[{"Name":"Calculating Current Density from Given Magnetic Field","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Deriving Equation for K","Duration":"5m 19s","ChapterTopicVideoID":21332,"CourseChapterTopicPlaylistID":246910,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/21332.jpeg","UploadDate":"2020-04-06T21:26:21.9370000","DurationForVideoObject":"PT5M19S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.815","Text":"Hello, in this lesson,"},{"Start":"00:01.815 ","End":"00:06.390","Text":"we\u0027re going to be deriving this equation for"},{"Start":"00:06.390 ","End":"00:15.100","Text":"the jump in the electric field that gives us what our current density is."},{"Start":"00:15.290 ","End":"00:17.310","Text":"What we\u0027re going to take is"},{"Start":"00:17.310 ","End":"00:23.890","Text":"a very thin infinite plane and we\u0027re going to look at it from the side."},{"Start":"00:23.930 ","End":"00:30.075","Text":"This is our infinite plane from the side and then as we know,"},{"Start":"00:30.075 ","End":"00:34.335","Text":"our k is current density per unit length."},{"Start":"00:34.335 ","End":"00:39.900","Text":"We have this current flowing through the plane like"},{"Start":"00:39.900 ","End":"00:45.639","Text":"so and then what we can do is we can take some length,"},{"Start":"00:45.639 ","End":"00:52.025","Text":"this over here, dl and then we see how much current is flowing through and that"},{"Start":"00:52.025 ","End":"00:59.090","Text":"is our k. We\u0027ve already seen that in this case,"},{"Start":"00:59.090 ","End":"01:05.810","Text":"we\u0027re going to have on top above this infinite plane,"},{"Start":"01:05.810 ","End":"01:09.875","Text":"a magnetic field in this direction and below,"},{"Start":"01:09.875 ","End":"01:13.670","Text":"the magnetic field is going to be in the opposite direction."},{"Start":"01:13.670 ","End":"01:16.940","Text":"However, what happens if I have"},{"Start":"01:16.940 ","End":"01:20.870","Text":"other objects that also have"},{"Start":"01:20.870 ","End":"01:25.634","Text":"current flowing through them so here I have a cylinder with current throwing through,"},{"Start":"01:25.634 ","End":"01:27.980","Text":"here I have some wire."},{"Start":"01:27.980 ","End":"01:34.510","Text":"Now how do I calculate my magnetic field and the jump in the magnetic field?"},{"Start":"01:34.510 ","End":"01:37.190","Text":"What I\u0027m going to say is,"},{"Start":"01:37.190 ","End":"01:44.195","Text":"I\u0027m going to imagine that I have above the plane a magnetic field B_1,"},{"Start":"01:44.195 ","End":"01:53.070","Text":"and below the plane in the same direction B_2 and this is just in a general way."},{"Start":"01:53.070 ","End":"01:55.220","Text":"It could be that one of these or"},{"Start":"01:55.220 ","End":"01:58.475","Text":"both of them are in the opposite direction or a negative,"},{"Start":"01:58.475 ","End":"02:02.990","Text":"but this is just the general way and now what I\u0027m going to do"},{"Start":"02:02.990 ","End":"02:08.565","Text":"is I\u0027m going to draw Ampere\u0027s loop around here."},{"Start":"02:08.565 ","End":"02:15.360","Text":"I draw this rectangular loop, like so."},{"Start":"02:15.360 ","End":"02:18.875","Text":"Now we\u0027re going to use Ampere\u0027s law."},{"Start":"02:18.875 ","End":"02:28.185","Text":"We know we have to multiply B by the distance that we\u0027ve traveled or Bdl."},{"Start":"02:28.185 ","End":"02:37.040","Text":"Let\u0027s say that the length of the loop over here is l. Then according to ampere,"},{"Start":"02:37.040 ","End":"02:46.190","Text":"so here we\u0027re going to have B_1(L) and then because here we\u0027re going to go down the loop,"},{"Start":"02:46.190 ","End":"02:52.530","Text":"so in the opposite direction to B_2 so we have -B_2L."},{"Start":"02:53.210 ","End":"02:57.905","Text":"This of course, is equal to Mu naught,"},{"Start":"02:57.905 ","End":"03:02.845","Text":"multiplied by the current enclosed, so in."},{"Start":"03:02.845 ","End":"03:06.450","Text":"What exactly is I_in?"},{"Start":"03:06.450 ","End":"03:10.685","Text":"I_in is equal to our current density per"},{"Start":"03:10.685 ","End":"03:16.430","Text":"unit length so that\u0027s k multiplied by the length dl,"},{"Start":"03:16.430 ","End":"03:20.638","Text":"where we already said our length is this length l over here."},{"Start":"03:20.638 ","End":"03:27.870","Text":"Now we can plug that in so this is equal to Mu naught multiplied by kl."},{"Start":"03:27.870 ","End":"03:33.410","Text":"Now we can divide both sides by l. Then what we\u0027re left"},{"Start":"03:33.410 ","End":"03:39.165","Text":"with is B_1-B_2,"},{"Start":"03:39.165 ","End":"03:45.505","Text":"which is equal to Mu naught k and what is B_1-B_2?"},{"Start":"03:45.505 ","End":"03:48.575","Text":"It\u0027s the jump in the magnetic field."},{"Start":"03:48.575 ","End":"03:50.930","Text":"This is exactly our Delta B."},{"Start":"03:50.930 ","End":"03:53.420","Text":"That\u0027s how we get the equation."},{"Start":"03:53.420 ","End":"03:58.100","Text":"Now what\u0027s important to note is that we\u0027re using the components of our B,"},{"Start":"03:58.100 ","End":"04:01.125","Text":"which is parallel to the plane."},{"Start":"04:01.125 ","End":"04:05.150","Text":"When we we were dealing with electrostatics, so in electricity,"},{"Start":"04:05.150 ","End":"04:09.500","Text":"we would look at the electric field which was perpendicular to"},{"Start":"04:09.500 ","End":"04:14.795","Text":"the plane and that\u0027s the electric field where we would calculate the jump."},{"Start":"04:14.795 ","End":"04:18.080","Text":"The jump in the electric field is perpendicular to the plane."},{"Start":"04:18.080 ","End":"04:19.970","Text":"However, here we\u0027re dealing with"},{"Start":"04:19.970 ","End":"04:24.020","Text":"the magnetic field and in the magnetic field we\u0027re calculating"},{"Start":"04:24.020 ","End":"04:31.000","Text":"the jump when it\u0027s parallel to the plane."},{"Start":"04:32.420 ","End":"04:39.230","Text":"What happens if we have a cylinder where also here we have"},{"Start":"04:39.230 ","End":"04:45.540","Text":"current flowing through only on the surface of the cylinder?"},{"Start":"04:45.540 ","End":"04:51.890","Text":"What happens if we look at 2 points that are very close to the surface of the cylinder,"},{"Start":"04:51.890 ","End":"04:53.675","Text":"and we zoom in on that,"},{"Start":"04:53.675 ","End":"04:58.020","Text":"then the surface of the cylinder just looks like an infinite plane."},{"Start":"04:59.060 ","End":"05:03.005","Text":"It will look like an infinite plane and then this equation"},{"Start":"05:03.005 ","End":"05:07.250","Text":"still works for this type of example."},{"Start":"05:07.250 ","End":"05:11.885","Text":"What\u0027s important to know is this equation over here"},{"Start":"05:11.885 ","End":"05:17.030","Text":"and also how we got it via using Ampere\u0027s Law."},{"Start":"05:17.030 ","End":"05:19.800","Text":"That\u0027s the end of this lesson."}],"ID":21412},{"Watched":false,"Name":"Exercise 1","Duration":"9m 22s","ChapterTopicVideoID":24815,"CourseChapterTopicPlaylistID":246910,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/24815.jpeg","UploadDate":"2021-02-14T18:23:21.7500000","DurationForVideoObject":"PT9M22S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.170","Text":"Hello, in this lesson we\u0027re going to be answering the following question."},{"Start":"00:04.170 ","End":"00:07.545","Text":"Calculate the current density throughout."},{"Start":"00:07.545 ","End":"00:11.835","Text":"Here we have an equation for the magnetic field,"},{"Start":"00:11.835 ","End":"00:13.680","Text":"and it\u0027s given in two regions,"},{"Start":"00:13.680 ","End":"00:17.505","Text":"where r is smaller than a and where r is greater than a."},{"Start":"00:17.505 ","End":"00:22.620","Text":"We can see that we have the radial coordinate and the z coordinate."},{"Start":"00:22.620 ","End":"00:24.515","Text":"We\u0027re going to be using,"},{"Start":"00:24.515 ","End":"00:27.835","Text":"of course, the cylindrical coordinate system."},{"Start":"00:27.835 ","End":"00:32.735","Text":"What we want to do is we want to find the current density given this magnetic field."},{"Start":"00:32.735 ","End":"00:38.015","Text":"First of all, we\u0027re going to be starting with finding our J,"},{"Start":"00:38.015 ","End":"00:42.305","Text":"where we know that our J is 1 divided by Mu naught"},{"Start":"00:42.305 ","End":"00:49.180","Text":"multiplied by the rotor of the magnetic field."},{"Start":"00:49.670 ","End":"00:56.595","Text":"Here we have the rotor of some vector field A and of course we\u0027re using B."},{"Start":"00:56.595 ","End":"01:06.490","Text":"We can see that B has variables r and it\u0027s in the z direction."},{"Start":"01:08.480 ","End":"01:16.555","Text":"Of course, Rho for us means our radius and Phi means just Theta."},{"Start":"01:16.555 ","End":"01:22.740","Text":"Here we have 1 divided by r multiplied by dA_z,"},{"Start":"01:23.110 ","End":"01:26.555","Text":"so the z component by d Phi."},{"Start":"01:26.555 ","End":"01:29.850","Text":"Of course we don\u0027t have a z component."}],"ID":25728}],"Thumbnail":null,"ID":246910}]

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