12:23.320","Text":"in the radial direction and what are bounds?"},{"Start":"12:23.320 ","End":"12:28.075","Text":"The radius goes from 0 to a radius of"},{"Start":"12:28.075 ","End":"12:34.705","Text":"r. Because I have lowercase r in my bounds so as not to get confused,"},{"Start":"12:34.705 ","End":"12:36.835","Text":"I\u0027ll add in tags over here."},{"Start":"12:36.835 ","End":"12:38.125","Text":"This is r tag,"},{"Start":"12:38.125 ","End":"12:42.985","Text":"r tag and the dr tag to not get confused with the bounce"},{"Start":"12:42.985 ","End":"12:49.060","Text":"and my Theta is going from 0 and a full circle around,"},{"Start":"12:49.060 ","End":"12:52.840","Text":"so from 0 until 2 Phi."},{"Start":"12:52.840 ","End":"12:58.190","Text":"Now what I want to do is I want to know what this integral is equal to."},{"Start":"12:59.010 ","End":"13:02.230","Text":"I wrote out the onset to the integral."},{"Start":"13:02.230 ","End":"13:07.960","Text":"Feel free to pause the video if you want to calculate it yourself but this is the onset."},{"Start":"13:07.960 ","End":"13:13.420","Text":"Now what I\u0027m going to do is I\u0027m going to equate both sides over here."},{"Start":"13:13.420 ","End":"13:16.240","Text":"I have over here my E.dl."},{"Start":"13:16.240 ","End":"13:19.990","Text":"If E multiplied by 2 Pi r,"},{"Start":"13:19.990 ","End":"13:24.970","Text":"that\u0027s my E.dl and this is equal to the negative double integral."},{"Start":"13:24.970 ","End":"13:32.680","Text":"This is equal to negative 2 Pi mu naught Rho Alpha"},{"Start":"13:32.680 ","End":"13:42.595","Text":"multiplied by 1.5 multiplied by R^2 squared r^2 divided by 2 minus r^4 divided by 4."},{"Start":"13:42.595 ","End":"13:48.490","Text":"Then we can see that the 2 Pi cancels out."},{"Start":"13:48.490 ","End":"13:55.510","Text":"Then I can also divide both sides by r. Here,"},{"Start":"13:55.510 ","End":"14:00.930","Text":"I\u0027ll divide by r and then here I can add an r over here."},{"Start":"14:00.930 ","End":"14:07.410","Text":"Then what we get is that our electric field is equal to the minus,"},{"Start":"14:07.410 ","End":"14:10.470","Text":"I\u0027ll put in to these brackets over here."},{"Start":"14:10.470 ","End":"14:17.622","Text":"So I have mu naught Rho Alpha divided by 2r."},{"Start":"14:17.622 ","End":"14:19.960","Text":"Then multiplied by,"},{"Start":"14:19.960 ","End":"14:21.625","Text":"so here I put in the minus,"},{"Start":"14:21.625 ","End":"14:31.580","Text":"so r^4 divided by 4 minus R^2 r^2 divided by 2."},{"Start":"14:32.850 ","End":"14:39.655","Text":"As we already saw, this is going to be in the Theta direction,"},{"Start":"14:39.655 ","End":"14:43.105","Text":"and let\u0027s just see if it\u0027s in the positive or negative Theta direction."},{"Start":"14:43.105 ","End":"14:45.520","Text":"Don\u0027t get confused with that here."},{"Start":"14:45.520 ","End":"14:47.680","Text":"We don\u0027t have a minus sign."},{"Start":"14:47.680 ","End":"14:50.570","Text":"Because if we look over here,"},{"Start":"14:50.670 ","End":"14:55.270","Text":"we have a smaller number minus a bigger number."},{"Start":"14:55.270 ","End":"15:02.290","Text":"So we can see that the electric field is in the opposite direction to what we drew it in."},{"Start":"15:02.290 ","End":"15:09.230","Text":"Instead, the electric field is going around like so."},{"Start":"15:11.310 ","End":"15:14.350","Text":"If we plug in the largest number,"},{"Start":"15:14.350 ","End":"15:21.400","Text":"we have capital R^4 and here it\u0027s only divided by 2."},{"Start":"15:21.400 ","End":"15:24.085","Text":"So this is a larger number than this"},{"Start":"15:24.085 ","End":"15:27.850","Text":"always and so we can see that it is in the opposite direction,"},{"Start":"15:27.850 ","End":"15:30.620","Text":"it\u0027s in the negative Theta direction."},{"Start":"15:30.620 ","End":"15:32.245","Text":"So don\u0027t get confused."},{"Start":"15:32.245 ","End":"15:35.635","Text":"The next thing is to note that this is when r is"},{"Start":"15:35.635 ","End":"15:39.865","Text":"smaller than the maximum radius of the cylinder,"},{"Start":"15:39.865 ","End":"15:42.940","Text":"what happens if we\u0027re in the region outside of the cylinder?"},{"Start":"15:42.940 ","End":"15:46.990","Text":"Of course, the electric field isn\u0027t going to be = 0."},{"Start":"15:46.990 ","End":"15:50.710","Text":"If we draw a loop,"},{"Start":"15:50.710 ","End":"15:58.967","Text":"so we can see that if we want to measure the electric field at this radius outside."},{"Start":"15:58.967 ","End":"16:01.165","Text":"So at this radius, r,"},{"Start":"16:01.165 ","End":"16:06.100","Text":"we can see that we going to do the exact same equation,"},{"Start":"16:06.100 ","End":"16:09.310","Text":"the exact same steps as we did before."},{"Start":"16:09.310 ","End":"16:14.182","Text":"We still have that E.dl is = E multiplied by 2Phi"},{"Start":"16:14.182 ","End":"16:18.970","Text":"r. We still have that B dot is equal to this and we still have to do this equation,"},{"Start":"16:18.970 ","End":"16:24.565","Text":"but the difference in this equation is the bounds for the radius."},{"Start":"16:24.565 ","End":"16:26.890","Text":"We\u0027re measuring outside of the cylinder,"},{"Start":"16:26.890 ","End":"16:31.390","Text":"which means that we have to integrate up until the maximal radius of the cylinder,"},{"Start":"16:31.390 ","End":"16:36.465","Text":"which is of course capital R. I\u0027ll write it in red."},{"Start":"16:36.465 ","End":"16:40.260","Text":"That means that here we\u0027re going to be integrating up"},{"Start":"16:40.260 ","End":"16:44.550","Text":"until capital R. Notice over here we\u0027re going to"},{"Start":"16:44.550 ","End":"16:52.870","Text":"get a different result and what we\u0027ll get is that E is equal to negative mu"},{"Start":"16:52.870 ","End":"16:56.050","Text":"naught Rho Alpha divided by"},{"Start":"16:56.050 ","End":"17:05.540","Text":"2r multiplied by R^4 divided by 4 in the Theta direction."},{"Start":"17:07.590 ","End":"17:11.665","Text":"One last thing for this section of the question."},{"Start":"17:11.665 ","End":"17:18.144","Text":"This is the electric field that we discovered due to the magnetic field but of course,"},{"Start":"17:18.144 ","End":"17:21.894","Text":"we have some charged cylinder,"},{"Start":"17:21.894 ","End":"17:23.950","Text":"which means that we\u0027re going to have"},{"Start":"17:23.950 ","End":"17:27.715","Text":"an electric field because of the charges and the cylinder themselves."},{"Start":"17:27.715 ","End":"17:30.070","Text":"Just Gauss\u0027s law."},{"Start":"17:30.070 ","End":"17:35.034","Text":"We know that we\u0027re going to have an E field"},{"Start":"17:35.034 ","End":"17:40.840","Text":"going like so in the radial direction just because of the charges."},{"Start":"17:40.840 ","End":"17:45.010","Text":"Then we just have to solve according to Gauss for a charged cylinder."},{"Start":"17:45.010 ","End":"17:49.420","Text":"That means that we have to add over here the radial component."},{"Start":"17:49.420 ","End":"17:53.515","Text":"Of course, the E field will be in the radial direction as we know with a cylinder."},{"Start":"17:53.515 ","End":"17:55.825","Text":"We have to add over here,"},{"Start":"17:55.825 ","End":"18:00.280","Text":"and over here the radial component"},{"Start":"18:00.280 ","End":"18:04.630","Text":"of the electric field due to the charged particles themselves."},{"Start":"18:04.630 ","End":"18:12.440","Text":"This section in black is the electric field due to the magnetic field and nothing else."},{"Start":"18:13.020 ","End":"18:18.640","Text":"Now, let\u0027s go to question Number 3."},{"Start":"18:18.640 ","End":"18:20.020","Text":"Let\u0027s do it over here."},{"Start":"18:20.020 ","End":"18:22.540","Text":"What force acts on charge q?"},{"Start":"18:22.540 ","End":"18:24.055","Text":"So this is very easy."},{"Start":"18:24.055 ","End":"18:32.035","Text":"F = q multiplied by E. Of course, for each region,"},{"Start":"18:32.035 ","End":"18:34.390","Text":"r is smaller than R,"},{"Start":"18:34.390 ","End":"18:43.430","Text":"or r is bigger than R. So we just multiply q by all of this and then we get the force."},{"Start":"18:44.670 ","End":"18:47.560","Text":"That\u0027s it. We\u0027ve answered all the questions,"},{"Start":"18:47.560 ","End":"18:49.870","Text":"but before the end of the lesson,"},{"Start":"18:49.870 ","End":"18:54.340","Text":"I want to go over what we see over here."},{"Start":"18:54.340 ","End":"19:00.400","Text":"What we get is that the electric field in both cases is in the negative Theta direction."},{"Start":"19:00.400 ","End":"19:03.415","Text":"Here we spoke about why this is the negative Theta direction"},{"Start":"19:03.415 ","End":"19:06.636","Text":"because what\u0027s inside the brackets over here is negative."},{"Start":"19:06.636 ","End":"19:08.080","Text":"Here, of course,"},{"Start":"19:08.080 ","End":"19:11.365","Text":"we have this negative coefficient."},{"Start":"19:11.365 ","End":"19:13.990","Text":"What we can see is that the electric field is in"},{"Start":"19:13.990 ","End":"19:18.655","Text":"the negative Theta direction or in other words,"},{"Start":"19:18.655 ","End":"19:24.190","Text":"the electric field is in the opposite direction to our angular velocity."},{"Start":"19:24.190 ","End":"19:27.265","Text":"It\u0027s in the opposite direction to Omega,"},{"Start":"19:27.265 ","End":"19:28.810","Text":"or in other words,"},{"Start":"19:28.810 ","End":"19:35.240","Text":"the opposite direction to our angular acceleration, Alpha."},{"Start":"19:35.760 ","End":"19:39.850","Text":"What we can see is that the electric field is in"},{"Start":"19:39.850 ","End":"19:44.125","Text":"the opposite direction to the angular acceleration."},{"Start":"19:44.125 ","End":"19:49.480","Text":"That is because if it was going in the same direction as the angular acceleration,"},{"Start":"19:49.480 ","End":"19:54.928","Text":"it would act together with the angular acceleration and increase the acceleration."},{"Start":"19:54.928 ","End":"20:00.565","Text":"Therefore, our cylinder would spin and spin faster and faster"},{"Start":"20:00.565 ","End":"20:06.910","Text":"until infinity which obviously isn\u0027t something that we know to be possible."},{"Start":"20:06.910 ","End":"20:10.990","Text":"In other words, it has to be always in the opposite direction to"},{"Start":"20:10.990 ","End":"20:15.434","Text":"the angular acceleration to act to slow down the acceleration."},{"Start":"20:15.434 ","End":"20:23.150","Text":"Therefore, we stay in the realm that we know is true in real life."},{"Start":"20:23.430 ","End":"20:28.360","Text":"The electric field is acting in the opposite direction to"},{"Start":"20:28.360 ","End":"20:35.035","Text":"the angular acceleration in order to negate it so that we don\u0027t accelerate into infinity."},{"Start":"20:35.035 ","End":"20:37.615","Text":"This is true for every question."},{"Start":"20:37.615 ","End":"20:43.210","Text":"Just remember that you want to get your direction for the electric field in"},{"Start":"20:43.210 ","End":"20:46.825","Text":"the opposite direction to your Omega and that\u0027s"},{"Start":"20:46.825 ","End":"20:51.520","Text":"a nice little trick to see if you got the directions, correct."},{"Start":"20:51.520 ","End":"20:54.590","Text":"That\u0027s the end of this lesson."}],"ID":21439},{"Watched":false,"Name":"Exercise 2","Duration":"27m 23s","ChapterTopicVideoID":21360,"CourseChapterTopicPlaylistID":99494,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.725","Text":"Hello. In this lesson,"},{"Start":"00:01.725 ","End":"00:04.260","Text":"we\u0027re going to be answering the following question,"},{"Start":"00:04.260 ","End":"00:10.005","Text":"which is dealing with an electric field which is dependent on time."},{"Start":"00:10.005 ","End":"00:13.410","Text":"Then through that, we\u0027re going to calculate the magnetic field."},{"Start":"00:13.410 ","End":"00:15.075","Text":"In order to do that,"},{"Start":"00:15.075 ","End":"00:20.120","Text":"we need to use this equation that we saw a few lessons ago."},{"Start":"00:20.120 ","End":"00:25.709","Text":"First, I\u0027m going to explain how to use this equation generally."},{"Start":"00:26.180 ","End":"00:30.315","Text":"This is Maxwell\u0027s fourth equation."},{"Start":"00:30.315 ","End":"00:35.865","Text":"What we can see is that this section is drawn in blue,"},{"Start":"00:35.865 ","End":"00:41.310","Text":"all this over here is Ampere\u0027s law."},{"Start":"00:41.480 ","End":"00:46.115","Text":"This, we\u0027ve already seen how to use it."},{"Start":"00:46.115 ","End":"00:54.060","Text":"This section over here is Maxwell\u0027s correction."},{"Start":"00:54.560 ","End":"00:59.270","Text":"What we can see from this equation as a whole is that if"},{"Start":"00:59.270 ","End":"01:03.395","Text":"we have an electric field which is time dependent,"},{"Start":"01:03.395 ","End":"01:07.280","Text":"that means that this whole integral is non-zero,"},{"Start":"01:07.280 ","End":"01:11.360","Text":"and that means that there must be a magnetic field present."},{"Start":"01:11.360 ","End":"01:18.155","Text":"If we have an electric field flowing in this direction,"},{"Start":"01:18.155 ","End":"01:20.555","Text":"so that will mean that we\u0027re going to have"},{"Start":"01:20.555 ","End":"01:24.980","Text":"some magnetic field going like so in the Theta direction."},{"Start":"01:24.980 ","End":"01:28.915","Text":"Whether this is the positive or negative Theta direction,"},{"Start":"01:28.915 ","End":"01:30.825","Text":"we\u0027ll find out later."},{"Start":"01:30.825 ","End":"01:34.760","Text":"But we know that it\u0027s in the general Theta direction."},{"Start":"01:34.760 ","End":"01:37.999","Text":"This happens much when we have current,"},{"Start":"01:37.999 ","End":"01:40.950","Text":"how it forms a magnetic field."},{"Start":"01:41.330 ","End":"01:44.835","Text":"Now let\u0027s go back over here."},{"Start":"01:44.835 ","End":"01:49.280","Text":"As we saw, this electric field is"},{"Start":"01:49.280 ","End":"01:55.070","Text":"acting like current in the fact that it forms this magnetic field."},{"Start":"01:55.070 ","End":"01:59.015","Text":"In other words, if I erase this,"},{"Start":"01:59.015 ","End":"02:06.605","Text":"so we already know that this from Ampere\u0027s law= I_in,"},{"Start":"02:06.605 ","End":"02:08.845","Text":"so this we\u0027ve already seen."},{"Start":"02:08.845 ","End":"02:14.405","Text":"Because this time dependent electric field acts like current"},{"Start":"02:14.405 ","End":"02:20.005","Text":"in the sense that it forms a magnetic field, we just saw that,"},{"Start":"02:20.005 ","End":"02:25.700","Text":"then we can say mathematically that it is also called"},{"Start":"02:25.700 ","End":"02:31.805","Text":"I and it is called I_d for displacement."},{"Start":"02:31.805 ","End":"02:35.300","Text":"Now, it\u0027s important to note this isn\u0027t actually current."},{"Start":"02:35.300 ","End":"02:38.540","Text":"It just mathematically behaves like current."},{"Start":"02:38.540 ","End":"02:44.525","Text":"I\u0027m going to put quotation marks over here so that you don\u0027t get confused."},{"Start":"02:44.525 ","End":"02:46.985","Text":"This is not current,"},{"Start":"02:46.985 ","End":"02:50.330","Text":"it just forms a magnetic field,"},{"Start":"02:50.330 ","End":"02:52.730","Text":"so it acts as if it is current."},{"Start":"02:52.730 ","End":"02:56.360","Text":"Mathematically, it works as a current,"},{"Start":"02:56.360 ","End":"02:57.860","Text":"but it isn\u0027t actually current,"},{"Start":"02:57.860 ","End":"02:59.960","Text":"it\u0027s an electric field."},{"Start":"02:59.960 ","End":"03:03.290","Text":"Then we know that this j ds,"},{"Start":"03:03.290 ","End":"03:07.685","Text":"so what we have I_in is the integral on j ds."},{"Start":"03:07.685 ","End":"03:14.760","Text":"Here we have that I displacement is the integral on something over here, ds."},{"Start":"03:14.760 ","End":"03:16.905","Text":"If I_in is j ds,"},{"Start":"03:16.905 ","End":"03:20.460","Text":"then I_d is j d ds."},{"Start":"03:20.460 ","End":"03:30.435","Text":"In other words, we can say that all of this over here is j displacement."},{"Start":"03:30.435 ","End":"03:34.025","Text":"Because it\u0027s acting like current density."},{"Start":"03:34.025 ","End":"03:39.660","Text":"However, there isn\u0027t any current over here as we\u0027ve already explained."},{"Start":"03:40.490 ","End":"03:44.235","Text":"Now let\u0027s answer the question."},{"Start":"03:44.235 ","End":"03:49.580","Text":"Here we have a parallel plate capacitor where each plate is circular"},{"Start":"03:49.580 ","End":"03:55.285","Text":"and has a radius of R and the plates are a distance d away from one another."},{"Start":"03:55.285 ","End":"04:00.750","Text":"Of course, we know that d is much smaller than"},{"Start":"04:00.750 ","End":"04:09.440","Text":"R. We\u0027re being told that a constant current I flows through the capacitor."},{"Start":"04:09.440 ","End":"04:13.460","Text":"Let\u0027s imagine that this is connected to some source,"},{"Start":"04:13.460 ","End":"04:19.110","Text":"and then here we have this current I flowing like so."},{"Start":"04:19.270 ","End":"04:25.625","Text":"Question number 1 is to calculate the charge on the capacitor as a function of time,"},{"Start":"04:25.625 ","End":"04:30.145","Text":"given that the initial charge is equal to 0."},{"Start":"04:30.145 ","End":"04:32.610","Text":"Let\u0027s answer this over here."},{"Start":"04:32.610 ","End":"04:40.090","Text":"As we know, I the current is equal to the time derivative of the charge."},{"Start":"04:40.090 ","End":"04:43.140","Text":"It\u0027s equal to dq by dt."},{"Start":"04:43.140 ","End":"04:46.350","Text":"Therefore, if we integrate along I,"},{"Start":"04:46.350 ","End":"04:49.155","Text":"then we\u0027ll get our charge q."},{"Start":"04:49.155 ","End":"04:54.770","Text":"We\u0027ll get that q is equal to I multiplied by t plus"},{"Start":"04:54.770 ","End":"05:02.045","Text":"the initial charge that we have at the beginning and the initial charge at t = 0."},{"Start":"05:02.045 ","End":"05:04.520","Text":"We add here 0, or in other words,"},{"Start":"05:04.520 ","End":"05:07.590","Text":"this is the answer to Question number 1."},{"Start":"05:08.240 ","End":"05:11.450","Text":"This is the charge as a function of time."},{"Start":"05:11.450 ","End":"05:14.240","Text":"Now, let\u0027s onset Question number 2."},{"Start":"05:14.240 ","End":"05:17.930","Text":"Calculate the electric field also as a function of time."},{"Start":"05:17.930 ","End":"05:21.425","Text":"Because we\u0027re dealing with a parallel plate capacitor,"},{"Start":"05:21.425 ","End":"05:22.730","Text":"you should remember this."},{"Start":"05:22.730 ","End":"05:29.945","Text":"The electric field only of a parallel plate capacitor is always equal to v divided by d,"},{"Start":"05:29.945 ","End":"05:37.200","Text":"or an equivalent equation is Sigma divided by Epsilon Naught."},{"Start":"05:38.120 ","End":"05:42.650","Text":"This is the electric field for a parallel plate capacitor."},{"Start":"05:42.650 ","End":"05:44.585","Text":"You can use one of these versions."},{"Start":"05:44.585 ","End":"05:47.060","Text":"Because in the previous question,"},{"Start":"05:47.060 ","End":"05:48.545","Text":"I calculated the charge,"},{"Start":"05:48.545 ","End":"05:54.180","Text":"so it\u0027s easier for me to use the Sigma divided by Epsilon Naught version."},{"Start":"05:54.180 ","End":"05:59.165","Text":"Sigma is simply the charge divided by the surface area."},{"Start":"05:59.165 ","End":"06:03.740","Text":"So my charge is q as a function of t,"},{"Start":"06:03.740 ","End":"06:05.945","Text":"which I just calculated in Question 1,"},{"Start":"06:05.945 ","End":"06:16.435","Text":"divided by Epsilon Naught and also divided by the area which is just Pie R^2."},{"Start":"06:16.435 ","End":"06:19.520","Text":"This is the magnitude of the electric field,"},{"Start":"06:19.520 ","End":"06:23.660","Text":"and now let\u0027s find the direction. Let\u0027s take a look."},{"Start":"06:23.660 ","End":"06:30.560","Text":"Over here, our current is flowing to the top plate. What does that mean?"},{"Start":"06:30.560 ","End":"06:32.525","Text":"That means that here we\u0027re getting"},{"Start":"06:32.525 ","End":"06:37.220","Text":"the positive charges and here the positive charges are leaving."},{"Start":"06:37.220 ","End":"06:40.340","Text":"So this is the positive plate and this is the negative plate,"},{"Start":"06:40.340 ","End":"06:43.790","Text":"which means that our electric field is"},{"Start":"06:43.790 ","End":"06:47.810","Text":"in this direction because the electric field goes from positive to negative."},{"Start":"06:47.810 ","End":"06:50.960","Text":"Now if we define a direction,"},{"Start":"06:50.960 ","End":"06:53.435","Text":"so let\u0027s say that this is the z-axis,"},{"Start":"06:53.435 ","End":"06:57.320","Text":"we can see that the electric field is in the negative z direction."},{"Start":"06:57.320 ","End":"07:02.835","Text":"It\u0027s in the negative z direction like so."},{"Start":"07:02.835 ","End":"07:06.660","Text":"This is the answer to Question 2."},{"Start":"07:06.660 ","End":"07:12.065","Text":"What we can see is we have an electric field which is dependent on time."},{"Start":"07:12.065 ","End":"07:14.375","Text":"Of course, we can also plug in instead of qt,"},{"Start":"07:14.375 ","End":"07:18.545","Text":"we can write It, but we can just leave it like this for now."},{"Start":"07:18.545 ","End":"07:20.435","Text":"The second that you will see this,"},{"Start":"07:20.435 ","End":"07:23.045","Text":"you should know that there\u0027s going to be a magnetic field."},{"Start":"07:23.045 ","End":"07:28.825","Text":"Indeed, Question number 3 asks us to calculate the magnetic field."},{"Start":"07:28.825 ","End":"07:33.660","Text":"We\u0027re going to be using this equation like so."},{"Start":"07:33.660 ","End":"07:40.285","Text":"The first thing that I\u0027m going to do is I\u0027m going to do like in Ampere\u0027s law."},{"Start":"07:40.285 ","End":"07:47.080","Text":"If I know that my electric field is going like so in the negative z direction,"},{"Start":"07:47.080 ","End":"07:51.580","Text":"then that means that my magnetic field is going to be"},{"Start":"07:51.580 ","End":"07:56.635","Text":"going in this Theta direction, like so."},{"Start":"07:56.635 ","End":"07:59.390","Text":"In a loop."},{"Start":"07:59.390 ","End":"08:05.770","Text":"What I can do is I can draw Ampere\u0027s loop like so,"},{"Start":"08:05.770 ","End":"08:08.140","Text":"and say that it has a radius of a"},{"Start":"08:08.140 ","End":"08:14.425","Text":"lowercase r. Now I can write out this side of the equation,"},{"Start":"08:14.425 ","End":"08:19.660","Text":"so my integral on B.dl."},{"Start":"08:19.660 ","End":"08:26.245","Text":"Because my magnetic field is constant across this length,"},{"Start":"08:26.245 ","End":"08:32.560","Text":"so I can just write that this is equal to B multiplied by the length of this loop,"},{"Start":"08:32.560 ","End":"08:35.125","Text":"which of course is the circumference of the loop,"},{"Start":"08:35.125 ","End":"08:43.073","Text":"B multiplied by 2 Pi r. This is of course equal to,"},{"Start":"08:43.073 ","End":"08:46.615","Text":"so let\u0027s take a look at this over here."},{"Start":"08:46.615 ","End":"08:48.925","Text":"This is of course,"},{"Start":"08:48.925 ","End":"08:51.595","Text":"Mu multiplied by I in,"},{"Start":"08:51.595 ","End":"09:00.440","Text":"Mu multiplied by the current flowing through this region over here."},{"Start":"09:00.450 ","End":"09:04.675","Text":"As we know, because we\u0027re dealing with capacitors,"},{"Start":"09:04.675 ","End":"09:09.340","Text":"then there\u0027s no current actually flowing between these 2 plates."},{"Start":"09:09.340 ","End":"09:16.135","Text":"Because remember there\u0027s space and so no charges can pass through this space over here."},{"Start":"09:16.135 ","End":"09:18.100","Text":"If no charges are passing through,"},{"Start":"09:18.100 ","End":"09:24.020","Text":"that means no current is passing through this area over here."},{"Start":"09:24.270 ","End":"09:34.015","Text":"That means that if we say that this is equal to Mu Naught multiplied by I in,"},{"Start":"09:34.015 ","End":"09:38.350","Text":"this is equal to 0 because of course no current is passing between the plates,"},{"Start":"09:38.350 ","End":"09:41.575","Text":"so no current his passing through this region."},{"Start":"09:41.575 ","End":"09:44.155","Text":"Now we have to add over here."},{"Start":"09:44.155 ","End":"09:52.510","Text":"We\u0027re adding plus Mu Naught multiplied by the integral of Epsilon Naught"},{"Start":"09:52.510 ","End":"10:01.060","Text":"multiplied by the time derivative of our electric field ds."},{"Start":"10:01.060 ","End":"10:05.710","Text":"First of all, let\u0027s calculate our E. our time derivative of our electric field,"},{"Start":"10:05.710 ","End":"10:08.840","Text":"or in other words our, dE by dt."},{"Start":"10:09.290 ","End":"10:13.560","Text":"What is dependent on time over here is our charge,"},{"Start":"10:13.560 ","End":"10:15.000","Text":"which is equal to It."},{"Start":"10:15.000 ","End":"10:17.675","Text":"If we take the time derivative of that,"},{"Start":"10:17.675 ","End":"10:21.940","Text":"we just get q. which is of course just equal to I."},{"Start":"10:21.940 ","End":"10:27.010","Text":"What we have is negative q. so negative I divided by"},{"Start":"10:27.010 ","End":"10:34.240","Text":"Epsilon Naught Pi R^2 and of course in the z direction."},{"Start":"10:34.240 ","End":"10:38.995","Text":"Now what we can do is we can plug this in over here."},{"Start":"10:38.995 ","End":"10:47.440","Text":"What we have is that B multiplied by 2Pir is equal to 0 plus Mu Naught"},{"Start":"10:47.440 ","End":"10:57.955","Text":"integral of Epsilon Naught multiplied by negative I divided by Epsilon Naught Pi R^2."},{"Start":"10:57.955 ","End":"11:07.600","Text":"Then all of this is z.ds or rather z hat.ds."},{"Start":"11:07.600 ","End":"11:10.765","Text":"First of all, we can see that"},{"Start":"11:10.765 ","End":"11:16.570","Text":"our time derivative of the electric field is constant in space,"},{"Start":"11:16.570 ","End":"11:18.895","Text":"okay we have no variables."},{"Start":"11:18.895 ","End":"11:22.150","Text":"The electric field is constant throughout this region."},{"Start":"11:22.150 ","End":"11:24.145","Text":"Which means that instead of integrating,"},{"Start":"11:24.145 ","End":"11:30.805","Text":"I can just multiply by the surface area of this Ampere loop that I put."},{"Start":"11:30.805 ","End":"11:39.265","Text":"We have B multiplied by 2 Pi r is equal to Mu Naught multiplied by Epsilon Naught."},{"Start":"11:39.265 ","End":"11:42.280","Text":"Case I\u0027m taking this out of the integral sign,"},{"Start":"11:42.280 ","End":"11:46.420","Text":"so multiplied by negative I divided by"},{"Start":"11:46.420 ","End":"11:54.100","Text":"Epsilon Naught Pi R^2 and multiplied by the surface area of this loop,"},{"Start":"11:54.100 ","End":"12:00.040","Text":"which is just equal to Pi r^2."},{"Start":"12:00.040 ","End":"12:03.340","Text":"Now I can see that Epsilon Naught and Epsilon Naught cancel out,"},{"Start":"12:03.340 ","End":"12:05.425","Text":"Pi and Pi cancel out."},{"Start":"12:05.425 ","End":"12:11.365","Text":"I can divide both sides by this lowercase r over here."},{"Start":"12:11.365 ","End":"12:15.610","Text":"Then I\u0027m going to divide both sides by 2 Pi to isolate out my B."},{"Start":"12:15.610 ","End":"12:21.110","Text":"My magnetic field is equal to negative Mu Naught"},{"Start":"12:21.110 ","End":"12:29.385","Text":"I multiplied by lowercase r"},{"Start":"12:29.385 ","End":"12:35.680","Text":"divided by 2 Pi R^2."},{"Start":"12:35.820 ","End":"12:39.100","Text":"This is the magnitude of the magnetic field."},{"Start":"12:39.100 ","End":"12:40.840","Text":"Of course this is a vector,"},{"Start":"12:40.840 ","End":"12:43.090","Text":"so we want to find the direction."},{"Start":"12:43.090 ","End":"12:47.710","Text":"As we said, the z direction is pointing upwards,"},{"Start":"12:47.710 ","End":"12:53.200","Text":"so the positive Theta direction is like so."},{"Start":"12:53.200 ","End":"12:56.530","Text":"This is the Theta direction."},{"Start":"12:56.530 ","End":"13:00.370","Text":"Now if I use the right-hand rule,"},{"Start":"13:00.370 ","End":"13:03.670","Text":"so I point my thumb in the direction of the electric field,"},{"Start":"13:03.670 ","End":"13:06.220","Text":"so my thumb is pointing downwards."},{"Start":"13:06.220 ","End":"13:12.820","Text":"Then I can see that my magnetic field is in fact going in the negative Theta direction."},{"Start":"13:12.820 ","End":"13:20.210","Text":"My magnetic field is going in this direction in fact, like so."},{"Start":"13:27.900 ","End":"13:33.715","Text":"I have to put that the direction is in the negative Theta direction."},{"Start":"13:33.715 ","End":"13:35.500","Text":"Here I already have a negative,"},{"Start":"13:35.500 ","End":"13:38.695","Text":"so I\u0027m just going to add in my Theta hat over here,"},{"Start":"13:38.695 ","End":"13:40.420","Text":"and there we have it."},{"Start":"13:40.420 ","End":"13:44.960","Text":"The magnetic field is in the negative Theta direction."},{"Start":"13:45.090 ","End":"13:48.790","Text":"This is the answer to question number 3."},{"Start":"13:48.790 ","End":"13:54.579","Text":"These are the main questions that you have a chance of being asked about in an exam,"},{"Start":"13:54.579 ","End":"13:57.415","Text":"they rarely follow on from one another."},{"Start":"13:57.415 ","End":"14:02.680","Text":"The next 2 sections of this question are added."},{"Start":"14:02.680 ","End":"14:06.057","Text":"You could be asked these in the exam,"},{"Start":"14:06.057 ","End":"14:08.995","Text":"so let\u0027s answer these as well."},{"Start":"14:08.995 ","End":"14:11.710","Text":"Now let\u0027s answer a question number 4"},{"Start":"14:11.710 ","End":"14:15.880","Text":"calculate the energy stored between the capacitor plates."},{"Start":"14:15.880 ","End":"14:21.400","Text":"We\u0027ve already seen that the energy is equal to the integral on the energy density."},{"Start":"14:21.400 ","End":"14:27.760","Text":"Over here, the energy density for an electric field is Mu Naught multiplied by"},{"Start":"14:27.760 ","End":"14:34.735","Text":"the electric field squared and divide it by 2 dv."},{"Start":"14:34.735 ","End":"14:37.330","Text":"That means we\u0027re summing up"},{"Start":"14:37.330 ","End":"14:40.810","Text":"the electric field squared multiplied by Epsilon Naught divided by"},{"Start":"14:40.810 ","End":"14:48.355","Text":"2 in the volume that fills up this area between the 2 capacitor plates."},{"Start":"14:48.355 ","End":"14:51.400","Text":"Then you might remember that we got that the energy of"},{"Start":"14:51.400 ","End":"14:55.480","Text":"the capacitor will after doing this equation,"},{"Start":"14:55.480 ","End":"15:02.020","Text":"this integral is equal to 1.5 multiplied by q squared divided by c,"},{"Start":"15:02.020 ","End":"15:05.320","Text":"where c is of course the capacitance."},{"Start":"15:05.320 ","End":"15:09.905","Text":"What we can see here is this is the energy due to the electric field."},{"Start":"15:09.905 ","End":"15:12.950","Text":"However, here we also know that we have a magnetic field."},{"Start":"15:12.950 ","End":"15:14.915","Text":"We already calculated that."},{"Start":"15:14.915 ","End":"15:19.747","Text":"What we\u0027re going to do is we have to add on to this equation a correction,"},{"Start":"15:19.747 ","End":"15:22.660","Text":"and this correction is for the magnetic field."},{"Start":"15:22.660 ","End":"15:28.024","Text":"We\u0027re integrating for the electric field energy density"},{"Start":"15:28.024 ","End":"15:32.730","Text":"and for the magnetic field energy density."},{"Start":"15:32.730 ","End":"15:40.480","Text":"Here we have the magnetic field squared and then it\u0027s divided by 2 Mu Naught."},{"Start":"15:40.480 ","End":"15:46.860","Text":"Then we integrate for the whole volume between the capacitor plates."},{"Start":"15:46.890 ","End":"15:50.900","Text":"Now what I\u0027m going to have to do is I\u0027m going to have to sum up for all of this,"},{"Start":"15:50.900 ","End":"15:53.750","Text":"for the whole volume of the capacitor."},{"Start":"15:53.750 ","End":"15:58.370","Text":"The units were obviously summing in cylindrical units."},{"Start":"15:58.370 ","End":"16:04.560","Text":"We have r dr d Theta dz."},{"Start":"16:04.560 ","End":"16:09.014","Text":"Of course, our integral becomes a triple integral,"},{"Start":"16:09.014 ","End":"16:13.220","Text":"so the bounds for r are from 0 up until a radius of"},{"Start":"16:13.220 ","End":"16:20.075","Text":"capital R. The bounds for theta is from 0 until 2 Pi because we\u0027re going all around."},{"Start":"16:20.075 ","End":"16:26.840","Text":"The bounds for z are from 0 up until the second plate,"},{"Start":"16:26.840 ","End":"16:28.730","Text":"which is a distance d away,"},{"Start":"16:28.730 ","End":"16:33.935","Text":"so from 0 until d. Then you do this integral,"},{"Start":"16:33.935 ","End":"16:35.870","Text":"I\u0027m not going to waste time solving the integral."},{"Start":"16:35.870 ","End":"16:37.940","Text":"Feel free to pause this video."},{"Start":"16:37.940 ","End":"16:43.010","Text":"But what we get in the end is that the energy is equal to"},{"Start":"16:43.010 ","End":"16:48.290","Text":"I^2 T^2 d divided"},{"Start":"16:48.290 ","End":"16:54.460","Text":"by 2 Epsilon Naught Pi R^2."},{"Start":"16:54.460 ","End":"17:00.460","Text":"This is the energy that comes from this term over here involving the electric field plus"},{"Start":"17:00.460 ","End":"17:10.080","Text":"Mu Naught I^2 d divided by 16 Pi."},{"Start":"17:10.080 ","End":"17:15.575","Text":"This term comes from this term over here involving the magnetic field."},{"Start":"17:15.575 ","End":"17:17.450","Text":"When you do this integral,"},{"Start":"17:17.450 ","End":"17:22.835","Text":"you just have to substitute in the E-field that we calculated up here,"},{"Start":"17:22.835 ","End":"17:28.895","Text":"remembering that qt is equal to It and of course,"},{"Start":"17:28.895 ","End":"17:32.079","Text":"to substitute in the b field into here,"},{"Start":"17:32.079 ","End":"17:34.650","Text":"and then you should get this answer."},{"Start":"17:34.890 ","End":"17:38.480","Text":"This is the answer to question number 4."},{"Start":"17:38.480 ","End":"17:43.340","Text":"Now, let\u0027s answer question number 5."},{"Start":"17:43.340 ","End":"17:47.915","Text":"We\u0027re being asked to calculate the Poynting vector on the edge of the capacitor,"},{"Start":"17:47.915 ","End":"17:53.767","Text":"find its flux through the Gaussian surface encasing the capacitor."},{"Start":"17:53.767 ","End":"17:56.740","Text":"First, let\u0027s calculate the Poynting vector,"},{"Start":"17:56.740 ","End":"17:58.870","Text":"and then we\u0027ll calculate the flux."},{"Start":"17:58.870 ","End":"18:05.320","Text":"The equation for the Poynting vector is S is equal to 1 divided"},{"Start":"18:05.320 ","End":"18:12.025","Text":"by Mu Naught multiplied by E cross B. I\u0027m just reminding you,"},{"Start":"18:12.025 ","End":"18:15.400","Text":"you might not have learned this already so if you don\u0027t know this,"},{"Start":"18:15.400 ","End":"18:17.890","Text":"you could also even skip this question and come back to it"},{"Start":"18:17.890 ","End":"18:21.190","Text":"later when we learn about waves."},{"Start":"18:21.190 ","End":"18:24.190","Text":"But you can look through this lesson now if you want to,"},{"Start":"18:24.190 ","End":"18:26.440","Text":"but if this isn\u0027t relevant, feel free to skip."},{"Start":"18:26.440 ","End":"18:30.565","Text":"Anyway, this is the equation for the Poynting vector."},{"Start":"18:30.565 ","End":"18:35.575","Text":"We\u0027re being asked to find the Poynting vector on the edge of the capacitor."},{"Start":"18:35.575 ","End":"18:41.990","Text":"We can imagine that this capacitor is covered by some cylindrical casing."},{"Start":"18:42.030 ","End":"18:45.670","Text":"Let\u0027s see how we\u0027re going to draw that."},{"Start":"18:45.670 ","End":"18:49.510","Text":"We have some cylindrical casing like so in"},{"Start":"18:49.510 ","End":"18:55.570","Text":"gray that is just covering the entire capacitor."},{"Start":"18:55.570 ","End":"19:02.440","Text":"Now what we want to do is we want to calculate the Poynting vector at some point."},{"Start":"19:02.440 ","End":"19:04.450","Text":"Let\u0027s look at it over here."},{"Start":"19:04.450 ","End":"19:07.540","Text":"First of all, if we\u0027re looking at this point over here,"},{"Start":"19:07.540 ","End":"19:11.305","Text":"we can see that the magnetic field is coming towards us."},{"Start":"19:11.305 ","End":"19:20.830","Text":"Because we\u0027re looking at the surface or the shell going around the capacitor,"},{"Start":"19:20.830 ","End":"19:24.265","Text":"so we know and we\u0027re being told to calculate it at the edge."},{"Start":"19:24.265 ","End":"19:27.453","Text":"We\u0027re looking at the magnetic field at radius R,"},{"Start":"19:27.453 ","End":"19:30.415","Text":"so capital R at the edge,"},{"Start":"19:30.415 ","End":"19:33.925","Text":"and it\u0027s coming towards us because it\u0027s in the negative Theta direction,"},{"Start":"19:33.925 ","End":"19:38.740","Text":"so we know that the magnetic field at this point is coming out towards us."},{"Start":"19:38.740 ","End":"19:45.190","Text":"If we look at our equation for the electric field,"},{"Start":"19:45.190 ","End":"19:50.875","Text":"we see that our electric field is in the negative z direction."},{"Start":"19:50.875 ","End":"19:54.910","Text":"We can come back down here and we see that the z direction is Poynting up."},{"Start":"19:54.910 ","End":"20:02.330","Text":"We can then say that the electric field over here is pointing down."},{"Start":"20:02.460 ","End":"20:10.165","Text":"Let\u0027s calculate first of all the magnitude of the Poynting vector."},{"Start":"20:10.165 ","End":"20:16.165","Text":"What I have is 1 divided by Mu Naught and then I have E cross B at"},{"Start":"20:16.165 ","End":"20:19.105","Text":"R. I\u0027m going to do"},{"Start":"20:19.105 ","End":"20:23.230","Text":"the cross-product as the magnitude of E multiplied by the magnitude of B,"},{"Start":"20:23.230 ","End":"20:27.170","Text":"multiplied by Sine of the angle between the two."},{"Start":"20:29.160 ","End":"20:32.635","Text":"I\u0027m just writing the magnitude without the direction,"},{"Start":"20:32.635 ","End":"20:36.535","Text":"I\u0027m paying no attention to any minus signs that they might be."},{"Start":"20:36.535 ","End":"20:42.205","Text":"The magnitude of E is equal to q as a function of"},{"Start":"20:42.205 ","End":"20:49.915","Text":"t divided by Epsilon Naught Pi R^2."},{"Start":"20:49.915 ","End":"20:52.690","Text":"This is written up top in the screen."},{"Start":"20:52.690 ","End":"20:58.150","Text":"Feel free to rewind the video if you can\u0027t remember if you didn\u0027t write this down."},{"Start":"20:58.150 ","End":"21:01.825","Text":"This multiplied by the magnitude of the B field."},{"Start":"21:01.825 ","End":"21:04.165","Text":"Again, not paying attention to the signs."},{"Start":"21:04.165 ","End":"21:10.090","Text":"That\u0027s Mu Naught I and the B field at R. Instead of lowercase r,"},{"Start":"21:10.090 ","End":"21:16.450","Text":"I substitute in capital R divided by 2 Pi R^2."},{"Start":"21:16.450 ","End":"21:20.185","Text":"This is multiplied by Sine of the angle between the two."},{"Start":"21:20.185 ","End":"21:25.210","Text":"Seeing as our B field is coming out towards us and our E field is pointing downwards,"},{"Start":"21:25.210 ","End":"21:28.660","Text":"the angle between the two is 90 degrees."},{"Start":"21:28.660 ","End":"21:33.770","Text":"As we know, this is Sine of 90 degrees is equal to 1."},{"Start":"21:34.980 ","End":"21:38.665","Text":"Now we want to know the direction of"},{"Start":"21:38.665 ","End":"21:43.555","Text":"S. All we have to do is we have to use the right-hand rule."},{"Start":"21:43.555 ","End":"21:47.380","Text":"Our thumb points in the direction of the E field,"},{"Start":"21:47.380 ","End":"21:48.565","Text":"so points down,"},{"Start":"21:48.565 ","End":"21:51.610","Text":"our fore finger points in the direction of the B field,"},{"Start":"21:51.610 ","End":"21:53.545","Text":"so it\u0027s pointing towards us."},{"Start":"21:53.545 ","End":"21:58.720","Text":"Then what we get is that our S is pointing inwards like this."},{"Start":"21:58.720 ","End":"22:00.190","Text":"Or in other words,"},{"Start":"22:00.190 ","End":"22:03.310","Text":"it\u0027s pointing in the negative radial direction."},{"Start":"22:03.310 ","End":"22:08.740","Text":"I can add here a negative and radial direction."},{"Start":"22:08.740 ","End":"22:11.185","Text":"Now we can play around and cancel stuff out."},{"Start":"22:11.185 ","End":"22:13.420","Text":"The Mu naught and Mu naught cancels out,"},{"Start":"22:13.420 ","End":"22:16.090","Text":"this R can cancel out with one of these"},{"Start":"22:16.090 ","End":"22:23.650","Text":"R. This is what you get in total for the pointing angle."},{"Start":"22:23.650 ","End":"22:27.120","Text":"Of course, remember that Qt is just It."},{"Start":"22:27.120 ","End":"22:30.660","Text":"So here you can have I^2 multiplied by t instead,"},{"Start":"22:30.660 ","End":"22:31.890","Text":"but it doesn\u0027t make a difference."},{"Start":"22:31.890 ","End":"22:34.365","Text":"It\u0027s of course and the negative I direction."},{"Start":"22:34.365 ","End":"22:41.650","Text":"That\u0027s the Poynting vector and now we want to find its flux through the Gaussian surface."},{"Start":"22:42.360 ","End":"22:45.130","Text":"I wrote here Gaussian surface,"},{"Start":"22:45.130 ","End":"22:47.500","Text":"this, of course, we\u0027re not using Gauss\u0027s law,"},{"Start":"22:47.500 ","End":"22:49.540","Text":"so this isn\u0027t a Gaussian surface,"},{"Start":"22:49.540 ","End":"22:51.249","Text":"but I just meant surface,"},{"Start":"22:51.249 ","End":"22:53.110","Text":"but it makes no difference."},{"Start":"22:53.110 ","End":"22:56.200","Text":"We have a surface that\u0027s acting the same way."},{"Start":"22:56.200 ","End":"23:00.505","Text":"Now we\u0027re calculating the flux of this Poynting vector."},{"Start":"23:00.505 ","End":"23:02.500","Text":"The equation for flux,"},{"Start":"23:02.500 ","End":"23:05.140","Text":"we\u0027re used to seeing it with Gauss\u0027s law"},{"Start":"23:05.140 ","End":"23:08.380","Text":"when dealing with the flux of the electric field,"},{"Start":"23:08.380 ","End":"23:15.055","Text":"and we also do a similar calculation when calculating the flux of the magnetic field."},{"Start":"23:15.055 ","End":"23:18.355","Text":"Here we\u0027re working out the flux of the Poynting vector."},{"Start":"23:18.355 ","End":"23:24.530","Text":"Of course, we\u0027re integrating with the Poynting vector.ds."},{"Start":"23:26.040 ","End":"23:30.010","Text":"The next thing that we can see is that we have a closed surface."},{"Start":"23:30.010 ","End":"23:32.860","Text":"I can just symbolize this with a circle over here."},{"Start":"23:32.860 ","End":"23:36.025","Text":"This is of course the flux for our Poynting vector."},{"Start":"23:36.025 ","End":"23:40.195","Text":"What we can see, because S,"},{"Start":"23:40.195 ","End":"23:44.635","Text":"our Poynting vector is always in the radial direction,"},{"Start":"23:44.635 ","End":"23:49.885","Text":"and we can see that it\u0027s pretty much a constant if time is constant,"},{"Start":"23:49.885 ","End":"23:55.510","Text":"but spatially it\u0027s uniform throughout."},{"Start":"23:55.510 ","End":"24:02.230","Text":"What we can see is that we\u0027re just going to multiply the Poynting vector by the area."},{"Start":"24:02.230 ","End":"24:08.080","Text":"When we\u0027re looking at the cylindrical parts the part that curves around the capacitor,"},{"Start":"24:08.080 ","End":"24:15.760","Text":"we can see that our vector is always perpendicular to the surface, which is great."},{"Start":"24:15.760 ","End":"24:18.640","Text":"Because then the dot product doesn\u0027t cancel out."},{"Start":"24:18.640 ","End":"24:20.860","Text":"Then that\u0027s great."},{"Start":"24:20.860 ","End":"24:25.689","Text":"However, we can see that our Poynting vector is parallel"},{"Start":"24:25.689 ","End":"24:31.675","Text":"to the plates themselves or to the disks on top and at the bottom of the cylinder."},{"Start":"24:31.675 ","End":"24:38.050","Text":"The dot product between two vectors that are parallel is equal to 0."},{"Start":"24:38.050 ","End":"24:42.160","Text":"What we see is that we\u0027re just taking the magnitude of"},{"Start":"24:42.160 ","End":"24:49.010","Text":"our Poynting vector and multiplying it by the total curved area."},{"Start":"24:49.140 ","End":"24:53.950","Text":"As we know, that\u0027s going to be the circumference of the cylinder."},{"Start":"24:53.950 ","End":"24:58.480","Text":"That\u0027s 2Pi multiplied by the radius of the cylinder and"},{"Start":"24:58.480 ","End":"25:03.460","Text":"then multiplied by the length to give us this surface area,"},{"Start":"25:03.460 ","End":"25:07.480","Text":"so multiplied by d. In other words,"},{"Start":"25:07.480 ","End":"25:12.820","Text":"the flux of the Poynting vector is equal to negative"},{"Start":"25:12.820 ","End":"25:19.195","Text":"q(t) multiplied by I divided by,"},{"Start":"25:19.195 ","End":"25:22.810","Text":"here we have this 2 will cancel out with this 2,"},{"Start":"25:22.810 ","End":"25:25.180","Text":"this Pi will cancel out with one of these Pi\u0027s,"},{"Start":"25:25.180 ","End":"25:28.870","Text":"this r will cancel out with one of these rs."},{"Start":"25:28.870 ","End":"25:37.795","Text":"What we get is negative qI for S multiplied by d from over here,"},{"Start":"25:37.795 ","End":"25:44.540","Text":"and then divided by Epsilon Naught PiR^2."},{"Start":"25:44.700 ","End":"25:49.300","Text":"Then I now just plug it in what q(t) was,"},{"Start":"25:49.300 ","End":"25:53.890","Text":"which was equal to I multiplied by t, I,"},{"Start":"25:53.890 ","End":"25:54.940","Text":"and the I from here,"},{"Start":"25:54.940 ","End":"26:00.680","Text":"I^2td and this is the answer."},{"Start":"26:00.930 ","End":"26:08.960","Text":"This is the flux of the Poynting vector through the surface encasing the capacitor."},{"Start":"26:10.050 ","End":"26:13.240","Text":"This what\u0027s here in this blue square is"},{"Start":"26:13.240 ","End":"26:15.790","Text":"an equation that you have to write in your equation books."},{"Start":"26:15.790 ","End":"26:18.100","Text":"Something that I want you to notice,"},{"Start":"26:18.100 ","End":"26:22.810","Text":"if we look over here at the flux of the Poynting vector,"},{"Start":"26:22.810 ","End":"26:25.060","Text":"I\u0027ll add an S over here."},{"Start":"26:25.060 ","End":"26:30.220","Text":"It is equal to the negative time derivative of the energy."},{"Start":"26:30.220 ","End":"26:33.459","Text":"As we can see, if we take the negative time derivative,"},{"Start":"26:33.459 ","End":"26:36.130","Text":"this time cancels out because it\u0027s independent of"},{"Start":"26:36.130 ","End":"26:40.615","Text":"t. And here of course we can take the derivative."},{"Start":"26:40.615 ","End":"26:45.260","Text":"If you see, you will get this exact onset."},{"Start":"26:45.270 ","End":"26:51.610","Text":"This is another equation that you should write in your equation books."},{"Start":"26:51.610 ","End":"26:54.730","Text":"Then it\u0027s a much easier way to get from"},{"Start":"26:54.730 ","End":"26:59.090","Text":"each answer if you don\u0027t want to do all of these calculations."},{"Start":"26:59.520 ","End":"27:02.140","Text":"Of course these equations,"},{"Start":"27:02.140 ","End":"27:03.880","Text":"the Poynting vector as a whole,"},{"Start":"27:03.880 ","End":"27:07.660","Text":"we\u0027re going to speak about much more in depth in later chapters."},{"Start":"27:07.660 ","End":"27:11.530","Text":"But in the meantime, I just wanted you to get a feel for this."},{"Start":"27:11.530 ","End":"27:15.355","Text":"Generally when you are answering questions dealing with the Poynting vector,"},{"Start":"27:15.355 ","End":"27:21.220","Text":"just knowing how to plug in the values into these equations will be good enough."},{"Start":"27:21.220 ","End":"27:24.110","Text":"That\u0027s the end of this lesson."}],"ID":21440},{"Watched":false,"Name":"Exercise 3","Duration":"20m 14s","ChapterTopicVideoID":21361,"CourseChapterTopicPlaylistID":99494,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.800","Text":"Hello. In this lesson,"},{"Start":"00:01.800 ","End":"00:04.350","Text":"we\u0027re going to be answering the following question."},{"Start":"00:04.350 ","End":"00:07.485","Text":"A parallel plate capacitor is given."},{"Start":"00:07.485 ","End":"00:12.330","Text":"Each plate is circular and has a radius of a."},{"Start":"00:12.330 ","End":"00:20.775","Text":"Over here, this length is a and the plates are a distance d from one another."},{"Start":"00:20.775 ","End":"00:25.980","Text":"The capacitor is connected to a conducting track where there\u0027s"},{"Start":"00:25.980 ","End":"00:33.150","Text":"no resistance and a non-resistant rod of length L is placed on the truck."},{"Start":"00:33.150 ","End":"00:38.490","Text":"The rod is pulled away from the capacitor at a velocity of v(t),"},{"Start":"00:38.490 ","End":"00:40.995","Text":"which is equal to a multiplied by t,"},{"Start":"00:40.995 ","End":"00:46.595","Text":"and a uniform magnetic field B acts into the page."},{"Start":"00:46.595 ","End":"00:48.260","Text":"Question Number 1 is,"},{"Start":"00:48.260 ","End":"00:55.445","Text":"calculate the charge on the capacitor and to also the signal which is the positive plate."},{"Start":"00:55.445 ","End":"00:57.485","Text":"First of all, let\u0027s see."},{"Start":"00:57.485 ","End":"01:01.670","Text":"We can calculate that EMF, which as we know,"},{"Start":"01:01.670 ","End":"01:09.995","Text":"is equal to either the negative time integral of the magnetic flux,"},{"Start":"01:09.995 ","End":"01:14.900","Text":"or you can say that the EMF when dealing with"},{"Start":"01:14.900 ","End":"01:22.050","Text":"the magnetic field is equal to B multiplied by L multiplied by V, the velocity."},{"Start":"01:22.130 ","End":"01:25.165","Text":"In our case, we\u0027re given B,"},{"Start":"01:25.165 ","End":"01:29.680","Text":"we\u0027re given L and we\u0027re given the velocity so let\u0027s just plug that in."},{"Start":"01:29.680 ","End":"01:37.400","Text":"We have BL in the velocity is equal to A multiplied by t."},{"Start":"01:37.730 ","End":"01:42.880","Text":"Now we can see that this rod or the movement of it is making it"},{"Start":"01:42.880 ","End":"01:47.740","Text":"act as if it is a voltage source."},{"Start":"01:47.740 ","End":"01:50.710","Text":"If it\u0027s a voltage source,"},{"Start":"01:50.710 ","End":"01:57.825","Text":"so we know that there\u0027s a connection between the charge on a capacitor and the voltage."},{"Start":"01:57.825 ","End":"02:00.940","Text":"We know that the equation is that the charge on a capacitor is"},{"Start":"02:00.940 ","End":"02:04.290","Text":"equal to the capacitance multiplied by"},{"Start":"02:04.290 ","End":"02:12.720","Text":"the voltage so in our case this is equal to the capacitance multiplied by the EMF."},{"Start":"02:13.160 ","End":"02:19.834","Text":"Let\u0027s plug in our values so the capacitance for a parallel plate capacitor"},{"Start":"02:19.834 ","End":"02:26.000","Text":"is equal to the epsilon node multiplied by the surface area of the plates."},{"Start":"02:26.000 ","End":"02:31.100","Text":"Here, we\u0027re told that they are circular plates of radius a,"},{"Start":"02:31.100 ","End":"02:37.070","Text":"so it\u0027s pi a squared and then divided by the distance between the 2 plates,"},{"Start":"02:37.070 ","End":"02:40.535","Text":"which is d, and then multiplied by our EMF,"},{"Start":"02:40.535 ","End":"02:43.110","Text":"which is equal to BLAt."},{"Start":"02:45.020 ","End":"02:48.875","Text":"Here we\u0027ve calculated the charge on"},{"Start":"02:48.875 ","End":"02:52.640","Text":"the capacitor and now we want to see which is the positive plate."},{"Start":"02:52.640 ","End":"02:55.505","Text":"If we would\u0027ve worked with this equation over here,"},{"Start":"02:55.505 ","End":"03:00.800","Text":"then we would have to use Lenz\u0027s law in order to answer this question,"},{"Start":"03:00.800 ","End":"03:08.090","Text":"which is a positive plate and with the version for the EMF that we\u0027ve used."},{"Start":"03:08.090 ","End":"03:11.195","Text":"We have to use the right-hand rule."},{"Start":"03:11.195 ","End":"03:16.070","Text":"We know that the force is equal to the charge multiplied"},{"Start":"03:16.070 ","End":"03:22.230","Text":"by the velocity cross product with the magnetic field."},{"Start":"03:23.170 ","End":"03:29.420","Text":"Our thumb points in the direction of the velocity so our thumb is pointing rightwords,"},{"Start":"03:29.420 ","End":"03:34.115","Text":"our forefinger points in the direction of the magnetic fields so inside the page."},{"Start":"03:34.115 ","End":"03:39.380","Text":"Then our middle finger represents the direction of the force and what we get is"},{"Start":"03:39.380 ","End":"03:45.345","Text":"that our force is upwards so this is our force."},{"Start":"03:45.345 ","End":"03:49.340","Text":"In that case, this is where the positive charges"},{"Start":"03:49.340 ","End":"03:54.590","Text":"are building up and this is where the negative charges are"},{"Start":"03:54.590 ","End":"04:04.580","Text":"building up so that means that the current is flowing like so."},{"Start":"04:04.580 ","End":"04:15.145","Text":"In other words, this is what the voltage source looks like."},{"Start":"04:15.145 ","End":"04:20.820","Text":"In other words, this is the positive plate and this is the negative plate."},{"Start":"04:21.280 ","End":"04:24.305","Text":"This is the answer to Question Number 1."},{"Start":"04:24.305 ","End":"04:25.610","Text":"This is the charge on"},{"Start":"04:25.610 ","End":"04:33.260","Text":"the capacitor and the top plate is positive and the bottom plate is negative."},{"Start":"04:33.260 ","End":"04:36.035","Text":"Now let\u0027s go on to answering Question Number 2,"},{"Start":"04:36.035 ","End":"04:40.040","Text":"calculate the electric field inside the capacitor."},{"Start":"04:40.040 ","End":"04:45.260","Text":"When we\u0027re dealing with a parallel plate capacitor only,"},{"Start":"04:45.260 ","End":"04:50.480","Text":"then we know that the electric field inside"},{"Start":"04:50.480 ","End":"04:53.765","Text":"a parallel plate capacitor is equal to"},{"Start":"04:53.765 ","End":"04:58.067","Text":"the voltage divided by the distance between the 2 plates."},{"Start":"04:58.067 ","End":"05:03.420","Text":"It\u0027s also equal to sigma divided by epsilon naught."},{"Start":"05:03.650 ","End":"05:07.515","Text":"What we can see is both answers will be the same."},{"Start":"05:07.515 ","End":"05:14.435","Text":"If we say that E is equal to the voltage divided by the distance between the 2 plates."},{"Start":"05:14.435 ","End":"05:16.040","Text":"The voltage is of course,"},{"Start":"05:16.040 ","End":"05:19.640","Text":"the EMF so then we can plug in the values."},{"Start":"05:19.640 ","End":"05:26.570","Text":"Our EMF is equal to BLAt from Question Number 1 divided by d and similarly,"},{"Start":"05:26.570 ","End":"05:31.715","Text":"if we said that E was equal to sigma divided by epsilon naught so sigma"},{"Start":"05:31.715 ","End":"05:37.205","Text":"is equal to the charge divided by the surface area."},{"Start":"05:37.205 ","End":"05:39.830","Text":"Let\u0027s write S for surface area."},{"Start":"05:39.830 ","End":"05:43.355","Text":"Then epsilon naught so the charge"},{"Start":"05:43.355 ","End":"05:47.720","Text":"is also from Question Number 1 so we have epsilon naught Pi"},{"Start":"05:47.720 ","End":"05:53.450","Text":"a squared BLAt divided"},{"Start":"05:53.450 ","End":"05:58.015","Text":"by d and then multiplied by the surface area."},{"Start":"05:58.015 ","End":"06:05.405","Text":"So the surface area of a circular plate is Pi a squared and of course,"},{"Start":"06:05.405 ","End":"06:07.543","Text":"here we also have an epsilon naught."},{"Start":"06:07.543 ","End":"06:09.440","Text":"The epsilon naught cancels out,"},{"Start":"06:09.440 ","End":"06:12.710","Text":"Pi a squared cancels out and then again,"},{"Start":"06:12.710 ","End":"06:17.010","Text":"we\u0027re left with the same onset BLAt divided by"},{"Start":"06:17.010 ","End":"06:22.130","Text":"d. You can see that you get the same answer in both methods and of course,"},{"Start":"06:22.130 ","End":"06:28.910","Text":"the electric field is going from the positive plate to the negative plate."},{"Start":"06:28.910 ","End":"06:32.660","Text":"If we say that this is the z direction,"},{"Start":"06:32.660 ","End":"06:39.060","Text":"we can give this in the z direction."},{"Start":"06:41.000 ","End":"06:44.600","Text":"This is the answer to Question Number 2."},{"Start":"06:44.600 ","End":"06:48.170","Text":"I\u0027m just going to erase this over here because it really gives"},{"Start":"06:48.170 ","End":"06:52.830","Text":"us the same answer to give us more space for Question Number 3."},{"Start":"06:53.030 ","End":"06:57.020","Text":"Question Number 3 is to calculate the magnetic field both"},{"Start":"06:57.020 ","End":"07:00.035","Text":"inside and outside of the capacitor."},{"Start":"07:00.035 ","End":"07:03.280","Text":"Ignore the fields produced by the rod."},{"Start":"07:03.280 ","End":"07:04.490","Text":"First of all,"},{"Start":"07:04.490 ","End":"07:14.014","Text":"we are dealing with an electric field over here that we can see is dependent on time."},{"Start":"07:14.014 ","End":"07:15.700","Text":"This is of course,"},{"Start":"07:15.700 ","End":"07:19.010","Text":"lowercase t. We can see that my electric field is dependent on"},{"Start":"07:19.010 ","End":"07:23.570","Text":"time and the following question is asking us for a magnetic field."},{"Start":"07:23.570 ","End":"07:26.360","Text":"In this Chapter and one of the previous lessons,"},{"Start":"07:26.360 ","End":"07:31.440","Text":"we learned that anytime we have an electric field as a function of time."},{"Start":"07:31.440 ","End":"07:34.775","Text":"That means we are going to have a magnetic field."},{"Start":"07:34.775 ","End":"07:38.735","Text":"We remember that in order to calculate the magnetic field,"},{"Start":"07:38.735 ","End":"07:41.660","Text":"given this time dependent electric field."},{"Start":"07:41.660 ","End":"07:46.345","Text":"We\u0027re going to be using Ampere\u0027s law with Maxwell\u0027s correction."},{"Start":"07:46.345 ","End":"07:53.180","Text":"We have this closed loop integral on b.dl,"},{"Start":"07:53.180 ","End":"07:57.050","Text":"which is equal to mu naught multiplied"},{"Start":"07:57.050 ","End":"08:01.130","Text":"by I in like in Ampere\u0027s law plus Maxwell\u0027s correction,"},{"Start":"08:01.130 ","End":"08:10.830","Text":"which is mu naught and the integral of epsilon naught dE Pi dt.ds."},{"Start":"08:14.570 ","End":"08:20.765","Text":"Before we begin, we see that we\u0027re calculating a magnetic field but of course,"},{"Start":"08:20.765 ","End":"08:23.471","Text":"we\u0027re also dealing with this external magnetic fields."},{"Start":"08:23.471 ","End":"08:25.010","Text":"Just so that we don\u0027t get confused,"},{"Start":"08:25.010 ","End":"08:28.435","Text":"let\u0027s call this external magnetic field B naught."},{"Start":"08:28.435 ","End":"08:29.900","Text":"Everywhere I\u0027ve used it,"},{"Start":"08:29.900 ","End":"08:37.230","Text":"I\u0027m going to plug in B naught so that we don\u0027t get confused."},{"Start":"08:37.240 ","End":"08:43.025","Text":"What we\u0027re doing is Ampere\u0027s law and we can see that inside"},{"Start":"08:43.025 ","End":"08:48.244","Text":"the capacitor our electric field is going from positive to negative,"},{"Start":"08:48.244 ","End":"08:49.490","Text":"or in other words,"},{"Start":"08:49.490 ","End":"08:52.370","Text":"in the z direction so that means that"},{"Start":"08:52.370 ","End":"08:57.960","Text":"our magnetic field is going to be in this theta direction."},{"Start":"08:57.980 ","End":"09:03.340","Text":"Somewhere along this theta axis via the positive or negative direction."},{"Start":"09:03.340 ","End":"09:08.970","Text":"Let\u0027s say that the radius of this green cycle is lowercase"},{"Start":"09:08.970 ","End":"09:13.940","Text":"r. What we can see is that in"},{"Start":"09:13.940 ","End":"09:19.925","Text":"space the electric field is constant inside the capacitor."},{"Start":"09:19.925 ","End":"09:27.180","Text":"That means that it doesn\u0027t change throughout this dl."},{"Start":"09:27.190 ","End":"09:32.690","Text":"We can just multiply the magnetic field by the length of"},{"Start":"09:32.690 ","End":"09:38.165","Text":"this trajectory or by the circumference of the circle."},{"Start":"09:38.165 ","End":"09:42.740","Text":"We get that the magnetic field that we\u0027re trying to calculate multiplied"},{"Start":"09:42.740 ","End":"09:44.160","Text":"by"},{"Start":"09:49.590 ","End":"09:53.780","Text":"2pi r. This"},{"Start":"09:53.780 ","End":"09:58.355","Text":"is the circumference of the circle is equal to,"},{"Start":"09:58.355 ","End":"09:59.790","Text":"so this comes from this."},{"Start":"09:59.790 ","End":"10:03.065","Text":"Then mu naught multiplied by I in."},{"Start":"10:03.065 ","End":"10:06.903","Text":"Because we\u0027re looking at the electric field inside the capacitor,"},{"Start":"10:06.903 ","End":"10:09.770","Text":"there is no current flowing through capacitor."},{"Start":"10:09.770 ","End":"10:13.190","Text":"Because of course, each plate is isolated from"},{"Start":"10:13.190 ","End":"10:17.144","Text":"the other so if the current passing through here is 0."},{"Start":"10:17.144 ","End":"10:22.520","Text":"So this term is equal to 0 and then we\u0027re adding over this so we have Mu naught."},{"Start":"10:22.520 ","End":"10:26.840","Text":"We can take out the epsilon naught because it\u0027s also a constant and then we\u0027re"},{"Start":"10:26.840 ","End":"10:31.715","Text":"integrating along the time derivative of the electric field."},{"Start":"10:31.715 ","End":"10:35.000","Text":"If we take the time derivative of the electric fields,"},{"Start":"10:35.000 ","End":"10:41.930","Text":"so we just cross out the t. So we have B naught LA divided by d,"},{"Start":"10:41.930 ","End":"10:47.270","Text":"and this is in the z direction dot ds."},{"Start":"10:47.270 ","End":"10:51.535","Text":"Again, we can see that this over here is constant."},{"Start":"10:51.535 ","End":"10:59.495","Text":"What we can do is we can simply multiply this term over here by the surface area."},{"Start":"10:59.495 ","End":"11:06.470","Text":"What we get is B multiplied by 2pi r is equal to mu naught,"},{"Start":"11:06.470 ","End":"11:11.720","Text":"epsilon naught and then we have B naught LA,"},{"Start":"11:11.720 ","End":"11:17.585","Text":"all of this divided by d. Then multiply it by the surface area,"},{"Start":"11:17.585 ","End":"11:21.230","Text":"which is the surface area of this green circle over here so"},{"Start":"11:21.230 ","End":"11:25.099","Text":"that\u0027s pi multiplied by the radius squared."},{"Start":"11:25.099 ","End":"11:31.690","Text":"Then we can cancel both pies out an r from here on and r from here."},{"Start":"11:31.810 ","End":"11:37.447","Text":"Then what we get is that our magnetic field."},{"Start":"11:37.447 ","End":"11:42.410","Text":"So the magnetic field caused by all of this that\u0027s going"},{"Start":"11:42.410 ","End":"11:47.620","Text":"on over here by this electric field;"},{"Start":"11:47.620 ","End":"11:49.120","Text":"which is dependent on time,"},{"Start":"11:49.120 ","End":"11:53.710","Text":"is equal to Mu naught Epsilon naught,"},{"Start":"11:53.710 ","End":"12:02.150","Text":"b_0 multiplied by LAr and divided by 2d."},{"Start":"12:04.020 ","End":"12:08.290","Text":"Just to remind you this is the induced magnetic field."},{"Start":"12:08.290 ","End":"12:14.440","Text":"This has got nothing to do with adding the external magnetic field on top of everything."},{"Start":"12:14.440 ","End":"12:17.470","Text":"This is the magnetic field, of course,"},{"Start":"12:17.470 ","End":"12:21.250","Text":"when we\u0027re located inside the capacitor."},{"Start":"12:21.250 ","End":"12:24.360","Text":"When r is smaller than a"},{"Start":"12:24.360 ","End":"12:30.280","Text":"but now let\u0027s look at what happens when we\u0027re outside of the capacitor."},{"Start":"12:30.280 ","End":"12:32.365","Text":"This is when r is bigger than a."},{"Start":"12:32.365 ","End":"12:36.685","Text":"The only thing that will change is that here"},{"Start":"12:36.685 ","End":"12:41.800","Text":"instead when we multiply by the ds, the surface area."},{"Start":"12:41.800 ","End":"12:47.305","Text":"So now the loop is going to look something like this,"},{"Start":"12:47.305 ","End":"12:50.020","Text":"coming out of the capacitor but of course,"},{"Start":"12:50.020 ","End":"12:53.905","Text":"the capacitor plate is only up until a radius a."},{"Start":"12:53.905 ","End":"12:56.140","Text":"That\u0027s a maximum radius."},{"Start":"12:56.140 ","End":"12:58.375","Text":"When we multiply by the surface area,"},{"Start":"12:58.375 ","End":"13:01.360","Text":"over here we\u0027re multiplying by a surface area of"},{"Start":"13:01.360 ","End":"13:07.090","Text":"Pi a^2 and all of the rest is exactly the same."},{"Start":"13:07.090 ","End":"13:10.750","Text":"What we get is that our B induced,"},{"Start":"13:10.750 ","End":"13:15.250","Text":"so our induced magnetic field outside of the capacitor is equal to"},{"Start":"13:15.250 ","End":"13:22.780","Text":"Mu naught Epsilon naught B_0 multiplied by LA."},{"Start":"13:22.780 ","End":"13:26.005","Text":"Then the Pi will still cancel out but here the a is squared."},{"Start":"13:26.005 ","End":"13:28.630","Text":"The lowercase a^2 will not cancel out."},{"Start":"13:28.630 ","End":"13:32.050","Text":"This r over here won\u0027t cancel out so of course,"},{"Start":"13:32.050 ","End":"13:35.180","Text":"we\u0027re dividing by 2dr."},{"Start":"13:37.020 ","End":"13:47.480","Text":"Then these are of course going to be vectors and they\u0027re in the Theta direction."},{"Start":"13:49.500 ","End":"13:53.110","Text":"Now let\u0027s answer Question Number 4."},{"Start":"13:53.110 ","End":"13:56.020","Text":"We\u0027re being told that the rod, sorry,"},{"Start":"13:56.020 ","End":"13:59.050","Text":"has mass M and to calculate the force"},{"Start":"13:59.050 ","End":"14:03.715","Text":"required in order to move the rod at the same velocity,"},{"Start":"14:03.715 ","End":"14:06.475","Text":"at that we\u0027re given in the question."},{"Start":"14:06.475 ","End":"14:08.080","Text":"In order to answer this question,"},{"Start":"14:08.080 ","End":"14:11.620","Text":"the only thing I need is my answer to Question Number 1."},{"Start":"14:11.620 ","End":"14:14.605","Text":"I\u0027m going to rub all of this out."},{"Start":"14:14.605 ","End":"14:18.410","Text":"Please write down the answers if you want."},{"Start":"14:19.350 ","End":"14:22.000","Text":"What we can see first of all,"},{"Start":"14:22.000 ","End":"14:28.630","Text":"we\u0027ve seen that we have this current I flowing through over here and of course,"},{"Start":"14:28.630 ","End":"14:31.780","Text":"also flowing through the rod."},{"Start":"14:31.780 ","End":"14:34.645","Text":"We have this I like so."},{"Start":"14:34.645 ","End":"14:37.030","Text":"Seeing as we have a current,"},{"Start":"14:37.030 ","End":"14:44.500","Text":"we know that therefore we must have a magnetic field and therefore a magnetic force."},{"Start":"14:44.500 ","End":"14:48.760","Text":"The magnetic force F_B is equal to"},{"Start":"14:48.760 ","End":"14:55.765","Text":"the current multiplied by the magnitude of the length over here,"},{"Start":"14:55.765 ","End":"15:01.420","Text":"and multiplied by the magnitude of the external magnetic field."},{"Start":"15:01.420 ","End":"15:05.770","Text":"B_0 multiplied by sine of the angle between the 2."},{"Start":"15:05.770 ","End":"15:11.470","Text":"We can see that B_0 is into the page and L is in this negative side direction,"},{"Start":"15:11.470 ","End":"15:13.645","Text":"so they\u0027re perpendicular to one another."},{"Start":"15:13.645 ","End":"15:16.330","Text":"Sine of 90 is equal to 1."},{"Start":"15:16.330 ","End":"15:18.100","Text":"This is what we get."},{"Start":"15:18.100 ","End":"15:21.355","Text":"If we want to know the direction,"},{"Start":"15:21.355 ","End":"15:25.150","Text":"we\u0027re going to use the right-hand rule."},{"Start":"15:25.150 ","End":"15:30.670","Text":"What we have over here is the direction of the current,"},{"Start":"15:30.670 ","End":"15:37.540","Text":"which is pointing in this z direction multiplied by the direction of the magnetic field,"},{"Start":"15:37.540 ","End":"15:39.010","Text":"which is pointing into the page."},{"Start":"15:39.010 ","End":"15:40.630","Text":"The thumb is pointing up,"},{"Start":"15:40.630 ","End":"15:43.585","Text":"your 4 fingers pointing into the page."},{"Start":"15:43.585 ","End":"15:49.720","Text":"Then you get that the force F_B is pointing leftwards,"},{"Start":"15:49.720 ","End":"15:51.445","Text":"so this is our F_B."},{"Start":"15:51.445 ","End":"15:53.770","Text":"This doesn\u0027t surprise us that"},{"Start":"15:53.770 ","End":"15:58.090","Text":"the magnetic force is trying to slow down this motion over here."},{"Start":"15:58.090 ","End":"16:03.625","Text":"It\u0027s acting in the opposite direction because if it wasn\u0027t acting in this direction,"},{"Start":"16:03.625 ","End":"16:08.290","Text":"then we would get some acceleration,"},{"Start":"16:08.290 ","End":"16:10.375","Text":"which we wouldn\u0027t expect."},{"Start":"16:10.375 ","End":"16:16.223","Text":"Let\u0027s imagine that this is the x-direction."},{"Start":"16:16.223 ","End":"16:18.785","Text":"We can say, well, you know what?"},{"Start":"16:18.785 ","End":"16:22.980","Text":"Let\u0027s say that this is the x-direction,"},{"Start":"16:22.980 ","End":"16:28.900","Text":"so in that case this is our F_B acting in the x-direction."},{"Start":"16:29.190 ","End":"16:33.685","Text":"Change the direction of the x to the rightwards direction again."},{"Start":"16:33.685 ","End":"16:38.320","Text":"That means that our F_B is in the negative x direction."},{"Start":"16:38.320 ","End":"16:43.195","Text":"Now what I\u0027m trying to find is my F exterior."},{"Start":"16:43.195 ","End":"16:47.290","Text":"This is the force required in order to move the rod at"},{"Start":"16:47.290 ","End":"16:51.250","Text":"the same velocity given that we\u0027re told that the rod is of"},{"Start":"16:51.250 ","End":"16:57.340","Text":"mass m. The direction"},{"Start":"16:57.340 ","End":"17:02.485","Text":"of F exterior is going to be in this direction."},{"Start":"17:02.485 ","End":"17:07.495","Text":"First of all, we can see that the sum of all of the forces is not equal to zero,"},{"Start":"17:07.495 ","End":"17:10.720","Text":"because we can see that our velocity is in constant."},{"Start":"17:10.720 ","End":"17:11.920","Text":"It\u0027s dependent on time,"},{"Start":"17:11.920 ","End":"17:13.840","Text":"which means it\u0027s always changing."},{"Start":"17:13.840 ","End":"17:22.165","Text":"We can say that the sum of all of the forces is equal to our F exterior,"},{"Start":"17:22.165 ","End":"17:25.225","Text":"which is in the positive x-direction,"},{"Start":"17:25.225 ","End":"17:27.250","Text":"plus our F_B,"},{"Start":"17:27.250 ","End":"17:30.100","Text":"which is in the negative x-direction."},{"Start":"17:30.100 ","End":"17:38.870","Text":"A magnetic force is equal to mass multiplied by the acceleration."},{"Start":"17:39.840 ","End":"17:45.205","Text":"Our mass is, we\u0027re told M,"},{"Start":"17:45.205 ","End":"17:49.390","Text":"and our acceleration is the time derivative of our velocity,"},{"Start":"17:49.390 ","End":"17:52.225","Text":"so acceleration is just as A."},{"Start":"17:52.225 ","End":"17:56.620","Text":"If we take the derivative of At with respect to time, we just get A."},{"Start":"17:56.620 ","End":"18:01.075","Text":"Then we have our F exterior which we\u0027re trying to find."},{"Start":"18:01.075 ","End":"18:02.905","Text":"Then we have plus our F_B,"},{"Start":"18:02.905 ","End":"18:11.065","Text":"which is minus because they\u0027re in opposite direction, so minus ILB_0."},{"Start":"18:11.065 ","End":"18:21.625","Text":"Now we can see that our F exterior is equal to MA plus ILB_0."},{"Start":"18:21.625 ","End":"18:24.730","Text":"Now we have a problem. What exactly is our current?"},{"Start":"18:24.730 ","End":"18:26.200","Text":"We figured that there is current,"},{"Start":"18:26.200 ","End":"18:27.535","Text":"but we don\u0027t know what it is."},{"Start":"18:27.535 ","End":"18:32.515","Text":"Now a lot of the times when we have the EMF or the electromotive force,"},{"Start":"18:32.515 ","End":"18:38.185","Text":"our current is equal to the EMF divided by the resistance."},{"Start":"18:38.185 ","End":"18:43.075","Text":"However, we\u0027re told that there is no resistance."},{"Start":"18:43.075 ","End":"18:45.115","Text":"We have non-resistant."},{"Start":"18:45.115 ","End":"18:49.270","Text":"That means that we cannot use this equation."},{"Start":"18:49.270 ","End":"18:53.110","Text":"However, we remember and this is an equation that we"},{"Start":"18:53.110 ","End":"18:56.515","Text":"use most of the time when we have a capacitor."},{"Start":"18:56.515 ","End":"19:03.865","Text":"The question is that our current is equal to the time derivative of our charge."},{"Start":"19:03.865 ","End":"19:07.780","Text":"Here we have our charge q."},{"Start":"19:07.780 ","End":"19:09.400","Text":"If we take the time derivative,"},{"Start":"19:09.400 ","End":"19:15.250","Text":"we just lose this t. We get that our current is equal to Epsilon naught"},{"Start":"19:15.250 ","End":"19:20.875","Text":"Pi A^2 B_0 LA"},{"Start":"19:20.875 ","End":"19:25.480","Text":"and all of this divided by d. In that case,"},{"Start":"19:25.480 ","End":"19:27.820","Text":"we can plug this into this equation."},{"Start":"19:27.820 ","End":"19:31.660","Text":"What we can see is that we have this capital A in both,"},{"Start":"19:31.660 ","End":"19:37.210","Text":"so we have a and then in brackets M plus I,"},{"Start":"19:37.210 ","End":"19:44.065","Text":"so that\u0027s Epsilon naught Pi a^2 B_0 L,"},{"Start":"19:44.065 ","End":"19:50.020","Text":"and we\u0027ve already taken the A out of the brackets divided by d. Of course,"},{"Start":"19:50.020 ","End":"19:57.895","Text":"we have to multiply this also by L and by B_0 again from here,"},{"Start":"19:57.895 ","End":"20:00.500","Text":"so these becomes squares."},{"Start":"20:01.350 ","End":"20:05.215","Text":"This is the answer to Question Number 4."},{"Start":"20:05.215 ","End":"20:10.840","Text":"The force required in order to move the rod at the same velocity."},{"Start":"20:10.840 ","End":"20:15.230","Text":"That\u0027s it. That is the end of this lesson."}],"ID":21441}],"Thumbnail":null,"ID":99494}]

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