Lorenz Transformations for Electromagnetic Fields
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[{"Name":"Lorenz Transformations for Electromagnetic Fields","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Electromagnetic Transformations","Duration":"11m 34s","ChapterTopicVideoID":21536,"CourseChapterTopicPlaylistID":99485,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/21536.jpeg","UploadDate":"2020-04-21T16:17:39.8000000","DurationForVideoObject":"PT11M34S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.785","Text":"Hello. In this lesson,"},{"Start":"00:01.785 ","End":"00:05.340","Text":"we\u0027re going to be speaking about transformations."},{"Start":"00:05.340 ","End":"00:08.310","Text":"In our unit, when we dealt with mechanics,"},{"Start":"00:08.310 ","End":"00:12.000","Text":"we saw that when dealing with special relativity,"},{"Start":"00:12.000 ","End":"00:16.303","Text":"that\u0027s cases where we\u0027re traveling close to the speed of light,"},{"Start":"00:16.303 ","End":"00:18.300","Text":"we saw that we have to use"},{"Start":"00:18.300 ","End":"00:23.950","Text":"Lorentz transformations when dealing with multiple frames of reference."},{"Start":"00:24.110 ","End":"00:30.315","Text":"What we\u0027re going to do now is we\u0027re going to look at these Lorentz transformations but"},{"Start":"00:30.315 ","End":"00:36.000","Text":"we\u0027re going to be looking at them when dealing with electric and magnetic fields."},{"Start":"00:36.000 ","End":"00:41.365","Text":"Let\u0027s imagine that we have 1 frame of reference over here."},{"Start":"00:41.365 ","End":"00:46.020","Text":"This is frame of reference S and there\u0027s someone"},{"Start":"00:46.020 ","End":"00:51.390","Text":"over here watching measuring the electromagnetic field."},{"Start":"00:51.390 ","End":"00:57.345","Text":"Then over here we have our S tag frame of reference."},{"Start":"00:57.345 ","End":"01:00.905","Text":"Where over here we also have someone standing."},{"Start":"01:00.905 ","End":"01:03.425","Text":"This is person tag."},{"Start":"01:03.425 ","End":"01:08.630","Text":"The S tag frame of reference is traveling in this direction with"},{"Start":"01:08.630 ","End":"01:15.670","Text":"a velocity of V relative to our S frame of reference."},{"Start":"01:16.120 ","End":"01:19.715","Text":"Now, in our S tag reference frame,"},{"Start":"01:19.715 ","End":"01:24.845","Text":"our person tag over here measures at this point"},{"Start":"01:24.845 ","End":"01:32.920","Text":"an electric field like so E tag and a magnetic field like so B tag."},{"Start":"01:33.140 ","End":"01:37.205","Text":"Then in our S reference frame,"},{"Start":"01:37.205 ","End":"01:42.500","Text":"this observer is going to measure at this same point."},{"Start":"01:42.500 ","End":"01:46.590","Text":"This is the same point as this."},{"Start":"01:46.590 ","End":"01:50.750","Text":"Our observer in our S reference frame will measure"},{"Start":"01:50.750 ","End":"01:55.025","Text":"a different electric and magnetic field"},{"Start":"01:55.025 ","End":"02:00.210","Text":"to what the observer in the S tag reference frame observed."},{"Start":"02:00.530 ","End":"02:06.155","Text":"What we\u0027re going to do in this lesson is we\u0027re going to see that if we have"},{"Start":"02:06.155 ","End":"02:12.125","Text":"the electromagnetic fields according to our S frame of reference,"},{"Start":"02:12.125 ","End":"02:20.270","Text":"then we can do some transformation in order to get what"},{"Start":"02:20.270 ","End":"02:24.230","Text":"the observer in the S tag frame of reference will measure at"},{"Start":"02:24.230 ","End":"02:28.220","Text":"that exact same point for the electromagnetic fields."},{"Start":"02:28.220 ","End":"02:33.415","Text":"Of course, there\u0027s also a transformation that goes the opposite way."},{"Start":"02:33.415 ","End":"02:39.935","Text":"That means that if we have the electromagnetic field in the S tag frame of reference,"},{"Start":"02:39.935 ","End":"02:43.580","Text":"we can do this transformation T tag and get"},{"Start":"02:43.580 ","End":"02:48.180","Text":"the electromagnetic field in the S frame of reference."},{"Start":"02:48.710 ","End":"02:51.990","Text":"These are the equations."},{"Start":"02:51.990 ","End":"02:54.495","Text":"If we know what E parallel is,"},{"Start":"02:54.495 ","End":"02:57.240","Text":"we know what E tag parallel is,"},{"Start":"02:57.240 ","End":"03:02.085","Text":"and B parallel is equal to B tag parallel."},{"Start":"03:02.085 ","End":"03:07.695","Text":"Then if we want to know what E tag perpendicular is equal to,"},{"Start":"03:07.695 ","End":"03:15.470","Text":"we have Gamma multiplied by the perpendicular component of the E field plus V,"},{"Start":"03:15.470 ","End":"03:21.560","Text":"the relativistic velocity cross-product with the perpendicular field."},{"Start":"03:21.560 ","End":"03:25.210","Text":"Our B tag perpendicular is equal to Gamma"},{"Start":"03:25.210 ","End":"03:30.480","Text":"multiplied by B perpendicular minus 1 divided by C^2,"},{"Start":"03:30.480 ","End":"03:36.740","Text":"where C is the velocity of light multiplied by V cross E perpendicular."},{"Start":"03:36.740 ","End":"03:41.390","Text":"What do we mean by the parallel component of the E and the B field?"},{"Start":"03:41.390 ","End":"03:48.475","Text":"The parallel component of the field is parallel to the relativistic velocity."},{"Start":"03:48.475 ","End":"03:53.375","Text":"Here we can see that B is parallel to V. Here,"},{"Start":"03:53.375 ","End":"03:56.645","Text":"our B field would be parallel."},{"Start":"03:56.645 ","End":"04:06.550","Text":"Whereas we can see that E over here is perpendicular to the velocity."},{"Start":"04:06.550 ","End":"04:10.580","Text":"Of course, we can use our cosine and sine"},{"Start":"04:10.580 ","End":"04:15.510","Text":"of the angle in order to calculate the different components."},{"Start":"04:16.400 ","End":"04:24.900","Text":"Here we can see that our velocity is in the x-direction."},{"Start":"04:25.370 ","End":"04:30.075","Text":"Therefore we can say that our E parallel is E"},{"Start":"04:30.075 ","End":"04:34.830","Text":"in the x-direction is going to be equal to E in the x-direction tag."},{"Start":"04:36.170 ","End":"04:38.595","Text":"With the perpendicular,"},{"Start":"04:38.595 ","End":"04:45.390","Text":"we also go according to if it\u0027s perpendicular to the relativistic velocity."},{"Start":"04:45.950 ","End":"04:49.535","Text":"In that case, our E perpendicular,"},{"Start":"04:49.535 ","End":"04:52.820","Text":"we can see it\u0027s in the y-direction."},{"Start":"04:52.820 ","End":"04:57.590","Text":"E tag_y is going to be equal to Gamma multiplied by"},{"Start":"04:57.590 ","End":"05:06.620","Text":"E_y plus our V cross product with the perpendicular component of the B field."},{"Start":"05:06.620 ","End":"05:12.035","Text":"Now what\u0027s important to note is that any way when we do V cross B,"},{"Start":"05:12.035 ","End":"05:14.450","Text":"we\u0027re going to get the perpendicular component."},{"Start":"05:14.450 ","End":"05:20.495","Text":"This section doesn\u0027t really matter but you could admit the perpendicular,"},{"Start":"05:20.495 ","End":"05:23.070","Text":"but that makes it a bit clearer."},{"Start":"05:23.450 ","End":"05:28.865","Text":"Of course, we can do it the same, another perpendicular direction."},{"Start":"05:28.865 ","End":"05:31.295","Text":"If the V is in the x-direction,"},{"Start":"05:31.295 ","End":"05:38.570","Text":"that\u0027s the electric field in the z direction because both y and z are perpendicular to x."},{"Start":"05:38.570 ","End":"05:41.150","Text":"Then we just write out the exact same thing just"},{"Start":"05:41.150 ","End":"05:44.150","Text":"instead of E tag_y we\u0027ll have E tag_z tag,"},{"Start":"05:44.150 ","End":"05:47.500","Text":"and here instead of E_y we\u0027ll have E_z."},{"Start":"05:47.500 ","End":"05:50.730","Text":"Then a reminder on what Gamma is."},{"Start":"05:50.730 ","End":"05:55.125","Text":"Gamma is equal to 1 divided the square root of 1 minus"},{"Start":"05:55.125 ","End":"06:01.050","Text":"the velocity divided by the speed of light^2."},{"Start":"06:01.050 ","End":"06:04.580","Text":"Now, another little thing that we need to note."},{"Start":"06:04.580 ","End":"06:08.870","Text":"Sometimes people find it difficult deciding which is"},{"Start":"06:08.870 ","End":"06:14.220","Text":"the S reference frame and which is the S tag reference frame."},{"Start":"06:14.840 ","End":"06:18.910","Text":"The trick is the S reference frame is"},{"Start":"06:18.910 ","End":"06:23.710","Text":"the one that is measuring the velocity of the other reference frame."},{"Start":"06:23.710 ","End":"06:31.710","Text":"For instance, here we were given that the velocity of this reference frame S tag is V"},{"Start":"06:31.710 ","End":"06:42.025","Text":"relative to S. So that means that this is our S reference frame and this is our S tag."},{"Start":"06:42.025 ","End":"06:45.380","Text":"If we\u0027re given some kind of velocity,"},{"Start":"06:45.380 ","End":"06:49.960","Text":"the frame of reference which measures that velocity is"},{"Start":"06:49.960 ","End":"06:57.140","Text":"the S frame of reference and the other frame of reference will be the S tag."},{"Start":"06:58.520 ","End":"07:02.380","Text":"The reference frame that measures the velocity of"},{"Start":"07:02.380 ","End":"07:07.100","Text":"the other reference frame is the S reference frame."},{"Start":"07:08.480 ","End":"07:14.980","Text":"This we can say is transformation T. What we saw now,"},{"Start":"07:14.980 ","End":"07:17.950","Text":"given E and B,"},{"Start":"07:17.950 ","End":"07:21.770","Text":"we can calculate E tag and B tag."},{"Start":"07:22.140 ","End":"07:26.475","Text":"Now, what happens if we want to know T tag?"},{"Start":"07:26.475 ","End":"07:29.895","Text":"The opposite transformation."},{"Start":"07:29.895 ","End":"07:36.170","Text":"As in we have E tag and B tag and we want to calculate E and B."},{"Start":"07:36.200 ","End":"07:41.239","Text":"A moment ago we were speaking about the reference frames where we have"},{"Start":"07:41.239 ","End":"07:46.400","Text":"the reference frame that measures the velocity is S,"},{"Start":"07:46.400 ","End":"07:53.290","Text":"and the reference frame which is moving relative to that reference frame is S tag."},{"Start":"07:53.290 ","End":"07:55.260","Text":"In order to get from S tag to S,"},{"Start":"07:55.260 ","End":"07:58.790","Text":"we don\u0027t have to do complicated algebra over here,"},{"Start":"07:58.790 ","End":"08:03.310","Text":"we just have to change around our velocity."},{"Start":"08:03.310 ","End":"08:06.860","Text":"If the S tag reference frame is moving in"},{"Start":"08:06.860 ","End":"08:13.355","Text":"the x-direction with a velocity of V in the x-direction,"},{"Start":"08:13.355 ","End":"08:19.350","Text":"the velocity of the S frame of reference relative to S tag"},{"Start":"08:19.350 ","End":"08:21.968","Text":"will have the exact same magnitude of velocity"},{"Start":"08:21.968 ","End":"08:24.815","Text":"but it will be traveling in the opposite direction,"},{"Start":"08:24.815 ","End":"08:27.750","Text":"so in the negative x-direction."},{"Start":"08:27.830 ","End":"08:30.185","Text":"Let\u0027s just speak about that."},{"Start":"08:30.185 ","End":"08:34.115","Text":"If we\u0027re talking about some velocity of some body"},{"Start":"08:34.115 ","End":"08:38.450","Text":"relative to another body or another reference frame,"},{"Start":"08:38.450 ","End":"08:46.015","Text":"we can switch between these 2 reference frames by just taking the negative velocity."},{"Start":"08:46.015 ","End":"08:50.835","Text":"If I have point A over here and B"},{"Start":"08:50.835 ","End":"08:55.750","Text":"starts like so and it\u0027s moving like so with a velocity of V,"},{"Start":"08:55.750 ","End":"09:03.650","Text":"it\u0027s the exact same thing as if I would have some point B."},{"Start":"09:04.250 ","End":"09:08.105","Text":"This exact point B over here."},{"Start":"09:08.105 ","End":"09:12.425","Text":"But instead of the velocity that we said was moving in this direction,"},{"Start":"09:12.425 ","End":"09:14.915","Text":"the velocity will be moving in"},{"Start":"09:14.915 ","End":"09:19.680","Text":"the opposite direction with the same velocity where B is stationary."},{"Start":"09:20.230 ","End":"09:29.760","Text":"What we can do is we can say that the velocity of S relative to S tag."},{"Start":"09:29.930 ","End":"09:38.225","Text":"The velocity of this relative to this is simply equal to the negative of the velocity"},{"Start":"09:38.225 ","End":"09:47.530","Text":"of S tag relative to S. What is the velocity of S tag relative to S?"},{"Start":"09:47.690 ","End":"09:51.755","Text":"We\u0027re at S and we\u0027re looking at the velocity of S tag."},{"Start":"09:51.755 ","End":"09:56.160","Text":"That was our V. That means it\u0027s equal to negative"},{"Start":"09:56.160 ","End":"10:01.785","Text":"V. This is the velocity of S tag relative to S,"},{"Start":"10:01.785 ","End":"10:05.849","Text":"as in S is looking at S tag and sees it with this velocity."},{"Start":"10:05.849 ","End":"10:12.065","Text":"That\u0027s negative V. Then we just change the equations accordingly."},{"Start":"10:12.065 ","End":"10:16.110","Text":"Wherever we see plus V or V,"},{"Start":"10:16.110 ","End":"10:20.950","Text":"we substitute in minus V. Let\u0027s do this in red."},{"Start":"10:21.650 ","End":"10:24.135","Text":"Here we have plus V,"},{"Start":"10:24.135 ","End":"10:29.520","Text":"we\u0027ll put in instead minus V. Here we have V,"},{"Start":"10:29.520 ","End":"10:32.200","Text":"we\u0027ll multiply it by, instead of V,"},{"Start":"10:32.200 ","End":"10:33.730","Text":"it will be minus V,"},{"Start":"10:33.730 ","End":"10:36.835","Text":"which means that over here the minus and the minus will cancel out."},{"Start":"10:36.835 ","End":"10:39.055","Text":"Here we\u0027ll have a plus."},{"Start":"10:39.055 ","End":"10:47.265","Text":"Then of course, because we have our E tag and our B tag and we want to find our E and B,"},{"Start":"10:47.265 ","End":"10:52.755","Text":"here we\u0027ll have E tag and B tag and here we won\u0027t have a tag."},{"Start":"10:52.755 ","End":"10:56.635","Text":"Here\u0027s the empty box where the tag would have been, but it isn\u0027t."},{"Start":"10:56.635 ","End":"11:01.055","Text":"Here again, we have E tag and B tag because we"},{"Start":"11:01.055 ","End":"11:06.250","Text":"have the information from our S tag frame of reference and we\u0027re trying to find B."},{"Start":"11:06.250 ","End":"11:12.930","Text":"This is B with an empty box of where the tag is missing because we\u0027re calculating B."},{"Start":"11:13.670 ","End":"11:20.435","Text":"What is in red gives us the transformation from S tag to S,"},{"Start":"11:20.435 ","End":"11:26.270","Text":"and what is in black gives us the transformation from S to S tag."},{"Start":"11:26.270 ","End":"11:31.970","Text":"Then you can just plug these values in like so."},{"Start":"11:31.970 ","End":"11:35.040","Text":"That\u0027s the end of this lesson."}],"ID":22390},{"Watched":false,"Name":"Exercise 1","Duration":"7m 34s","ChapterTopicVideoID":21338,"CourseChapterTopicPlaylistID":99485,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/21338.jpeg","UploadDate":"2020-04-06T21:40:46.4900000","DurationForVideoObject":"PT7M34S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.551","Text":"Hello. In this lesson we\u0027re going to be answering the following question,"},{"Start":"00:04.551 ","End":"00:10.155","Text":"an observer travels at a velocity v in the x-direction relative to the lab."},{"Start":"00:10.155 ","End":"00:13.065","Text":"The observer measures an electric field E naught"},{"Start":"00:13.065 ","End":"00:16.590","Text":"in the z direction and the magnetic field of 0."},{"Start":"00:16.590 ","End":"00:22.050","Text":"What electric and magnetic field will the observer in the lab measure?"},{"Start":"00:22.050 ","End":"00:30.750","Text":"What we have is over here we have our lab and we have some observer over here."},{"Start":"00:30.750 ","End":"00:33.630","Text":"We can call this frame of reference the S frame of"},{"Start":"00:33.630 ","End":"00:37.140","Text":"reference and relative to the S frame of reference we"},{"Start":"00:37.140 ","End":"00:43.910","Text":"have this S tag frame of reference moving with a velocity of v in the x-direction."},{"Start":"00:43.910 ","End":"00:48.030","Text":"Here, of course, we also have an observer."},{"Start":"00:48.320 ","End":"00:51.930","Text":"In the S tag frame of reference,"},{"Start":"00:51.930 ","End":"00:57.180","Text":"the observer here measures an electric field in the lab."},{"Start":"00:59.330 ","End":"01:04.050","Text":"What we have is E tag,"},{"Start":"01:04.050 ","End":"01:10.610","Text":"which is equal to E_0 in the z direction,"},{"Start":"01:10.610 ","End":"01:14.030","Text":"and of course, our B tag."},{"Start":"01:14.030 ","End":"01:20.285","Text":"So the magnetic field in our S tag frame of reference we were told is equal to 0."},{"Start":"01:20.285 ","End":"01:28.680","Text":"Then we\u0027re also told that we have our velocity which has some v in the x direction."},{"Start":"01:28.730 ","End":"01:33.680","Text":"If we remember our equation for transformations,"},{"Start":"01:33.680 ","End":"01:39.650","Text":"we know that we have a parallel component and a perpendicular component."},{"Start":"01:39.650 ","End":"01:44.840","Text":"The parallel component for our electric field."},{"Start":"01:44.840 ","End":"01:46.265","Text":"What\u0027s a parallel component?"},{"Start":"01:46.265 ","End":"01:51.330","Text":"It means the component parallel to the direction of the velocity."},{"Start":"01:51.330 ","End":"01:54.485","Text":"Because we know that the velocity is in the x direction,"},{"Start":"01:54.485 ","End":"01:59.100","Text":"so we\u0027re looking at the electric field in the x-direction."},{"Start":"01:59.180 ","End":"02:03.755","Text":"Of course, we only have a z component for the electric field,"},{"Start":"02:03.755 ","End":"02:06.205","Text":"so this is equal to 0."},{"Start":"02:06.205 ","End":"02:11.879","Text":"Now let\u0027s take a look at our perpendicular component."},{"Start":"02:11.879 ","End":"02:17.120","Text":"E perpendicular is the perpendicular field"},{"Start":"02:17.120 ","End":"02:20.540","Text":"when it\u0027s perpendicular to the direction of the velocity."},{"Start":"02:20.540 ","End":"02:25.260","Text":"That means that because our velocity is in the x direction,"},{"Start":"02:25.260 ","End":"02:31.805","Text":"so we\u0027re either dealing with E_y or E_z and specifically,"},{"Start":"02:31.805 ","End":"02:36.680","Text":"we see that we only have a z component for our electric field."},{"Start":"02:36.680 ","End":"02:41.490","Text":"Therefore, our E perpendicular is just our E_z."},{"Start":"02:43.280 ","End":"02:52.270","Text":"Of course, that is just equal to E_0 in the z direction."},{"Start":"02:52.270 ","End":"02:56.910","Text":"Then we can add a vector sign over here."},{"Start":"02:57.620 ","End":"03:03.185","Text":"Now what we\u0027re going to do is we\u0027re going to use our Lorentz transformations."},{"Start":"03:03.185 ","End":"03:08.900","Text":"First of all, what we\u0027re being told in the question is that we\u0027re looking"},{"Start":"03:08.900 ","End":"03:15.020","Text":"from the S tag frame of reference to the S. What we have is E tag and B tag,"},{"Start":"03:15.020 ","End":"03:17.320","Text":"and what we want is E and B."},{"Start":"03:17.320 ","End":"03:18.860","Text":"As we said over here,"},{"Start":"03:18.860 ","End":"03:20.960","Text":"we just have to change this around."},{"Start":"03:20.960 ","End":"03:23.039","Text":"First of all, these are fine."},{"Start":"03:23.039 ","End":"03:26.000","Text":"We don\u0027t really have to change the order over here."},{"Start":"03:26.000 ","End":"03:28.475","Text":"We can just flip around these equal signs."},{"Start":"03:28.475 ","End":"03:31.755","Text":"But over here, instead of E tag,"},{"Start":"03:31.755 ","End":"03:35.450","Text":"we want E and instead of B tag, we want B."},{"Start":"03:35.450 ","End":"03:39.080","Text":"That means here instead of E we\u0027re having E tag and instead of B,"},{"Start":"03:39.080 ","End":"03:40.400","Text":"we have B tag."},{"Start":"03:40.400 ","End":"03:43.890","Text":"Here, B tag and E tag."},{"Start":"03:43.890 ","End":"03:51.560","Text":"Because now we\u0027re looking from the S tag frame of reference,"},{"Start":"03:51.560 ","End":"03:55.010","Text":"so if the S tag frame of reference is moving at"},{"Start":"03:55.010 ","End":"03:59.510","Text":"a velocity v in the x direction relative to S,"},{"Start":"03:59.510 ","End":"04:05.700","Text":"then the S frame of reference is moving at a velocity of v in the negative x"},{"Start":"04:05.700 ","End":"04:12.125","Text":"direction relative to S. We can even just write it over here."},{"Start":"04:12.125 ","End":"04:19.010","Text":"V tag is always going to just be equal to negative v. That"},{"Start":"04:19.010 ","End":"04:25.940","Text":"means that over here we have a minus sign and over here we also have a minus sign,"},{"Start":"04:25.940 ","End":"04:31.010","Text":"which is of course going to cancel out with this minus sign."},{"Start":"04:31.190 ","End":"04:36.110","Text":"Now let\u0027s plug everything into the equations and of course,"},{"Start":"04:36.110 ","End":"04:39.560","Text":"don\u0027t forget the tags over here which I forgot to add in."},{"Start":"04:39.560 ","End":"04:42.250","Text":"We\u0027re dealing just with tags."},{"Start":"04:43.040 ","End":"04:48.120","Text":"Our E tag parallel is this."},{"Start":"04:48.120 ","End":"04:51.440","Text":"Our E tag parallel, as we said,"},{"Start":"04:51.440 ","End":"04:54.335","Text":"was equal to E_x tag,"},{"Start":"04:54.335 ","End":"04:56.045","Text":"which is equal to 0."},{"Start":"04:56.045 ","End":"04:57.539","Text":"We already saw that."},{"Start":"04:57.539 ","End":"05:02.760","Text":"Of course, our B tag parallel"},{"Start":"05:04.460 ","End":"05:10.820","Text":"is equal to our B parallel over here,"},{"Start":"05:10.820 ","End":"05:14.730","Text":"which is, as we saw, equal to 0."},{"Start":"05:15.410 ","End":"05:20.010","Text":"Our E_x tag is equal to our E_x,"},{"Start":"05:20.010 ","End":"05:21.915","Text":"which is equal to 0."},{"Start":"05:21.915 ","End":"05:26.080","Text":"Now let\u0027s look at our E perpendicular."},{"Start":"05:26.210 ","End":"05:29.440","Text":"We\u0027re just plugging into the equation."},{"Start":"05:29.440 ","End":"05:38.560","Text":"We have E perpendicular is equal to Gamma multiplied by E tag perpendicular,"},{"Start":"05:38.560 ","End":"05:41.285","Text":"which is equal to this."},{"Start":"05:41.285 ","End":"05:45.369","Text":"E_0 in the z direction"},{"Start":"05:45.860 ","End":"05:52.045","Text":"minus our v cross B tag perpendicular,"},{"Start":"05:52.045 ","End":"05:54.850","Text":"where we know that our B tag perpendicular is"},{"Start":"05:54.850 ","End":"05:57.955","Text":"equal to 0 because our magnetic field is equal to 0,"},{"Start":"05:57.955 ","End":"06:01.825","Text":"so v cross 0 is going to equal to 0."},{"Start":"06:01.825 ","End":"06:10.000","Text":"In total, our E perpendicular is equal to Gamma E_0 in the z direction."},{"Start":"06:10.010 ","End":"06:15.705","Text":"Then let\u0027s do our B perpendicular."},{"Start":"06:15.705 ","End":"06:21.190","Text":"This is equal to Gamma multiplied by B tag perpendicular,"},{"Start":"06:21.190 ","End":"06:27.225","Text":"which we know is equal to 0 because B tag is equal to 0."},{"Start":"06:27.225 ","End":"06:32.273","Text":"Minus and a minus and plus,1 divided by c^2,"},{"Start":"06:32.273 ","End":"06:38.380","Text":"and then we have v which is v in the x direction."},{"Start":"06:38.380 ","End":"06:43.580","Text":"Cross-product with E tag perpendicular which,"},{"Start":"06:43.580 ","End":"06:45.400","Text":"as we said over here, is this."},{"Start":"06:45.400 ","End":"06:50.360","Text":"Cross multiplied by E_0 in the z-direction."},{"Start":"06:50.360 ","End":"07:00.690","Text":"Then what we\u0027re left with is Gamma multiplied by 1 divided by c^2 multiplied by vE_0."},{"Start":"07:00.690 ","End":"07:02.274","Text":"Then we have x hat,"},{"Start":"07:02.274 ","End":"07:07.680","Text":"cross z hat, which is just equal to negative y hat."},{"Start":"07:08.750 ","End":"07:14.285","Text":"This is the perpendicular electric field that"},{"Start":"07:14.285 ","End":"07:18.570","Text":"the observer in the lab will"},{"Start":"07:18.570 ","End":"07:20.000","Text":"measure and this is"},{"Start":"07:20.000 ","End":"07:24.845","Text":"the perpendicular magnetic field that the observer in the lab will measure."},{"Start":"07:24.845 ","End":"07:30.725","Text":"Regarding the parallel fields for electric field and magnetic field,"},{"Start":"07:30.725 ","End":"07:32.660","Text":"it will measure 0."},{"Start":"07:32.660 ","End":"07:35.520","Text":"That\u0027s the end of the lesson."}],"ID":21418},{"Watched":false,"Name":"Transformation of Charge Distribution","Duration":"8m 47s","ChapterTopicVideoID":21537,"CourseChapterTopicPlaylistID":99485,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/21537.jpeg","UploadDate":"2020-04-21T16:18:55.8530000","DurationForVideoObject":"PT8M47S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.940","Text":"Hello. In this lesson we\u0027re going to be speaking about"},{"Start":"00:02.940 ","End":"00:06.330","Text":"the transformation of charge distribution."},{"Start":"00:06.330 ","End":"00:10.845","Text":"If we have some frame of reference over here,"},{"Start":"00:10.845 ","End":"00:15.225","Text":"S and another frame of reference which is moving"},{"Start":"00:15.225 ","End":"00:21.670","Text":"relative to our S frame of reference with some velocity v in this direction."},{"Start":"00:21.670 ","End":"00:30.515","Text":"If our S frame of reference measures some charge distribution Sigma,"},{"Start":"00:30.515 ","End":"00:38.800","Text":"our S tag frame of reference will measure a different charge distribution, Sigma tag."},{"Start":"00:39.100 ","End":"00:42.395","Text":"The charge distributions are different."},{"Start":"00:42.395 ","End":"00:49.505","Text":"However, the total charge that each reference frame will measure is the same,"},{"Start":"00:49.505 ","End":"00:53.700","Text":"just the distribution of the charge is different."},{"Start":"00:53.950 ","End":"00:57.200","Text":"Let\u0027s talk about why this happens."},{"Start":"00:57.200 ","End":"01:01.640","Text":"Let\u0027s imagine that we have some rod over here,"},{"Start":"01:01.640 ","End":"01:06.830","Text":"and it has a charge distribution of Lambda naught."},{"Start":"01:06.830 ","End":"01:08.525","Text":"What is Lambda naught?"},{"Start":"01:08.525 ","End":"01:13.930","Text":"Lambda naught means that we\u0027re in the rod\u0027s frame of reference."},{"Start":"01:13.930 ","End":"01:17.440","Text":"The rod has its own frame of reference and we\u0027re looking there."},{"Start":"01:17.440 ","End":"01:20.218","Text":"Relative to its own frame of reference,"},{"Start":"01:20.218 ","End":"01:23.600","Text":"it has a charge distribution of Lambda naught and we can"},{"Start":"01:23.600 ","End":"01:28.015","Text":"say that it has a length of L naught."},{"Start":"01:28.015 ","End":"01:34.285","Text":"The length of the rod not in its own frame of reference,"},{"Start":"01:34.285 ","End":"01:35.780","Text":"the length of the rod,"},{"Start":"01:35.780 ","End":"01:41.585","Text":"either in the S or S tag reference frame"},{"Start":"01:41.585 ","End":"01:47.870","Text":"is going to be equal to its length in its own reference frame,"},{"Start":"01:47.870 ","End":"01:50.840","Text":"so L naught divided by Gamma."},{"Start":"01:50.840 ","End":"01:54.350","Text":"This we saw in the chapter where we spoke about"},{"Start":"01:54.350 ","End":"01:59.310","Text":"special relativity in the mechanics section."},{"Start":"01:59.950 ","End":"02:04.270","Text":"The same total charge is measured"},{"Start":"02:04.270 ","End":"02:09.170","Text":"throughout because there\u0027s a certain number of charges in this rod,"},{"Start":"02:09.170 ","End":"02:12.910","Text":"so we\u0027re going to measure the same number of charges."},{"Start":"02:12.910 ","End":"02:15.665","Text":"The total charge, as we know,"},{"Start":"02:15.665 ","End":"02:17.580","Text":"is equal to Q."},{"Start":"02:17.580 ","End":"02:20.180","Text":"This Q is equal to,"},{"Start":"02:20.180 ","End":"02:22.655","Text":"when we\u0027re looking at the rods frame of reference,"},{"Start":"02:22.655 ","End":"02:24.860","Text":"is equal to its charge distribution,"},{"Start":"02:24.860 ","End":"02:29.195","Text":"Lambda naught multiplied by its length, L naught."},{"Start":"02:29.195 ","End":"02:32.915","Text":"This is of course equal to,"},{"Start":"02:32.915 ","End":"02:37.495","Text":"if we\u0027re looking at 1 of the other frames of reference, the Lambda."},{"Start":"02:37.495 ","End":"02:41.615","Text":"The charge distribution measured in that specific frame of reference"},{"Start":"02:41.615 ","End":"02:47.590","Text":"multiplied by the length measured in that specific frame of reference."},{"Start":"02:48.620 ","End":"02:52.080","Text":"Now let\u0027s isolate out our Lambda."},{"Start":"02:52.080 ","End":"03:00.290","Text":"Lambda is simply equal to Lambda naught multiplied by L naught divided by"},{"Start":"03:00.290 ","End":"03:03.350","Text":"L. All I\u0027ve done is divided both sides over here by"},{"Start":"03:03.350 ","End":"03:08.960","Text":"L. L naught divided by L. If we look at this equation,"},{"Start":"03:08.960 ","End":"03:12.110","Text":"if we isolate out my Gamma over here,"},{"Start":"03:12.110 ","End":"03:19.400","Text":"what I\u0027ll get is that Gamma is equal to L naught divided by L. I can see that"},{"Start":"03:19.400 ","End":"03:23.060","Text":"my Lambda measured in 1 of my reference frames is equal"},{"Start":"03:23.060 ","End":"03:28.780","Text":"to Lambda naught multiplied by Gamma."},{"Start":"03:30.140 ","End":"03:32.580","Text":"This is an equation."},{"Start":"03:32.580 ","End":"03:35.785","Text":"The equation for Lambda,"},{"Start":"03:35.785 ","End":"03:41.090","Text":"which is the charge distribution in a certain reference frame."},{"Start":"03:42.410 ","End":"03:46.300","Text":"The charges region and a certain reference frame,"},{"Start":"03:46.300 ","End":"03:51.815","Text":"Lambda naught is the charge distribution in its own reference frame."},{"Start":"03:51.815 ","End":"03:54.330","Text":"Gamma over here is, as we know,"},{"Start":"03:54.330 ","End":"03:58.150","Text":"1 divided by the square root of 1 minus,"},{"Start":"03:58.150 ","End":"04:03.260","Text":"this is a minus v divided by c squared."},{"Start":"04:04.190 ","End":"04:08.605","Text":"We\u0027ve looked at charge distribution per unit length."},{"Start":"04:08.605 ","End":"04:13.755","Text":"What about charge distribution per unit area?"},{"Start":"04:13.755 ","End":"04:23.965","Text":"Again, we have our S reference frame and our S tag reference frame with some observer."},{"Start":"04:23.965 ","End":"04:29.680","Text":"But this time we have some plane with a total charge Q."},{"Start":"04:29.680 ","End":"04:34.865","Text":"Its dimensions are b naught and a naught."},{"Start":"04:34.865 ","End":"04:41.080","Text":"The S tag reference frame is traveling with this velocity v in this direction."},{"Start":"04:41.080 ","End":"04:45.200","Text":"First of all, because our length over here,"},{"Start":"04:45.200 ","End":"04:49.415","Text":"a naught is perpendicular to the direction of travel,"},{"Start":"04:49.415 ","End":"04:53.270","Text":"we can say that its length is uniform."},{"Start":"04:53.270 ","End":"04:57.335","Text":"It isn\u0027t changing because it\u0027s perpendicular to the travel."},{"Start":"04:57.335 ","End":"05:06.210","Text":"We have a and the S or S tag reference frame is equal to just a naught."},{"Start":"05:06.210 ","End":"05:12.760","Text":"The length in this plains own reference frame."},{"Start":"05:14.510 ","End":"05:18.670","Text":"However, b is going to get shorter because b is"},{"Start":"05:18.670 ","End":"05:22.930","Text":"parallel to the direction of travel of our S tag reference frame."},{"Start":"05:22.930 ","End":"05:26.540","Text":"The length of b is going to change."},{"Start":"05:26.540 ","End":"05:30.520","Text":"B in 1of the other reference frames,"},{"Start":"05:30.520 ","End":"05:38.525","Text":"S or S tag reference frame is going to be equal to b naught."},{"Start":"05:38.525 ","End":"05:46.730","Text":"The size or length of b naught in its own reference frame and divide it by Gamma."},{"Start":"05:46.730 ","End":"05:50.200","Text":"Now if we look at the total charge, we know it\u0027s Q,"},{"Start":"05:50.200 ","End":"05:56.540","Text":"and that is simply going to be equal to Sigma naught."},{"Start":"05:56.630 ","End":"06:01.355","Text":"That\u0027s the charge distribution over here."},{"Start":"06:01.355 ","End":"06:05.060","Text":"Then that is multiplied by the dimensions of the plane,"},{"Start":"06:05.060 ","End":"06:08.045","Text":"which is a naught, b naught."},{"Start":"06:08.045 ","End":"06:09.560","Text":"But on the other hand,"},{"Start":"06:09.560 ","End":"06:10.760","Text":"in the other reference frame,"},{"Start":"06:10.760 ","End":"06:13.775","Text":"it\u0027s going to be equal to Sigma"},{"Start":"06:13.775 ","End":"06:18.755","Text":"that we measure in the other reference frame multiplied by a,"},{"Start":"06:18.755 ","End":"06:21.095","Text":"which as we said,"},{"Start":"06:21.095 ","End":"06:22.440","Text":"is equal to a naught."},{"Start":"06:22.440 ","End":"06:26.330","Text":"We can just put it here because a naught isn\u0027t changing and"},{"Start":"06:26.330 ","End":"06:31.640","Text":"multiply it by some dimension b over here because we saw that that is changing."},{"Start":"06:31.640 ","End":"06:39.735","Text":"We can divide both sides by a naught and then we can isolate out Sigma."},{"Start":"06:39.735 ","End":"06:41.610","Text":"We divide both sides by b."},{"Start":"06:41.610 ","End":"06:50.750","Text":"What we get is that Sigma is equal to Sigma naught multiplied by b naught divided by b."},{"Start":"06:50.750 ","End":"06:55.100","Text":"Then what we can do is we can overhear isolate out our Gamma."},{"Start":"06:55.100 ","End":"07:00.515","Text":"We get that our Gamma is equal to b naught divided by b."},{"Start":"07:00.515 ","End":"07:06.855","Text":"In other words, our Sigma is equal to Sigma naught multiplied by Gamma."},{"Start":"07:06.855 ","End":"07:12.635","Text":"What we get is that Sigma is equal to Gamma"},{"Start":"07:12.635 ","End":"07:19.440","Text":"multiplied by Sigma naught."},{"Start":"07:19.580 ","End":"07:24.470","Text":"Of course, if we do the exact same calculation, but for row,"},{"Start":"07:24.470 ","End":"07:27.275","Text":"so charge density per unit volume,"},{"Start":"07:27.275 ","End":"07:29.585","Text":"we will get the same relationship."},{"Start":"07:29.585 ","End":"07:32.705","Text":"Of course this is also true for if"},{"Start":"07:32.705 ","End":"07:35.450","Text":"our charge distribution is not uniform because"},{"Start":"07:35.450 ","End":"07:38.725","Text":"then what we\u0027ll do is we\u0027ll just look at small squares."},{"Start":"07:38.725 ","End":"07:43.060","Text":"Then the charge on each small square is dq,"},{"Start":"07:43.060 ","End":"07:45.585","Text":"which is equal to in the own reference frame,"},{"Start":"07:45.585 ","End":"07:46.850","Text":"let\u0027s say over here,"},{"Start":"07:46.850 ","End":"07:53.545","Text":"Sigma naught multiplied by dx naught multiplied by dy naught."},{"Start":"07:53.545 ","End":"07:56.899","Text":"Relative to another frame of reference,"},{"Start":"07:56.899 ","End":"08:00.650","Text":"it will be equal to some charge distribution,"},{"Start":"08:00.650 ","End":"08:07.770","Text":"Sigma multiplied by some dx,"},{"Start":"08:07.770 ","End":"08:14.300","Text":"which is changing multiplied by dy naught because our y is"},{"Start":"08:14.300 ","End":"08:17.480","Text":"perpendicular to the direction of travel of v. We can"},{"Start":"08:17.480 ","End":"08:21.105","Text":"cancel both sides over here with the dy naught."},{"Start":"08:21.105 ","End":"08:31.499","Text":"Then what we\u0027re left with is that Sigma dx is equal to Sigma naught dx naught."},{"Start":"08:31.810 ","End":"08:35.690","Text":"Then we can just integrate and follow these rules."},{"Start":"08:35.690 ","End":"08:39.185","Text":"But what we can see is that the same relationship even"},{"Start":"08:39.185 ","End":"08:43.595","Text":"when the charge density is non-uniform is kept."},{"Start":"08:43.595 ","End":"08:46.770","Text":"That\u0027s the end of this lesson."}],"ID":22391},{"Watched":false,"Name":"Transformation of Current Density","Duration":"2m 18s","ChapterTopicVideoID":21339,"CourseChapterTopicPlaylistID":99485,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/21339.jpeg","UploadDate":"2020-04-06T21:41:55.9130000","DurationForVideoObject":"PT2M18S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.760","Text":"Hello. In this lesson,"},{"Start":"00:01.760 ","End":"00:06.185","Text":"we\u0027re going to be learning about the transformation of current density."},{"Start":"00:06.185 ","End":"00:10.025","Text":"Here we have some kind of frame of reference."},{"Start":"00:10.025 ","End":"00:14.790","Text":"Then what we have is some kind of charged cylinder with"},{"Start":"00:14.790 ","End":"00:20.145","Text":"charged density per unit volume, Rho,"},{"Start":"00:20.145 ","End":"00:24.210","Text":"and it is traveling in this direction with some kind of velocity,"},{"Start":"00:24.210 ","End":"00:31.635","Text":"V. The current density is going to be equal to Rho multiplied by V,"},{"Start":"00:31.635 ","End":"00:40.450","Text":"where Rho is the charge density in the reference frame where we are calculating J."},{"Start":"00:41.390 ","End":"00:47.210","Text":"This Rho is in the frame of reference where we are calculating J."},{"Start":"00:47.210 ","End":"00:49.340","Text":"In other words, in the question,"},{"Start":"00:49.340 ","End":"00:58.095","Text":"we\u0027ll usually be given the charge density in the cylinder\u0027s own reference frame,"},{"Start":"00:58.095 ","End":"01:01.380","Text":"or in other words, in the rest frame."},{"Start":"01:01.380 ","End":"01:03.245","Text":"That will be equal to,"},{"Start":"01:03.245 ","End":"01:05.650","Text":"as we saw in the previous lesson,"},{"Start":"01:05.650 ","End":"01:08.625","Text":"Gamma multiplied by Rho naught."},{"Start":"01:08.625 ","End":"01:15.895","Text":"In other words, J will be equal to Gamma multiplied by Rho naught multiplied by V,"},{"Start":"01:15.895 ","End":"01:23.250","Text":"where this is the charge density in the cylinder\u0027s rest frame."},{"Start":"01:23.250 ","End":"01:28.360","Text":"Then similarly, if we\u0027re calculating k"},{"Start":"01:28.360 ","End":"01:35.030","Text":"and the k that we\u0027re calculating isn\u0027t in the cylinder\u0027s rest frame."},{"Start":"01:35.030 ","End":"01:39.045","Text":"Our k, instead of being equal to Sigma V,"},{"Start":"01:39.045 ","End":"01:44.430","Text":"it will be equal to Gamma Sigma naught V, where of course,"},{"Start":"01:44.430 ","End":"01:48.005","Text":"Sigma naught is the charge density per"},{"Start":"01:48.005 ","End":"01:55.550","Text":"unit area in the object\u0027s own reference frame or in its rest frame."},{"Start":"01:55.550 ","End":"02:03.140","Text":"In the same way, I will be equal to either gamma V if we\u0027re in the objects rest frame."},{"Start":"02:03.140 ","End":"02:05.090","Text":"But if we\u0027re calculating I in"},{"Start":"02:05.090 ","End":"02:09.500","Text":"a different reference frame to the object\u0027s rest frame, so similarly,"},{"Start":"02:09.500 ","End":"02:16.010","Text":"it will be Gamma multiplied by Lambda naught multiplied by V, and that\u0027s it."},{"Start":"02:16.010 ","End":"02:18.810","Text":"That is the end of this lesson."}],"ID":21419},{"Watched":false,"Name":"Exercise 2","Duration":"20m 14s","ChapterTopicVideoID":24816,"CourseChapterTopicPlaylistID":99485,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/24816.jpeg","UploadDate":"2021-12-29T07:33:10.8400000","DurationForVideoObject":"PT20M14S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.590","Text":"Hello. In this lesson,"},{"Start":"00:01.590 ","End":"00:04.245","Text":"we\u0027re going to be answering the following question."},{"Start":"00:04.245 ","End":"00:09.375","Text":"An infinite plane has a charge density of Sigma."},{"Start":"00:09.375 ","End":"00:16.740","Text":"The plane travels at a constant velocity of v in the x direction relative to the lab."},{"Start":"00:16.740 ","End":"00:21.960","Text":"What we have here is our S frame of reference. This is our lab."},{"Start":"00:21.960 ","End":"00:23.480","Text":"Then over here,"},{"Start":"00:23.480 ","End":"00:27.195","Text":"we have our S tag frame of reference where we have"},{"Start":"00:27.195 ","End":"00:33.555","Text":"some infinite plane with charge density per unit area Sigma."},{"Start":"00:33.555 ","End":"00:38.010","Text":"If we say that this is the x-direction,"},{"Start":"00:38.010 ","End":"00:47.980","Text":"so we can see that the plane is traveling at a velocity v in the x direction."},{"Start":"00:47.980 ","End":"00:52.070","Text":"Now we\u0027re being told to calculate the electric and magnetic fields,"},{"Start":"00:52.070 ","End":"00:57.595","Text":"as well as the charge and current densities in the lab\u0027s reference frame."},{"Start":"00:57.595 ","End":"00:59.960","Text":"We\u0027re being asked to use both methods."},{"Start":"00:59.960 ","End":"01:05.100","Text":"But also fast, let\u0027s use the first method."},{"Start":"01:05.180 ","End":"01:07.820","Text":"First, we\u0027re going to calculate"},{"Start":"01:07.820 ","End":"01:12.230","Text":"the electric and magnetic fields in the plane\u0027s rest frame,"},{"Start":"01:12.230 ","End":"01:16.369","Text":"where we\u0027re in this S tag frame of reference,"},{"Start":"01:16.369 ","End":"01:22.935","Text":"using the charge distribution over here, so Sigma."},{"Start":"01:22.935 ","End":"01:26.680","Text":"If we\u0027re in the plane\u0027s reference frame,"},{"Start":"01:26.680 ","End":"01:34.715","Text":"that means that its velocity relative to itself is equal to 0."},{"Start":"01:34.715 ","End":"01:42.190","Text":"Let\u0027s just write over here V tag because we\u0027re in it\u0027s own frame of reference."},{"Start":"01:42.190 ","End":"01:44.345","Text":"It\u0027s equal to 0."},{"Start":"01:44.345 ","End":"01:49.270","Text":"Now let\u0027s see what the electric field is equal to."},{"Start":"01:49.270 ","End":"01:52.750","Text":"From Gauss\u0027s Law, we know that E tag,"},{"Start":"01:52.750 ","End":"01:55.555","Text":"the electric field in this frame of reference,"},{"Start":"01:55.555 ","End":"01:59.395","Text":"is the electric field of an infinite plane,"},{"Start":"01:59.395 ","End":"02:08.085","Text":"which is equal to Sigma divided by 2 Epsilon naught in the z direction."},{"Start":"02:08.085 ","End":"02:13.530","Text":"This is, of course, above the plane."},{"Start":"02:14.690 ","End":"02:18.665","Text":"Then the magnetic field, of course,"},{"Start":"02:18.665 ","End":"02:21.250","Text":"is going to be equal to 0,"},{"Start":"02:21.250 ","End":"02:26.535","Text":"because there\u0027s nothing here telling us that we have a magnetic field."},{"Start":"02:26.535 ","End":"02:30.565","Text":"This was a. Now,"},{"Start":"02:30.565 ","End":"02:32.835","Text":"let\u0027s take a look at b."},{"Start":"02:32.835 ","End":"02:40.170","Text":"First of all, these are the values that we need to keep an eye on."},{"Start":"02:40.170 ","End":"02:45.120","Text":"First of all, let\u0027s look at our E parallel."},{"Start":"02:45.120 ","End":"02:51.390","Text":"Our E tag parallel is equal to our E parallel."},{"Start":"02:51.390 ","End":"02:53.800","Text":"What is this parallel?"},{"Start":"02:53.800 ","End":"03:01.070","Text":"It means the electric field parallel to the direction of travel of the plane."},{"Start":"03:01.070 ","End":"03:02.510","Text":"The direction of travel,"},{"Start":"03:02.510 ","End":"03:04.040","Text":"although the velocity is equal to 0,"},{"Start":"03:04.040 ","End":"03:08.135","Text":"it\u0027s moving along the x-axis."},{"Start":"03:08.135 ","End":"03:12.635","Text":"The parallel direction to the x axis is the x-axis."},{"Start":"03:12.635 ","End":"03:15.645","Text":"Let\u0027s just write that actually."},{"Start":"03:15.645 ","End":"03:21.705","Text":"That\u0027s the E tag in the x-direction."},{"Start":"03:21.705 ","End":"03:28.370","Text":"This is of course equal to 0 because we don\u0027t have an x component for our electric field."},{"Start":"03:28.370 ","End":"03:30.020","Text":"This is equal to 0."},{"Start":"03:30.020 ","End":"03:36.920","Text":"Then of course, B tag parallel is equal to B parallel."},{"Start":"03:36.920 ","End":"03:41.150","Text":"Of course, B tag=0 in all the directions."},{"Start":"03:41.150 ","End":"03:44.220","Text":"So this is also equal to 0."},{"Start":"03:44.590 ","End":"03:51.000","Text":"Now let\u0027s look at E perpendicular."},{"Start":"03:51.050 ","End":"03:58.474","Text":"This is going to be equal to Gamma multiplied by"},{"Start":"03:58.474 ","End":"04:07.290","Text":"E perpendicular tag minus V cross B tag."},{"Start":"04:07.290 ","End":"04:10.115","Text":"This is going to be equal to Gamma,"},{"Start":"04:10.115 ","End":"04:12.894","Text":"E tag perpendicular is this."},{"Start":"04:12.894 ","End":"04:21.935","Text":"So Sigma divided by 2 Epsilon_0 in the z direction minus V cross V. V,"},{"Start":"04:21.935 ","End":"04:26.686","Text":"as we said, is equal to 0 in the rest frame of the plane."},{"Start":"04:26.686 ","End":"04:29.560","Text":"So 0 multiplied by something is equal to 0."},{"Start":"04:29.560 ","End":"04:31.665","Text":"So we have minus 0."},{"Start":"04:31.665 ","End":"04:38.210","Text":"What we get is that our E perpendicular is equal to Gamma multiplied"},{"Start":"04:38.210 ","End":"04:45.090","Text":"by Sigma divided by 2 Epsilon_0 in the z direction."},{"Start":"04:45.740 ","End":"04:51.295","Text":"Because this is the only component of the electric field,"},{"Start":"04:51.295 ","End":"04:55.630","Text":"because the parallel component is equal to 0,"},{"Start":"04:55.630 ","End":"05:00.780","Text":"so this is also equal to E perpendicular."},{"Start":"05:00.780 ","End":"05:07.405","Text":"It\u0027s of course, also equal to the electric field because there\u0027s no parallel component."},{"Start":"05:07.405 ","End":"05:11.200","Text":"Then let\u0027s take a look at what our B field is."},{"Start":"05:11.200 ","End":"05:14.605","Text":"We already said that B parallel is equal to 0."},{"Start":"05:14.605 ","End":"05:16.915","Text":"But what about B perpendicular?"},{"Start":"05:16.915 ","End":"05:25.725","Text":"This is equal to Gamma multiplied by B perpendicular tag"},{"Start":"05:25.725 ","End":"05:31.950","Text":"plus 1 divided by c^2 multiplied"},{"Start":"05:31.950 ","End":"05:38.910","Text":"by V cross E perpendicular tag."},{"Start":"05:38.910 ","End":"05:46.670","Text":"We have Gamma multiplied by B perpendicular is equal to 0,"},{"Start":"05:46.670 ","End":"05:52.040","Text":"because the total B tag magnetic field is equal to 0,"},{"Start":"05:52.040 ","End":"05:55.920","Text":"so the perpendicular component is also equal to 0."},{"Start":"05:55.920 ","End":"05:59.775","Text":"So 0+1 divided by c^2."},{"Start":"05:59.775 ","End":"06:03.460","Text":"Then we have our V,"},{"Start":"06:03.560 ","End":"06:09.580","Text":"which is V in the x direction."},{"Start":"06:11.210 ","End":"06:15.275","Text":"Remember, here we were meant to have a minus."},{"Start":"06:15.275 ","End":"06:23.765","Text":"But then this V is because we\u0027re going from the tag field to the non-tag field,"},{"Start":"06:23.765 ","End":"06:25.430","Text":"from S tag to S,"},{"Start":"06:25.430 ","End":"06:29.300","Text":"so the velocity becomes a negative over here."},{"Start":"06:29.300 ","End":"06:33.685","Text":"Then we have a negative and a negative and that becomes a plus."},{"Start":"06:33.685 ","End":"06:35.690","Text":"That\u0027s why there\u0027s a plus here."},{"Start":"06:35.690 ","End":"06:40.175","Text":"Vx cross the E perpendicular."},{"Start":"06:40.175 ","End":"06:42.275","Text":"Let\u0027s just scroll over here."},{"Start":"06:42.275 ","End":"06:44.030","Text":"E perpendicular tags, sorry,"},{"Start":"06:44.030 ","End":"06:50.910","Text":"so that\u0027s Sigma divided by 2 Epsilon_0 in the z direction."},{"Start":"06:50.910 ","End":"06:55.910","Text":"This is simply going to be equal to Gamma divided"},{"Start":"06:55.910 ","End":"07:02.835","Text":"by c^2 multiplied by V,"},{"Start":"07:02.835 ","End":"07:08.055","Text":"multiplied by Sigma divided by 2 Epsilon_0."},{"Start":"07:08.055 ","End":"07:11.330","Text":"Then we have x cross z,"},{"Start":"07:11.330 ","End":"07:14.165","Text":"which is equal to negative y."},{"Start":"07:14.165 ","End":"07:17.330","Text":"All of this is in the negative y-direction."},{"Start":"07:17.330 ","End":"07:20.765","Text":"This is equal to B perpendicular."},{"Start":"07:20.765 ","End":"07:23.870","Text":"But also because there\u0027s no parallel component,"},{"Start":"07:23.870 ","End":"07:28.590","Text":"this is also equal to the entire B field."},{"Start":"07:29.390 ","End":"07:33.980","Text":"The next stage is to calculate the charge and current densities in"},{"Start":"07:33.980 ","End":"07:40.325","Text":"the lab\u0027s reference frame using the fields calculated in the previous step."},{"Start":"07:40.325 ","End":"07:44.235","Text":"Let\u0027s just scroll a little bit."},{"Start":"07:44.235 ","End":"07:54.575","Text":"Now what we\u0027re doing c. The charge density in the lab,"},{"Start":"07:54.575 ","End":"07:56.285","Text":"so I\u0027ll write L,"},{"Start":"07:56.285 ","End":"08:03.045","Text":"is simply equal to Epsilon_0 multiplied by Delta E,"},{"Start":"08:03.045 ","End":"08:05.265","Text":"the jump in the E field."},{"Start":"08:05.265 ","End":"08:07.415","Text":"What is the jump in the E field?"},{"Start":"08:07.415 ","End":"08:11.590","Text":"Just like here, we said that the E field was above the plane."},{"Start":"08:11.590 ","End":"08:17.090","Text":"Also over here, the E field and the B fields are above the plane."},{"Start":"08:17.090 ","End":"08:19.055","Text":"If here we have the plane,"},{"Start":"08:19.055 ","End":"08:29.855","Text":"so our E over here is equal to Gamma Sigma divided by 2 Epsilon_0 in the z direction."},{"Start":"08:29.855 ","End":"08:37.605","Text":"The B field is equal to Gamma V Sigma divided by"},{"Start":"08:37.605 ","End":"08:41.595","Text":"2 Epsilon_0 c^2 in"},{"Start":"08:41.595 ","End":"08:48.410","Text":"the negative y direction."},{"Start":"08:48.410 ","End":"08:54.750","Text":"Below the plane, our E over here is equal to negative this,"},{"Start":"08:54.750 ","End":"08:59.480","Text":"so negative Gamma Sigma divided by 2 Epsilon_0."},{"Start":"08:59.480 ","End":"09:04.220","Text":"The B field below the plane is equal to negative,"},{"Start":"09:04.220 ","End":"09:06.095","Text":"so this negative cancels out."},{"Start":"09:06.095 ","End":"09:10.635","Text":"It\u0027s just Gamma V Sigma divided by"},{"Start":"09:10.635 ","End":"09:16.980","Text":"2 Epsilon_0 c^2 and just in the y-direction this time."},{"Start":"09:16.980 ","End":"09:19.730","Text":"The jump in the electric field."},{"Start":"09:19.730 ","End":"09:21.140","Text":"Let\u0027s go back to this."},{"Start":"09:21.140 ","End":"09:29.780","Text":"Sigma in the lab is equal to Epsilon_0 multiplied by the electric field at the top,"},{"Start":"09:29.780 ","End":"09:36.485","Text":"which is Gamma Sigma divided by 2 Epsilon naught minus the electric field on the bottom,"},{"Start":"09:36.485 ","End":"09:42.465","Text":"which is negative Gamma Sigma divided by 2 Epsilon_0."},{"Start":"09:42.465 ","End":"09:47.400","Text":"Then this is simply equal to."},{"Start":"09:47.400 ","End":"09:55.455","Text":"So we have gamma sigma divided by 2 Epsilon_0 plus Gamma Sigma divided by 2 Epsilon_0."},{"Start":"09:55.455 ","End":"09:58.110","Text":"This is simply going to be equal"},{"Start":"09:58.110 ","End":"10:05.115","Text":"to Gamma Sigma divided by Epsilon_0 multiplied by Epsilon_0."},{"Start":"10:05.115 ","End":"10:08.824","Text":"This is just equal to Gamma Sigma,"},{"Start":"10:08.824 ","End":"10:12.410","Text":"which is exactly what we would have expected to get from"},{"Start":"10:12.410 ","End":"10:16.145","Text":"our previous lesson where we learned about this."},{"Start":"10:16.145 ","End":"10:20.180","Text":"Then let\u0027s take a look at our current density."},{"Start":"10:20.180 ","End":"10:23.150","Text":"The current density, because we\u0027re dealing with a plane."},{"Start":"10:23.150 ","End":"10:28.260","Text":"So we\u0027re using k. So k=1 divided"},{"Start":"10:28.260 ","End":"10:34.440","Text":"by Mu_0 multiplied by the jump in the magnetic fields."},{"Start":"10:34.670 ","End":"10:39.369","Text":"That means that this is equal to 1 divided by Mu_0"},{"Start":"10:39.369 ","End":"10:44.015","Text":"multiplied by the magnetic field on top,"},{"Start":"10:44.015 ","End":"10:46.250","Text":"which is over here we have a minus."},{"Start":"10:46.250 ","End":"10:54.490","Text":"So it\u0027s minus Gamma V Sigma divided by 2 Epsilon_0 c^2."},{"Start":"10:54.530 ","End":"11:00.800","Text":"Then we have minus the magnetic field at the bottom,"},{"Start":"11:00.800 ","End":"11:05.190","Text":"which is Gamma V Sigma divided"},{"Start":"11:05.190 ","End":"11:11.010","Text":"by 2 Epsilon_0 c^2."},{"Start":"11:11.010 ","End":"11:18.130","Text":"Then what we have is the 2s cancel out like so."},{"Start":"11:19.340 ","End":"11:27.950","Text":"Then what we\u0027re left with is 1 divided by Mu_0 multiplied by"},{"Start":"11:27.950 ","End":"11:37.680","Text":"negative Gamma Sigma V divided by c^2 Epsilon naught."},{"Start":"11:37.680 ","End":"11:40.350","Text":"Then as a note,"},{"Start":"11:40.350 ","End":"11:50.470","Text":"we know that 1 divided by C^2=Mu_0 Epsilon_0."},{"Start":"11:51.020 ","End":"11:55.845","Text":"1 divided by c^2 is Mu_0 Epsilon_0."},{"Start":"11:55.845 ","End":"11:58.115","Text":"Therefore, once we plug that in,"},{"Start":"11:58.115 ","End":"12:00.890","Text":"we get that k is equal to"},{"Start":"12:00.890 ","End":"12:09.125","Text":"negative Gamma Sigma V. This is the magnitude of k. But,"},{"Start":"12:09.125 ","End":"12:10.850","Text":"of course, k is a vector."},{"Start":"12:10.850 ","End":"12:14.285","Text":"So we want to know the direction."},{"Start":"12:14.285 ","End":"12:22.280","Text":"As we know, k is in a direction which is perpendicular to"},{"Start":"12:22.280 ","End":"12:29.465","Text":"the jump in the magnetic field"},{"Start":"12:29.465 ","End":"12:34.380","Text":"and perpendicular to the direction of the magnetic field."},{"Start":"12:35.510 ","End":"12:37.725","Text":"Let\u0027s take a look."},{"Start":"12:37.725 ","End":"12:43.940","Text":"If we see, we\u0027ve already said that this is"},{"Start":"12:43.940 ","End":"12:50.285","Text":"the x-direction and that this is the z direction."},{"Start":"12:50.285 ","End":"12:57.270","Text":"That means that this direction over here is the y-direction."},{"Start":"12:57.650 ","End":"13:04.830","Text":"What we can see with the B field is that here, it\u0027s negative y-direction."}],"ID":25729}],"Thumbnail":null,"ID":99485}]

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