Amperes Law
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[{"Name":"Amperes Law","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Intro to Ameperes Law","Duration":"10m 3s","ChapterTopicVideoID":21512,"CourseChapterTopicPlaylistID":99468,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/21512.jpeg","UploadDate":"2020-04-21T15:17:35.7200000","DurationForVideoObject":"PT10M3S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:05.115","Text":"Hello, in this lesson we\u0027re going to be learning about Ampere\u0027s law."},{"Start":"00:05.115 ","End":"00:07.950","Text":"This over here is Ampere\u0027s law,"},{"Start":"00:07.950 ","End":"00:13.375","Text":"and what we can see is that if we have some current flowing through something,"},{"Start":"00:13.375 ","End":"00:18.925","Text":"we are going to obtain or there will be present a magnetic field."},{"Start":"00:18.925 ","End":"00:23.330","Text":"This equation should always jump into your mind whenever there\u0027s"},{"Start":"00:23.330 ","End":"00:27.365","Text":"current because then you know that there\u0027s going to be a magnetic field,"},{"Start":"00:27.365 ","End":"00:31.500","Text":"and here we can see that this integral is a closed loop integral."},{"Start":"00:31.500 ","End":"00:32.600","Text":"So soon in this lesson,"},{"Start":"00:32.600 ","End":"00:33.985","Text":"we\u0027re going to speak about that."},{"Start":"00:33.985 ","End":"00:41.573","Text":"We can see that the integral on B.dl is equal to Mu naught multiplied by I_in,"},{"Start":"00:41.573 ","End":"00:45.058","Text":"and often times we won\u0027t be given what I_in,"},{"Start":"00:45.058 ","End":"00:47.552","Text":"the current inside is equal to."},{"Start":"00:47.552 ","End":"00:51.350","Text":"So in order to calculate that we do the integral on J ds,"},{"Start":"00:51.350 ","End":"00:55.950","Text":"where of course J is our current density."},{"Start":"00:59.750 ","End":"01:06.295","Text":"Any closed loop integral is sometimes denoted C for circulation,"},{"Start":"01:06.295 ","End":"01:10.735","Text":"so C_B for circulation of the magnetic field,"},{"Start":"01:10.735 ","End":"01:14.950","Text":"and this is just a term you don\u0027t really need to remember that."},{"Start":"01:14.950 ","End":"01:16.390","Text":"If you see it mentioned somewhere,"},{"Start":"01:16.390 ","End":"01:19.180","Text":"just know that it means there specifically C_B"},{"Start":"01:19.180 ","End":"01:22.690","Text":"refers to this closed loop integral and the magnetic field."},{"Start":"01:22.690 ","End":"01:25.165","Text":"But of course, we can have other fields."},{"Start":"01:25.165 ","End":"01:27.662","Text":"Anyway, back to this equation."},{"Start":"01:27.662 ","End":"01:32.470","Text":"We can see it looks a lot like Gauss\u0027s law for electric fields."},{"Start":"01:32.470 ","End":"01:34.180","Text":"Just like in Gauss\u0027s law,"},{"Start":"01:34.180 ","End":"01:38.795","Text":"we\u0027re first going to calculate the left side of the equation."},{"Start":"01:38.795 ","End":"01:42.005","Text":"Aside from this integral,"},{"Start":"01:42.005 ","End":"01:45.160","Text":"as we said, this whole equation looking like Gauss\u0027s law,"},{"Start":"01:45.160 ","End":"01:50.150","Text":"we can also notice something about this integral on B.dl,"},{"Start":"01:50.150 ","End":"01:52.970","Text":"we can see that this is an integral along"},{"Start":"01:52.970 ","End":"01:56.240","Text":"some path and because we have this circle over here,"},{"Start":"01:56.240 ","End":"01:59.470","Text":"we\u0027re being told that it\u0027s a closed path."},{"Start":"01:59.470 ","End":"02:02.180","Text":"What does that mean? This integral,"},{"Start":"02:02.180 ","End":"02:06.230","Text":"we\u0027ve seen this type of integral in many different places,"},{"Start":"02:06.230 ","End":"02:09.455","Text":"for instance, when we want to calculate voltage,"},{"Start":"02:09.455 ","End":"02:16.500","Text":"we already saw that this is the negative path interval on E.dr,"},{"Start":"02:16.500 ","End":"02:21.525","Text":"and it\u0027s important to note that dl and dr basically mean the exact same thing."},{"Start":"02:21.525 ","End":"02:24.310","Text":"There are along some line."},{"Start":"02:24.310 ","End":"02:26.705","Text":"We saw this when calculating voltage,"},{"Start":"02:26.705 ","End":"02:30.545","Text":"and we saw this also when calculating work,"},{"Start":"02:30.545 ","End":"02:36.370","Text":"which is equal to the negative line integral of F.dr."},{"Start":"02:36.370 ","End":"02:40.155","Text":"This is the exact same thing, but with B."},{"Start":"02:40.155 ","End":"02:45.185","Text":"What we can see is we can have some route,"},{"Start":"02:45.185 ","End":"02:47.533","Text":"like so, some path."},{"Start":"02:47.533 ","End":"02:53.820","Text":"Then let\u0027s imagine that our magnetic field is in this direction over here."},{"Start":"02:54.050 ","End":"03:01.505","Text":"What this integral, B.dl or something.dr means,"},{"Start":"03:01.505 ","End":"03:03.635","Text":"we\u0027re not looking at this closed route yet,"},{"Start":"03:03.635 ","End":"03:08.000","Text":"means that we\u0027re taking the root parallel to"},{"Start":"03:08.000 ","End":"03:15.970","Text":"this path and we\u0027re summing up the magnetic field along this path like so."},{"Start":"03:17.180 ","End":"03:23.420","Text":"Now finally, we see that we have this circle over here and the integral sign,"},{"Start":"03:23.420 ","End":"03:25.655","Text":"which means that this is a closed path."},{"Start":"03:25.655 ","End":"03:29.944","Text":"It\u0027s circular, so that means that the ends connect."},{"Start":"03:29.944 ","End":"03:32.105","Text":"It doesn\u0027t matter what the shape looks like,"},{"Start":"03:32.105 ","End":"03:37.439","Text":"it just matters that the ends connect."},{"Start":"03:37.439 ","End":"03:41.399","Text":"Therefore, we just go like this and we carry"},{"Start":"03:41.399 ","End":"03:46.470","Text":"on in a circle summing up the magnetic field at each point."},{"Start":"03:47.360 ","End":"03:50.940","Text":"This is the left side of the equation,"},{"Start":"03:50.940 ","End":"03:56.090","Text":"this is where we sum up the magnetic field along this closed loop."},{"Start":"03:56.090 ","End":"03:59.705","Text":"Now let\u0027s look at the right side of the equation,"},{"Start":"03:59.705 ","End":"04:06.144","Text":"what we can see is that this closed loop encompasses some surface."},{"Start":"04:06.144 ","End":"04:15.029","Text":"Inside over here we have this surface that is defined by the loop."},{"Start":"04:15.029 ","End":"04:18.590","Text":"Or rather it\u0027s not a surface,"},{"Start":"04:18.590 ","End":"04:23.910","Text":"but it\u0027s some surface area that is enclosed within this loop."},{"Start":"04:23.910 ","End":"04:28.370","Text":"What we\u0027re doing on the right side of the equation is we\u0027re"},{"Start":"04:28.370 ","End":"04:33.485","Text":"seeing the current that flows through this loop,"},{"Start":"04:33.485 ","End":"04:38.940","Text":"I_in, the current inside this loop."},{"Start":"04:39.260 ","End":"04:43.640","Text":"What we\u0027re doing is we want to see how many charges"},{"Start":"04:43.640 ","End":"04:47.480","Text":"per second are passing through this area."},{"Start":"04:47.480 ","End":"04:53.240","Text":"Let\u0027s say here we have a wire and we have current,"},{"Start":"04:53.240 ","End":"04:56.900","Text":"and the current is going inside the page and there\u0027s a current I,"},{"Start":"04:56.900 ","End":"04:58.952","Text":"so I going into the page."},{"Start":"04:58.952 ","End":"05:03.545","Text":"Here we also have another wire with current I going inside the page,"},{"Start":"05:03.545 ","End":"05:08.825","Text":"and here we have another wire with current I coming out of the page,"},{"Start":"05:08.825 ","End":"05:12.570","Text":"and here another wire with I coming out of the page."},{"Start":"05:12.570 ","End":"05:14.921","Text":"What is the total I_in?"},{"Start":"05:14.921 ","End":"05:17.171","Text":"Let\u0027s write it here."},{"Start":"05:17.171 ","End":"05:20.300","Text":"So I_in, we\u0027re looking,"},{"Start":"05:20.300 ","End":"05:23.285","Text":"remember just in the area inside the loop."},{"Start":"05:23.285 ","End":"05:27.883","Text":"This wire over here doesn\u0027t interest me because it\u0027s outside of the loop."},{"Start":"05:27.883 ","End":"05:31.410","Text":"Only these three wires interest me."},{"Start":"05:31.410 ","End":"05:36.850","Text":"We have I into the page plus I into the page, that\u0027s 2I,"},{"Start":"05:36.850 ","End":"05:39.560","Text":"and then this current is flowing in"},{"Start":"05:39.560 ","End":"05:43.280","Text":"the opposite direction because it\u0027s coming out of the page, so minus I."},{"Start":"05:43.280 ","End":"05:45.095","Text":"We have 2I minus I,"},{"Start":"05:45.095 ","End":"05:47.486","Text":"which leaves us just with I."},{"Start":"05:47.486 ","End":"05:52.380","Text":"Our total I_in inside this loop is I."},{"Start":"05:53.300 ","End":"05:58.940","Text":"Now, what happens if we don\u0027t have wires with current flowing through,"},{"Start":"05:58.940 ","End":"06:03.245","Text":"but we have some current density J that we\u0027re given,"},{"Start":"06:03.245 ","End":"06:08.795","Text":"that means that we have current going"},{"Start":"06:08.795 ","End":"06:14.570","Text":"at every single point in the surface area going into the page,"},{"Start":"06:14.570 ","End":"06:18.185","Text":"or charges rather going into the page at every single point."},{"Start":"06:18.185 ","End":"06:21.890","Text":"What we\u0027ll do is we\u0027ll find the current density"},{"Start":"06:21.890 ","End":"06:25.905","Text":"for each little square or each small surface,"},{"Start":"06:25.905 ","End":"06:29.778","Text":"the surface area of the square is ds,"},{"Start":"06:29.778 ","End":"06:32.160","Text":"and then we find the current density."},{"Start":"06:32.160 ","End":"06:36.125","Text":"Let\u0027s say we know it, we\u0027re given it in the question and then we just integrate,"},{"Start":"06:36.125 ","End":"06:44.970","Text":"or we sum up the current densities for the total surface area of this area over here."},{"Start":"06:45.320 ","End":"06:48.270","Text":"Now we can see this equation,"},{"Start":"06:48.270 ","End":"06:51.320","Text":"we\u0027re back to this side of the equals sign,"},{"Start":"06:51.320 ","End":"06:52.693","Text":"the left side,"},{"Start":"06:52.693 ","End":"06:55.925","Text":"and we can see that we have to integrate along some path."},{"Start":"06:55.925 ","End":"07:00.873","Text":"It\u0027s very difficult to integrate if we don\u0027t have the function."},{"Start":"07:00.873 ","End":"07:02.030","Text":"A lot of time,"},{"Start":"07:02.030 ","End":"07:05.780","Text":"they will give us a case of symmetry,"},{"Start":"07:05.780 ","End":"07:07.895","Text":"so we have symmetry."},{"Start":"07:07.895 ","End":"07:11.870","Text":"In that case, the closed loop integral of"},{"Start":"07:11.870 ","End":"07:18.170","Text":"B.dl will be equal"},{"Start":"07:18.170 ","End":"07:24.600","Text":"to b multiplied by the length of the path,"},{"Start":"07:24.600 ","End":"07:26.744","Text":"multiplied by the path length."},{"Start":"07:26.744 ","End":"07:30.800","Text":"What is very, very important to remember over here,"},{"Start":"07:30.800 ","End":"07:32.885","Text":"in order to do this jump,"},{"Start":"07:32.885 ","End":"07:38.482","Text":"the magnetic field has to be constant along the path."},{"Start":"07:38.482 ","End":"07:40.830","Text":"Well, it has to be uniform along the path,"},{"Start":"07:40.830 ","End":"07:44.900","Text":"it can be different in other areas in space,"},{"Start":"07:44.900 ","End":"07:48.695","Text":"but it\u0027s super, super important that long the specific path,"},{"Start":"07:48.695 ","End":"07:51.600","Text":"the B field is uniform."},{"Start":"07:52.730 ","End":"07:57.740","Text":"Because of this, there\u0027s only a certain type of question"},{"Start":"07:57.740 ","End":"08:02.010","Text":"that we can be asked in order to calculate the magnetic field,"},{"Start":"08:02.010 ","End":"08:05.459","Text":"that\u0027s usually what will be asked in these types of questions."},{"Start":"08:05.459 ","End":"08:13.505","Text":"That is, when we\u0027re dealing with either a wire or a thick wire,"},{"Start":"08:13.505 ","End":"08:14.765","Text":"what is a thick wire?"},{"Start":"08:14.765 ","End":"08:17.812","Text":"It\u0027s a cylinder."},{"Start":"08:17.812 ","End":"08:19.205","Text":"When we\u0027re dealing with that,"},{"Start":"08:19.205 ","End":"08:25.600","Text":"or we\u0027re dealing with a cylindrical shell,"},{"Start":"08:25.600 ","End":"08:27.450","Text":"and of course they have to be infinite."},{"Start":"08:27.450 ","End":"08:29.465","Text":"If we\u0027re dealing with an infinite wire,"},{"Start":"08:29.465 ","End":"08:32.615","Text":"or a cylinder, or cylindrical shell,"},{"Start":"08:32.615 ","End":"08:34.250","Text":"and all of these are infinite,"},{"Start":"08:34.250 ","End":"08:39.525","Text":"then we can use this because we have some case of symmetry."},{"Start":"08:39.525 ","End":"08:48.058","Text":"Also, we can deal with these calculations when we have an infinite plane,"},{"Start":"08:48.058 ","End":"08:54.130","Text":"and the third case is when we have an infinite coil."},{"Start":"08:54.130 ","End":"09:00.810","Text":"In this chapter, we\u0027re going to be learning how to deal with each one of these cases."},{"Start":"09:00.890 ","End":"09:08.360","Text":"We can really do some flow chart for how to find the magnetic field."},{"Start":"09:08.360 ","End":"09:13.640","Text":"If we have a case of symmetry,"},{"Start":"09:13.640 ","End":"09:20.280","Text":"in other words, all of these options over here."},{"Start":"09:20.280 ","End":"09:22.310","Text":"If this is the case,"},{"Start":"09:22.310 ","End":"09:31.450","Text":"then what we\u0027re going to use is Ampere\u0027s law."},{"Start":"09:31.450 ","End":"09:33.860","Text":"If we don\u0027t have symmetry,"},{"Start":"09:33.860 ","End":"09:36.695","Text":"not one of these cases,"},{"Start":"09:36.695 ","End":"09:44.995","Text":"then what we\u0027re going to do is we\u0027re going to use Bio-Savart\u0027s law,"},{"Start":"09:44.995 ","End":"09:48.395","Text":"where we split up the shape into lots of different pieces,"},{"Start":"09:48.395 ","End":"09:52.100","Text":"and then we have to also integrate there."},{"Start":"09:52.100 ","End":"09:55.290","Text":"In this chapter, we\u0027re going to be dealing with Ampere\u0027s law,"},{"Start":"09:55.290 ","End":"09:57.425","Text":"so in the case when we have symmetry,"},{"Start":"09:57.425 ","End":"10:00.575","Text":"which means that we have one of these options."},{"Start":"10:00.575 ","End":"10:03.450","Text":"That\u0027s the end of this lesson."}],"ID":22272},{"Watched":false,"Name":"Exercise 1","Duration":"4m 57s","ChapterTopicVideoID":21513,"CourseChapterTopicPlaylistID":99468,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/21513.jpeg","UploadDate":"2020-04-21T15:18:45.8770000","DurationForVideoObject":"PT4M57S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.915","Text":"Hello. In this lesson we\u0027re going to be answering the following question."},{"Start":"00:03.915 ","End":"00:07.290","Text":"Here we have a wire with a current I flowing through"},{"Start":"00:07.290 ","End":"00:11.205","Text":"it and we\u0027re being asked to calculate the magnetic field."},{"Start":"00:11.205 ","End":"00:15.795","Text":"First of all, from the chapter dealing with BO_r."},{"Start":"00:15.795 ","End":"00:18.645","Text":"We saw that if we have"},{"Start":"00:18.645 ","End":"00:24.410","Text":"a current carrying wire that we\u0027re going to have a magnetic field in"},{"Start":"00:24.410 ","End":"00:28.640","Text":"some circular trajectory around the wire that is"},{"Start":"00:28.640 ","End":"00:33.620","Text":"traveling like so in this Theta direction around it,"},{"Start":"00:33.620 ","End":"00:36.805","Text":"according to the right-hand rule."},{"Start":"00:36.805 ","End":"00:40.130","Text":"The version that I\u0027m"},{"Start":"00:40.130 ","End":"00:43.609","Text":"speaking about right now is the version where you take your right-hand,"},{"Start":"00:43.609 ","End":"00:46.700","Text":"your thumb points in the direction of the current,"},{"Start":"00:46.700 ","End":"00:52.290","Text":"and then your fingers curl around in the direction of the magnetic field."},{"Start":"00:52.490 ","End":"00:57.379","Text":"We can see that the magnetic field is going like this and this Theta direction."},{"Start":"00:57.379 ","End":"01:01.969","Text":"If we are to look from this angle,"},{"Start":"01:01.969 ","End":"01:05.165","Text":"we\u0027re looking down the wire."},{"Start":"01:05.165 ","End":"01:07.690","Text":"In that case, we have a wire,"},{"Start":"01:07.690 ","End":"01:11.860","Text":"we\u0027re looking down so we can see the current is traveling away from us or"},{"Start":"01:11.860 ","End":"01:16.380","Text":"into the page and this is current I,"},{"Start":"01:16.380 ","End":"01:20.035","Text":"then we can see that we have"},{"Start":"01:20.035 ","End":"01:27.625","Text":"some circular path over here where our magnetic field is traveling like so."},{"Start":"01:27.625 ","End":"01:38.415","Text":"We can define that this is of some radius I and this is of course going down the z-axis,"},{"Start":"01:38.415 ","End":"01:40.785","Text":"as we can see from here."},{"Start":"01:40.785 ","End":"01:43.650","Text":"I\u0027ve just chosen a random radius,"},{"Start":"01:43.650 ","End":"01:47.560","Text":"I could have chosen a larger circle or not."},{"Start":"01:47.560 ","End":"01:50.435","Text":"I want to calculate the magnetic field,"},{"Start":"01:50.435 ","End":"01:53.810","Text":"so we\u0027ve already seen from Ampere\u0027s law that we have this closed loop"},{"Start":"01:53.810 ","End":"01:58.770","Text":"integral on B dot dl and that this is equal"},{"Start":"01:58.770 ","End":"02:02.525","Text":"to Mu_naught multiplied by I_in"},{"Start":"02:02.525 ","End":"02:08.400","Text":"reminding you this as an equation for your equation sheets."},{"Start":"02:08.440 ","End":"02:13.655","Text":"Let\u0027s see what our Bdl is equal to."},{"Start":"02:13.655 ","End":"02:17.270","Text":"When we\u0027re dealing with this infinite wire,"},{"Start":"02:17.270 ","End":"02:19.490","Text":"this is an infinitely long wire."},{"Start":"02:19.490 ","End":"02:24.665","Text":"We know that the magnetic field around it is dependent on the radius."},{"Start":"02:24.665 ","End":"02:29.255","Text":"If I look at this radius over here,"},{"Start":"02:29.255 ","End":"02:30.680","Text":"which is much larger,"},{"Start":"02:30.680 ","End":"02:33.395","Text":"the magnetic field will be different to"},{"Start":"02:33.395 ","End":"02:36.650","Text":"the magnetic field of the smaller radius, however,"},{"Start":"02:36.650 ","End":"02:41.404","Text":"along this path, the magnetic field is constant,"},{"Start":"02:41.404 ","End":"02:45.035","Text":"which as we said in the previous lesson, is very important."},{"Start":"02:45.035 ","End":"02:53.850","Text":"Because there is a constant B field along the path,"},{"Start":"02:55.550 ","End":"02:58.425","Text":"so I\u0027ll just rub out the second one."},{"Start":"02:58.425 ","End":"03:05.035","Text":"The magnetic field is uniform or constant along this path at this constant radius."},{"Start":"03:05.035 ","End":"03:09.365","Text":"Therefore, our B dot dl integral,"},{"Start":"03:09.365 ","End":"03:16.130","Text":"we can say that it\u0027s just B multiplied by the path length,"},{"Start":"03:16.130 ","End":"03:19.905","Text":"which because we\u0027re dealing with a perfect circle,"},{"Start":"03:19.905 ","End":"03:24.260","Text":"we just multiply by the circumference of the circle,"},{"Start":"03:24.260 ","End":"03:29.345","Text":"which is equal to 2 Pi multiplied by the radius."},{"Start":"03:29.345 ","End":"03:31.550","Text":"This is this side,"},{"Start":"03:31.550 ","End":"03:36.005","Text":"then if we want to look at our M_naught I_in,"},{"Start":"03:36.005 ","End":"03:40.865","Text":"so we can see that I_in the only current inside this loop."},{"Start":"03:40.865 ","End":"03:42.545","Text":"There could be currents outside,"},{"Start":"03:42.545 ","End":"03:43.970","Text":"but that doesn\u0027t interest me."},{"Start":"03:43.970 ","End":"03:48.305","Text":"The only current inside this loop is this current I."},{"Start":"03:48.305 ","End":"03:52.870","Text":"Therefore I_in is just I."},{"Start":"03:52.910 ","End":"03:58.800","Text":"Therefore, we get Mu_naught multiplied by I."},{"Start":"03:58.800 ","End":"04:03.740","Text":"What we have is B multiplied by 2Pir."},{"Start":"04:03.740 ","End":"04:05.965","Text":"This is our integral over here,"},{"Start":"04:05.965 ","End":"04:10.470","Text":"Bdl is equal to Mu_naught I."},{"Start":"04:10.470 ","End":"04:13.300","Text":"Therefore, we can isolate out our B,"},{"Start":"04:13.300 ","End":"04:21.770","Text":"and we get that our B is equal to Mu naught multiplied by I divided by 2Pir."},{"Start":"04:21.770 ","End":"04:26.750","Text":"We can see if this into the page is the z-axis,"},{"Start":"04:26.750 ","End":"04:28.760","Text":"so this is the Theta direction,"},{"Start":"04:28.760 ","End":"04:32.530","Text":"so we can even add the direction."},{"Start":"04:32.530 ","End":"04:35.000","Text":"This is the answer to the question,"},{"Start":"04:35.000 ","End":"04:39.890","Text":"but this is also an equation for the magnetic field of an infinite wire."},{"Start":"04:39.890 ","End":"04:44.020","Text":"Anytime you have an infinite current carrying wire,"},{"Start":"04:44.020 ","End":"04:47.115","Text":"and it has to have a current,"},{"Start":"04:47.115 ","End":"04:50.420","Text":"so anytime we have an infinite current carrying wire,"},{"Start":"04:50.420 ","End":"04:54.105","Text":"this will be the magnetic field around it."},{"Start":"04:54.105 ","End":"04:58.100","Text":"That is the end of this lesson."}],"ID":22273},{"Watched":false,"Name":"Exercise 2","Duration":"24m 56s","ChapterTopicVideoID":21514,"CourseChapterTopicPlaylistID":99468,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/21514.jpeg","UploadDate":"2020-04-21T15:28:17.9770000","DurationForVideoObject":"PT24M56S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.224","Text":"Hello, in this lesson we\u0027re going to be answering"},{"Start":"00:03.224 ","End":"00:06.810","Text":"a question dealing with a coaxial cable,"},{"Start":"00:06.810 ","End":"00:08.930","Text":"or a coax cable."},{"Start":"00:08.930 ","End":"00:10.790","Text":"What is a coax cable?"},{"Start":"00:10.790 ","End":"00:15.450","Text":"It\u0027s some cable that has different layers inside of it,"},{"Start":"00:15.450 ","End":"00:19.440","Text":"and all of the layers share the same axis."},{"Start":"00:19.440 ","End":"00:23.925","Text":"That\u0027s where the name comes from, coaxial cable."},{"Start":"00:23.925 ","End":"00:29.400","Text":"What we have is over here a conducting cylinder."},{"Start":"00:29.400 ","End":"00:33.130","Text":"This is a full cylinder of radius R,"},{"Start":"00:33.130 ","End":"00:34.564","Text":"and it conducts,"},{"Start":"00:34.564 ","End":"00:41.320","Text":"and it has this current I naught flowing through it in the direction into the page,"},{"Start":"00:41.320 ","End":"00:46.705","Text":"and surrounding it is a thin insulating layer."},{"Start":"00:46.705 ","End":"00:54.665","Text":"We\u0027re told that inside over here the current density is uniform."},{"Start":"00:54.665 ","End":"00:59.100","Text":"Then from this radius R until 2R,"},{"Start":"00:59.100 ","End":"01:04.215","Text":"we have a thick conducting tube of width 2R minus"},{"Start":"01:04.215 ","End":"01:10.385","Text":"R. It is surrounding this thin insulating layer,"},{"Start":"01:10.385 ","End":"01:15.090","Text":"and the inner conducting cylinder."},{"Start":"01:15.090 ","End":"01:19.405","Text":"Now, the same current I naught is flowing through it."},{"Start":"01:19.405 ","End":"01:23.560","Text":"However, this time the direction of the current is out of the page,"},{"Start":"01:23.560 ","End":"01:25.690","Text":"as we can see over here."},{"Start":"01:25.690 ","End":"01:32.320","Text":"We\u0027re told that also in this region the current density is also uniform."},{"Start":"01:32.320 ","End":"01:37.060","Text":"Now, it\u0027s important to note that even though the current density in the outer tube,"},{"Start":"01:37.060 ","End":"01:39.385","Text":"and in the inner tube is uniform,"},{"Start":"01:39.385 ","End":"01:42.040","Text":"it doesn\u0027t mean that they are both equal."},{"Start":"01:42.040 ","End":"01:48.410","Text":"They\u0027re just both uniform throughout but they could be different values."},{"Start":"01:48.830 ","End":"01:52.780","Text":"The first question that we\u0027re going to be answering,"},{"Start":"01:52.780 ","End":"01:58.460","Text":"question Number 1 is what is the current density J throughout?"},{"Start":"01:58.820 ","End":"02:04.015","Text":"As we know, there are a few equations for J."},{"Start":"02:04.015 ","End":"02:07.535","Text":"But 1 of the equations we know,"},{"Start":"02:07.535 ","End":"02:11.780","Text":"the most basic equation maybe is that I is equal"},{"Start":"02:11.780 ","End":"02:18.435","Text":"to the integral of J.ds."},{"Start":"02:18.435 ","End":"02:24.920","Text":"In the case where J is uniform or is a constant value,"},{"Start":"02:24.920 ","End":"02:29.050","Text":"we can say that this is just equal to J.S."},{"Start":"02:29.050 ","End":"02:31.865","Text":"We know that here it\u0027s a uniform because we\u0027re being"},{"Start":"02:31.865 ","End":"02:35.905","Text":"told that the current density is uniform."},{"Start":"02:35.905 ","End":"02:40.410","Text":"You can only do this if J is uniform."},{"Start":"02:40.410 ","End":"02:43.690","Text":"In that case, we can isolate out our J,"},{"Start":"02:43.690 ","End":"02:47.390","Text":"so first of all, let\u0027s work out this inner region first."},{"Start":"02:47.390 ","End":"02:50.356","Text":"Here r is smaller than R,"},{"Start":"02:50.356 ","End":"02:54.050","Text":"so we\u0027re in the region of this inner cylinder."},{"Start":"02:54.050 ","End":"02:59.555","Text":"J is equal to I divided by S,"},{"Start":"02:59.555 ","End":"03:03.460","Text":"which our current over here is I naught."},{"Start":"03:03.460 ","End":"03:10.265","Text":"In which direction? Let\u0027s say that the z direction is going into the page."},{"Start":"03:10.265 ","End":"03:13.460","Text":"Okay, so the z is into the page."},{"Start":"03:13.460 ","End":"03:19.415","Text":"We have I naught divided by the surface area or the cross-sectional surface area"},{"Start":"03:19.415 ","End":"03:25.600","Text":"of the cylinder which as we know is equal to Pi R^2."},{"Start":"03:25.600 ","End":"03:27.830","Text":"We said that this is going to be in"},{"Start":"03:27.830 ","End":"03:31.235","Text":"the direction of I of the current which is into the page,"},{"Start":"03:31.235 ","End":"03:34.920","Text":"which is the positive z direction."},{"Start":"03:35.330 ","End":"03:40.010","Text":"This is the current density in the inner cylinder."},{"Start":"03:40.010 ","End":"03:45.395","Text":"Now let\u0027s look at the current density when we\u0027re located in this outer tube."},{"Start":"03:45.395 ","End":"03:50.235","Text":"That means that R is bigger than r. However,"},{"Start":"03:50.235 ","End":"03:53.445","Text":"it\u0027s smaller than 2R."},{"Start":"03:53.445 ","End":"04:00.365","Text":"Over here again we\u0027re being told that the current density is uniform."},{"Start":"04:00.365 ","End":"04:04.775","Text":"Again, we can say that J is equal to I divided by"},{"Start":"04:04.775 ","End":"04:11.449","Text":"S. First of all our current is again I naught,"},{"Start":"04:11.449 ","End":"04:14.960","Text":"but our surface area is this over here."},{"Start":"04:14.960 ","End":"04:17.210","Text":"It\u0027s just this ring going around over here,"},{"Start":"04:17.210 ","End":"04:18.995","Text":"the cross-sectional surface area."},{"Start":"04:18.995 ","End":"04:24.520","Text":"What it\u0027s going to be is the surface area if this whole thing was full."},{"Start":"04:24.520 ","End":"04:28.470","Text":"That would be Pi multiplied by the radius^2,"},{"Start":"04:28.470 ","End":"04:32.395","Text":"so here the radius is 2R, so 2R^2."},{"Start":"04:32.395 ","End":"04:35.600","Text":"But then we have this hole inside,"},{"Start":"04:35.600 ","End":"04:40.055","Text":"so minus the cross-sectional surface area of this hole,"},{"Start":"04:40.055 ","End":"04:43.610","Text":"which as we know is equal to Pi multiplied by the radius ^2,"},{"Start":"04:43.610 ","End":"04:50.270","Text":"where the radius is R. What we get is I naught divided by,"},{"Start":"04:50.270 ","End":"04:55.485","Text":"so here we have 4Pi R^2 minus Pi R^2,"},{"Start":"04:55.485 ","End":"05:00.360","Text":"so we have divided by 3Pi R^2,"},{"Start":"05:00.360 ","End":"05:03.130","Text":"and this is also in the z-direction."},{"Start":"05:03.130 ","End":"05:05.180","Text":"But as it\u0027s coming out of the page,"},{"Start":"05:05.180 ","End":"05:08.700","Text":"it\u0027s in the negative z direction."},{"Start":"05:09.040 ","End":"05:13.460","Text":"Of course, if we want to find the current density outside of"},{"Start":"05:13.460 ","End":"05:17.680","Text":"our coax cable we\u0027re just going to do the same thing but here,"},{"Start":"05:17.680 ","End":"05:21.460","Text":"so we see that J is equal to I divided by S,"},{"Start":"05:21.460 ","End":"05:25.670","Text":"but the current outside of the coax cable is equal to 0."},{"Start":"05:25.670 ","End":"05:31.160","Text":"We have J is equal to 0 divided by some surface area which is equal to 0."},{"Start":"05:31.160 ","End":"05:36.230","Text":"We have no current density outside which is what we were expecting."},{"Start":"05:36.230 ","End":"05:39.230","Text":"Now let\u0027s move on to question Number 2."},{"Start":"05:39.230 ","End":"05:43.590","Text":"What is the magnetic field B throughout?"},{"Start":"05:43.640 ","End":"05:46.655","Text":"In order to find the magnetic field,"},{"Start":"05:46.655 ","End":"05:51.620","Text":"we can use either Biot Savart law or Ampere\u0027s law."},{"Start":"05:51.620 ","End":"05:56.210","Text":"Because we\u0027re dealing with this infinite cable,"},{"Start":"05:56.210 ","End":"05:59.053","Text":"so this infinitely long cylinder,"},{"Start":"05:59.053 ","End":"06:01.475","Text":"so we\u0027re going to use Ampere\u0027s law."},{"Start":"06:01.475 ","End":"06:04.830","Text":"Also because there\u0027s symmetry over here."},{"Start":"06:05.660 ","End":"06:08.075","Text":"This is Ampere\u0027s law,"},{"Start":"06:08.075 ","End":"06:15.850","Text":"we have the closed circuit integral of B.dl."},{"Start":"06:15.850 ","End":"06:20.965","Text":"The magnetic field multiplied by its route or its trajectory,"},{"Start":"06:20.965 ","End":"06:24.145","Text":"which is equal to Mu naught which is a constant,"},{"Start":"06:24.145 ","End":"06:31.905","Text":"multiplied by I_in which is the current inside this closed circuit."},{"Start":"06:31.905 ","End":"06:36.070","Text":"Now let\u0027s take a look at what we\u0027ve got over here so that we can understand"},{"Start":"06:36.070 ","End":"06:40.520","Text":"how to solve the left side of this equation."},{"Start":"06:40.790 ","End":"06:47.110","Text":"Let\u0027s look first of all in this inner region over here."},{"Start":"06:47.690 ","End":"06:53.380","Text":"We\u0027re looking at the region where r is smaller than R,"},{"Start":"06:53.380 ","End":"06:59.290","Text":"and we can see that we have the current going in the positive z direction."},{"Start":"06:59.290 ","End":"07:02.110","Text":"If we use our right-hand rule,"},{"Start":"07:02.110 ","End":"07:06.775","Text":"so we point our thumb in the direction of the current."},{"Start":"07:06.775 ","End":"07:15.115","Text":"What we get that the direction of the magnetic field is in the theta direction."},{"Start":"07:15.115 ","End":"07:22.360","Text":"In this, it\u0027s going in this direction which is also the positive theta direction."},{"Start":"07:22.360 ","End":"07:24.820","Text":"Let\u0027s just go like this,"},{"Start":"07:24.820 ","End":"07:28.420","Text":"and this is the theta direction, like so."},{"Start":"07:28.420 ","End":"07:30.890","Text":"The positive theta direction."},{"Start":"07:31.340 ","End":"07:39.055","Text":"Now what we\u0027re going to do is we\u0027re going to say that this is at a radius of lowercase r."},{"Start":"07:39.055 ","End":"07:48.130","Text":"Then in that case over here we have the closed loop integral on B.dl."},{"Start":"07:48.130 ","End":"07:54.728","Text":"Now, because our current over here is uniform,"},{"Start":"07:54.728 ","End":"07:57.580","Text":"so we know that the magnetic field over here is"},{"Start":"07:57.580 ","End":"08:01.060","Text":"also going to be uniform because if the current\u0027s uniform,"},{"Start":"08:01.060 ","End":"08:03.400","Text":"the magnetic field is going to be uniform."},{"Start":"08:03.400 ","End":"08:06.700","Text":"There\u0027s nothing in our question that will lead us to"},{"Start":"08:06.700 ","End":"08:11.110","Text":"believe that our magnetic field is fluctuating throughout this region."},{"Start":"08:11.110 ","End":"08:14.090","Text":"Therefore, so over here,"},{"Start":"08:14.090 ","End":"08:19.170","Text":"because we have uniform B,"},{"Start":"08:19.960 ","End":"08:29.705","Text":"we can say that this integral is simply equal to B.l where l is of course,"},{"Start":"08:29.705 ","End":"08:32.150","Text":"the length of this trajectory,"},{"Start":"08:32.150 ","End":"08:33.965","Text":"which over here is just a circle."},{"Start":"08:33.965 ","End":"08:37.685","Text":"The circumference of the circle."},{"Start":"08:37.685 ","End":"08:41.420","Text":"What we get that this is equal to the magnetic field"},{"Start":"08:41.420 ","End":"08:45.275","Text":"multiplied by the circumference of the circle,"},{"Start":"08:45.275 ","End":"08:52.570","Text":"which is equal to 2Pi multiplied by lowercase r."},{"Start":"08:54.060 ","End":"08:57.040","Text":"Now we want to do the right side."},{"Start":"08:57.040 ","End":"08:59.935","Text":"We see that we have this I_in over here."},{"Start":"08:59.935 ","End":"09:05.455","Text":"What we want to do is we want to find the current inside this loop,"},{"Start":"09:05.455 ","End":"09:08.755","Text":"this Ampere\u0027s loop of radius i."},{"Start":"09:08.755 ","End":"09:18.117","Text":"Notice i is smaller than R. We\u0027re trying to find I_in;"},{"Start":"09:18.117 ","End":"09:20.680","Text":"so I_in as we saw,"},{"Start":"09:20.680 ","End":"09:26.860","Text":"is equal to the integral of J.ds."},{"Start":"09:26.860 ","End":"09:28.810","Text":"As we\u0027ve already been told,"},{"Start":"09:28.810 ","End":"09:30.835","Text":"and also as we\u0027ve calculated,"},{"Start":"09:30.835 ","End":"09:32.770","Text":"our J is uniform."},{"Start":"09:32.770 ","End":"09:37.960","Text":"We can just say that this is equal to J.S."},{"Start":"09:37.960 ","End":"09:42.460","Text":"Our J in this region where r is smaller than R is equal"},{"Start":"09:42.460 ","End":"09:50.450","Text":"to I_naught divided by PiR^2."},{"Start":"09:51.420 ","End":"09:58.015","Text":"This is of course, in the positives at direction and this is multiplied by"},{"Start":"09:58.015 ","End":"10:05.545","Text":"the cross-sectional area of this Ampere\u0027s loop of radius r. Of course,"},{"Start":"10:05.545 ","End":"10:11.230","Text":"that is going to be equal to Pir^2."},{"Start":"10:11.230 ","End":"10:14.050","Text":"We can cross out these Pis."},{"Start":"10:14.050 ","End":"10:19.210","Text":"Let\u0027s just scroll down a little bit to give us some space."},{"Start":"10:19.210 ","End":"10:25.765","Text":"What we have is that I_in is simply equal to I_naught R^2"},{"Start":"10:25.765 ","End":"10:33.235","Text":"in the z-direction divided by R^2."},{"Start":"10:33.235 ","End":"10:36.190","Text":"Therefore, we can equate these 2 sides."},{"Start":"10:36.190 ","End":"10:39.325","Text":"Don\u0027t forget to multiply by Mu_naught."},{"Start":"10:39.325 ","End":"10:42.040","Text":"What we have is B.dl,"},{"Start":"10:42.040 ","End":"10:43.510","Text":"which we got over here,"},{"Start":"10:43.510 ","End":"10:49.519","Text":"is equal to 2BPir"},{"Start":"10:49.519 ","End":"10:55.345","Text":"is equal to Mu_naught multiplied by I_in,"},{"Start":"10:55.345 ","End":"11:03.310","Text":"which is I naught_ r^2 divided by R^2."},{"Start":"11:03.310 ","End":"11:08.410","Text":"Then what we can do is we can divide both sides by this r,"},{"Start":"11:08.410 ","End":"11:11.785","Text":"and now we just want to isolate out"},{"Start":"11:11.785 ","End":"11:16.180","Text":"our B because that\u0027s our question, the magnetic field."},{"Start":"11:16.180 ","End":"11:21.685","Text":"Magnetic field B, and don\u0027t forget that this is in the z-direction."},{"Start":"11:21.685 ","End":"11:30.130","Text":"Our magnetic field is going to be equal to Mu_naught I_naught R divided"},{"Start":"11:30.130 ","End":"11:39.580","Text":"by 2PiR^2 in the positive z-direction."},{"Start":"11:39.580 ","End":"11:45.460","Text":"This is of course, the region where r is smaller than R. Sorry, I made a mistake,"},{"Start":"11:45.460 ","End":"11:47.950","Text":"here because we\u0027re not dealing with vectors on"},{"Start":"11:47.950 ","End":"11:51.580","Text":"this side so we don\u0027t write the z-direction over here."},{"Start":"11:51.580 ","End":"11:53.770","Text":"But here, the magnetic field,"},{"Start":"11:53.770 ","End":"11:56.230","Text":"as we already saw through the right-hand rule,"},{"Start":"11:56.230 ","End":"11:59.470","Text":"is in the positive Theta direction."},{"Start":"11:59.470 ","End":"12:02.725","Text":"Just don\u0027t make the same mistake that I made."},{"Start":"12:02.725 ","End":"12:05.540","Text":"We already calculated the direction of the magnetic field,"},{"Start":"12:05.540 ","End":"12:10.015","Text":"it\u0027s in the Theta direction so just write that in."},{"Start":"12:10.015 ","End":"12:17.800","Text":"Now what we want to do is we want to calculate the magnetic field at a different radius."},{"Start":"12:17.800 ","End":"12:22.945","Text":"Now we\u0027re looking at this radius over here."},{"Start":"12:22.945 ","End":"12:26.575","Text":"This is our r right now."},{"Start":"12:26.575 ","End":"12:35.180","Text":"We\u0027re in the region where r is between R and 2R again."},{"Start":"12:35.250 ","End":"12:41.030","Text":"Let\u0027s just say again that the magnetic field is going this way."},{"Start":"12:41.040 ","End":"12:43.600","Text":"The magnetic field is some,"},{"Start":"12:43.600 ","End":"12:47.395","Text":"where along the Theta axis be it positive or negative,"},{"Start":"12:47.395 ","End":"12:49.405","Text":"we will soon discover."},{"Start":"12:49.405 ","End":"12:51.250","Text":"Again, we\u0027re doing this."},{"Start":"12:51.250 ","End":"12:57.850","Text":"We know that our magnetic field is going to be constant, it\u0027s uniform throughout."},{"Start":"12:57.850 ","End":"13:06.880","Text":"That means that this side we can straight away write B.l and as we can see, the l,"},{"Start":"13:06.880 ","End":"13:10.000","Text":"the roots is the circumference of this circle,"},{"Start":"13:10.000 ","End":"13:16.795","Text":"which is still 2Pi multiplied by r. Just the same R is in a different region."},{"Start":"13:16.795 ","End":"13:25.900","Text":"So B.dl becomes 2Pi B r"},{"Start":"13:25.900 ","End":"13:31.265","Text":"and this is equal to Mu naught multiplied by I_in."},{"Start":"13:31.265 ","End":"13:33.705","Text":"What exactly is I_in?"},{"Start":"13:33.705 ","End":"13:40.315","Text":"I_in is the current in this entire region over here."},{"Start":"13:40.315 ","End":"13:44.245","Text":"What we can see is that we have the I_in,"},{"Start":"13:44.245 ","End":"13:49.090","Text":"in this region over here between R and 2R."},{"Start":"13:49.090 ","End":"13:56.080","Text":"But we also have the I_in inside over here, the inner cylinder."},{"Start":"13:56.080 ","End":"14:02.930","Text":"What we\u0027re going to do is we\u0027re going to split this into 2 equations to find the I_in."},{"Start":"14:03.240 ","End":"14:06.505","Text":"We have Mu_naught which multiplies"},{"Start":"14:06.505 ","End":"14:11.020","Text":"everything and then we have I_in for this inner cylinder."},{"Start":"14:11.020 ","End":"14:17.110","Text":"First of all, we know that I_in is equal to the integral of J.ds,"},{"Start":"14:17.110 ","End":"14:21.730","Text":"which we calculated J up here."},{"Start":"14:21.730 ","End":"14:25.900","Text":"However, if we see the way that we calculated it,"},{"Start":"14:25.900 ","End":"14:29.785","Text":"because the full cylinder is inside over here."},{"Start":"14:29.785 ","End":"14:38.960","Text":"We\u0027re multiplying J by the cross-sectional area of the full cylinder, which is PiI^2."},{"Start":"14:39.090 ","End":"14:46.780","Text":"If we take I_naught divided by PiI^2 multiplied by PiR^2,"},{"Start":"14:46.780 ","End":"14:48.670","Text":"we\u0027re going to get our I_in,"},{"Start":"14:48.670 ","End":"14:50.724","Text":"which is just I_naught."},{"Start":"14:50.724 ","End":"14:54.430","Text":"Or we can also see from the question we were told that we"},{"Start":"14:54.430 ","End":"14:58.225","Text":"have a current I_naught flowing through the inner cylinder."},{"Start":"14:58.225 ","End":"15:04.850","Text":"The total I in this entire inner cylinder is I_naught."},{"Start":"15:05.310 ","End":"15:10.180","Text":"In the previous region, over here,"},{"Start":"15:10.180 ","End":"15:13.870","Text":"we calculated we had to use J.S because"},{"Start":"15:13.870 ","End":"15:19.180","Text":"the cross-sectional area was different because we took this radius of r,"},{"Start":"15:19.180 ","End":"15:28.840","Text":"which was smaller than R. We had a smaller internal current in this region over here,"},{"Start":"15:28.840 ","End":"15:33.430","Text":"in this region of r. However,"},{"Start":"15:33.430 ","End":"15:36.520","Text":"now we\u0027re in this whole region over here,"},{"Start":"15:36.520 ","End":"15:43.300","Text":"which means that this entire region up until R radius is enclosed."},{"Start":"15:43.300 ","End":"15:47.020","Text":"We\u0027re taking all the current that is in the inner cylinder,"},{"Start":"15:47.020 ","End":"15:49.135","Text":"which just happens to be I_naught."},{"Start":"15:49.135 ","End":"15:52.045","Text":"We can understand that from the question and even if you didn\u0027t,"},{"Start":"15:52.045 ","End":"15:55.360","Text":"once you do J.S with a J that we"},{"Start":"15:55.360 ","End":"15:58.840","Text":"previously calculated multiplied by this full surface area,"},{"Start":"15:58.840 ","End":"16:00.685","Text":"we will get I_naught."},{"Start":"16:00.685 ","End":"16:07.330","Text":"Then we add on the I_in in this region over here."},{"Start":"16:07.330 ","End":"16:11.960","Text":"This region over here is going to be J."},{"Start":"16:13.140 ","End":"16:15.340","Text":"It\u0027s this over here,"},{"Start":"16:15.340 ","End":"16:20.020","Text":"so this j so we have negative"},{"Start":"16:20.020 ","End":"16:26.725","Text":"I naught divided by 3PiR^2."},{"Start":"16:26.725 ","End":"16:30.010","Text":"This is J.S,"},{"Start":"16:30.010 ","End":"16:32.005","Text":"the surface area over here."},{"Start":"16:32.005 ","End":"16:34.105","Text":"Our surface area, as we saw,"},{"Start":"16:34.105 ","End":"16:40.600","Text":"is this width multiplied by the circumference. What is this?"},{"Start":"16:40.600 ","End":"16:43.990","Text":"The surface area of a circle is of course,"},{"Start":"16:43.990 ","End":"16:49.585","Text":"PiR^2 so we have this radius over here,"},{"Start":"16:49.585 ","End":"16:52.210","Text":"which is r,"},{"Start":"16:52.210 ","End":"17:00.295","Text":"minus this radius over here,"},{"Start":"17:00.295 ","End":"17:03.580","Text":"which is R and of course,"},{"Start":"17:03.580 ","End":"17:06.230","Text":"each one is squared."},{"Start":"17:06.870 ","End":"17:12.565","Text":"Then what we can do is we can just isolate out our B,"},{"Start":"17:12.565 ","End":"17:22.650","Text":"so what we have is that B is equal to Mu naught divided by 2Pi r and then"},{"Start":"17:22.650 ","End":"17:32.100","Text":"multiplied by I naught minus I naught Pi divided by"},{"Start":"17:32.100 ","End":"17:42.950","Text":"3Pi r^2 multiplied by i^2 minus R^2."},{"Start":"17:43.620 ","End":"17:48.130","Text":"Of course we can cross out these 2 Pi\u0027s,"},{"Start":"17:48.130 ","End":"17:53.950","Text":"they cancel out so we have divided by 3r^2 instead."},{"Start":"17:53.950 ","End":"17:58.015","Text":"I\u0027m not going to cancel this out anymore right now."},{"Start":"17:58.015 ","End":"18:00.895","Text":"This is the onset, but of course,"},{"Start":"18:00.895 ","End":"18:04.045","Text":"our B is a magnetic field and what we want to know is,"},{"Start":"18:04.045 ","End":"18:07.015","Text":"is it in the positive Theta direction,"},{"Start":"18:07.015 ","End":"18:10.309","Text":"or the negative Theta direction?"},{"Start":"18:10.350 ","End":"18:12.910","Text":"Let\u0027s take a look at it."},{"Start":"18:12.910 ","End":"18:17.800","Text":"First of all we know that in the inner cylinder we have a current of I"},{"Start":"18:17.800 ","End":"18:23.260","Text":"naught and in the thick outer tube,"},{"Start":"18:23.260 ","End":"18:26.440","Text":"we also have a current of I naught."},{"Start":"18:26.440 ","End":"18:31.750","Text":"What we can see here is when we summed to find the magnetic field,"},{"Start":"18:31.750 ","End":"18:35.660","Text":"the whole inner cylinder is included,"},{"Start":"18:35.660 ","End":"18:39.820","Text":"so all of I naught is included."},{"Start":"18:39.820 ","End":"18:45.805","Text":"Of course this I naught is forming a magnetic field in the positive Theta direction,"},{"Start":"18:45.805 ","End":"18:48.190","Text":"as we\u0027ve already seen."},{"Start":"18:48.190 ","End":"18:52.960","Text":"Then of course, we\u0027re also summing in this outer cylinder,"},{"Start":"18:52.960 ","End":"18:54.685","Text":"but the outer cylinder,"},{"Start":"18:54.685 ","End":"19:01.000","Text":"which has a total current of I naught but we\u0027re not summing the whole outer cylinder."},{"Start":"19:01.000 ","End":"19:06.205","Text":"We\u0027re just summing this region over here in green,"},{"Start":"19:06.205 ","End":"19:08.815","Text":"which is less than the whole outer cylinder."},{"Start":"19:08.815 ","End":"19:12.790","Text":"If we\u0027re summing on less than the whole region,"},{"Start":"19:12.790 ","End":"19:18.265","Text":"that means that our total summation is going to be less than I naught,"},{"Start":"19:18.265 ","End":"19:23.620","Text":"so the total current I_in,"},{"Start":"19:23.620 ","End":"19:27.800","Text":"in the region over here,"},{"Start":"19:28.560 ","End":"19:33.310","Text":"so in this thick outer tube is going to be"},{"Start":"19:33.310 ","End":"19:38.440","Text":"less than I naught because we haven\u0027t summed on the whole outer tube."},{"Start":"19:38.440 ","End":"19:41.470","Text":"Only if we were summing in the whole outer tube,"},{"Start":"19:41.470 ","End":"19:46.150","Text":"then we will have the maximal current flowing,"},{"Start":"19:46.150 ","End":"19:47.650","Text":"which is I naught."},{"Start":"19:47.650 ","End":"19:53.470","Text":"What we have is we have a magnetic field which is equal to something,"},{"Start":"19:53.470 ","End":"19:57.590","Text":"some constant multiplied by I naught,"},{"Start":"19:57.870 ","End":"20:04.190","Text":"subtracting something which is smaller than I-naught."},{"Start":"20:04.650 ","End":"20:08.755","Text":"In that case, what we\u0027re going to have is something bigger"},{"Start":"20:08.755 ","End":"20:13.310","Text":"minus something smaller which means,"},{"Start":"20:13.440 ","End":"20:17.440","Text":"let\u0027s call this I_1,"},{"Start":"20:17.440 ","End":"20:19.645","Text":"and let\u0027s call this over here,"},{"Start":"20:19.645 ","End":"20:26.860","Text":"I_2 and we know that I_1 is bigger than I_2, so therefore,"},{"Start":"20:26.860 ","End":"20:29.680","Text":"when we subtract I_2 from both sides,"},{"Start":"20:29.680 ","End":"20:33.730","Text":"what we have is I_1 minus I_2,"},{"Start":"20:33.730 ","End":"20:36.955","Text":"which is bigger than 0."},{"Start":"20:36.955 ","End":"20:39.160","Text":"If something is bigger than 0,"},{"Start":"20:39.160 ","End":"20:42.350","Text":"that means that it is positive."},{"Start":"20:42.350 ","End":"20:49.490","Text":"In that case, we know what that is in the positive Theta direction."},{"Start":"20:49.490 ","End":"20:51.430","Text":"I\u0027ll even write the positive over here."},{"Start":"20:51.430 ","End":"20:55.630","Text":"You don\u0027t have to, but just to make it clear that this is also in"},{"Start":"20:55.630 ","End":"21:01.940","Text":"the positive Theta direction, like so."},{"Start":"21:02.490 ","End":"21:06.520","Text":"What we can see the difference between"},{"Start":"21:06.520 ","End":"21:10.870","Text":"the 2 regions with a magnetic field is in the first region,"},{"Start":"21:10.870 ","End":"21:13.015","Text":"so inside the inner cylinder,"},{"Start":"21:13.015 ","End":"21:15.220","Text":"we have this r,"},{"Start":"21:15.220 ","End":"21:17.965","Text":"which is of course what\u0027s changing."},{"Start":"21:17.965 ","End":"21:23.570","Text":"It\u0027s slowly increasing and the r is located in the numerator."},{"Start":"21:23.570 ","End":"21:27.210","Text":"In other words, as r increases,"},{"Start":"21:27.210 ","End":"21:34.150","Text":"so as we move away from the center towards the outer radius of the cylinder,"},{"Start":"21:34.170 ","End":"21:38.350","Text":"this whole fraction is going to increase,"},{"Start":"21:38.350 ","End":"21:41.965","Text":"which means that our magnetic field is getting stronger."},{"Start":"21:41.965 ","End":"21:45.160","Text":"As we move away from the origin,"},{"Start":"21:45.160 ","End":"21:51.100","Text":"the magnetic field in this region of the inner cylinder increases until it reaches a"},{"Start":"21:51.100 ","End":"21:57.670","Text":"maximum over here at radius R. Then as we move into the next region,"},{"Start":"21:57.670 ","End":"22:00.700","Text":"so the thick outer tube,"},{"Start":"22:00.700 ","End":"22:03.910","Text":"we see that this r,"},{"Start":"22:03.910 ","End":"22:05.604","Text":"which is what\u0027s changing,"},{"Start":"22:05.604 ","End":"22:08.800","Text":"is in the denominator,"},{"Start":"22:08.800 ","End":"22:14.780","Text":"so we can see that as we get further and further away from the center,"},{"Start":"22:14.880 ","End":"22:20.275","Text":"r is going to be increasing and because it\u0027s in the denominator,"},{"Start":"22:20.275 ","End":"22:24.820","Text":"this whole fraction is going to be decreasing,"},{"Start":"22:24.820 ","End":"22:30.200","Text":"which means that the magnetic field in this region is going to be decreasing."},{"Start":"22:30.660 ","End":"22:35.380","Text":"If we want to draw this as a graph,"},{"Start":"22:35.380 ","End":"22:37.450","Text":"so let\u0027s draw it here."},{"Start":"22:37.450 ","End":"22:40.135","Text":"Here we have the radius R,"},{"Start":"22:40.135 ","End":"22:44.110","Text":"and here we have the radius 2R,"},{"Start":"22:44.110 ","End":"22:50.960","Text":"and this is r and here we have our magnetic field."},{"Start":"22:51.330 ","End":"22:59.050","Text":"In this region, we see that r can increase in a linear way,"},{"Start":"22:59.050 ","End":"23:01.780","Text":"and it affects the magnetic field linearly."},{"Start":"23:01.780 ","End":"23:05.470","Text":"It\u0027s multiplied by some constant,"},{"Start":"23:05.470 ","End":"23:13.780","Text":"so it will go something like so and then it reaches this maximum value at radius R,"},{"Start":"23:13.780 ","End":"23:15.745","Text":"and then over here,"},{"Start":"23:15.745 ","End":"23:20.080","Text":"it decreases as a function of this."},{"Start":"23:20.080 ","End":"23:22.330","Text":"Here we can look at it."},{"Start":"23:22.330 ","End":"23:23.710","Text":"It makes it a bit more complicated,"},{"Start":"23:23.710 ","End":"23:26.455","Text":"but let\u0027s just look at this area over here."},{"Start":"23:26.455 ","End":"23:31.630","Text":"It\u0027s functioning the same as 1 divided by r multiplied by something."},{"Start":"23:31.630 ","End":"23:35.425","Text":"I know that here we have this factor of r^2,"},{"Start":"23:35.425 ","End":"23:37.340","Text":"which changes a little bit,"},{"Start":"23:37.340 ","End":"23:39.265","Text":"but all in all,"},{"Start":"23:39.265 ","End":"23:45.440","Text":"this is decreasing as a function of 1 divided by r so it will go like so."},{"Start":"23:46.770 ","End":"23:52.555","Text":"I\u0027ve just opened up this equation just to make it a bit clearer."},{"Start":"23:52.555 ","End":"24:01.540","Text":"What we see is here we have a decrease of 1 divided by r. Here we have a linear increase,"},{"Start":"24:01.540 ","End":"24:04.405","Text":"but it\u0027s so small because if we look at the denominator,"},{"Start":"24:04.405 ","End":"24:07.525","Text":"the denominator is extremely small,"},{"Start":"24:07.525 ","End":"24:11.890","Text":"it\u0027s 3R^2 times as small as this,"},{"Start":"24:11.890 ","End":"24:18.580","Text":"so we have a very slight increase and also there\u0027s a minus here,"},{"Start":"24:18.580 ","End":"24:21.430","Text":"so it\u0027s actually still a decrease."},{"Start":"24:21.430 ","End":"24:26.170","Text":"Then over here, we also are going down as a function of 1"},{"Start":"24:26.170 ","End":"24:31.690","Text":"divided by r. All in all we have this pattern."},{"Start":"24:31.690 ","End":"24:34.720","Text":"It\u0027s not exactly as a function of 1 divided by r,"},{"Start":"24:34.720 ","End":"24:39.070","Text":"but it can be approximated to that."},{"Start":"24:39.070 ","End":"24:47.560","Text":"That is what we can see as we reach the radius of R over here so the magnetic field"},{"Start":"24:47.560 ","End":"24:51.370","Text":"begins decreasing as an approximate function of 1"},{"Start":"24:51.370 ","End":"24:56.480","Text":"divided by r. That\u0027s the end of this lesson."}],"ID":22274},{"Watched":false,"Name":"Exercise 3","Duration":"25m 14s","ChapterTopicVideoID":21340,"CourseChapterTopicPlaylistID":99468,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/21340.jpeg","UploadDate":"2020-04-06T21:24:33.4200000","DurationForVideoObject":"PT25M14S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.620","Text":"Hello. In this lesson,"},{"Start":"00:01.620 ","End":"00:04.485","Text":"we\u0027re going to be answering the following question."},{"Start":"00:04.485 ","End":"00:07.590","Text":"We\u0027re being asked to calculate the magnetic field of"},{"Start":"00:07.590 ","End":"00:14.775","Text":"an infinite thin plane with a uniform charge density of Sigma."},{"Start":"00:14.775 ","End":"00:17.895","Text":"This is of course the charge density per unit area,"},{"Start":"00:17.895 ","End":"00:21.960","Text":"which is moving with a velocity of v naught."},{"Start":"00:21.960 ","End":"00:25.335","Text":"It\u0027s a constant velocity in the x direction."},{"Start":"00:25.335 ","End":"00:28.320","Text":"Let\u0027s add in our axes."},{"Start":"00:28.320 ","End":"00:32.085","Text":"Let\u0027s say that this is the y direction,"},{"Start":"00:32.085 ","End":"00:37.980","Text":"this is the z direction and over here,"},{"Start":"00:37.980 ","End":"00:39.615","Text":"coming out of the page,"},{"Start":"00:39.615 ","End":"00:43.821","Text":"we have our x direction,"},{"Start":"00:43.821 ","End":"00:48.043","Text":"and our velocity is of course also in this direction."},{"Start":"00:48.043 ","End":"00:56.875","Text":"Our plane is moving like so in this direction with a velocity of v naught."},{"Start":"00:56.875 ","End":"01:01.520","Text":"Now let\u0027s look at this from the side view where we\u0027re"},{"Start":"01:01.520 ","End":"01:05.975","Text":"looking right over here at this edge of this infinite plane."},{"Start":"01:05.975 ","End":"01:09.710","Text":"What we\u0027ll see is a line."},{"Start":"01:11.150 ","End":"01:17.870","Text":"This is still the y direction,"},{"Start":"01:17.870 ","End":"01:22.835","Text":"this direction is still the z direction,"},{"Start":"01:22.835 ","End":"01:27.095","Text":"and the x direction is coming out of the page,"},{"Start":"01:27.095 ","End":"01:30.509","Text":"so this is the x direction."},{"Start":"01:31.300 ","End":"01:37.237","Text":"As we know, we have this charge density on the plane,"},{"Start":"01:37.237 ","End":"01:41.360","Text":"and the plane is traveling towards us in the x-direction,"},{"Start":"01:41.360 ","End":"01:44.090","Text":"so towards us, with a velocity."},{"Start":"01:44.090 ","End":"01:47.690","Text":"That means that all of the charges on the plane,"},{"Start":"01:47.690 ","End":"01:52.020","Text":"so we can draw them in green."},{"Start":"01:52.020 ","End":"01:53.600","Text":"Let\u0027s draw them over here."},{"Start":"01:53.600 ","End":"01:56.990","Text":"We have these charges over here like so,"},{"Start":"01:56.990 ","End":"02:00.680","Text":"they\u0027re all also traveling towards us in"},{"Start":"02:00.680 ","End":"02:05.360","Text":"this direction with the same velocity of v naught."},{"Start":"02:05.360 ","End":"02:08.194","Text":"If the charges are moving,"},{"Start":"02:08.194 ","End":"02:11.195","Text":"that means that we have a current."},{"Start":"02:11.195 ","End":"02:14.300","Text":"In other words, we have a current and"},{"Start":"02:14.300 ","End":"02:18.875","Text":"the direction of the current is the direction that the charges are moving in,"},{"Start":"02:18.875 ","End":"02:23.615","Text":"which is towards us or in the positive x direction."},{"Start":"02:23.615 ","End":"02:30.530","Text":"What we can imagine is that this plane is made up of lots of different"},{"Start":"02:30.530 ","End":"02:38.610","Text":"current carrying wires where the current is flowing towards us."},{"Start":"02:38.610 ","End":"02:43.719","Text":"Now what I\u0027m going to do is I\u0027m going to pick a random area in space,"},{"Start":"02:43.719 ","End":"02:50.880","Text":"and what I want to do is I want to find out what the magnetic field is at this point."},{"Start":"02:51.170 ","End":"02:56.270","Text":"In this case, what we do is let\u0027s choose"},{"Start":"02:56.270 ","End":"03:01.370","Text":"this wire over here and from the right hand rule,"},{"Start":"03:01.370 ","End":"03:05.030","Text":"we point our thumb in the direction that the current is flowing."},{"Start":"03:05.030 ","End":"03:10.944","Text":"We point our thumb that it\u0027s pointing towards us out of the page,"},{"Start":"03:10.944 ","End":"03:15.080","Text":"and then our fingers curl in the direction of the magnetic field."},{"Start":"03:15.080 ","End":"03:17.720","Text":"If you point your thumb outwards,"},{"Start":"03:17.720 ","End":"03:19.160","Text":"you\u0027ll see that first of all,"},{"Start":"03:19.160 ","End":"03:23.090","Text":"the magnetic field is some kind of circle, as we know."},{"Start":"03:23.090 ","End":"03:25.355","Text":"Imagine that this is a good circle,"},{"Start":"03:25.355 ","End":"03:30.865","Text":"and we\u0027ll see that it\u0027s traveling in this direction."},{"Start":"03:30.865 ","End":"03:35.280","Text":"This is the direction of the magnetic field."},{"Start":"03:35.920 ","End":"03:39.500","Text":"Specifically at this point over here,"},{"Start":"03:39.500 ","End":"03:45.556","Text":"the blue point where we\u0027re trying to see what the magnetic field is at this point,"},{"Start":"03:45.556 ","End":"03:50.150","Text":"so the magnetic field is of course tangent to"},{"Start":"03:50.150 ","End":"03:56.340","Text":"the circle like so and it is pointing in this direction."},{"Start":"03:56.650 ","End":"03:59.060","Text":"Now, let\u0027s choose"},{"Start":"03:59.060 ","End":"04:06.020","Text":"another wire on the other side of this point that we\u0027re trying to measure."},{"Start":"04:06.020 ","End":"04:07.520","Text":"Again, from the right hand rule,"},{"Start":"04:07.520 ","End":"04:09.680","Text":"we point our thumb in the direction of the current,"},{"Start":"04:09.680 ","End":"04:19.225","Text":"so it\u0027s pointing out of the page and we get this circular magnetic fields going like so."},{"Start":"04:19.225 ","End":"04:22.580","Text":"Of course, our fingers curl around in the direction of"},{"Start":"04:22.580 ","End":"04:27.920","Text":"the magnetic field so they curl in this direction."},{"Start":"04:27.920 ","End":"04:37.074","Text":"But this time, I\u0027ll draw this actually in pink to show that it\u0027s for this wire."},{"Start":"04:37.074 ","End":"04:42.350","Text":"This time, the magnetic field over here is"},{"Start":"04:42.350 ","End":"04:49.175","Text":"pointing in this direction tangent of course to this circle."},{"Start":"04:49.175 ","End":"04:57.179","Text":"What we can see is if we want to see the net magnetic field,"},{"Start":"04:57.179 ","End":"05:03.275","Text":"so the pink is made up of a component in the z direction,"},{"Start":"05:03.275 ","End":"05:08.710","Text":"and a component in the negative y direction and the gray,"},{"Start":"05:08.710 ","End":"05:14.290","Text":"so for this wire is made up of a component in the negative z direction,"},{"Start":"05:14.290 ","End":"05:17.195","Text":"and a component in the negative y direction."},{"Start":"05:17.195 ","End":"05:20.570","Text":"We can see that these 2 components are equal and"},{"Start":"05:20.570 ","End":"05:24.860","Text":"opposite because the currents are exactly the same,"},{"Start":"05:24.860 ","End":"05:27.920","Text":"and they\u0027re equidistant away from this point."},{"Start":"05:27.920 ","End":"05:31.775","Text":"That means that they\u0027re equal and opposite."},{"Start":"05:31.775 ","End":"05:36.650","Text":"So we have 1 component which we\u0027re going to add,"},{"Start":"05:36.650 ","End":"05:37.670","Text":"and that\u0027s this one."},{"Start":"05:37.670 ","End":"05:41.015","Text":"Let\u0027s draw it in black."},{"Start":"05:41.015 ","End":"05:44.555","Text":"This is the component of the magnetic fields"},{"Start":"05:44.555 ","End":"05:47.720","Text":"that we\u0027re trying to calculate at this point."},{"Start":"05:47.720 ","End":"05:56.560","Text":"The first thing that we can notice is that it is in the negative y direction."},{"Start":"05:58.040 ","End":"06:01.310","Text":"Because we have this infinite plane,"},{"Start":"06:01.310 ","End":"06:08.550","Text":"the magnetic field above the plane is always going to be in this negative y direction,"},{"Start":"06:08.550 ","End":"06:11.030","Text":"and if I do the exact same thing,"},{"Start":"06:11.030 ","End":"06:15.080","Text":"but for a point below the plane over here,"},{"Start":"06:15.080 ","End":"06:20.030","Text":"what I\u0027ll get is that the magnetic field below the plane is"},{"Start":"06:20.030 ","End":"06:26.190","Text":"in the positive y direction for the exact same reason."},{"Start":"06:26.240 ","End":"06:30.650","Text":"What we can see is that because the plane is infinite,"},{"Start":"06:30.650 ","End":"06:34.588","Text":"we know that our magnetic field is independent of y."},{"Start":"06:34.588 ","End":"06:39.605","Text":"Because it doesn\u0027t make a difference if I say that the origin is over here,"},{"Start":"06:39.605 ","End":"06:43.565","Text":"or over here, or over here because the plane is infinite."},{"Start":"06:43.565 ","End":"06:47.555","Text":"But the magnetic field might be dependent on z."},{"Start":"06:47.555 ","End":"06:48.620","Text":"We\u0027ll see that soon."},{"Start":"06:48.620 ","End":"06:52.865","Text":"We\u0027re going to calculate what the magnetic field is and we\u0027ll see if it\u0027s dependent on z."},{"Start":"06:52.865 ","End":"06:54.800","Text":"But in the meantime,"},{"Start":"06:54.800 ","End":"06:56.825","Text":"we can assume that it is."},{"Start":"06:56.825 ","End":"07:01.280","Text":"What we can see is that we have the direction of"},{"Start":"07:01.280 ","End":"07:03.485","Text":"the magnetic field above and below"},{"Start":"07:03.485 ","End":"07:06.860","Text":"the plane and what we want to do is we want to see its value."},{"Start":"07:06.860 ","End":"07:12.120","Text":"Let\u0027s say that it\u0027s some certain height,"},{"Start":"07:12.120 ","End":"07:14.960","Text":"or at some certain value of z."},{"Start":"07:14.960 ","End":"07:19.080","Text":"Let\u0027s say that this height over here is z."},{"Start":"07:20.660 ","End":"07:24.914","Text":"Also, let\u0027s say that on this side,"},{"Start":"07:24.914 ","End":"07:28.110","Text":"we\u0027re looking at the magnetic field at a height of z,"},{"Start":"07:28.110 ","End":"07:31.170","Text":"below the z axis,"},{"Start":"07:31.170 ","End":"07:34.275","Text":"below the 0s, so it\u0027s just negative z."},{"Start":"07:34.275 ","End":"07:39.180","Text":"What we can see is that if we flip our plane around,"},{"Start":"07:39.180 ","End":"07:43.155","Text":"we have symmetry in the z axis."},{"Start":"07:43.155 ","End":"07:45.210","Text":"Because if I take this plane,"},{"Start":"07:45.210 ","End":"07:48.000","Text":"and they flip it around 180 degrees so that what\u0027s"},{"Start":"07:48.000 ","End":"07:51.240","Text":"on top is now in the bottom and vice versa,"},{"Start":"07:51.240 ","End":"07:54.540","Text":"we\u0027ll be left with the exact same questions."},{"Start":"07:54.540 ","End":"07:59.040","Text":"The same magnetic field pointing in the same direction on both sides,"},{"Start":"07:59.040 ","End":"08:00.630","Text":"on the top and on the bottom."},{"Start":"08:00.630 ","End":"08:05.325","Text":"Therefore, if we have this symmetry in the z axis,"},{"Start":"08:05.325 ","End":"08:09.465","Text":"we can say that the magnetic field at this point z"},{"Start":"08:09.465 ","End":"08:15.310","Text":"is equal to the magnetic field at this point, negative z."},{"Start":"08:16.400 ","End":"08:19.965","Text":"In order to find what the magnetic field is,"},{"Start":"08:19.965 ","End":"08:22.200","Text":"we\u0027re going to use Ampere\u0027s law."},{"Start":"08:22.200 ","End":"08:30.855","Text":"As we know, Ampere\u0027s law needs a closed loop integral along B.dl."},{"Start":"08:30.855 ","End":"08:33.765","Text":"We know that this is equal to mu naught,"},{"Start":"08:33.765 ","End":"08:41.260","Text":"a constant multiplied by the current inside this closed loop integral."},{"Start":"08:41.260 ","End":"08:45.825","Text":"Now what we\u0027re going to do is we\u0027re going to draw this loop."},{"Start":"08:45.825 ","End":"08:48.250","Text":"Let\u0027s draw it in gray."},{"Start":"08:48.740 ","End":"08:56.985","Text":"If I have that my magnetic field is going in a circle or in the Theta direction,"},{"Start":"08:56.985 ","End":"08:59.865","Text":"I\u0027ll choose my amperes loop to be in a circle."},{"Start":"08:59.865 ","End":"09:03.720","Text":"But if my magnetic field is traveling in a straight line,"},{"Start":"09:03.720 ","End":"09:12.030","Text":"then I\u0027ll choose my amperes loop to be some kind of rectangle or square."},{"Start":"09:12.030 ","End":"09:17.505","Text":"Over here I can see that my magnetic field is going in a straight line."},{"Start":"09:17.505 ","End":"09:27.300","Text":"So I\u0027m going to choose the loop to be this rectangle. Going like so."},{"Start":"09:27.300 ","End":"09:29.610","Text":"I\u0027m going to say that this length,"},{"Start":"09:29.610 ","End":"09:32.070","Text":"the length of my rectangle is some length that I choose,"},{"Start":"09:32.070 ","End":"09:35.415","Text":"let\u0027s say lowercase l,"},{"Start":"09:35.415 ","End":"09:41.830","Text":"and of course, my magnetic field is traveling like so."},{"Start":"09:42.080 ","End":"09:46.019","Text":"What we want to do is we want to calculate first"},{"Start":"09:46.019 ","End":"09:50.415","Text":"the left side of the equation, so B. dl."},{"Start":"09:50.415 ","End":"09:54.210","Text":"What we can see is that this side of the equation is"},{"Start":"09:54.210 ","End":"09:58.035","Text":"dependent on the vectors B of the magnetic field,"},{"Start":"09:58.035 ","End":"10:00.780","Text":"so the size and direction of the magnetic field,"},{"Start":"10:00.780 ","End":"10:06.045","Text":"and it\u0027s dependent on the size and the direction of dl,"},{"Start":"10:06.045 ","End":"10:11.550","Text":"where dl is the trajectory or the root that I\u0027m taking."},{"Start":"10:11.550 ","End":"10:15.630","Text":"Let\u0027s say that the route that I\u0027m taking is going to"},{"Start":"10:15.630 ","End":"10:19.890","Text":"go over here in the same direction as the magnetic field."},{"Start":"10:19.890 ","End":"10:22.875","Text":"I\u0027m starting at this corner over here,"},{"Start":"10:22.875 ","End":"10:25.410","Text":"and I\u0027m traveling in this direction,"},{"Start":"10:25.410 ","End":"10:31.050","Text":"then it will go down here and we\u0027ll just carry on in the direction of the loop."},{"Start":"10:31.050 ","End":"10:37.500","Text":"Over here we\u0027re also going in this direction and then back up to the starting point."},{"Start":"10:37.500 ","End":"10:41.055","Text":"What we can see is that on this edge,"},{"Start":"10:41.055 ","End":"10:47.430","Text":"the route taken is in the same direction as the magnetic field."},{"Start":"10:47.430 ","End":"10:50.070","Text":"They\u0027re both in the negative y-direction."},{"Start":"10:50.070 ","End":"10:53.565","Text":"Also over here, below the plane,"},{"Start":"10:53.565 ","End":"10:57.300","Text":"the route taken is in the same direction as the magnetic field,"},{"Start":"10:57.300 ","End":"10:59.760","Text":"in the positive y-direction."},{"Start":"10:59.760 ","End":"11:03.345","Text":"That\u0027s good. We have that B and our dl,"},{"Start":"11:03.345 ","End":"11:05.835","Text":"when we\u0027re doing this closed loop integral,"},{"Start":"11:05.835 ","End":"11:08.655","Text":"are always traveling in the same direction."},{"Start":"11:08.655 ","End":"11:11.400","Text":"We can see that at this level,"},{"Start":"11:11.400 ","End":"11:13.455","Text":"as long as z is constant,"},{"Start":"11:13.455 ","End":"11:16.905","Text":"our magnetic field is going to be constant."},{"Start":"11:16.905 ","End":"11:19.635","Text":"We have a uniform magnetic field over here."},{"Start":"11:19.635 ","End":"11:22.935","Text":"Maybe later we\u0027ll see that B is independent of z."},{"Start":"11:22.935 ","End":"11:25.590","Text":"We\u0027ll get to that potentially later."},{"Start":"11:25.590 ","End":"11:27.810","Text":"It might not be, we\u0027ll see."},{"Start":"11:27.810 ","End":"11:29.790","Text":"But at least for now,"},{"Start":"11:29.790 ","End":"11:34.695","Text":"we can see that as long as we\u0027re at the same height of the z,"},{"Start":"11:34.695 ","End":"11:36.615","Text":"due to the symmetry of the question,"},{"Start":"11:36.615 ","End":"11:39.060","Text":"the magnetic field is uniform."},{"Start":"11:39.060 ","End":"11:41.880","Text":"Therefore, we can say,"},{"Start":"11:41.880 ","End":"11:44.520","Text":"because the magnetic field is uniform,"},{"Start":"11:44.520 ","End":"11:50.535","Text":"we can write this equation as B.dl,"},{"Start":"11:50.535 ","End":"11:56.560","Text":"the length of the trajectory is equal to Mu_ I_n."},{"Start":"11:56.690 ","End":"11:59.715","Text":"Now let\u0027s do this equation."},{"Start":"11:59.715 ","End":"12:07.725","Text":"First, let\u0027s take B.dl of this upper side of this rectangle."},{"Start":"12:07.725 ","End":"12:13.650","Text":"What we get is the magnetic field B multiplied by the length of the loop over here,"},{"Start":"12:13.650 ","End":"12:15.690","Text":"or the length of this section of the loop,"},{"Start":"12:15.690 ","End":"12:22.125","Text":"which we said was equal to lowercase l. Then we carry on in our loop."},{"Start":"12:22.125 ","End":"12:24.870","Text":"This is perpendicular to the magnetic field,"},{"Start":"12:24.870 ","End":"12:27.015","Text":"so we don\u0027t add anything over here."},{"Start":"12:27.015 ","End":"12:31.545","Text":"Then we\u0027re at the bottom side of the loop."},{"Start":"12:31.545 ","End":"12:36.840","Text":"Again, we have a plus because we are traveling in the direction of"},{"Start":"12:36.840 ","End":"12:40.065","Text":"the loop trajectory and it is"},{"Start":"12:40.065 ","End":"12:44.325","Text":"in the positive direction of the magnetic field in this region."},{"Start":"12:44.325 ","End":"12:47.940","Text":"We add plus B,"},{"Start":"12:47.940 ","End":"12:53.055","Text":"the field, which is again also over here uniform for the same reason explained above,"},{"Start":"12:53.055 ","End":"12:55.890","Text":"multiplied by the length of this section of the loop,"},{"Start":"12:55.890 ","End":"12:59.640","Text":"which is lowercase l. Then we travel up here,"},{"Start":"12:59.640 ","End":"13:01.320","Text":"which is again,"},{"Start":"13:01.320 ","End":"13:02.865","Text":"just like the other side,"},{"Start":"13:02.865 ","End":"13:05.700","Text":"perpendicular to the magnetic field,"},{"Start":"13:05.700 ","End":"13:10.493","Text":"which means that we have a 0 magnetic field over here,"},{"Start":"13:10.493 ","End":"13:17.685","Text":"so we\u0027ll have 0 multiplied by 2z plus 0 multiplied by 2z, which is 0."},{"Start":"13:17.685 ","End":"13:24.435","Text":"This is equal to Mu_0 multiplied by I_n."},{"Start":"13:24.435 ","End":"13:29.640","Text":"Now what we want to do is we want to calculate the current I."},{"Start":"13:29.640 ","End":"13:39.615","Text":"In general, we\u0027ve seen that I_n is equal to the integral on J.ds."},{"Start":"13:39.615 ","End":"13:49.080","Text":"However, what we have over here is some current that\u0027s flowing over here lengthwise."},{"Start":"13:49.080 ","End":"13:52.020","Text":"We have some not J,"},{"Start":"13:52.020 ","End":"13:57.780","Text":"but we have some K. Because if we take J,"},{"Start":"13:57.780 ","End":"13:59.385","Text":"if we want to calculate J,"},{"Start":"13:59.385 ","End":"14:01.980","Text":"we need Rho,"},{"Start":"14:01.980 ","End":"14:09.345","Text":"charged density per unit volume multiplied by the velocity."},{"Start":"14:09.345 ","End":"14:15.840","Text":"K is equal to Sigma multiplied by the velocity."},{"Start":"14:15.840 ","End":"14:22.605","Text":"We have Sigma, so we need to find K. If we want to go straight to I,"},{"Start":"14:22.605 ","End":"14:26.550","Text":"where we don\u0027t have to do this whole integral and equation,"},{"Start":"14:26.550 ","End":"14:28.065","Text":"then we need Lambda,"},{"Start":"14:28.065 ","End":"14:33.990","Text":"charged density per unit length multiplied by the velocity."},{"Start":"14:33.990 ","End":"14:37.860","Text":"This is 1 option to calculate the current."},{"Start":"14:37.860 ","End":"14:47.505","Text":"Another option is knowing that I is equal to dq by dt, then calculating it."},{"Start":"14:47.505 ","End":"14:51.068","Text":"First we\u0027re going to calculate the current using this method,"},{"Start":"14:51.068 ","End":"14:52.490","Text":"and at the end of the lesson,"},{"Start":"14:52.490 ","End":"14:54.560","Text":"we\u0027ll use this method."},{"Start":"14:54.560 ","End":"14:56.129","Text":"But both methods work,"},{"Start":"14:56.129 ","End":"15:00.330","Text":"you can choose whichever 1 you feel more comfortable with."},{"Start":"15:00.910 ","End":"15:04.460","Text":"Let\u0027s use this method."},{"Start":"15:04.460 ","End":"15:13.770","Text":"dq is equal to either Rho multiplied by dv,"},{"Start":"15:13.770 ","End":"15:20.505","Text":"in volume, Sigma ds or Lambda dl."},{"Start":"15:20.505 ","End":"15:22.330","Text":"Of course, we have Sigma,"},{"Start":"15:22.330 ","End":"15:27.300","Text":"that means that dq is equal to Sigma ds."},{"Start":"15:27.300 ","End":"15:28.965","Text":"What is the ds over here?"},{"Start":"15:28.965 ","End":"15:32.205","Text":"ds is a unit of area."},{"Start":"15:32.205 ","End":"15:36.560","Text":"Seeing as our plane is along the x and y axes,"},{"Start":"15:36.560 ","End":"15:39.565","Text":"or the plane is on the x, y plane,"},{"Start":"15:39.565 ","End":"15:44.630","Text":"our unit of area is going to be dx, dy,"},{"Start":"15:44.630 ","End":"15:46.999","Text":"a unit of x,"},{"Start":"15:46.999 ","End":"15:50.820","Text":"and a unit of y, like so."},{"Start":"15:51.070 ","End":"15:55.100","Text":"Now we\u0027ll say that our I,"},{"Start":"15:55.100 ","End":"15:58.980","Text":"which is equal to dq by dt,"},{"Start":"15:59.230 ","End":"16:04.110","Text":"we have dq by dt,"},{"Start":"16:04.110 ","End":"16:12.060","Text":"where dq is equal to Sigma dxdy divided by dt."},{"Start":"16:12.060 ","End":"16:14.295","Text":"At this stage,"},{"Start":"16:14.295 ","End":"16:19.870","Text":"1 of the differentials in the numerator will join"},{"Start":"16:19.870 ","End":"16:25.495","Text":"with the differential in the denominator to equal the velocity."},{"Start":"16:25.495 ","End":"16:29.935","Text":"In our case, our velocity is in the x-direction."},{"Start":"16:29.935 ","End":"16:36.610","Text":"In other words, it\u0027s equal to the change in x with respect to the change in time,"},{"Start":"16:36.610 ","End":"16:40.030","Text":"that is the velocity in the x-direction."},{"Start":"16:40.030 ","End":"16:47.380","Text":"What we get is that this is equal to Sigma multiplied by the velocity,"},{"Start":"16:47.380 ","End":"16:54.375","Text":"which we were told is v_0 in the x direction, multiplied by dy."},{"Start":"16:54.375 ","End":"16:57.090","Text":"Because we have a differential here dy,"},{"Start":"16:57.090 ","End":"17:00.100","Text":"this is equal to dI."},{"Start":"17:00.660 ","End":"17:04.105","Text":"What we have over here is our dI,"},{"Start":"17:04.105 ","End":"17:09.430","Text":"which is of course the current per unit width over here,"},{"Start":"17:09.430 ","End":"17:15.940","Text":"so it\u0027s how much current is in this interval over here, dy."},{"Start":"17:15.940 ","End":"17:19.135","Text":"This is what our dI is equal to."},{"Start":"17:19.135 ","End":"17:22.120","Text":"If we look back at the original equation,"},{"Start":"17:22.120 ","End":"17:24.198","Text":"we\u0027re trying to find I in,"},{"Start":"17:24.198 ","End":"17:29.260","Text":"so the current inside this closed circuit loop,"},{"Start":"17:29.260 ","End":"17:33.340","Text":"so the current inside this region over here,"},{"Start":"17:33.340 ","End":"17:38.035","Text":"which we defined as being of a width in the y-direction of"},{"Start":"17:38.035 ","End":"17:44.740","Text":"l. What I\u0027m going to do is I\u0027m going to write over here I in,"},{"Start":"17:44.740 ","End":"17:49.825","Text":"which is equal to the integral on dI."},{"Start":"17:49.825 ","End":"17:52.765","Text":"That is the integral from,"},{"Start":"17:52.765 ","End":"17:56.065","Text":"let\u0027s say that this is 0 and this over here is l,"},{"Start":"17:56.065 ","End":"17:59.200","Text":"so from 0 to l of the dI,"},{"Start":"17:59.200 ","End":"18:03.280","Text":"which is Sigma v naught dy,"},{"Start":"18:03.280 ","End":"18:06.610","Text":"which is of course going to give us"},{"Start":"18:06.610 ","End":"18:15.820","Text":"Sigma v naught multiplied by y and we substitute in l minus 0,"},{"Start":"18:15.820 ","End":"18:24.250","Text":"which is just l, so Sigma v naught multiplied by l. Now let\u0027s just scroll down."},{"Start":"18:24.250 ","End":"18:26.440","Text":"We\u0027re trying to find the magnetic field."},{"Start":"18:26.440 ","End":"18:35.125","Text":"I\u0027ll remind you we have 2Bl is equal to Mu naught multiplied by I in,"},{"Start":"18:35.125 ","End":"18:43.135","Text":"where in is Sigma v naught l. Now we can isolate out our B."},{"Start":"18:43.135 ","End":"18:48.580","Text":"We\u0027ll get that our magnetic field is equal to Mu naught"},{"Start":"18:48.580 ","End":"18:55.045","Text":"Sigma v naught l divided by 2l."},{"Start":"18:55.045 ","End":"18:59.740","Text":"Of course, these 2l\u0027s cancel out,"},{"Start":"18:59.740 ","End":"19:03.560","Text":"and we\u0027re left with this."},{"Start":"19:04.440 ","End":"19:07.270","Text":"Here we have the magnitude of RB,"},{"Start":"19:07.270 ","End":"19:09.865","Text":"but of course we want it as a vector,"},{"Start":"19:09.865 ","End":"19:13.185","Text":"so that means with direction as well."},{"Start":"19:13.185 ","End":"19:15.164","Text":"Let\u0027s look at our diagram."},{"Start":"19:15.164 ","End":"19:17.265","Text":"We have above the plane,"},{"Start":"19:17.265 ","End":"19:20.100","Text":"if the plane is at z is equal to naught."},{"Start":"19:20.100 ","End":"19:24.420","Text":"We have above the plane where z is greater than naught."},{"Start":"19:24.420 ","End":"19:29.790","Text":"We have below the plane where z is less than naught."},{"Start":"19:29.790 ","End":"19:35.670","Text":"Greater than naught, we see that the direction of our magnetic field is going leftwards,"},{"Start":"19:35.670 ","End":"19:43.165","Text":"which is in the negative y direction and below the plane where z is less than naught,"},{"Start":"19:43.165 ","End":"19:46.480","Text":"we see that our magnetic field is traveling rightwards,"},{"Start":"19:46.480 ","End":"19:50.635","Text":"which is in the same direction as the y-direction,"},{"Start":"19:50.635 ","End":"19:55.128","Text":"or in other words, it\u0027s in the positive y-direction."},{"Start":"19:55.128 ","End":"19:57.850","Text":"This is what our magnetic field is,"},{"Start":"19:57.850 ","End":"20:03.055","Text":"depending on if we\u0027re above the plane or below the plane."},{"Start":"20:03.055 ","End":"20:09.710","Text":"As we can see, our magnetic field is independent of z."},{"Start":"20:10.020 ","End":"20:14.950","Text":"To know the direction of the magnetic field,"},{"Start":"20:14.950 ","End":"20:17.290","Text":"we have to know if we\u0027re above or below the plane,"},{"Start":"20:17.290 ","End":"20:20.875","Text":"which is of course dependent on z, the direction."},{"Start":"20:20.875 ","End":"20:25.600","Text":"However, the magnitude of the magnetic field itself is constant for z."},{"Start":"20:25.600 ","End":"20:28.735","Text":"Whether we\u0027re located at this height,"},{"Start":"20:28.735 ","End":"20:32.875","Text":"z over here, or this height over here,"},{"Start":"20:32.875 ","End":"20:34.480","Text":"or at this height over here,"},{"Start":"20:34.480 ","End":"20:38.073","Text":"we\u0027re going to have the exact same magnetic field,"},{"Start":"20:38.073 ","End":"20:41.305","Text":"and if we\u0027re located 2z below,"},{"Start":"20:41.305 ","End":"20:43.915","Text":"or 3z below the magnetic field,"},{"Start":"20:43.915 ","End":"20:47.590","Text":"we\u0027re going to have the exact same magnetic field."},{"Start":"20:47.590 ","End":"20:50.830","Text":"The magnetic field is uniform in z."},{"Start":"20:50.830 ","End":"20:55.555","Text":"The only difference is if it\u0027s in the positive y-direction"},{"Start":"20:55.555 ","End":"21:02.060","Text":"below the plane or in the negative y-direction above the plane."},{"Start":"21:02.460 ","End":"21:07.255","Text":"This is the answer to the question."},{"Start":"21:07.255 ","End":"21:10.540","Text":"I did that using this method and they said that I\u0027m going to go"},{"Start":"21:10.540 ","End":"21:13.935","Text":"over using this method afterwards."},{"Start":"21:13.935 ","End":"21:17.215","Text":"If you don\u0027t want to see this method or you\u0027re ready know it,"},{"Start":"21:17.215 ","End":"21:18.880","Text":"then you can stop the video now."},{"Start":"21:18.880 ","End":"21:21.670","Text":"However, if you want to see how I do this method,"},{"Start":"21:21.670 ","End":"21:24.225","Text":"so I\u0027m going to show it right now."},{"Start":"21:24.225 ","End":"21:28.455","Text":"Now we\u0027re using this method and we\u0027re using,"},{"Start":"21:28.455 ","End":"21:35.665","Text":"of course k because we\u0027re given that our plane is charged with this charge Sigma."},{"Start":"21:35.665 ","End":"21:43.150","Text":"What we have is that k is equal to Sigma multiplied by the velocity,"},{"Start":"21:43.150 ","End":"21:48.400","Text":"where the velocity is some v naught in the x-direction."},{"Start":"21:48.400 ","End":"21:53.575","Text":"This is k and now we want to calculate our I in."},{"Start":"21:53.575 ","End":"22:00.260","Text":"We know that our In is the integral on kdl,"},{"Start":"22:00.260 ","End":"22:02.170","Text":"so we have the integral on k,"},{"Start":"22:02.170 ","End":"22:06.025","Text":"which is Sigma v naughts."},{"Start":"22:06.025 ","End":"22:12.520","Text":"Now we just have to do is we have to understand what the dl is equal to."},{"Start":"22:12.520 ","End":"22:17.170","Text":"In this method, this is really what is difficult,"},{"Start":"22:17.170 ","End":"22:19.735","Text":"it\u0027s to understand what the dl is."},{"Start":"22:19.735 ","End":"22:25.170","Text":"Now note, dl is what we\u0027re summing on."},{"Start":"22:25.170 ","End":"22:28.170","Text":"We are not summing in the x-direction."},{"Start":"22:28.170 ","End":"22:34.970","Text":"As we know, the x-direction is the direction of the current."},{"Start":"22:34.970 ","End":"22:39.205","Text":"But what we want to know is how much current is flowing"},{"Start":"22:39.205 ","End":"22:43.030","Text":"in this region of our gray dotted line"},{"Start":"22:43.030 ","End":"22:50.484","Text":"in this length l. What we can see is that we want to sum up each of these,"},{"Start":"22:50.484 ","End":"22:55.165","Text":"what we said where these wires carrying current that make up the plane."},{"Start":"22:55.165 ","End":"22:58.390","Text":"We want to sum up all of these. What does that mean?"},{"Start":"22:58.390 ","End":"23:00.955","Text":"We\u0027re summing up along this length l,"},{"Start":"23:00.955 ","End":"23:04.720","Text":"which happens to be along the y-axis."},{"Start":"23:04.720 ","End":"23:08.950","Text":"What we\u0027re trying to do over here in this integral is"},{"Start":"23:08.950 ","End":"23:13.750","Text":"we\u0027re calculating the current per unit length."},{"Start":"23:13.750 ","End":"23:17.560","Text":"We\u0027re trying to see if we take some length,"},{"Start":"23:17.560 ","End":"23:21.039","Text":"how much current is flowing through here."},{"Start":"23:21.039 ","End":"23:22.525","Text":"In order to do that,"},{"Start":"23:22.525 ","End":"23:25.773","Text":"we\u0027re going to take this and see how much current is here,"},{"Start":"23:25.773 ","End":"23:29.830","Text":"and add on this and this and this and so on."},{"Start":"23:29.830 ","End":"23:34.030","Text":"We can see very clearly once we sit to think"},{"Start":"23:34.030 ","End":"23:38.035","Text":"about what this integral actually means and we look at our diagram,"},{"Start":"23:38.035 ","End":"23:40.855","Text":"that\u0027s why it\u0027s very important to draw the diagram."},{"Start":"23:40.855 ","End":"23:43.915","Text":"It\u0027s very easy to understand that in this case,"},{"Start":"23:43.915 ","End":"23:49.730","Text":"our dl is just dy in the y-direction."},{"Start":"23:50.160 ","End":"23:52.872","Text":"It\u0027s very easy to get confused."},{"Start":"23:52.872 ","End":"23:55.990","Text":"Just remember, we\u0027re trying to find the total current in,"},{"Start":"23:55.990 ","End":"23:59.650","Text":"so you want to choose one of the sides,"},{"Start":"23:59.650 ","End":"24:04.750","Text":"dl is going to represent one of the sides which is perpendicular to our k vector."},{"Start":"24:04.750 ","End":"24:06.895","Text":"That\u0027s another way to think about it."},{"Start":"24:06.895 ","End":"24:10.090","Text":"Here our k vector is in the x-direction."},{"Start":"24:10.090 ","End":"24:13.240","Text":"The directions which are perpendicular to"},{"Start":"24:13.240 ","End":"24:18.175","Text":"the x-direction is the y-direction or the z-direction."},{"Start":"24:18.175 ","End":"24:19.960","Text":"If we sum across z,"},{"Start":"24:19.960 ","End":"24:24.670","Text":"we see that we only have current over here, z is equal to 0."},{"Start":"24:24.670 ","End":"24:27.115","Text":"Above that any other value of z,"},{"Start":"24:27.115 ","End":"24:29.020","Text":"we have no current."},{"Start":"24:29.020 ","End":"24:32.530","Text":"However, if we go along the y-axis,"},{"Start":"24:32.530 ","End":"24:34.525","Text":"we can see that we have current."},{"Start":"24:34.525 ","End":"24:38.350","Text":"Therefore, we\u0027re going to want to sum up with y."},{"Start":"24:38.350 ","End":"24:41.950","Text":"We\u0027re going to want to integrate with respect to y."},{"Start":"24:41.950 ","End":"24:48.640","Text":"What we can see is that over here is actually the equivalent of our dI,"},{"Start":"24:48.640 ","End":"24:50.800","Text":"which we saw over here."},{"Start":"24:50.800 ","End":"24:54.220","Text":"This is exactly what we got and of course we put in our bounds from"},{"Start":"24:54.220 ","End":"24:59.110","Text":"0 until the full length of our loop."},{"Start":"24:59.110 ","End":"25:02.395","Text":"Then of course we just do the integral."},{"Start":"25:02.395 ","End":"25:06.550","Text":"We plug it in to this equation that we had over here for B,"},{"Start":"25:06.550 ","End":"25:09.130","Text":"and then you isolated out B,"},{"Start":"25:09.130 ","End":"25:12.310","Text":"and you\u0027ll get the exact same answer."},{"Start":"25:12.310 ","End":"25:15.290","Text":"That\u0027s the end of this lesson."}],"ID":21421},{"Watched":false,"Name":"Exercise 4","Duration":"19m 45s","ChapterTopicVideoID":21341,"CourseChapterTopicPlaylistID":99468,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/21341.jpeg","UploadDate":"2020-04-06T21:32:21.5670000","DurationForVideoObject":"PT19M45S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.065","Text":"Hello. In this lesson we\u0027re going to be answering the following question."},{"Start":"00:04.065 ","End":"00:08.130","Text":"Calculate the magnetic field of an infinite plane of width d with"},{"Start":"00:08.130 ","End":"00:15.825","Text":"a uniform charge density of Rho moving with a velocity of v _0 in the x direction."},{"Start":"00:15.825 ","End":"00:20.165","Text":"Because we have an infinite plane on the x-y plane,"},{"Start":"00:20.165 ","End":"00:23.130","Text":"that means that we can place this origin"},{"Start":"00:23.130 ","End":"00:28.420","Text":"anywhere along the x-y plane because of this symmetry."},{"Start":"00:28.970 ","End":"00:33.800","Text":"However, in the z axis we can see that we have"},{"Start":"00:33.800 ","End":"00:38.705","Text":"this width d so the origin has to be right in the middle of this width d,"},{"Start":"00:38.705 ","End":"00:43.985","Text":"which is of course, at 1/2 d. As long as we\u0027re in the middle of the width,"},{"Start":"00:43.985 ","End":"00:49.055","Text":"we can place the origin anywhere else with respect to x and y."},{"Start":"00:49.055 ","End":"00:53.870","Text":"This question is very similar to the previous question that we dealt with."},{"Start":"00:53.870 ","End":"00:58.220","Text":"The only difference is now we\u0027re dealing with an infinite plane with width."},{"Start":"00:58.220 ","End":"01:04.543","Text":"What we can do is we can split up the plane into lots of thin planes,"},{"Start":"01:04.543 ","End":"01:09.060","Text":"and then through them calculate the magnetic field."},{"Start":"01:09.080 ","End":"01:11.750","Text":"We\u0027re going to do that soon."},{"Start":"01:11.750 ","End":"01:17.015","Text":"But let\u0027s first of all start with just the direction of the magnetic field."},{"Start":"01:17.015 ","End":"01:18.740","Text":"From the previous lesson,"},{"Start":"01:18.740 ","End":"01:20.720","Text":"it\u0027s the exact same explanation."},{"Start":"01:20.720 ","End":"01:26.810","Text":"We can see that the magnetic field above the plane is in the negative y direction,"},{"Start":"01:26.810 ","End":"01:34.400","Text":"and the magnetic fields below the plane is in the positive y direction."},{"Start":"01:34.400 ","End":"01:36.890","Text":"If you want to understand the explanation,"},{"Start":"01:36.890 ","End":"01:40.060","Text":"please go back to the previous video."},{"Start":"01:40.060 ","End":"01:43.565","Text":"Now we know the direction of the magnetic field."},{"Start":"01:43.565 ","End":"01:48.590","Text":"What we want to do is we want to work out the magnetic field itself, the magnitude."},{"Start":"01:48.590 ","End":"01:50.825","Text":"We\u0027re going to use Ampere\u0027s law,"},{"Start":"01:50.825 ","End":"01:54.760","Text":"which means that we have a closed loop integral on"},{"Start":"01:54.760 ","End":"02:00.005","Text":"B so the magnetic field dot-product with dl,"},{"Start":"02:00.005 ","End":"02:05.105","Text":"the closed loop, that is of course a vector so it has a direction."},{"Start":"02:05.105 ","End":"02:10.125","Text":"This is equal to Mu naught multiplied by I_in,"},{"Start":"02:10.125 ","End":"02:13.220","Text":"where Mu naught is a constant and I_in is the"},{"Start":"02:13.220 ","End":"02:17.820","Text":"current contained in this closed loop integral."},{"Start":"02:18.230 ","End":"02:21.065","Text":"Just like in the previous video,"},{"Start":"02:21.065 ","End":"02:27.995","Text":"we\u0027re going to take this rectangular closed loop like so."},{"Start":"02:27.995 ","End":"02:32.690","Text":"We\u0027re going to say that it\u0027s direction is like so,"},{"Start":"02:32.690 ","End":"02:35.345","Text":"so it\u0027s going anticlockwise."},{"Start":"02:35.345 ","End":"02:39.530","Text":"What we can see is that the magnetic field is traveling"},{"Start":"02:39.530 ","End":"02:45.030","Text":"like so on the sides we\u0027ll deal with in a second."},{"Start":"02:45.030 ","End":"02:51.710","Text":"Let\u0027s just give this an arbitrary length of l. This length over here is"},{"Start":"02:51.710 ","End":"03:00.380","Text":"l. Because the magnetic field"},{"Start":"03:00.380 ","End":"03:02.970","Text":"is uniform across this height,"},{"Start":"03:02.970 ","End":"03:13.060","Text":"and let\u0027s say that this height is the height z away from where z is equal to 0."},{"Start":"03:13.130 ","End":"03:16.655","Text":"We know for sure that this height z,"},{"Start":"03:16.655 ","End":"03:20.245","Text":"the magnetic field is going to be uniform."},{"Start":"03:20.245 ","End":"03:25.930","Text":"What we can do is we can take B multiplied by the length of this side of the loop,"},{"Start":"03:25.930 ","End":"03:31.690","Text":"which is l. Then when we go down this side,"},{"Start":"03:31.690 ","End":"03:38.750","Text":"so our magnetic field is either in the positive y direction or the negative y direction."},{"Start":"03:38.810 ","End":"03:41.035","Text":"That\u0027s one thing, however,"},{"Start":"03:41.035 ","End":"03:43.175","Text":"that doesn\u0027t make a difference."},{"Start":"03:43.175 ","End":"03:50.110","Text":"Our dl vector for this region is in the z or negative z direction,"},{"Start":"03:50.110 ","End":"03:54.080","Text":"and our b vector is either in the positive or negative y direction,"},{"Start":"03:54.080 ","End":"03:55.415","Text":"but it doesn\u0027t make a difference."},{"Start":"03:55.415 ","End":"03:59.800","Text":"We have a vector in the y direction,"},{"Start":"03:59.800 ","End":"04:03.935","Text":"dot product with a vector in the z direction."},{"Start":"04:03.935 ","End":"04:07.430","Text":"Of course, these 2 vectors on this region are"},{"Start":"04:07.430 ","End":"04:12.095","Text":"perpendicular to 1 another so their dot products is going to be equal to 0."},{"Start":"04:12.095 ","End":"04:16.715","Text":"Then we\u0027re carrying on over on this section,"},{"Start":"04:16.715 ","End":"04:25.078","Text":"where we see that our magnetic field is again in the same direction as the loop,"},{"Start":"04:25.078 ","End":"04:27.880","Text":"so our B is in the same direction as our dl."},{"Start":"04:27.880 ","End":"04:34.530","Text":"They\u0027re parallel. We can just add on another Bl so we\u0027re left with 2Bl."},{"Start":"04:34.530 ","End":"04:38.315","Text":"As we go back up to our starting point, again,"},{"Start":"04:38.315 ","End":"04:43.860","Text":"we have that our B vector\u0027s somewhere in 1 of the y direction is positive"},{"Start":"04:43.860 ","End":"04:49.310","Text":"or negative and that our dl vector is along the z axis,"},{"Start":"04:49.310 ","End":"04:54.185","Text":"which is perpendicular to the y-axis so their dot products is going to be equal to 0."},{"Start":"04:54.185 ","End":"04:59.550","Text":"That is it. This is the left side of the equation."},{"Start":"04:59.550 ","End":"05:06.075","Text":"This is, of course, equal to Mu naught multiplied by I_in."},{"Start":"05:06.075 ","End":"05:12.360","Text":"Now what we want to do is we want to calculate what I_in is equal to."},{"Start":"05:12.800 ","End":"05:17.930","Text":"There\u0027s 2 ways rather that we can calculate this."},{"Start":"05:17.930 ","End":"05:23.090","Text":"Either we can use the equation where I is equal to dq by dt,"},{"Start":"05:23.090 ","End":"05:28.700","Text":"so the change in charge per change in time or we can use the equation that"},{"Start":"05:28.700 ","End":"05:34.910","Text":"J is equal to Rho multiplied by the velocity."},{"Start":"05:34.910 ","End":"05:38.210","Text":"In our case, we\u0027re going to use this equation because"},{"Start":"05:38.210 ","End":"05:42.130","Text":"it\u0027s slightly easier to use when we have width."},{"Start":"05:42.130 ","End":"05:44.390","Text":"When dealing with Rho,"},{"Start":"05:44.390 ","End":"05:46.430","Text":"charge density per unit volume,"},{"Start":"05:46.430 ","End":"05:48.860","Text":"it\u0027s easier to use this method."},{"Start":"05:48.860 ","End":"05:52.835","Text":"In the previous video where we had a thin infinite plane,"},{"Start":"05:52.835 ","End":"05:55.700","Text":"we use this method as it was slightly easier."},{"Start":"05:55.700 ","End":"05:57.230","Text":"But at the end of the lesson,"},{"Start":"05:57.230 ","End":"06:01.370","Text":"we also showed how to do it where we use the equivalent of J,"},{"Start":"06:01.370 ","End":"06:06.560","Text":"which is k, which was equal to Sigma multiplied by the velocity."},{"Start":"06:06.560 ","End":"06:08.270","Text":"If you want to go over that,"},{"Start":"06:08.270 ","End":"06:10.550","Text":"please go back to the previous lesson."},{"Start":"06:10.550 ","End":"06:19.300","Text":"Rho we\u0027re given and our velocity we were told is equal to v_0 in the x direction."},{"Start":"06:19.300 ","End":"06:27.760","Text":"As we know, I_in is equal to integral of Jds,"},{"Start":"06:27.760 ","End":"06:30.635","Text":"where ds is a unit of area."},{"Start":"06:30.635 ","End":"06:35.675","Text":"Now notice that when we\u0027re calculating this,"},{"Start":"06:35.675 ","End":"06:39.230","Text":"what we can see is that we have like"},{"Start":"06:39.230 ","End":"06:45.815","Text":"so charges for Rho that are moving in this x direction,"},{"Start":"06:45.815 ","End":"06:48.680","Text":"so that are moving out of the page."},{"Start":"06:48.680 ","End":"06:56.720","Text":"What we can see is that we have lots of little wires that are carrying current."},{"Start":"06:56.720 ","End":"06:58.955","Text":"That\u0027s what we can say that it looks like,"},{"Start":"06:58.955 ","End":"07:03.958","Text":"current-carrying wires where the current is traveling in the x direction,"},{"Start":"07:03.958 ","End":"07:06.895","Text":"and this fills this whole area."},{"Start":"07:06.895 ","End":"07:14.835","Text":"What we can see is if I take the highlighter,"},{"Start":"07:14.835 ","End":"07:18.980","Text":"what I want to do is I want to sum up this whole area,"},{"Start":"07:18.980 ","End":"07:25.085","Text":"all of the current in this area that I\u0027ve marked out with this loop."},{"Start":"07:25.085 ","End":"07:30.515","Text":"However, in this region over here and in this region up here,"},{"Start":"07:30.515 ","End":"07:34.505","Text":"there is no current because there\u0027s no charge."},{"Start":"07:34.505 ","End":"07:38.540","Text":"Rho is only on this width d. In other words,"},{"Start":"07:38.540 ","End":"07:44.930","Text":"what I want to do is I want to calculate the current that"},{"Start":"07:44.930 ","End":"07:52.120","Text":"is in this region where I have charges so within this region."},{"Start":"07:52.120 ","End":"08:00.365","Text":"From that is equal to 0 until z is equal to d divided by 2."},{"Start":"08:00.365 ","End":"08:05.050","Text":"This whole region in yellow."},{"Start":"08:05.050 ","End":"08:13.655","Text":"Now because we\u0027re not told that Rho is changing with respect to something,"},{"Start":"08:13.655 ","End":"08:18.605","Text":"we can say that the charges are uniformly distributed throughout,"},{"Start":"08:18.605 ","End":"08:23.490","Text":"which means that J is going to be uniform."},{"Start":"08:23.490 ","End":"08:30.815","Text":"While J is a constant,"},{"Start":"08:30.815 ","End":"08:39.450","Text":"I can say that this integral is simply equal to J multiplied by the area."},{"Start":"08:40.010 ","End":"08:46.685","Text":"The area is simply equal to length times width."},{"Start":"08:46.685 ","End":"08:52.220","Text":"The length is l and the width is d. We have J,"},{"Start":"08:52.220 ","End":"08:57.540","Text":"which is Rho multiplied by v_0 multiplied by the area"},{"Start":"08:57.540 ","End":"09:04.580","Text":"l multiplied by d. Then we can plug this into our equation."},{"Start":"09:04.580 ","End":"09:11.030","Text":"We get that 2Bl is equal to Mu naught multiplied by our I_in,"},{"Start":"09:11.030 ","End":"09:16.315","Text":"so Rho v_0 ld."},{"Start":"09:16.315 ","End":"09:22.140","Text":"The ls cancel out and we get that our magnetic field, B,"},{"Start":"09:22.140 ","End":"09:30.940","Text":"is equal to Mu naught Rho v_0 d divided by 2."},{"Start":"09:32.660 ","End":"09:38.300","Text":"Then if we want to add in directions so when we\u0027re in"},{"Start":"09:38.300 ","End":"09:45.390","Text":"the region where we\u0027re at z is greater than d divided by 2."},{"Start":"09:45.390 ","End":"09:51.675","Text":"We\u0027re above the plane so z is greater than d divided by 2."},{"Start":"09:51.675 ","End":"09:57.240","Text":"We can see that the magnetic field is in the negative y direction."},{"Start":"09:57.240 ","End":"09:59.070","Text":"When we\u0027re below the plane,"},{"Start":"09:59.070 ","End":"10:08.890","Text":"so in the region where z is smaller than negative d divided by 2,"},{"Start":"10:08.890 ","End":"10:11.570","Text":"this is negative d divided by 2,"},{"Start":"10:11.570 ","End":"10:17.505","Text":"then the magnetic field is in the positive y direction."},{"Start":"10:17.505 ","End":"10:22.150","Text":"This is the magnetic field above and below the plane,"},{"Start":"10:22.150 ","End":"10:26.740","Text":"but what about the magnetic field inside the plane?"},{"Start":"10:26.740 ","End":"10:29.275","Text":"Inside the plane,"},{"Start":"10:29.275 ","End":"10:32.245","Text":"we\u0027re choosing some kind of height,"},{"Start":"10:32.245 ","End":"10:35.650","Text":"z, where we\u0027re located inside."},{"Start":"10:35.650 ","End":"10:39.625","Text":"We can choose this height over here, z."},{"Start":"10:39.625 ","End":"10:46.300","Text":"We\u0027re in the region where z is smaller than d divided by 2,"},{"Start":"10:46.300 ","End":"10:48.505","Text":"so it\u0027s smaller than this upper edge,"},{"Start":"10:48.505 ","End":"10:52.600","Text":"but it\u0027s larger or it\u0027s located above the lower edge,"},{"Start":"10:52.600 ","End":"10:58.420","Text":"where the lower edge is of course at negative d divided by 2."},{"Start":"10:58.830 ","End":"11:06.940","Text":"Again, we build this amperes loop, like so."},{"Start":"11:06.940 ","End":"11:13.270","Text":"This length again is of length l. Again,"},{"Start":"11:13.270 ","End":"11:17.575","Text":"we can say that the direction of the loop is in this anticlockwise direction,"},{"Start":"11:17.575 ","End":"11:22.090","Text":"and our magnetic field is also going something like that,"},{"Start":"11:22.090 ","End":"11:24.130","Text":"in the same direction."},{"Start":"11:24.130 ","End":"11:31.900","Text":"Just like before, the left side of the equation will leave us with to 2Bl,"},{"Start":"11:31.900 ","End":"11:36.160","Text":"for the exact same reason as when we were looking at the larger loop."},{"Start":"11:36.160 ","End":"11:43.070","Text":"Now we just want to see what Mu_naught multiplied by I_in is equal to."},{"Start":"11:43.380 ","End":"11:48.565","Text":"First of all, let\u0027s just see what our I_in is."},{"Start":"11:48.565 ","End":"11:52.120","Text":"Our I_in let\u0027s see."},{"Start":"11:52.120 ","End":"12:00.470","Text":"As we know, it\u0027s the integral of our Jds, just like here."},{"Start":"12:00.750 ","End":"12:04.030","Text":"But again, our J,"},{"Start":"12:04.030 ","End":"12:06.940","Text":"because our J is constant, this integral,"},{"Start":"12:06.940 ","End":"12:10.360","Text":"we can just multiply J.s."},{"Start":"12:10.360 ","End":"12:13.165","Text":"J multiplied by s,"},{"Start":"12:13.165 ","End":"12:16.690","Text":"where here, our J is the exact same,"},{"Start":"12:16.690 ","End":"12:24.850","Text":"so Rhov_naught and our s is the area of this entire rectangle."},{"Start":"12:24.850 ","End":"12:30.730","Text":"Here we have current throughout the whole ampere loop that we did,"},{"Start":"12:30.730 ","End":"12:34.420","Text":"like so, so we\u0027re summing up for the whole loop."},{"Start":"12:34.420 ","End":"12:37.510","Text":"We have this length l,"},{"Start":"12:37.510 ","End":"12:42.430","Text":"and the width is twice z."},{"Start":"12:42.430 ","End":"12:47.320","Text":"We have z above the origin and z below the origin,"},{"Start":"12:47.320 ","End":"12:49.630","Text":"another z over here,"},{"Start":"12:49.630 ","End":"12:53.120","Text":"so the length is 2z."},{"Start":"12:54.270 ","End":"13:01.539","Text":"In other words, 2Bl is equal to Mu_naught multiplied"},{"Start":"13:01.539 ","End":"13:08.245","Text":"by I_in which is Rhov_naught l multiplied by 2z."},{"Start":"13:08.245 ","End":"13:12.565","Text":"We can cancel out the l\u0027s from both side and the 2,"},{"Start":"13:12.565 ","End":"13:22.700","Text":"and what we\u0027re left with is that B is equal to Mu_naught Rho v_naught z."},{"Start":"13:23.670 ","End":"13:26.440","Text":"Now, because of course,"},{"Start":"13:26.440 ","End":"13:28.720","Text":"B is a vector,"},{"Start":"13:28.720 ","End":"13:31.465","Text":"so we want to add in directions."},{"Start":"13:31.465 ","End":"13:38.125","Text":"I\u0027m just going to say that the direction is in the negative y direction."},{"Start":"13:38.125 ","End":"13:43.960","Text":"I don\u0027t have to split it up into above or below the origin this time,"},{"Start":"13:43.960 ","End":"13:45.205","Text":"and why is this?"},{"Start":"13:45.205 ","End":"13:47.815","Text":"Because my z will cancel it out."},{"Start":"13:47.815 ","End":"13:50.950","Text":"If I\u0027m looking above the origin,"},{"Start":"13:50.950 ","End":"13:53.140","Text":"that means that my z is positive,"},{"Start":"13:53.140 ","End":"13:59.215","Text":"and then my magnetic field in total is in the negative y direction."},{"Start":"13:59.215 ","End":"14:03.925","Text":"Which is what we said originally in the negative y direction."},{"Start":"14:03.925 ","End":"14:06.700","Text":"If I\u0027m located below the origin,"},{"Start":"14:06.700 ","End":"14:12.550","Text":"then I\u0027m located at a value where z is a negative value."},{"Start":"14:12.550 ","End":"14:14.980","Text":"Z is smaller than 0."},{"Start":"14:14.980 ","End":"14:16.420","Text":"It\u0027s a negative value,"},{"Start":"14:16.420 ","End":"14:19.000","Text":"which means that here I\u0027ll have a negative,"},{"Start":"14:19.000 ","End":"14:21.826","Text":"multiplied by negative y direction,"},{"Start":"14:21.826 ","End":"14:24.460","Text":"will give me the positive y direction,"},{"Start":"14:24.460 ","End":"14:26.605","Text":"which is exactly what we said."},{"Start":"14:26.605 ","End":"14:31.100","Text":"The B field in this region is in the positive y direction."},{"Start":"14:31.800 ","End":"14:38.350","Text":"This is the magnetic field in the region"},{"Start":"14:38.350 ","End":"14:45.520","Text":"where z is between negative d divided by 2 and positive d divided by 2."},{"Start":"14:45.520 ","End":"14:48.445","Text":"That\u0027s the end of this lesson."},{"Start":"14:48.445 ","End":"14:49.660","Text":"Now, I\u0027m just going to give"},{"Start":"14:49.660 ","End":"14:57.760","Text":"a slightly more mathematical proof of why the magnetic field has to be along the y-axis,"},{"Start":"14:57.760 ","End":"15:01.120","Text":"be it in the positive or negative direction,"},{"Start":"15:01.120 ","End":"15:03.505","Text":"but in the y-direction in general."},{"Start":"15:03.505 ","End":"15:05.800","Text":"That\u0027s what I\u0027m going to do now."},{"Start":"15:05.800 ","End":"15:09.025","Text":"What you need to happen,"},{"Start":"15:09.025 ","End":"15:12.820","Text":"you need the rotor of"},{"Start":"15:12.820 ","End":"15:19.860","Text":"the magnetic field to be equal to Mu_naught multiplied by J,"},{"Start":"15:19.860 ","End":"15:22.875","Text":"which is of course the charge density."},{"Start":"15:22.875 ","End":"15:32.230","Text":"We saw that the charge density was equal to the charge multiplied by the velocity,"},{"Start":"15:32.230 ","End":"15:36.415","Text":"which was v_naught in the x direction."},{"Start":"15:36.415 ","End":"15:40.465","Text":"What we want is to show that this has to happen."},{"Start":"15:40.465 ","End":"15:43.990","Text":"We need that the rotor of the magnetic field is equal"},{"Start":"15:43.990 ","End":"15:47.710","Text":"to Mu_naught Rho v_naught in the x direction."},{"Start":"15:47.710 ","End":"15:55.780","Text":"In other words, the rotor of the magnetic field has to be in the x direction."},{"Start":"15:55.780 ","End":"16:03.850","Text":"This is how you have a magnetic field according to this equation. Let\u0027s do this rotor."},{"Start":"16:03.850 ","End":"16:08.530","Text":"In order to calculate the rotor of B,"},{"Start":"16:08.530 ","End":"16:10.420","Text":"we just take the determinant."},{"Start":"16:10.420 ","End":"16:16.465","Text":"We have the x, y, and z direction."},{"Start":"16:16.465 ","End":"16:19.029","Text":"In the x direction,"},{"Start":"16:19.029 ","End":"16:21.910","Text":"so we have d by dx,"},{"Start":"16:21.910 ","End":"16:24.910","Text":"then here we have d by dy,"},{"Start":"16:24.910 ","End":"16:27.955","Text":"and here d by dz."},{"Start":"16:27.955 ","End":"16:33.145","Text":"Then here we have the x component of the magnetic field,"},{"Start":"16:33.145 ","End":"16:35.770","Text":"the y component of the magnetic field,"},{"Start":"16:35.770 ","End":"16:41.320","Text":"and the z component of the magnetic field."},{"Start":"16:41.320 ","End":"16:43.540","Text":"Once we do this,"},{"Start":"16:43.540 ","End":"16:50.110","Text":"because we want a rotor in this example to be only in the x direction,"},{"Start":"16:50.110 ","End":"16:57.670","Text":"that means that we want the parts of the rotor in the y and z direction to be equal to 0."},{"Start":"16:57.670 ","End":"17:00.460","Text":"In the x direction,"},{"Start":"17:00.460 ","End":"17:06.400","Text":"we\u0027re going to have d by dy of Bz,"},{"Start":"17:06.400 ","End":"17:10.570","Text":"so we\u0027ll have dBz by"},{"Start":"17:10.570 ","End":"17:16.825","Text":"dy minus d by dz of By,"},{"Start":"17:16.825 ","End":"17:22.400","Text":"so d of By by dz."},{"Start":"17:23.010 ","End":"17:25.795","Text":"This is in the x direction."},{"Start":"17:25.795 ","End":"17:32.500","Text":"Then of course we have the components in the y and the z directions,"},{"Start":"17:32.500 ","End":"17:34.585","Text":"which we want it to be equal to 0."},{"Start":"17:34.585 ","End":"17:40.585","Text":"Now, something to note is that we\u0027re dealing with an infinite plane."},{"Start":"17:40.585 ","End":"17:42.460","Text":"If the plane is infinite,"},{"Start":"17:42.460 ","End":"17:49.225","Text":"that means that a change in y or a change in x isn\u0027t meant to change anything over here,"},{"Start":"17:49.225 ","End":"17:52.360","Text":"because if I look at a point over here,"},{"Start":"17:52.360 ","End":"17:54.085","Text":"or a point over here,"},{"Start":"17:54.085 ","End":"17:57.565","Text":"nothing with respect to y is meant to change because it\u0027s infinite."},{"Start":"17:57.565 ","End":"17:59.440","Text":"Because I can just shift it along,"},{"Start":"17:59.440 ","End":"18:03.490","Text":"it makes no difference because the plane is infinite in y."},{"Start":"18:03.490 ","End":"18:09.820","Text":"In that case, if the plane has this infinity in the y direction,"},{"Start":"18:09.820 ","End":"18:16.630","Text":"that has to mean that my d by dy of anything has to be equal to 0."},{"Start":"18:16.630 ","End":"18:18.700","Text":"There has to be no change in y,"},{"Start":"18:18.700 ","End":"18:22.765","Text":"so my function can\u0027t be dependent on y."},{"Start":"18:22.765 ","End":"18:24.160","Text":"That\u0027s what this means."},{"Start":"18:24.160 ","End":"18:28.615","Text":"Here I can just cross this off because this is equal to 0."},{"Start":"18:28.615 ","End":"18:33.970","Text":"Because our function for our magnetic field is independent of y,"},{"Start":"18:33.970 ","End":"18:37.825","Text":"which means that if I take the differential with respect to y, I\u0027ll get 0."},{"Start":"18:37.825 ","End":"18:42.430","Text":"What I\u0027m left with is that negative dBy,"},{"Start":"18:42.430 ","End":"18:46.030","Text":"so the y component of the magnetic field by"},{"Start":"18:46.030 ","End":"18:54.370","Text":"dz is equal to this,"},{"Start":"18:54.370 ","End":"18:57.760","Text":"Rhov_naught."},{"Start":"18:57.760 ","End":"19:03.445","Text":"That means that I have a y component to my magnetic field,"},{"Start":"19:03.445 ","End":"19:06.330","Text":"which is with respect to z."},{"Start":"19:06.330 ","End":"19:09.320","Text":"It has z as a variable."},{"Start":"19:09.320 ","End":"19:15.100","Text":"In other words, my magnetic field has to be in the y direction because this is"},{"Start":"19:15.100 ","End":"19:18.490","Text":"the y component of B. I\u0027m going"},{"Start":"19:18.490 ","End":"19:22.105","Text":"to have some kind of magnetic field which is dependent on z,"},{"Start":"19:22.105 ","End":"19:27.370","Text":"it has a z as a variable and it\u0027s in the y direction,"},{"Start":"19:27.370 ","End":"19:29.050","Text":"which is exactly what we found."},{"Start":"19:29.050 ","End":"19:30.505","Text":"If you go back in the video,"},{"Start":"19:30.505 ","End":"19:33.805","Text":"we see that the magnetic field is in the y direction,"},{"Start":"19:33.805 ","End":"19:39.753","Text":"and at least in-between or insides of the plane,"},{"Start":"19:39.753 ","End":"19:42.205","Text":"it really is dependent on z."},{"Start":"19:42.205 ","End":"19:43.525","Text":"Okay, that\u0027s it."},{"Start":"19:43.525 ","End":"19:45.920","Text":"That\u0027s the end of this lesson."}],"ID":21422},{"Watched":false,"Name":"Exercise 5","Duration":"10m 55s","ChapterTopicVideoID":21342,"CourseChapterTopicPlaylistID":99468,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/21342.jpeg","UploadDate":"2020-04-06T21:34:23.7370000","DurationForVideoObject":"PT10M55S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:07.065","Text":"Hello. In this lesson we\u0027re going to calculate the magnetic field of a coil of length l,"},{"Start":"00:07.065 ","End":"00:11.490","Text":"so this is the length of l,"},{"Start":"00:11.490 ","End":"00:14.775","Text":"that has a current I flowing through."},{"Start":"00:14.775 ","End":"00:17.640","Text":"We\u0027re told that the radius of the coil is a."},{"Start":"00:17.640 ","End":"00:21.180","Text":"That means that this distance over here is"},{"Start":"00:21.180 ","End":"00:27.000","Text":"a and the total number of turns in the coil is n."},{"Start":"00:27.000 ","End":"00:31.980","Text":"First thing is that we can say that the cross-sectional area S of"},{"Start":"00:31.980 ","End":"00:38.020","Text":"the coil is going to be this cross-sectional area of this circle of radius a."},{"Start":"00:38.020 ","End":"00:41.690","Text":"That\u0027s going to be equal to Pi a^2."},{"Start":"00:42.020 ","End":"00:46.325","Text":"Then we can also say that the turn density,"},{"Start":"00:46.325 ","End":"00:52.715","Text":"so the number of turns that we have per unit length, so this is n,"},{"Start":"00:52.715 ","End":"00:56.210","Text":"the turn density is equal to"},{"Start":"00:56.210 ","End":"01:02.150","Text":"the total number of turns in the entire coil divided by the total length,"},{"Start":"01:02.150 ","End":"01:07.655","Text":"so divided by l. Now what we can see is that the current starts over here."},{"Start":"01:07.655 ","End":"01:09.440","Text":"It goes down here,"},{"Start":"01:09.440 ","End":"01:12.575","Text":"up over here and then at this point over here,"},{"Start":"01:12.575 ","End":"01:18.145","Text":"we can see that it\u0027s coming out of the page."},{"Start":"01:18.145 ","End":"01:24.005","Text":"We can draw an x over here symbolizing that it\u0027s coming out of the page."},{"Start":"01:24.005 ","End":"01:28.415","Text":"Then we continue with the coil and we can see that over here at this point,"},{"Start":"01:28.415 ","End":"01:32.330","Text":"the current is going inside the page."},{"Start":"01:32.330 ","End":"01:35.840","Text":"Then again, we follow and here again it\u0027s"},{"Start":"01:35.840 ","End":"01:39.500","Text":"coming out of the page and we keep on following."},{"Start":"01:39.500 ","End":"01:44.580","Text":"Again here it\u0027s coming inside of the page."},{"Start":"01:44.580 ","End":"01:48.360","Text":"What we can do is we can also draw it like this."},{"Start":"01:48.400 ","End":"01:56.240","Text":"Here we have the top over here where the current is coming out of the page"},{"Start":"01:56.240 ","End":"02:04.240","Text":"and here we have the bottom where the current is coming inside from the page."},{"Start":"02:04.240 ","End":"02:07.670","Text":"Now what we want to do is we want to imagine that this coil is"},{"Start":"02:07.670 ","End":"02:10.670","Text":"an infinite Lelong or an infinite coil."},{"Start":"02:10.670 ","End":"02:12.200","Text":"What does that mean?"},{"Start":"02:12.200 ","End":"02:14.810","Text":"That means that the turn density is very,"},{"Start":"02:14.810 ","End":"02:16.850","Text":"very great or in other words,"},{"Start":"02:16.850 ","End":"02:22.375","Text":"the number of turns is much greater than the length of the wire."},{"Start":"02:22.375 ","End":"02:26.060","Text":"In other words, we can say that we have lots and lots of"},{"Start":"02:26.060 ","End":"02:31.940","Text":"turns that are very, very close together."},{"Start":"02:31.940 ","End":"02:37.820","Text":"Then we can see this as an infinite wire with lots of different points where"},{"Start":"02:37.820 ","End":"02:43.975","Text":"we have the current going out of the page and the current going into the page."},{"Start":"02:43.975 ","End":"02:49.070","Text":"Now what we can see is that we have symmetry in the z direction."},{"Start":"02:49.070 ","End":"02:55.254","Text":"Let\u0027s say that this is the z direction going right through our coil."},{"Start":"02:55.254 ","End":"02:56.990","Text":"Because this is infinite,"},{"Start":"02:56.990 ","End":"03:02.300","Text":"we have the symmetry in the z direction and because we\u0027re dealing with coils,"},{"Start":"03:02.300 ","End":"03:08.290","Text":"so loops, then we also are going to have symmetry in the Theta direction."},{"Start":"03:08.290 ","End":"03:11.270","Text":"Now, if we use the right-hand rule,"},{"Start":"03:11.270 ","End":"03:17.990","Text":"we can see that the magnetic field is going in this direction,"},{"Start":"03:17.990 ","End":"03:21.680","Text":"in the negative z direction."},{"Start":"03:21.680 ","End":"03:28.680","Text":"Now what I\u0027m going to do is I\u0027m going to build my Ampere\u0027s loop on this same line."},{"Start":"03:28.720 ","End":"03:32.585","Text":"It just goes down like so,"},{"Start":"03:32.585 ","End":"03:36.775","Text":"and it finishes over here at infinity."},{"Start":"03:36.775 ","End":"03:42.620","Text":"What I want is I always want my Ampere\u0027s loop to look similar to the magnetic field."},{"Start":"03:42.620 ","End":"03:48.110","Text":"I have it over here on the lines of the magnetic field where"},{"Start":"03:48.110 ","End":"03:53.780","Text":"they\u0027re parallel and then they come out of the coil like so."},{"Start":"03:53.780 ","End":"04:01.105","Text":"We\u0027ve already seen that Ampere\u0027s law is equal to the closed loop integral of"},{"Start":"04:01.105 ","End":"04:10.590","Text":"B.dl and this is equal to Mu naught multiplied by I_in,"},{"Start":"04:10.590 ","End":"04:14.530","Text":"the current flowing through this loop."},{"Start":"04:14.750 ","End":"04:18.200","Text":"Let\u0027s first do the left side of this equation."},{"Start":"04:18.200 ","End":"04:21.770","Text":"We see that we have this magnetic field B,"},{"Start":"04:21.770 ","End":"04:26.090","Text":"which is parallel to the loop,"},{"Start":"04:26.090 ","End":"04:29.635","Text":"and it\u0027s going in the negative z direction."},{"Start":"04:29.635 ","End":"04:34.324","Text":"Let\u0027s say that the direction of the loop is also the negative z direction."},{"Start":"04:34.324 ","End":"04:37.640","Text":"We have B multiplied by,"},{"Start":"04:37.640 ","End":"04:41.330","Text":"let\u0027s say that the length of this loop is L,"},{"Start":"04:41.330 ","End":"04:43.160","Text":"so the length of the coil."},{"Start":"04:43.160 ","End":"04:47.270","Text":"It\u0027d be B.L, then the magnetic field along with these sides,"},{"Start":"04:47.270 ","End":"04:51.460","Text":"which are perpendicular to the direction of the magnetic field."},{"Start":"04:51.460 ","End":"04:54.940","Text":"The magnetic field is like so in the z direction."},{"Start":"04:54.940 ","End":"05:00.159","Text":"We can see that this side of the loop is perpendicular to the z direction."},{"Start":"05:00.159 ","End":"05:02.845","Text":"When we do the dot-product, we\u0027ll get zero."},{"Start":"05:02.845 ","End":"05:06.640","Text":"Of course the side that\u0027s way down here is an infinity."},{"Start":"05:06.640 ","End":"05:10.540","Text":"The magnetic field over there is also equal to zero."},{"Start":"05:10.540 ","End":"05:12.250","Text":"From this side of the equation,"},{"Start":"05:12.250 ","End":"05:14.520","Text":"we\u0027re just left with B.L."},{"Start":"05:14.520 ","End":"05:19.665","Text":"This is equal to Mu naught multiplied by I_in."},{"Start":"05:19.665 ","End":"05:24.190","Text":"Now our question is, what is I_in?"},{"Start":"05:25.520 ","End":"05:33.110","Text":"As we can see, each of these circles represents 1 side of"},{"Start":"05:33.110 ","End":"05:40.145","Text":"the wire where our current is coming out at that exact point or going in."},{"Start":"05:40.145 ","End":"05:43.955","Text":"In that case, we can see that we have current I."},{"Start":"05:43.955 ","End":"05:46.820","Text":"Here we have I, I, I."},{"Start":"05:46.820 ","End":"05:50.325","Text":"We have this current I going through each."},{"Start":"05:50.325 ","End":"05:54.320","Text":"Now what we want to do is count how many of these turns"},{"Start":"05:54.320 ","End":"05:58.730","Text":"or how many of these points of currents there are."},{"Start":"05:58.730 ","End":"06:00.920","Text":"We\u0027re going to use this n,"},{"Start":"06:00.920 ","End":"06:03.235","Text":"so this is the turns density,"},{"Start":"06:03.235 ","End":"06:08.750","Text":"n. Then we want to multiply it by the length,"},{"Start":"06:08.750 ","End":"06:11.225","Text":"so the length of the coil,"},{"Start":"06:11.225 ","End":"06:16.635","Text":"which is equal to L. Then we can plug this in."},{"Start":"06:16.635 ","End":"06:26.690","Text":"Therefore what we get is that B.L is equal to Mu naught multiplied by In. That\u0027s InL."},{"Start":"06:26.690 ","End":"06:33.610","Text":"Then we can divide both sides by L and therefore we get that the magnetic field is equal"},{"Start":"06:33.610 ","End":"06:41.390","Text":"to Mu naught I multiplied by n, the turn density."},{"Start":"06:42.470 ","End":"06:45.840","Text":"This is in the z direction."},{"Start":"06:45.840 ","End":"06:48.970","Text":"This is not just the answer to this question,"},{"Start":"06:48.970 ","End":"06:53.410","Text":"but this is also the equation for the magnetic field of a coil."},{"Start":"06:53.410 ","End":"06:56.620","Text":"You can write this in your equation sheets."},{"Start":"06:56.620 ","End":"07:01.100","Text":"Just remember that this lowercase n is the turns density and"},{"Start":"07:01.100 ","End":"07:06.400","Text":"that means the number of turns divided by the length of the coil."},{"Start":"07:06.400 ","End":"07:13.550","Text":"The next thing to notice is that it is independent of where you are within the coils."},{"Start":"07:13.550 ","End":"07:18.155","Text":"If you\u0027re located over here, or here,"},{"Start":"07:18.155 ","End":"07:19.835","Text":"or here, or here,"},{"Start":"07:19.835 ","End":"07:24.335","Text":"you\u0027re going to have the exact same magnetic field."},{"Start":"07:24.335 ","End":"07:27.980","Text":"Sorry, just specifically over here because we define"},{"Start":"07:27.980 ","End":"07:32.120","Text":"the z direction in the opposite direction to the magnetic fields,"},{"Start":"07:32.120 ","End":"07:34.595","Text":"so you can add a minus over here."},{"Start":"07:34.595 ","End":"07:39.710","Text":"But if we would have just defined the z direction in the leftward direction,"},{"Start":"07:39.710 ","End":"07:42.395","Text":"which is of course an arbitrary decision,"},{"Start":"07:42.395 ","End":"07:44.690","Text":"then we would have gotten it in the z direction."},{"Start":"07:44.690 ","End":"07:49.400","Text":"Just note that. As we said before,"},{"Start":"07:49.400 ","End":"07:57.020","Text":"using the right-hand rule is 1 way to calculate the direction of the magnetic fields,"},{"Start":"07:57.020 ","End":"08:00.080","Text":"but another way is to use it\u0027s rotor."},{"Start":"08:00.080 ","End":"08:02.815","Text":"Let\u0027s just scroll down here."},{"Start":"08:02.815 ","End":"08:06.320","Text":"In other words, we also spoke about this in a previous lesson."},{"Start":"08:06.320 ","End":"08:12.840","Text":"We have the rotor of the magnetic field is equal to Mu naught multiplied by j,"},{"Start":"08:13.330 ","End":"08:17.360","Text":"which is the direction of the current."},{"Start":"08:17.360 ","End":"08:24.655","Text":"Over here, we can see that the direction of the current is in the Theta direction."},{"Start":"08:24.655 ","End":"08:28.350","Text":"We just want the Theta component of"},{"Start":"08:28.350 ","End":"08:32.235","Text":"our j and of course that\u0027s going to be in the Theta direction."},{"Start":"08:32.235 ","End":"08:34.310","Text":"Now what we want to do is we want to write out"},{"Start":"08:34.310 ","End":"08:38.700","Text":"the equation for the rotor of the magnetic field."},{"Start":"08:38.930 ","End":"08:44.090","Text":"This is what will probably be in your equation sheet."},{"Start":"08:44.090 ","End":"08:47.330","Text":"You\u0027ll have the rotor of some vector a."},{"Start":"08:47.330 ","End":"08:51.210","Text":"Here our vector a is just the magnetic field."},{"Start":"08:51.460 ","End":"08:53.870","Text":"As we\u0027ve already said,"},{"Start":"08:53.870 ","End":"09:00.520","Text":"our current or our j component that we\u0027re using is in the Theta direction."},{"Start":"09:00.520 ","End":"09:05.280","Text":"We can use r_Theta and z or Rho Phi and z,"},{"Start":"09:05.280 ","End":"09:06.750","Text":"it doesn\u0027t make a difference."},{"Start":"09:06.750 ","End":"09:09.350","Text":"We\u0027re of course, using cylindrical coordinates,"},{"Start":"09:09.350 ","End":"09:12.425","Text":"because the coil is in the shape of a cylinder."},{"Start":"09:12.425 ","End":"09:17.785","Text":"Theta and Phi in these equations are of course interchangeable."},{"Start":"09:17.785 ","End":"09:22.204","Text":"We just want the j component in the Theta direction."},{"Start":"09:22.204 ","End":"09:26.600","Text":"Here we have something in the z direction so we can cross off everything."},{"Start":"09:26.600 ","End":"09:28.790","Text":"In here we have something in the Rho direction,"},{"Start":"09:28.790 ","End":"09:30.620","Text":"so we can cross off everything."},{"Start":"09:30.620 ","End":"09:32.270","Text":"Then as we said before,"},{"Start":"09:32.270 ","End":"09:37.340","Text":"we have symmetry in the z direction and also in the Theta direction."},{"Start":"09:37.340 ","End":"09:45.575","Text":"If you wanted to, you could have crossed off the components alone or just according to,"},{"Start":"09:45.575 ","End":"09:48.455","Text":"if we\u0027re taking the derivative with respect to Theta,"},{"Start":"09:48.455 ","End":"09:50.630","Text":"because we have symmetry in the theta direction,"},{"Start":"09:50.630 ","End":"09:53.135","Text":"this would have crossed out because of that."},{"Start":"09:53.135 ","End":"09:56.570","Text":"We also have symmetry in the z direction, so this,"},{"Start":"09:56.570 ","End":"10:00.260","Text":"let\u0027s say it would have crossed out because of that and of course so will this,"},{"Start":"10:00.260 ","End":"10:02.720","Text":"because if we take d by dz,"},{"Start":"10:02.720 ","End":"10:05.870","Text":"when we have symmetry in the z direction, this will cross out."},{"Start":"10:05.870 ","End":"10:13.040","Text":"Then what we\u0027re left with is our vector with the z components."},{"Start":"10:13.040 ","End":"10:16.820","Text":"What we can see is that we\u0027re going to have a vector,"},{"Start":"10:16.820 ","End":"10:19.730","Text":"which here is the magnetic field and"},{"Start":"10:19.730 ","End":"10:23.120","Text":"we\u0027re only going to be speaking about its z components."},{"Start":"10:23.120 ","End":"10:28.250","Text":"In that case, it\u0027s going to be a magnetic field of some magnitude,"},{"Start":"10:28.250 ","End":"10:30.049","Text":"but in the z direction,"},{"Start":"10:30.049 ","End":"10:32.920","Text":"which is what we got in the z direction."},{"Start":"10:32.920 ","End":"10:36.560","Text":"That is of course another way by using"},{"Start":"10:36.560 ","End":"10:41.795","Text":"the rotor to see in what direction the magnetic field will be in."},{"Start":"10:41.795 ","End":"10:44.645","Text":"Remember, if you use the rotor way,"},{"Start":"10:44.645 ","End":"10:48.050","Text":"you\u0027re looking at what letter is over here."},{"Start":"10:48.050 ","End":"10:51.300","Text":"Here, we are left with z."},{"Start":"10:52.030 ","End":"10:55.440","Text":"That\u0027s the end of this lesson."}],"ID":21423},{"Watched":false,"Name":"Toroid","Duration":"6m 47s","ChapterTopicVideoID":21343,"CourseChapterTopicPlaylistID":99468,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/21343.jpeg","UploadDate":"2020-04-06T21:35:22.3830000","DurationForVideoObject":"PT6M47S","Description":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 learning about a toroid."},{"Start":"00:04.350 ","End":"00:09.675","Text":"A toroid is like a solenoid that is bent into a circle."},{"Start":"00:09.675 ","End":"00:14.820","Text":"We just dealt with a coil where we had"},{"Start":"00:14.820 ","End":"00:21.210","Text":"this type of situation and we had a current flowing through it I,"},{"Start":"00:21.210 ","End":"00:26.295","Text":"and then we saw that we\u0027re going to have a magnetic field"},{"Start":"00:26.295 ","End":"00:32.040","Text":"flowing through like so through the center of the coil."},{"Start":"00:32.040 ","End":"00:37.155","Text":"What happens if we then bend this coil so that"},{"Start":"00:37.155 ","End":"00:43.205","Text":"this end over here attaches to this end over here?"},{"Start":"00:43.205 ","End":"00:49.475","Text":"What we\u0027ll have is a circular coil or something that looks exactly like this."},{"Start":"00:49.475 ","End":"00:53.405","Text":"In that case, what will the magnetic field be doing?"},{"Start":"00:53.405 ","End":"00:58.620","Text":"Itself is going to bend in the direction of the coil,"},{"Start":"00:58.620 ","End":"01:05.945","Text":"so we\u0027re going to have a magnetic field traveling in this circular form."},{"Start":"01:05.945 ","End":"01:11.240","Text":"Here, we can see that we have a current going in in this direction,"},{"Start":"01:11.240 ","End":"01:15.395","Text":"which means that at this point over here,"},{"Start":"01:15.395 ","End":"01:20.510","Text":"the current is going inside the page and then here on the other side,"},{"Start":"01:20.510 ","End":"01:22.375","Text":"it\u0027s coming out of the page."},{"Start":"01:22.375 ","End":"01:23.680","Text":"Then it keeps traveling,"},{"Start":"01:23.680 ","End":"01:25.480","Text":"so here it goes inside the page,"},{"Start":"01:25.480 ","End":"01:26.935","Text":"so I\u0027ll draw it over here,"},{"Start":"01:26.935 ","End":"01:29.975","Text":"inside and then over here,"},{"Start":"01:29.975 ","End":"01:31.400","Text":"it\u0027s coming out of the page."},{"Start":"01:31.400 ","End":"01:36.470","Text":"This carries on and on and on until we see something like so."},{"Start":"01:36.470 ","End":"01:40.220","Text":"We have here, the current is coming out of"},{"Start":"01:40.220 ","End":"01:45.425","Text":"the page on the outskirts of the toroid and inside the toroid,"},{"Start":"01:45.425 ","End":"01:49.084","Text":"right through over here on the radius,"},{"Start":"01:49.084 ","End":"01:55.020","Text":"we can see that the current is going in the direction inside the page."},{"Start":"01:55.400 ","End":"02:00.695","Text":"What we see is we have this magnetic field going in a circular direction,"},{"Start":"02:00.695 ","End":"02:06.010","Text":"or in other words, we have a magnetic field in the Theta direction."},{"Start":"02:06.010 ","End":"02:09.185","Text":"Before when we were dealing with a simple coil,"},{"Start":"02:09.185 ","End":"02:11.629","Text":"the magnetic field was in the z direction,"},{"Start":"02:11.629 ","End":"02:13.460","Text":"and now it\u0027s in the Theta direction,"},{"Start":"02:13.460 ","End":"02:15.900","Text":"and we want to calculate it."},{"Start":"02:16.160 ","End":"02:19.305","Text":"From the center of the toroid,"},{"Start":"02:19.305 ","End":"02:24.745","Text":"we have this radius r to the center of the coil that we just bent,"},{"Start":"02:24.745 ","End":"02:30.920","Text":"and over here, we have this magnetic field going like so in this direction."},{"Start":"02:30.920 ","End":"02:32.360","Text":"Okay, so ignore the sketch,"},{"Start":"02:32.360 ","End":"02:34.100","Text":"it\u0027s in the opposite direction."},{"Start":"02:34.100 ","End":"02:39.440","Text":"What we want to do is we want to calculate the magnetic field at this point,"},{"Start":"02:39.440 ","End":"02:44.980","Text":"so the magnetic field at point r. We"},{"Start":"02:44.980 ","End":"02:51.260","Text":"already have this Ampere loop over here going along the magnetic field lines,"},{"Start":"02:51.260 ","End":"02:55.775","Text":"so it\u0027s going to be the circular loop Ampere\u0027s loop."},{"Start":"02:55.775 ","End":"02:58.370","Text":"As we know, Ampere\u0027s law,"},{"Start":"02:58.370 ","End":"03:02.930","Text":"we have the closed loop integral of B.dl,"},{"Start":"03:02.930 ","End":"03:10.270","Text":"which is equal to Mu_0 multiplied by I_in."},{"Start":"03:10.270 ","End":"03:12.060","Text":"What is our B.dl?"},{"Start":"03:12.060 ","End":"03:17.195","Text":"It\u0027s just our magnetic fields B multiplied by the length of the loop,"},{"Start":"03:17.195 ","End":"03:20.750","Text":"which is just going to be the circumference of the circle,"},{"Start":"03:20.750 ","End":"03:22.860","Text":"which is just 2Pir,"},{"Start":"03:24.080 ","End":"03:29.260","Text":"and this is equal to Mu_0 multiplied by I_in."},{"Start":"03:29.780 ","End":"03:36.940","Text":"The I_in is the total current which is located within my Ampere\u0027s loop,"},{"Start":"03:36.940 ","End":"03:39.370","Text":"which we said over here, is a circle,"},{"Start":"03:39.370 ","End":"03:43.220","Text":"so it\u0027s all the current that\u0027s located here."},{"Start":"03:43.490 ","End":"03:50.305","Text":"The total current is just over here where we see these circles with xs."},{"Start":"03:50.305 ","End":"03:54.220","Text":"I can say that this is equal to Mu_0 multiplied"},{"Start":"03:54.220 ","End":"03:58.293","Text":"by the total circles with xs, which, of course,"},{"Start":"03:58.293 ","End":"04:02.750","Text":"corresponds to the total turns in the wire,"},{"Start":"04:02.750 ","End":"04:04.285","Text":"so that\u0027s equal to,"},{"Start":"04:04.285 ","End":"04:06.415","Text":"as we said in the previous lesson,"},{"Start":"04:06.415 ","End":"04:09.995","Text":"N and, of course,"},{"Start":"04:09.995 ","End":"04:11.765","Text":"multiplied by the current,"},{"Start":"04:11.765 ","End":"04:16.005","Text":"which is I, and that\u0027s it."},{"Start":"04:16.005 ","End":"04:19.805","Text":"Now, we can isolate out our magnetic field B,"},{"Start":"04:19.805 ","End":"04:26.295","Text":"so we get that it\u0027s equal to Mu_0 NI divide it by 2Pir,"},{"Start":"04:26.295 ","End":"04:28.970","Text":"and, of course, as we said before,"},{"Start":"04:28.970 ","End":"04:32.385","Text":"it\u0027s in the Theta direction."},{"Start":"04:32.385 ","End":"04:35.010","Text":"What we can see is that this time,"},{"Start":"04:35.010 ","End":"04:38.135","Text":"our magnetic field is dependent on I."},{"Start":"04:38.135 ","End":"04:41.075","Text":"When we were dealing with the magnetic field of a coil,"},{"Start":"04:41.075 ","End":"04:47.480","Text":"we saw that the magnetic field was independent of where we were within the coil."},{"Start":"04:47.480 ","End":"04:50.585","Text":"We could be located over here,"},{"Start":"04:50.585 ","End":"04:53.945","Text":"or over here, or anywhere we wanted,"},{"Start":"04:53.945 ","End":"04:55.625","Text":"and we\u0027d have the same magnetic field."},{"Start":"04:55.625 ","End":"04:59.660","Text":"Here, however, it is dependent on r and we can see that"},{"Start":"04:59.660 ","End":"05:04.780","Text":"as we get further away from the center of the toroid,"},{"Start":"05:04.780 ","End":"05:07.265","Text":"so as r increases,"},{"Start":"05:07.265 ","End":"05:11.555","Text":"our magnetic field is going to decrease."},{"Start":"05:11.555 ","End":"05:17.370","Text":"Our greatest value for r is going to be when we\u0027re located somewhere over here."},{"Start":"05:17.900 ","End":"05:24.070","Text":"Of course, this is the magnetic field when we\u0027re located between the radius R_in,"},{"Start":"05:24.070 ","End":"05:27.185","Text":"so that\u0027s over here,"},{"Start":"05:27.185 ","End":"05:32.420","Text":"and when r is between the outer radius,"},{"Start":"05:32.420 ","End":"05:35.630","Text":"so R_out, which is this line over here,"},{"Start":"05:35.630 ","End":"05:37.775","Text":"so that\u0027s when we\u0027re located in this area."},{"Start":"05:37.775 ","End":"05:42.450","Text":"If, however, we took a radius over here,"},{"Start":"05:42.450 ","End":"05:47.630","Text":"so, of course, here we have no current."},{"Start":"05:47.630 ","End":"05:51.860","Text":"If r was smaller than R_in,"},{"Start":"05:51.860 ","End":"05:56.885","Text":"we would have a magnetic field equal to 0 because there\u0027s no current over here."},{"Start":"05:56.885 ","End":"06:00.799","Text":"Similarly, if we were to take an Ampere\u0027s loop over here,"},{"Start":"06:00.799 ","End":"06:02.400","Text":"and this was our r,"},{"Start":"06:02.400 ","End":"06:06.765","Text":"so if r was greater than R_out,"},{"Start":"06:06.765 ","End":"06:09.045","Text":"then here, the total current,"},{"Start":"06:09.045 ","End":"06:15.620","Text":"so we have current coming out of the page and current going into the page,"},{"Start":"06:15.620 ","End":"06:19.820","Text":"and the total number of current going out of"},{"Start":"06:19.820 ","End":"06:24.215","Text":"the page is equal to the total number of currents going into the page,"},{"Start":"06:24.215 ","End":"06:26.435","Text":"so they will cancel each other out."},{"Start":"06:26.435 ","End":"06:27.770","Text":"Also, over here,"},{"Start":"06:27.770 ","End":"06:33.510","Text":"the total magnetic field is equal to 0."},{"Start":"06:35.000 ","End":"06:39.860","Text":"This is the equation for the magnetic field of"},{"Start":"06:39.860 ","End":"06:45.125","Text":"a toroid and please write this in your equation sheets."},{"Start":"06:45.125 ","End":"06:48.660","Text":"Okay. That\u0027s the end of this lesson."}],"ID":21424},{"Watched":false,"Name":"Exercise 6","Duration":"18m 28s","ChapterTopicVideoID":21344,"CourseChapterTopicPlaylistID":99468,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/21344.jpeg","UploadDate":"2020-04-06T21:40:53.2100000","DurationForVideoObject":"PT18M28S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.875","Text":"Hello. In this lesson,"},{"Start":"00:01.875 ","End":"00:04.560","Text":"we\u0027re going to be answering the following question."},{"Start":"00:04.560 ","End":"00:11.250","Text":"An infinite plane of width d has a changing charge density per unit volume,"},{"Start":"00:11.250 ","End":"00:15.390","Text":"given by the charge density per unit volume is equal to"},{"Start":"00:15.390 ","End":"00:20.865","Text":"an initial charge density multiplied by e to the power of alpha z,"},{"Start":"00:20.865 ","End":"00:23.425","Text":"where alpha is a constant."},{"Start":"00:23.425 ","End":"00:29.750","Text":"The plane is parallel to the x-y plane and the origin is at its center,"},{"Start":"00:29.750 ","End":"00:31.370","Text":"so it d divided by 2."},{"Start":"00:31.370 ","End":"00:34.535","Text":"The plane begins moving in the x-direction,"},{"Start":"00:34.535 ","End":"00:36.620","Text":"so out of the page,"},{"Start":"00:36.620 ","End":"00:38.915","Text":"at a velocity of v naught,"},{"Start":"00:38.915 ","End":"00:44.580","Text":"calculate the magnetic field both inside and outside the plane."},{"Start":"00:44.690 ","End":"00:50.750","Text":"In this question, we have a charged density per unit volume,"},{"Start":"00:50.750 ","End":"00:53.981","Text":"which is different at every single point."},{"Start":"00:53.981 ","End":"01:00.590","Text":"So if we look at a graph and this is a graph of e to the power of alpha z."},{"Start":"01:00.590 ","End":"01:04.029","Text":"So take e to the power of z, just alpha,"},{"Start":"01:04.029 ","End":"01:09.665","Text":"will just change it\u0027s sharpness depending on its value it will flatten out the function"},{"Start":"01:09.665 ","End":"01:16.520","Text":"or narrow it down but what we have is that when z is greater than 0."},{"Start":"01:16.520 ","End":"01:19.915","Text":"We get a function that looks like this,"},{"Start":"01:19.915 ","End":"01:22.885","Text":"and when z is less than 0,"},{"Start":"01:22.885 ","End":"01:26.575","Text":"our function approaches 0 value."},{"Start":"01:26.575 ","End":"01:34.840","Text":"We can see that on either side of this line of symmetry so our y-axis,"},{"Start":"01:34.840 ","End":"01:38.240","Text":"we get above the y-axis,"},{"Start":"01:38.240 ","End":"01:47.395","Text":"our charge is approaching infinity and below the y-axis the charge is approaching 0."},{"Start":"01:47.395 ","End":"01:53.415","Text":"Given that, if I was to draw an unpaired loop,"},{"Start":"01:53.415 ","End":"02:00.590","Text":"like so and then let\u0027s say that here I have my magnetic field."},{"Start":"02:00.590 ","End":"02:04.310","Text":"I\u0027m clearly going to have 1 magnetic field at the top here"},{"Start":"02:04.310 ","End":"02:09.275","Text":"and a completely different magnetic field down here."},{"Start":"02:09.275 ","End":"02:16.310","Text":"Then in that case, if I do the closed loop integral on b.dl,"},{"Start":"02:16.310 ","End":"02:17.705","Text":"as we\u0027ve seen before."},{"Start":"02:17.705 ","End":"02:22.130","Text":"Then what I\u0027ll have is b_1 multiplied by,"},{"Start":"02:22.130 ","End":"02:29.945","Text":"let\u0027s say the length of that edge is L plus b_2 multiplied by the same length."},{"Start":"02:29.945 ","End":"02:32.390","Text":"Then what I have is I have 2 unknowns,"},{"Start":"02:32.390 ","End":"02:34.100","Text":"both b_1 and b_2,"},{"Start":"02:34.100 ","End":"02:40.260","Text":"and that means that I cannot use this rule or this equation."},{"Start":"02:40.600 ","End":"02:43.760","Text":"What do we do in this type of case?"},{"Start":"02:43.760 ","End":"02:50.330","Text":"So what do we do is we split up our thick plane into lots of thin plane."},{"Start":"02:50.330 ","End":"02:57.015","Text":"Let\u0027s erase all of this. We split this up."},{"Start":"02:57.015 ","End":"03:01.280","Text":"Let\u0027s say I take a thin plane,"},{"Start":"03:01.280 ","End":"03:06.485","Text":"like so and let\u0027s say that its width is"},{"Start":"03:06.485 ","End":"03:12.140","Text":"dz tag and that it is"},{"Start":"03:12.140 ","End":"03:19.395","Text":"located a height of z tag above the 0."},{"Start":"03:19.395 ","End":"03:21.080","Text":"What we\u0027re trying to do is we\u0027re trying to"},{"Start":"03:21.080 ","End":"03:24.695","Text":"find the magnetic field at this point over here."},{"Start":"03:24.695 ","End":"03:28.570","Text":"Let\u0027s just say that this point over here is at a height set."},{"Start":"03:28.570 ","End":"03:32.195","Text":"That\u0027s why I did these head tag and said tag."},{"Start":"03:32.195 ","End":"03:39.790","Text":"Now, all we\u0027re doing is we\u0027re calculating the magnetic field just for this over here."},{"Start":"03:39.790 ","End":"03:43.575","Text":"Just for this thin plane,"},{"Start":"03:43.575 ","End":"03:46.820","Text":"so what we know is that the plane is"},{"Start":"03:46.820 ","End":"03:51.335","Text":"moving towards us so out of the plane in the x-direction,"},{"Start":"03:51.335 ","End":"03:57.980","Text":"which means that the current is coming out in our direction, like so."},{"Start":"03:57.980 ","End":"04:06.480","Text":"Now what we\u0027re going to do is we\u0027re going to build our Ampere\u0027s loop. Like so."},{"Start":"04:06.480 ","End":"04:11.630","Text":"Where, of course, just taking into account this plane,"},{"Start":"04:11.630 ","End":"04:14.495","Text":"so ignore the thick plain,"},{"Start":"04:14.495 ","End":"04:17.645","Text":"ignore these lines were just looking at this."},{"Start":"04:17.645 ","End":"04:20.165","Text":"So I\u0027ll just draw it in red."},{"Start":"04:20.165 ","End":"04:27.100","Text":"We\u0027re just looking at this thin plane."},{"Start":"04:27.100 ","End":"04:30.930","Text":"Now here we\u0027re going to have a magnetic field"},{"Start":"04:30.930 ","End":"04:34.595","Text":"and I know that if the current is coming out of the page,"},{"Start":"04:34.595 ","End":"04:38.720","Text":"so my magnetic field on top is going to be in this direction and"},{"Start":"04:38.720 ","End":"04:44.291","Text":"my magnetic fields at the bottom is going to be in the opposite direction like so."},{"Start":"04:44.291 ","End":"04:48.500","Text":"Let\u0027s say that the length of this is"},{"Start":"04:48.500 ","End":"04:56.870","Text":"l. Now we\u0027re going to use Ampere\u0027s law just for this thin plane that we chose."},{"Start":"04:56.870 ","End":"05:02.440","Text":"We have the closed loop integral of b.dl,"},{"Start":"05:02.440 ","End":"05:05.695","Text":"which is going to be equal 2."},{"Start":"05:05.695 ","End":"05:08.785","Text":"Because we\u0027re dealing with a thin plane,"},{"Start":"05:08.785 ","End":"05:12.830","Text":"so it\u0027s just like the regular thin plane that we saw before."},{"Start":"05:12.830 ","End":"05:18.050","Text":"We know that the magnetic field over here is parallel and is equal to"},{"Start":"05:18.050 ","End":"05:23.450","Text":"b multiplied by l. Then over here on these 2 sides,"},{"Start":"05:23.450 ","End":"05:26.225","Text":"the magnetic field is perpendicular to the sides."},{"Start":"05:26.225 ","End":"05:29.300","Text":"When we do b.dl will get 0."},{"Start":"05:29.300 ","End":"05:32.330","Text":"Then again here in the direction of the loop,"},{"Start":"05:32.330 ","End":"05:37.650","Text":"we have b multiplied by l. So what we have is 2,"},{"Start":"05:37.650 ","End":"05:39.615","Text":"b multiplied by l,"},{"Start":"05:39.615 ","End":"05:44.160","Text":"b multiplied by l plus b multiplied by l and that\u0027s it."},{"Start":"05:44.160 ","End":"05:48.030","Text":"This is equal 2 mu naught multiplied by I"},{"Start":"05:48.030 ","End":"05:53.215","Text":"n. Now what we want to do is we want to find out what i n is equal to."},{"Start":"05:53.215 ","End":"06:02.865","Text":"We\u0027re going to use the equation that it\u0027s equal to the integral of j.ds. Let\u0027s see."},{"Start":"06:02.865 ","End":"06:04.250","Text":"What is our j?"},{"Start":"06:04.250 ","End":"06:07.255","Text":"So our j is equal to"},{"Start":"06:07.255 ","End":"06:16.350","Text":"the charge density multiplied by the velocity that it\u0027s traveling in."},{"Start":"06:16.350 ","End":"06:21.505","Text":"I charged density is equal to rho naught e to the power of alpha,"},{"Start":"06:21.505 ","End":"06:29.310","Text":"z and our velocity is v naught in the x direction."},{"Start":"06:29.310 ","End":"06:32.909","Text":"This is our j so therefore,"},{"Start":"06:32.909 ","End":"06:39.590","Text":"we get that our i n is equal to the integral of j ds. What is our ds?"},{"Start":"06:39.590 ","End":"06:43.070","Text":"We can see that a plane is over here."},{"Start":"06:43.070 ","End":"06:45.230","Text":"The line is in the y-direction,"},{"Start":"06:45.230 ","End":"06:47.910","Text":"and it has a small width of dz."},{"Start":"06:49.550 ","End":"06:54.000","Text":"We have this length l in the y direction."},{"Start":"06:54.000 ","End":"07:02.900","Text":"We\u0027re going from 0 to l. Then we have a j,"},{"Start":"07:02.900 ","End":"07:06.725","Text":"so rho naught e to the power of alpha z,"},{"Start":"07:06.725 ","End":"07:13.745","Text":"v naught and then we\u0027re integrating with respect to dy."},{"Start":"07:13.745 ","End":"07:17.060","Text":"Of course, we have to take in this width and"},{"Start":"07:17.060 ","End":"07:19.760","Text":"because this width is so small, we could just already,"},{"Start":"07:19.760 ","End":"07:22.130","Text":"instead of multiplying from 0 until dz,"},{"Start":"07:22.130 ","End":"07:26.048","Text":"we\u0027re not integrating from 0 into dz."},{"Start":"07:26.048 ","End":"07:31.295","Text":"We can just add in or just multiply over here with dz tag."},{"Start":"07:31.295 ","End":"07:34.085","Text":"Then of course, this is just dy."},{"Start":"07:34.085 ","End":"07:37.540","Text":"We don\u0027t have any y variables in here,"},{"Start":"07:37.540 ","End":"07:44.195","Text":"so we just multiply everything by l. We have rho naught e to the power of alpha z,"},{"Start":"07:44.195 ","End":"07:48.710","Text":"v naught and then when we integrate with respect to y,"},{"Start":"07:48.710 ","End":"07:53.845","Text":"we\u0027re just multiplying by l. Then of course, dz tag."},{"Start":"07:53.845 ","End":"07:57.980","Text":"I could have done a double integral here from 0 until dz tag,"},{"Start":"07:57.980 ","End":"08:01.790","Text":"and we\u0027ve gotten the exact same thing."},{"Start":"08:01.790 ","End":"08:06.180","Text":"Now we can plug this into this equation."},{"Start":"08:06.180 ","End":"08:14.630","Text":"What we have is 2bl is equal to mu naught multiplied by i n,"},{"Start":"08:14.630 ","End":"08:18.125","Text":"which is rho naught e to the power of alpha z,"},{"Start":"08:18.125 ","End":"08:22.110","Text":"v naught l dz tag."},{"Start":"08:22.110 ","End":"08:26.270","Text":"The l\u0027s can cancel out and then we can isolate out the b"},{"Start":"08:26.270 ","End":"08:30.410","Text":"and we get that b is equal to half mu naught,"},{"Start":"08:30.410 ","End":"08:39.580","Text":"rho naught, v naught e to the power of alpha z, dz tag."},{"Start":"08:39.580 ","End":"08:42.495","Text":"This is our b,"},{"Start":"08:42.495 ","End":"08:47.190","Text":"and in that case this is our db and similarly,"},{"Start":"08:47.190 ","End":"08:49.990","Text":"this is going to be our dIn."},{"Start":"08:49.990 ","End":"08:55.880","Text":"Why is that? Because we have this dz over here because we have a differential on 1 side."},{"Start":"08:55.880 ","End":"08:59.095","Text":"We have to make it the same on the other side."},{"Start":"08:59.095 ","End":"09:02.875","Text":"If we\u0027re left with dz, this becomes dIn."},{"Start":"09:02.875 ","End":"09:05.915","Text":"Here again, we\u0027re left with this differential over here,"},{"Start":"09:05.915 ","End":"09:10.050","Text":"dz tag, so this becomes db."},{"Start":"09:10.050 ","End":"09:12.410","Text":"Remember if it\u0027s on 1 side,"},{"Start":"09:12.410 ","End":"09:15.480","Text":"it has to also appear on the other side."},{"Start":"09:16.750 ","End":"09:20.755","Text":"Now what we want to do is we want to integrate on b."},{"Start":"09:20.755 ","End":"09:24.850","Text":"The first thing is we can say that db is a vector,"},{"Start":"09:24.850 ","End":"09:26.920","Text":"so it has directions."},{"Start":"09:26.920 ","End":"09:29.740","Text":"We multiply all of this."},{"Start":"09:29.740 ","End":"09:33.690","Text":"If we\u0027re above the piece,"},{"Start":"09:33.690 ","End":"09:35.890","Text":"above our thin strip,"},{"Start":"09:35.890 ","End":"09:39.805","Text":"this is going to be in the negative y direction, as we saw,"},{"Start":"09:39.805 ","End":"09:42.520","Text":"the b is going in the leftward direction,"},{"Start":"09:42.520 ","End":"09:45.677","Text":"so this is in the negative y-direction."},{"Start":"09:45.677 ","End":"09:47.290","Text":"If we\u0027re below the strip,"},{"Start":"09:47.290 ","End":"09:54.633","Text":"we can see that we\u0027re going in the positive y direction."},{"Start":"09:54.633 ","End":"09:58.680","Text":"This is above the strip and this is below the strip."},{"Start":"09:58.680 ","End":"10:04.080","Text":"Now we\u0027re going to work out the magnetic fields inside and outside the plane,"},{"Start":"10:04.080 ","End":"10:09.885","Text":"so right now we\u0027re looking at the entire thick plane that we have."},{"Start":"10:09.885 ","End":"10:12.490","Text":"Let\u0027s take a look."},{"Start":"10:14.570 ","End":"10:17.415","Text":"Let\u0027s take a look above the plane,"},{"Start":"10:17.415 ","End":"10:19.110","Text":"so above the thick plane,"},{"Start":"10:19.110 ","End":"10:23.085","Text":"that means that z is greater than d divided by 2,"},{"Start":"10:23.085 ","End":"10:29.490","Text":"because this whole width is d. This is 0 and this is positive d divided by 2,"},{"Start":"10:29.490 ","End":"10:32.385","Text":"and this is negative d divided by 2."},{"Start":"10:32.385 ","End":"10:35.505","Text":"If we\u0027re greater than d divided by 2,"},{"Start":"10:35.505 ","End":"10:38.415","Text":"we\u0027re above all of the planes."},{"Start":"10:38.415 ","End":"10:44.550","Text":"All of the thin planes that we can push in-between here were above them."},{"Start":"10:44.550 ","End":"10:53.370","Text":"We can say therefore that B = the integral between all of the thin plane,"},{"Start":"10:53.370 ","End":"10:55.545","Text":"so between this line over here,"},{"Start":"10:55.545 ","End":"11:00.000","Text":"negative d divided by 2 until the top of the plane over here,"},{"Start":"11:00.000 ","End":"11:03.280","Text":"positive d divided by 2."},{"Start":"11:05.480 ","End":"11:11.010","Text":"Here we know it\u0027s already in the negative y direction,"},{"Start":"11:11.010 ","End":"11:14.865","Text":"so we can add in the negative over here and then we have"},{"Start":"11:14.865 ","End":"11:23.385","Text":"Rho naught Mu naught v naught e^ Alpha z."},{"Start":"11:23.385 ","End":"11:29.400","Text":"All of this is divided by 2 and of course,"},{"Start":"11:29.400 ","End":"11:38.100","Text":"these add tag and we\u0027re integrating with respect to z tag and of course,"},{"Start":"11:38.100 ","End":"11:42.820","Text":"this is in the negative y direction."},{"Start":"11:45.620 ","End":"11:54.000","Text":"What we get i we get that this is equal to negative Mu naught Rho naught v naught divided"},{"Start":"11:54.000 ","End":"12:02.460","Text":"by 2 multiplied by the derivative of e^ Alpha z,"},{"Start":"12:02.460 ","End":"12:05.865","Text":"which is 1 divided by Alpha."},{"Start":"12:05.865 ","End":"12:10.440","Text":"Then we have e^ Alphas z,"},{"Start":"12:10.440 ","End":"12:16.020","Text":"y hat between the bounds and this is of course z tag within"},{"Start":"12:16.020 ","End":"12:22.305","Text":"the bounds of negative d divided by 2 to positive d divided by 2."},{"Start":"12:22.305 ","End":"12:29.670","Text":"Which is going to just leave us with therefore that the magnetic field in this region"},{"Start":"12:29.670 ","End":"12:38.480","Text":"is = Mu naught Rho naught v naught divided by 2 Alpha."},{"Start":"12:38.480 ","End":"12:43.790","Text":"Then I\u0027m just going to add in the negative into here."},{"Start":"12:43.790 ","End":"12:48.300","Text":"This is multiplied by e to"},{"Start":"12:48.300 ","End":"12:57.840","Text":"the negative Alpha d divided by 2,"},{"Start":"12:57.840 ","End":"13:00.570","Text":"and then minus because I put it in here,"},{"Start":"13:00.570 ","End":"13:04.995","Text":"e^ Alpha d divided by 2."},{"Start":"13:04.995 ","End":"13:09.090","Text":"Of course, this is in the y direction."},{"Start":"13:09.090 ","End":"13:15.510","Text":"However, if we then look at the magnetic field in the area where z is"},{"Start":"13:15.510 ","End":"13:19.260","Text":"smaller than negative d divided by"},{"Start":"13:19.260 ","End":"13:24.585","Text":"2 is anywhere below all of the thin of planes that we fit in here."},{"Start":"13:24.585 ","End":"13:27.135","Text":"It\u0027s going to be, as we saw before,"},{"Start":"13:27.135 ","End":"13:29.550","Text":"the exact same magnetic field,"},{"Start":"13:29.550 ","End":"13:32.070","Text":"just in the opposite direction."},{"Start":"13:32.070 ","End":"13:35.535","Text":"In that case, we can multiply this,"},{"Start":"13:35.535 ","End":"13:38.520","Text":"so here by plus 1."},{"Start":"13:38.520 ","End":"13:42.360","Text":"We\u0027re multiplying and here by negative 1."},{"Start":"13:42.360 ","End":"13:47.310","Text":"Plus 1 is when z is bigger than d divided by 2,"},{"Start":"13:47.310 ","End":"13:54.780","Text":"so we\u0027re above all of the thin planes or we\u0027re above the whole of the thick plane."},{"Start":"13:54.780 ","End":"13:56.700","Text":"Though plane with the width."},{"Start":"13:56.700 ","End":"14:02.940","Text":"We\u0027re multiplying by negative 1 when z is smaller than negative d divided by 2."},{"Start":"14:02.940 ","End":"14:06.150","Text":"As we\u0027re located below here."},{"Start":"14:06.150 ","End":"14:10.650","Text":"We\u0027re below the plane with thickness,"},{"Start":"14:10.650 ","End":"14:14.760","Text":"the thick plain slash in the same way where"},{"Start":"14:14.760 ","End":"14:20.950","Text":"below all of the thin planes that we squeezed in here."},{"Start":"14:21.650 ","End":"14:25.410","Text":"Now we found the magnetic field."},{"Start":"14:25.410 ","End":"14:27.810","Text":"Remember the question was to find"},{"Start":"14:27.810 ","End":"14:30.915","Text":"the magnetic field both inside and outside of the plane."},{"Start":"14:30.915 ","End":"14:36.270","Text":"What we\u0027ve done up until now is we\u0027ve found the magnetic field above and below the plane."},{"Start":"14:36.270 ","End":"14:39.810","Text":"This is the magnetic field outside of the plane."},{"Start":"14:39.810 ","End":"14:43.485","Text":"Now what we want to do is we want to calculate it inside of the plane."},{"Start":"14:43.485 ","End":"14:47.320","Text":"I\u0027m going to rub out this Ampere\u0027s loop."},{"Start":"14:48.200 ","End":"14:51.539","Text":"We\u0027re still using these thin strips,"},{"Start":"14:51.539 ","End":"14:54.840","Text":"just like we\u0027ve done before but as we\u0027ve said,"},{"Start":"14:54.840 ","End":"14:58.740","Text":"we can\u0027t do this Ampere\u0027s loop inside over"},{"Start":"14:58.740 ","End":"15:03.390","Text":"here because we have this changing charge density."},{"Start":"15:03.390 ","End":"15:06.930","Text":"What we\u0027re going to do is we\u0027re going to superimpose"},{"Start":"15:06.930 ","End":"15:11.265","Text":"the magnetic field of each one of these thin strips."},{"Start":"15:11.265 ","End":"15:14.520","Text":"Here we have the magnetic field and of course,"},{"Start":"15:14.520 ","End":"15:19.490","Text":"we have to remember if we\u0027re located above the thin strip,"},{"Start":"15:19.490 ","End":"15:22.370","Text":"or if we\u0027re located below the thin strip,"},{"Start":"15:22.370 ","End":"15:26.550","Text":"the direction of the magnetic field is going to change."},{"Start":"15:26.810 ","End":"15:30.525","Text":"Now let\u0027s take a random height over here,"},{"Start":"15:30.525 ","End":"15:34.035","Text":"so this point over here,"},{"Start":"15:34.035 ","End":"15:37.300","Text":"and let\u0027s say that this is at a height z."},{"Start":"15:37.790 ","End":"15:43.740","Text":"What I\u0027m going to do is I\u0027m going to calculate the magnetic field that results from being"},{"Start":"15:43.740 ","End":"15:49.575","Text":"above all of the thin strips that are below this point over here."},{"Start":"15:49.575 ","End":"15:52.815","Text":"A thin strip here, this thin strip,"},{"Start":"15:52.815 ","End":"15:54.480","Text":"all of the ones below,"},{"Start":"15:54.480 ","End":"15:57.705","Text":"up until the bottom over here."},{"Start":"15:57.705 ","End":"16:04.920","Text":"We can say that the magnetic field is going to be = the integral"},{"Start":"16:04.920 ","End":"16:12.495","Text":"from negative d divided by 2 up until this height z."},{"Start":"16:12.495 ","End":"16:15.450","Text":"We\u0027re taking into account all of these thin strips,"},{"Start":"16:15.450 ","End":"16:19.725","Text":"so we\u0027re looking at the magnetic field above all of the thin strips."},{"Start":"16:19.725 ","End":"16:21.840","Text":"This over here."},{"Start":"16:21.840 ","End":"16:31.095","Text":"We\u0027re integrating half Mu naught Rho naught v naught"},{"Start":"16:31.095 ","End":"16:41.550","Text":"e^ Alpha z tag and so this is also z tag over here in the negative y direction."},{"Start":"16:41.550 ","End":"16:45.900","Text":"Then we\u0027re adding the magnetic field that"},{"Start":"16:45.900 ","End":"16:51.450","Text":"results from being below all the thin strips over here."},{"Start":"16:51.450 ","End":"16:58.964","Text":"We\u0027re integrating from our height z until positive d divided by 2 but here,"},{"Start":"16:58.964 ","End":"17:02.650","Text":"we\u0027re located below these same strips."},{"Start":"17:04.190 ","End":"17:14.550","Text":"Then again, we have a half of Mu naught Rho naught v naught e^Alpha z tag,"},{"Start":"17:14.550 ","End":"17:20.400","Text":"dz tag but this time we\u0027re multiplying"},{"Start":"17:20.400 ","End":"17:28.080","Text":"by positive y hat because we\u0027re located below the strips."},{"Start":"17:28.080 ","End":"17:31.755","Text":"Then once you do the integral,"},{"Start":"17:31.755 ","End":"17:34.800","Text":"you get that the magnetic field is"},{"Start":"17:34.800 ","End":"17:41.820","Text":"= Mu naught Rho naught v naught divided by 2 Alpha like"},{"Start":"17:41.820 ","End":"17:48.810","Text":"before and then we have multiplied by e^ negative Alpha d divided by"},{"Start":"17:48.810 ","End":"17:55.500","Text":"2 plus e^Alpha d divided by 2 minus"},{"Start":"17:55.500 ","End":"18:05.145","Text":"2 e^ Alpha z in the y direction."},{"Start":"18:05.145 ","End":"18:10.380","Text":"This is of course, when we\u0027re located within the thick plane,"},{"Start":"18:10.380 ","End":"18:14.760","Text":"so z is bigger than negative d divided by 2,"},{"Start":"18:14.760 ","End":"18:17.730","Text":"but smaller than positive d divided by 2,"},{"Start":"18:17.730 ","End":"18:19.320","Text":"so we\u0027re located here."},{"Start":"18:19.320 ","End":"18:26.745","Text":"This is the magnetic field for inside the plane and outside the plane."},{"Start":"18:26.745 ","End":"18:29.680","Text":"That\u0027s the end of this lesson."}],"ID":21425},{"Watched":false,"Name":"Exercise 7","Duration":"9m 2s","ChapterTopicVideoID":21345,"CourseChapterTopicPlaylistID":99468,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/21345.jpeg","UploadDate":"2020-04-06T21:43:30.1230000","DurationForVideoObject":"PT9M2S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.620","Text":"Hello. In this lesson,"},{"Start":"00:01.620 ","End":"00:03.930","Text":"we\u0027re answering the following question."},{"Start":"00:03.930 ","End":"00:09.240","Text":"We have a current flowing through an infinite cylinder of inner radius a and outer"},{"Start":"00:09.240 ","End":"00:14.910","Text":"radius b and the current density j is as a function of the radius,"},{"Start":"00:14.910 ","End":"00:20.730","Text":"and it\u0027s equal to a constant multiplied by i^3 in the Theta direction."},{"Start":"00:20.730 ","End":"00:23.040","Text":"If this is the positive Theta direction,"},{"Start":"00:23.040 ","End":"00:27.835","Text":"that means the z-axis is coming out towards us."},{"Start":"00:27.835 ","End":"00:31.875","Text":"We\u0027re being asked to calculate the magnetic fields throughout."},{"Start":"00:31.875 ","End":"00:37.005","Text":"What we can do is we can look at this as if this is like a coil,"},{"Start":"00:37.005 ","End":"00:42.420","Text":"which is a question that we calculated a few lessons ago."},{"Start":"00:43.130 ","End":"00:46.035","Text":"When we dealt with the coils,"},{"Start":"00:46.035 ","End":"00:51.680","Text":"we can also imagine that there are lots of different coils wrapped around one another."},{"Start":"00:51.680 ","End":"01:00.005","Text":"What we did is we saw that the magnetic field outside of the coils is equal to 0."},{"Start":"01:00.005 ","End":"01:03.125","Text":"We can see that the magnetic field in this region,"},{"Start":"01:03.125 ","End":"01:06.875","Text":"so where r is greater than b,"},{"Start":"01:06.875 ","End":"01:09.665","Text":"so the magnetic field is equal to 0."},{"Start":"01:09.665 ","End":"01:15.930","Text":"What we\u0027ll see is that the magnetic field in general is going to be,"},{"Start":"01:15.930 ","End":"01:18.380","Text":"from the right-hand rule, in the z-direction."},{"Start":"01:18.380 ","End":"01:23.015","Text":"Our fingers curl in the direction of the current or of j,"},{"Start":"01:23.015 ","End":"01:27.290","Text":"and then our thumb is pointing in the direction of the magnetic field,"},{"Start":"01:27.290 ","End":"01:31.440","Text":"which is in the positive z-direction."},{"Start":"01:32.360 ","End":"01:38.120","Text":"Now what we want to do is we want to calculate the magnetic field inside."},{"Start":"01:38.120 ","End":"01:40.655","Text":"Now what we want to do is we want to see"},{"Start":"01:40.655 ","End":"01:47.110","Text":"the magnetic fields in this region between A and B."},{"Start":"01:47.110 ","End":"01:51.920","Text":"As we remember, we are going to take an ampere\u0027s loop,"},{"Start":"01:51.920 ","End":"01:57.775","Text":"and remember the ampere\u0027s loop has to be perpendicular to the direction of current."},{"Start":"01:57.775 ","End":"02:02.385","Text":"If the current is coming from the Theta direction,"},{"Start":"02:02.385 ","End":"02:09.285","Text":"we want to take a loop that looks something like this."},{"Start":"02:09.285 ","End":"02:11.370","Text":"This drawing isn\u0027t very clear,"},{"Start":"02:11.370 ","End":"02:13.020","Text":"let\u0027s just redraw this."},{"Start":"02:13.020 ","End":"02:17.240","Text":"Here we have our inner radius a,"},{"Start":"02:17.240 ","End":"02:22.445","Text":"and here we have our outer radius b,"},{"Start":"02:22.445 ","End":"02:27.735","Text":"and this is our cylinder."},{"Start":"02:27.735 ","End":"02:37.380","Text":"As we saw, we have our current or our current density j flowing in this direction."},{"Start":"02:38.080 ","End":"02:45.550","Text":"This is our z-axis that goes right through the center."},{"Start":"02:45.550 ","End":"02:51.995","Text":"In that case, if our current density is flowing like this,"},{"Start":"02:51.995 ","End":"02:56.840","Text":"as we said, we want our loop to be perpendicular to that."},{"Start":"02:56.840 ","End":"02:59.015","Text":"It\u0027s exactly like this."},{"Start":"02:59.015 ","End":"03:03.300","Text":"This is a much clearer drawing than the previous one."},{"Start":"03:03.300 ","End":"03:07.090","Text":"I don\u0027t know if you can see this."},{"Start":"03:07.220 ","End":"03:09.455","Text":"It\u0027s something like this."},{"Start":"03:09.455 ","End":"03:13.265","Text":"This is basically this loop. I\u0027ll draw it here."},{"Start":"03:13.265 ","End":"03:15.860","Text":"What we want is that our j,"},{"Start":"03:15.860 ","End":"03:16.910","Text":"our current density,"},{"Start":"03:16.910 ","End":"03:20.345","Text":"is going to flow through here."},{"Start":"03:20.345 ","End":"03:23.150","Text":"Now we have our amperes loop,"},{"Start":"03:23.150 ","End":"03:25.260","Text":"so we\u0027re going to use ampere\u0027s law."},{"Start":"03:25.260 ","End":"03:30.090","Text":"The closed loop integral of b.dl is"},{"Start":"03:30.090 ","End":"03:36.210","Text":"equal to Mu naught multiplied by in."},{"Start":"03:36.210 ","End":"03:38.165","Text":"As we\u0027ve already said,"},{"Start":"03:38.165 ","End":"03:42.400","Text":"our magnetic field is in the z-direction,"},{"Start":"03:42.400 ","End":"03:43.980","Text":"so this is our B."},{"Start":"03:43.980 ","End":"03:49.640","Text":"Let\u0027s say that this length of this loop is l. In that case,"},{"Start":"03:49.640 ","End":"03:53.015","Text":"this is going to be equal to b.l."},{"Start":"03:53.015 ","End":"03:56.700","Text":"Why is that? Here we have B along this edge,"},{"Start":"03:56.700 ","End":"04:00.560","Text":"then here our magnetic field is perpendicular to the loop,"},{"Start":"04:00.560 ","End":"04:03.650","Text":"which means when we do the dot-product between"},{"Start":"04:03.650 ","End":"04:07.535","Text":"b and dl because they\u0027re perpendicular will equal to 0."},{"Start":"04:07.535 ","End":"04:11.120","Text":"Here, of course, we\u0027ve already seen that we have no magnetic field"},{"Start":"04:11.120 ","End":"04:15.490","Text":"outside of our cylinder or outside of our coil."},{"Start":"04:15.490 ","End":"04:18.510","Text":"Here, our b is equal to 0."},{"Start":"04:18.510 ","End":"04:21.995","Text":"We have 0 multiplied by l, which is of course 0,"},{"Start":"04:21.995 ","End":"04:26.885","Text":"and this is equal to Mu naught multiplied by Iin."},{"Start":"04:26.885 ","End":"04:30.005","Text":"What exactly is Iin?"},{"Start":"04:30.005 ","End":"04:40.575","Text":"As we know, Iin is equal to the integral of j.ds."},{"Start":"04:40.575 ","End":"04:42.825","Text":"This is the integral,"},{"Start":"04:42.825 ","End":"04:45.225","Text":"because we have ds, it\u0027s a double integral."},{"Start":"04:45.225 ","End":"04:46.830","Text":"So rj, we were given,"},{"Start":"04:46.830 ","End":"04:50.700","Text":"it\u0027s equal to ar^3,"},{"Start":"04:50.700 ","End":"04:53.740","Text":"and of course we\u0027ll add in a dash because soon we\u0027re going to be"},{"Start":"04:53.740 ","End":"04:56.455","Text":"integrating along r. Then,"},{"Start":"04:56.455 ","End":"04:59.215","Text":"this is along this axis."},{"Start":"04:59.215 ","End":"05:05.555","Text":"Let\u0027s say that we put our loop a distance away from the center of the cylinder,"},{"Start":"05:05.555 ","End":"05:07.460","Text":"a distance of r away,"},{"Start":"05:07.460 ","End":"05:14.840","Text":"so we\u0027re going to go integrating along dr tag,"},{"Start":"05:14.840 ","End":"05:23.585","Text":"where the bounds for a dr tag integral are from this r until the edge of the cylinder."},{"Start":"05:23.585 ","End":"05:27.005","Text":"Remember, we\u0027re trying to see how much current"},{"Start":"05:27.005 ","End":"05:31.865","Text":"is flowing through this section over here in gray."},{"Start":"05:31.865 ","End":"05:35.135","Text":"All of this area over here."},{"Start":"05:35.135 ","End":"05:41.915","Text":"In that case, we\u0027re going from r until the outer radius of the cylinder, which is b."},{"Start":"05:41.915 ","End":"05:44.135","Text":"So from r until b."},{"Start":"05:44.135 ","End":"05:49.455","Text":"Then, we\u0027re also integrating along the length of this loop."},{"Start":"05:49.455 ","End":"05:52.050","Text":"That\u0027s of length l, total length."},{"Start":"05:52.050 ","End":"05:58.845","Text":"So from 0 until l. We\u0027re just going length times width for the area."},{"Start":"05:58.845 ","End":"06:05.150","Text":"The width is from r until b and the length is from 0 until l,"},{"Start":"06:05.150 ","End":"06:11.025","Text":"and this is of course dz."},{"Start":"06:11.025 ","End":"06:15.075","Text":"Then, what we\u0027re going to get is dz."},{"Start":"06:15.075 ","End":"06:17.010","Text":"We\u0027re just going to multiply by l,"},{"Start":"06:17.010 ","End":"06:18.825","Text":"so we have la."},{"Start":"06:18.825 ","End":"06:22.905","Text":"Then by integrating the di tag,"},{"Start":"06:22.905 ","End":"06:27.990","Text":"so we\u0027re going to have r to the r tag to the power of"},{"Start":"06:27.990 ","End":"06:34.260","Text":"4 divided by 4 between the bounds of r and b."},{"Start":"06:34.260 ","End":"06:39.090","Text":"This is going to be equal to al and then we have b to"},{"Start":"06:39.090 ","End":"06:44.655","Text":"the power 4 minus r to the power 4 divided by 4."},{"Start":"06:44.655 ","End":"06:46.770","Text":"That\u0027s our Iin."},{"Start":"06:46.770 ","End":"06:52.440","Text":"Now we have that b multiplied by l is equal to Mu naught multiplied by in."},{"Start":"06:52.440 ","End":"06:59.340","Text":"So multiplied by alb to the power 4 minus r to the power 4 divided by 4,"},{"Start":"06:59.340 ","End":"07:05.190","Text":"divide both sides by l. Then we get that b is equal to Mu naught"},{"Start":"07:05.190 ","End":"07:12.195","Text":"a multiplied by b to the power 4 minus r to the power 4 divided by 4."},{"Start":"07:12.195 ","End":"07:19.535","Text":"We already said that our b field is in the z-direction from the right-hand rule,"},{"Start":"07:19.535 ","End":"07:21.565","Text":"so in the z-direction."},{"Start":"07:21.565 ","End":"07:28.624","Text":"Now, the final region that we want to look at is this region inside over here,"},{"Start":"07:28.624 ","End":"07:34.185","Text":"when r is located when it\u0027s smaller than a."},{"Start":"07:34.185 ","End":"07:36.880","Text":"Again, we\u0027re using an ampere\u0027s loop,"},{"Start":"07:36.880 ","End":"07:41.515","Text":"but this time it goes like so into this region."},{"Start":"07:41.515 ","End":"07:43.850","Text":"Then we can also draw it over here."},{"Start":"07:43.850 ","End":"07:48.585","Text":"It\u0027s just going to be something like so, going like this."},{"Start":"07:48.585 ","End":"07:54.470","Text":"What we can see is that now it\u0027s independent of r,"},{"Start":"07:54.470 ","End":"07:57.605","Text":"because here, there\u0027s no current flowing through."},{"Start":"07:57.605 ","End":"08:01.450","Text":"The current is only flowing through between a and b."},{"Start":"08:01.450 ","End":"08:04.355","Text":"In that case, we just want to see"},{"Start":"08:04.355 ","End":"08:08.010","Text":"how much current is flowing through all the time a and b."},{"Start":"08:08.010 ","End":"08:10.865","Text":"It doesn\u0027t make a difference about"},{"Start":"08:10.865 ","End":"08:15.050","Text":"our r so long as it\u0027s in the region that\u0027s less than a."},{"Start":"08:15.050 ","End":"08:17.495","Text":"Now r is less than a."},{"Start":"08:17.495 ","End":"08:22.370","Text":"We\u0027re going to be using the exact same equation as we did over here."},{"Start":"08:22.370 ","End":"08:30.530","Text":"But the only difference is that instead of having our bounds over here for r,"},{"Start":"08:30.530 ","End":"08:32.150","Text":"between r and b,"},{"Start":"08:32.150 ","End":"08:35.465","Text":"it\u0027s now going to be between a to b."},{"Start":"08:35.465 ","End":"08:38.750","Text":"Then, we can just already write this out."},{"Start":"08:38.750 ","End":"08:44.585","Text":"We\u0027ll just write it\u0027s equal to Mu naught a multiplied by B to the power 4 minus."},{"Start":"08:44.585 ","End":"08:46.550","Text":"Then instead of r to the power 4,"},{"Start":"08:46.550 ","End":"08:54.810","Text":"it\u0027s going to be a to the power 4 divided by 4 in the z-axis."},{"Start":"08:55.670 ","End":"08:57.855","Text":"This is the answer,"},{"Start":"08:57.855 ","End":"09:00.395","Text":"this is the magnetic field throughout,"},{"Start":"09:00.395 ","End":"09:03.210","Text":"and that is the end of this lesson."}],"ID":21426}],"Thumbnail":null,"ID":99468}]

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