Relativistic Transformations for Electric and Magnetic Fields
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[{"Name":"Relativistic Transformations for Electric and Magnetic Fields","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Transformations of Electric and Magnetic Fields","Duration":"7m 12s","ChapterTopicVideoID":12095,"CourseChapterTopicPlaylistID":152630,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/12095.jpeg","UploadDate":"2018-06-28T03:12:39.2300000","DurationForVideoObject":"PT7M12S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.695","Text":"Hello. In this lesson,"},{"Start":"00:01.695 ","End":"00:06.285","Text":"we\u0027re going to be learning about the transformation of electric and magnetic fields."},{"Start":"00:06.285 ","End":"00:08.580","Text":"First of all, we\u0027re going to go over"},{"Start":"00:08.580 ","End":"00:13.870","Text":"some intuition and then we\u0027ll take a look at the equations."},{"Start":"00:14.570 ","End":"00:20.805","Text":"Let\u0027s imagine that we have a uniform electric field"},{"Start":"00:20.805 ","End":"00:29.490","Text":"pointing so in this direction and then we have a charged particle q."},{"Start":"00:29.490 ","End":"00:36.550","Text":"We know that there\u0027s going to be an electrical force F,"},{"Start":"00:36.550 ","End":"00:41.930","Text":"which is simply equal to the charge of the particle multiplied"},{"Start":"00:41.930 ","End":"00:47.630","Text":"by the electric field E. Of course,"},{"Start":"00:47.630 ","End":"00:52.710","Text":"the force is acting in the same direction as the electric field."},{"Start":"00:53.620 ","End":"01:04.530","Text":"Now let\u0027s assume that we are an observer and we are located in this S tag reference frame"},{"Start":"01:04.530 ","End":"01:10.040","Text":"and this S tag reference frame is moving rightwards with a velocity of"},{"Start":"01:10.040 ","End":"01:17.310","Text":"V. This is our laboratory"},{"Start":"01:17.310 ","End":"01:21.010","Text":"so this is where an observer in the lab,"},{"Start":"01:21.010 ","End":"01:24.340","Text":"they will just see this configuration and if"},{"Start":"01:24.340 ","End":"01:27.850","Text":"we\u0027re the observer in this S tag frame of reference,"},{"Start":"01:27.850 ","End":"01:30.295","Text":"the lab can be the S frame of reference"},{"Start":"01:30.295 ","End":"01:33.190","Text":"and we\u0027re moving rightwards with a velocity of V,"},{"Start":"01:33.190 ","End":"01:39.370","Text":"so what we will see is that the charge is moving leftwards with"},{"Start":"01:39.370 ","End":"01:42.460","Text":"a velocity of V. If we\u0027re moving"},{"Start":"01:42.460 ","End":"01:46.600","Text":"rightwards with a velocity of V and where in this frame of reference,"},{"Start":"01:46.600 ","End":"01:50.335","Text":"so according to us in this frame of reference, we\u0027re stationary,"},{"Start":"01:50.335 ","End":"01:55.610","Text":"but this is moving in the opposite direction leftwards with the same velocity."},{"Start":"01:55.610 ","End":"01:58.470","Text":"This we\u0027ve already seen,"},{"Start":"01:58.470 ","End":"02:03.040","Text":"and this is from Galileo\u0027s transformation and also from intuition,"},{"Start":"02:03.040 ","End":"02:06.190","Text":"we can see that this is what will happen."},{"Start":"02:06.190 ","End":"02:08.990","Text":"What we will see,"},{"Start":"02:08.990 ","End":"02:10.885","Text":"let\u0027s draw this again,"},{"Start":"02:10.885 ","End":"02:14.755","Text":"is that we\u0027ll have this charged particle q,"},{"Start":"02:14.755 ","End":"02:19.870","Text":"which is moving leftwards with a velocity of V and we\u0027ll"},{"Start":"02:19.870 ","End":"02:25.615","Text":"have this force F pointing upwards over here."},{"Start":"02:25.615 ","End":"02:30.905","Text":"What we can notice is that F and V are perpendicular to one another."},{"Start":"02:30.905 ","End":"02:34.765","Text":"This is exactly what we get in a magnetic field."},{"Start":"02:34.765 ","End":"02:41.690","Text":"A magnetic field produces a force which is perpendicular to the velocity."},{"Start":"02:42.710 ","End":"02:49.115","Text":"What we can see is that if we have a charged particle q in an electric field,"},{"Start":"02:49.115 ","End":"02:54.820","Text":"and we are moving in a reference frame in some direction with a velocity of V,"},{"Start":"02:54.820 ","End":"03:00.500","Text":"then what we will see from this reference frame is that we are stationary,"},{"Start":"03:00.500 ","End":"03:03.845","Text":"but this charged particle q is"},{"Start":"03:03.845 ","End":"03:08.750","Text":"moving with a velocity of V in the opposite direction and of course,"},{"Start":"03:08.750 ","End":"03:10.205","Text":"because of the electric fields,"},{"Start":"03:10.205 ","End":"03:12.950","Text":"there\u0027s an electric force acting."},{"Start":"03:12.950 ","End":"03:16.670","Text":"Then what we get is that the force is perpendicular to the velocity,"},{"Start":"03:16.670 ","End":"03:19.650","Text":"just like in the magnetic field."},{"Start":"03:20.420 ","End":"03:27.175","Text":"What is happening that the person in the lab or the S frame of reference sees"},{"Start":"03:27.175 ","End":"03:30.930","Text":"an electric force but the person in"},{"Start":"03:30.930 ","End":"03:35.275","Text":"the S tag frame of reference that\u0027s moving with some velocity,"},{"Start":"03:35.275 ","End":"03:39.470","Text":"is going to see a magnetic force."},{"Start":"03:39.710 ","End":"03:45.775","Text":"We can see that for the exact same system in one reference frame,"},{"Start":"03:45.775 ","End":"03:49.030","Text":"we\u0027re seeing an electric force and"},{"Start":"03:49.030 ","End":"03:52.360","Text":"then another reference frame for the exact same system,"},{"Start":"03:52.360 ","End":"03:53.695","Text":"the exact same setup,"},{"Start":"03:53.695 ","End":"03:56.780","Text":"we\u0027re seeing a magnetic force."},{"Start":"03:57.530 ","End":"04:02.695","Text":"What we can see is that depending on where we are as an observer,"},{"Start":"04:02.695 ","End":"04:05.675","Text":"we\u0027ll see an electric force become"},{"Start":"04:05.675 ","End":"04:11.060","Text":"a magnetic force or an electric field become a magnetic fields and vice versa."},{"Start":"04:11.060 ","End":"04:14.615","Text":"Or we can see a magnetic field if this was our lab,"},{"Start":"04:14.615 ","End":"04:18.980","Text":"and then here we would see an electric field."},{"Start":"04:18.980 ","End":"04:24.920","Text":"The proof of this is a very complicated proof so I\u0027m not going to go over it,"},{"Start":"04:24.920 ","End":"04:29.510","Text":"but we\u0027re going to now look at the equations which are very important to know."},{"Start":"04:29.510 ","End":"04:32.660","Text":"If you do want to read up a little bit more about this,"},{"Start":"04:32.660 ","End":"04:38.540","Text":"this comes from the theory of relativity so I won\u0027t go into it now,"},{"Start":"04:38.540 ","End":"04:42.060","Text":"but just know that this is due to relativity."},{"Start":"04:42.560 ","End":"04:52.430","Text":"If we have a case of Observer 1 measuring an electric field E and a magnetic field B,"},{"Start":"04:52.430 ","End":"04:54.860","Text":"so this is in the lab, for instance."},{"Start":"04:54.860 ","End":"04:57.540","Text":"Then Observer 2 in the S tag,"},{"Start":"04:57.540 ","End":"04:58.940","Text":"let\u0027s say frame of reference,"},{"Start":"04:58.940 ","End":"05:05.240","Text":"is moving with a velocity of V relative to Observer 1."},{"Start":"05:05.690 ","End":"05:14.735","Text":"This observer is going to measure this E tag electric field and B tag magnetic field."},{"Start":"05:14.735 ","End":"05:18.920","Text":"E tag will be equal to the electric field"},{"Start":"05:18.920 ","End":"05:24.440","Text":"measured by Observer 1 plus the velocity of Observer 2"},{"Start":"05:24.440 ","End":"05:29.365","Text":"cross multiplied with the magnetic field measured by"},{"Start":"05:29.365 ","End":"05:34.835","Text":"Observer 1 and the B field measured by Observer 2,"},{"Start":"05:34.835 ","End":"05:39.440","Text":"so B tag is equal to the magnetic field measured by Observer"},{"Start":"05:39.440 ","End":"05:45.620","Text":"1 minus the velocity at which Observer 2 is traveling with"},{"Start":"05:45.620 ","End":"05:52.460","Text":"cross multiplied by the E-field that Observer 1 measures and divide it by c^2 where"},{"Start":"05:52.460 ","End":"06:01.200","Text":"c is the velocity of light and it is equal to 3 times 10 to the 8 meters per second."},{"Start":"06:03.440 ","End":"06:05.945","Text":"This is the rule."},{"Start":"06:05.945 ","End":"06:10.805","Text":"Now the only thing we have to take into account is that V,"},{"Start":"06:10.805 ","End":"06:15.275","Text":"the velocity at which observer Number 2 is traveling at,"},{"Start":"06:15.275 ","End":"06:18.795","Text":"is much smaller than c,"},{"Start":"06:18.795 ","End":"06:20.850","Text":"the speed of light."},{"Start":"06:20.850 ","End":"06:25.935","Text":"This is true in most cases and most questions."},{"Start":"06:25.935 ","End":"06:32.300","Text":"Not many things can travel even close to the speed of light but it\u0027s important to know,"},{"Start":"06:32.300 ","End":"06:35.780","Text":"so if you\u0027re traveling at a velocity which is close to the speed of light,"},{"Start":"06:35.780 ","End":"06:38.825","Text":"then you cannot use these equations."},{"Start":"06:38.825 ","End":"06:42.360","Text":"This is very important to remember."},{"Start":"06:42.710 ","End":"06:48.170","Text":"This is the condition for using the equations and because we\u0027re taking"},{"Start":"06:48.170 ","End":"06:51.260","Text":"this approximation that the velocity that"},{"Start":"06:51.260 ","End":"06:54.725","Text":"Observer 2 is traveling at is significantly smaller than the speed of light,"},{"Start":"06:54.725 ","End":"06:59.540","Text":"you can also say that E tag is approximately equal to this and"},{"Start":"06:59.540 ","End":"07:05.940","Text":"that B tag is approximately equal to this because we\u0027re taking this approximation."},{"Start":"07:05.940 ","End":"07:09.245","Text":"Remember these equations, remember the condition for"},{"Start":"07:09.245 ","End":"07:12.900","Text":"it and that is the end of this lesson."}],"ID":23060},{"Watched":false,"Name":"Exercise 1","Duration":"8m 16s","ChapterTopicVideoID":12096,"CourseChapterTopicPlaylistID":152630,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/12096.jpeg","UploadDate":"2018-06-28T03:17:10.1530000","DurationForVideoObject":"PT8M16S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.875","Text":"Hello. In this lesson,"},{"Start":"00:01.875 ","End":"00:04.425","Text":"we\u0027re going to be answering the following question."},{"Start":"00:04.425 ","End":"00:07.710","Text":"A charged particle Q is placed on"},{"Start":"00:07.710 ","End":"00:12.885","Text":"an infinite plane of charge density per unit area of Sigma."},{"Start":"00:12.885 ","End":"00:17.445","Text":"The charged plane is located on the xy plane."},{"Start":"00:17.445 ","End":"00:22.320","Text":"Find the force acting on the particle relative to"},{"Start":"00:22.320 ","End":"00:28.035","Text":"an observer moving with velocity v in the y direction."},{"Start":"00:28.035 ","End":"00:31.590","Text":"Assume that the particle is stationary relative to the plane."},{"Start":"00:31.590 ","End":"00:36.410","Text":"What we have here is that here we have our S frame of reference,"},{"Start":"00:36.410 ","End":"00:40.330","Text":"and here we have our S tag frame of reference."},{"Start":"00:40.330 ","End":"00:44.300","Text":"The S tag frame of reference is moving with a velocity v in"},{"Start":"00:44.300 ","End":"00:50.065","Text":"the y direction relative to our charged particle."},{"Start":"00:50.065 ","End":"00:53.090","Text":"Then we have this charged particle which is subject to"},{"Start":"00:53.090 ","End":"00:56.345","Text":"an electric field from this infinite plane."},{"Start":"00:56.345 ","End":"01:01.070","Text":"Therefore, we\u0027re going to see a force acting"},{"Start":"01:01.070 ","End":"01:06.750","Text":"on this charged particle and we\u0027re being asked to find what this force is."},{"Start":"01:07.420 ","End":"01:16.095","Text":"The first thing we know is that we have this electric field over here being applied to Q,"},{"Start":"01:16.095 ","End":"01:18.480","Text":"due to this infinite plane."},{"Start":"01:18.480 ","End":"01:24.290","Text":"We know that the electric field of an infinite plane from Gauss\u0027s law is"},{"Start":"01:24.290 ","End":"01:26.000","Text":"equal to Sigma divided by"},{"Start":"01:26.000 ","End":"01:29.810","Text":"2 epsilon naught and we can see that it\u0027s going in this direction,"},{"Start":"01:29.810 ","End":"01:33.610","Text":"which here we\u0027ve said is the z direction."},{"Start":"01:34.310 ","End":"01:36.705","Text":"Of course, this is E,"},{"Start":"01:36.705 ","End":"01:39.920","Text":"so this is in the S reference frame."},{"Start":"01:39.920 ","End":"01:45.155","Text":"Then we know that the magnetic field in this reference frame,"},{"Start":"01:45.155 ","End":"01:48.530","Text":"so B is going to be equal to 0."},{"Start":"01:48.530 ","End":"01:50.580","Text":"We have no magnetic field here,"},{"Start":"01:50.580 ","End":"01:53.750","Text":"we have a charged infinite plane,"},{"Start":"01:53.750 ","End":"01:57.090","Text":"which is in which will give us this electric field,"},{"Start":"01:57.090 ","End":"02:01.100","Text":"and we have a charged particle also giving this electric field."},{"Start":"02:01.100 ","End":"02:05.110","Text":"We have a 0 B field."},{"Start":"02:05.180 ","End":"02:08.235","Text":"This is our S reference frame,"},{"Start":"02:08.235 ","End":"02:11.240","Text":"and now let\u0027s look into S tag reference frame."},{"Start":"02:11.240 ","End":"02:14.895","Text":"Let\u0027s take a look at E tag."},{"Start":"02:14.895 ","End":"02:16.560","Text":"E tag is equal to,"},{"Start":"02:16.560 ","End":"02:26.400","Text":"so the equation is E plus V cross product with B."},{"Start":"02:26.650 ","End":"02:31.070","Text":"Our E field we just saw is equal to Sigma divided by"},{"Start":"02:31.070 ","End":"02:35.165","Text":"2 epsilon naught and this is in the z direction,"},{"Start":"02:35.165 ","End":"02:37.010","Text":"and then we have v,"},{"Start":"02:37.010 ","End":"02:41.920","Text":"so velocity in the y-direction cross-product with 0."},{"Start":"02:41.920 ","End":"02:44.790","Text":"That\u0027s just going to be equal to 0."},{"Start":"02:44.790 ","End":"02:48.875","Text":"Then our B field in the reference frame, so B tag,"},{"Start":"02:48.875 ","End":"02:50.435","Text":"so the equation is"},{"Start":"02:50.435 ","End":"03:00.340","Text":"B minus v cross E divided by c^2."},{"Start":"03:00.340 ","End":"03:03.025","Text":"Our B is equal to 0."},{"Start":"03:03.025 ","End":"03:04.995","Text":"Then we have minus,"},{"Start":"03:04.995 ","End":"03:11.330","Text":"so our v we\u0027re being told that the velocity is v in the y-direction."},{"Start":"03:11.330 ","End":"03:17.600","Text":"We have v in the y-direction, cross-product with our E field,"},{"Start":"03:17.600 ","End":"03:28.110","Text":"so that is Sigma divided by 2 epsilon naught in the z-direction, divided by c^2."},{"Start":"03:28.820 ","End":"03:33.240","Text":"Let\u0027s just simplify this over here."},{"Start":"03:33.240 ","End":"03:35.030","Text":"We have 0 minus,"},{"Start":"03:35.030 ","End":"03:37.190","Text":"so we\u0027ll add in the minus in a second,"},{"Start":"03:37.190 ","End":"03:40.050","Text":"but let\u0027s just put all the constants in one place."},{"Start":"03:40.050 ","End":"03:49.950","Text":"We have v multiplied by Sigma divided by 2 epsilon naught multiplied by c^2."},{"Start":"03:49.950 ","End":"03:53.495","Text":"Then we have y cross z,"},{"Start":"03:53.495 ","End":"03:55.565","Text":"which is going to give us x."},{"Start":"03:55.565 ","End":"03:57.650","Text":"Then we have this minus over here,"},{"Start":"03:57.650 ","End":"04:00.660","Text":"so we\u0027ll write minus x."},{"Start":"04:00.660 ","End":"04:06.125","Text":"Then we can see that our B field or our B tag field is this."},{"Start":"04:06.125 ","End":"04:12.270","Text":"All of this is a constant and it\u0027s acting in the negative x direction."},{"Start":"04:13.700 ","End":"04:17.300","Text":"Now we\u0027re being asked to find the force acting on"},{"Start":"04:17.300 ","End":"04:22.715","Text":"the particle relative to this observer that\u0027s moving."},{"Start":"04:22.715 ","End":"04:25.040","Text":"Let\u0027s start over here."},{"Start":"04:25.040 ","End":"04:34.920","Text":"F tag, because we\u0027re in this S tag reference frame is equal to qE,"},{"Start":"04:38.390 ","End":"04:42.640","Text":"and of course, because when S tag frame of reference,"},{"Start":"04:42.640 ","End":"04:47.395","Text":"so this is the electric field that we\u0027ll see in the S tag frame of reference."},{"Start":"04:47.395 ","End":"04:53.740","Text":"Then plus qv where of course,"},{"Start":"04:53.740 ","End":"04:58.580","Text":"this is the velocity of the charge."},{"Start":"04:59.210 ","End":"05:02.175","Text":"It\u0027s for our charge q,"},{"Start":"05:02.175 ","End":"05:06.190","Text":"and then our qv is cross multiplied"},{"Start":"05:06.190 ","End":"05:11.080","Text":"by the magnetic field that will be seen in the S tag frame of reference,"},{"Start":"05:11.080 ","End":"05:13.570","Text":"which is B tag."},{"Start":"05:14.450 ","End":"05:17.785","Text":"What is the velocity for the charge q?"},{"Start":"05:17.785 ","End":"05:21.295","Text":"We know that in the S reference frame,"},{"Start":"05:21.295 ","End":"05:24.740","Text":"our q over here it\u0027s stationary."},{"Start":"05:24.740 ","End":"05:27.875","Text":"However, we\u0027re looking from this reference frame"},{"Start":"05:27.875 ","End":"05:33.145","Text":"where it\u0027s moving with a velocity of v in the y direction."},{"Start":"05:33.145 ","End":"05:38.615","Text":"For an observer in this reference frame, it is stationary."},{"Start":"05:38.615 ","End":"05:42.170","Text":"The observer relative to itself is stationary."},{"Start":"05:42.170 ","End":"05:47.555","Text":"What we\u0027ll see is that this particle q is moving with a velocity v,"},{"Start":"05:47.555 ","End":"05:50.400","Text":"but in the opposite direction."},{"Start":"05:50.960 ","End":"05:56.180","Text":"It will be moving with this direction over here."},{"Start":"05:56.180 ","End":"06:03.420","Text":"V will be equal to v in the negative y direction."},{"Start":"06:03.740 ","End":"06:06.330","Text":"Then we can write that over here."},{"Start":"06:06.330 ","End":"06:12.125","Text":"V for the charge q,"},{"Start":"06:12.125 ","End":"06:16.800","Text":"can write that over here is equal to negative vy."},{"Start":"06:18.530 ","End":"06:22.980","Text":"Now let\u0027s plug in everything."},{"Start":"06:22.980 ","End":"06:26.280","Text":"Q is just as capital Q,"},{"Start":"06:26.280 ","End":"06:30.540","Text":"and then we can see that Q multiplies both."},{"Start":"06:30.540 ","End":"06:34.500","Text":"Our E tag is equal to this over here,"},{"Start":"06:34.500 ","End":"06:38.210","Text":"so that\u0027s Sigma divided by 2 epsilon naught in"},{"Start":"06:38.210 ","End":"06:42.440","Text":"these that direction plus so we\u0027ve already multiplied by Q,"},{"Start":"06:42.440 ","End":"06:45.845","Text":"but we\u0027re going to have the v of our particle Q,"},{"Start":"06:45.845 ","End":"06:52.100","Text":"which is equal to v in the negative y direction."},{"Start":"06:52.100 ","End":"06:56.435","Text":"This is cross multiplied by a B tag field,"},{"Start":"06:56.435 ","End":"06:58.295","Text":"which is this over here,"},{"Start":"06:58.295 ","End":"07:04.005","Text":"v Sigma divided by 2 epsilon naught c^2,"},{"Start":"07:04.005 ","End":"07:08.620","Text":"and this is in the negative x direction."},{"Start":"07:11.030 ","End":"07:14.870","Text":"Negative and a negative when we multiply them together,"},{"Start":"07:14.870 ","End":"07:16.915","Text":"they become a positive."},{"Start":"07:16.915 ","End":"07:21.720","Text":"Then y cross x is equal to z."},{"Start":"07:21.720 ","End":"07:25.415","Text":"We\u0027re going to have everything in the z-direction."},{"Start":"07:25.415 ","End":"07:28.940","Text":"What we\u0027re going to end up with is equal to Q"},{"Start":"07:28.940 ","End":"07:33.445","Text":"multiplied by Sigma divided by 2 epsilon naught"},{"Start":"07:33.445 ","End":"07:38.450","Text":"minus v^2 multiplied by"},{"Start":"07:38.450 ","End":"07:44.465","Text":"Sigma divided by 2 epsilon naught c^2,"},{"Start":"07:44.465 ","End":"07:49.170","Text":"and all of this is in the z-direction."},{"Start":"07:49.170 ","End":"07:51.870","Text":"This is the final answer."},{"Start":"07:51.870 ","End":"07:55.880","Text":"We can see that the force that will be recorded in"},{"Start":"07:55.880 ","End":"07:58.745","Text":"this S tag frame of reference"},{"Start":"07:58.745 ","End":"08:03.365","Text":"is the same force that we would see in the S frame of reference."},{"Start":"08:03.365 ","End":"08:07.865","Text":"Then we have this extra little bit over here added"},{"Start":"08:07.865 ","End":"08:13.730","Text":"because we\u0027re dealing with relativity that we already said we won\u0027t go into."},{"Start":"08:13.730 ","End":"08:16.710","Text":"That is the end of this lesson."}],"ID":23061},{"Watched":false,"Name":"Deriving Biot-Savart Law from Transformation Equations","Duration":"10m 8s","ChapterTopicVideoID":12097,"CourseChapterTopicPlaylistID":152630,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/12097.jpeg","UploadDate":"2018-06-28T03:22:08.8930000","DurationForVideoObject":"PT10M8S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.770","Text":"Hello. In this lesson,"},{"Start":"00:01.770 ","End":"00:05.175","Text":"we\u0027re going to be deriving Biot-Savart\u0027s law from"},{"Start":"00:05.175 ","End":"00:09.690","Text":"the transformation equations that we saw in the previous lesson."},{"Start":"00:09.690 ","End":"00:14.175","Text":"Let\u0027s say that over here we have a point charge,"},{"Start":"00:14.175 ","End":"00:19.145","Text":"and it\u0027s located along this coordinate system,"},{"Start":"00:19.145 ","End":"00:26.400","Text":"and it\u0027s called the S coordinate system and it looks like so."},{"Start":"00:26.400 ","End":"00:29.510","Text":"Then we know that from this point charge,"},{"Start":"00:29.510 ","End":"00:37.570","Text":"we\u0027re going to have an E field in the radial direction going out in all directions."},{"Start":"00:37.570 ","End":"00:40.715","Text":"This E field from Coulomb\u0027s law,"},{"Start":"00:40.715 ","End":"00:47.345","Text":"we know is equal to kq divided by r^2 in the r hat direction."},{"Start":"00:47.345 ","End":"00:56.490","Text":"Or we can also write this as kq divided by r^3 in the r vector direction."},{"Start":"00:58.080 ","End":"01:03.730","Text":"In this situation, the B field would of course be 0."},{"Start":"01:03.730 ","End":"01:09.325","Text":"Now let\u0027s look at another case where we have the same point charge q,"},{"Start":"01:09.325 ","End":"01:14.810","Text":"and it\u0027s located along this axis."},{"Start":"01:14.870 ","End":"01:18.383","Text":"It\u0027s still located at the origin,"},{"Start":"01:18.383 ","End":"01:24.185","Text":"but the only difference is that now it has a velocity in this direction,"},{"Start":"01:24.185 ","End":"01:26.515","Text":"say this is the x-direction."},{"Start":"01:26.515 ","End":"01:29.125","Text":"It is a velocity in this direction."},{"Start":"01:29.125 ","End":"01:36.200","Text":"Now what we want to know is what the electric and magnetic fields are in this case."},{"Start":"01:36.890 ","End":"01:39.860","Text":"Here the particle starts off at"},{"Start":"01:39.860 ","End":"01:43.685","Text":"the origin and it\u0027s traveling in this direction with a velocity V,"},{"Start":"01:43.685 ","End":"01:47.060","Text":"and this is our S reference frame."},{"Start":"01:47.060 ","End":"01:49.550","Text":"Then let\u0027s imagine that over here,"},{"Start":"01:49.550 ","End":"01:54.065","Text":"we have another reference frame and this is S\u0027."},{"Start":"01:54.065 ","End":"01:58.010","Text":"Here, this whole reference frame is moving in"},{"Start":"01:58.010 ","End":"02:02.225","Text":"the rightwards direction with a velocity of"},{"Start":"02:02.225 ","End":"02:10.445","Text":"V. We can see that according to an observer in the S\u0027 reference frame,"},{"Start":"02:10.445 ","End":"02:16.115","Text":"q our charge is stationary because our charge is moving like so in"},{"Start":"02:16.115 ","End":"02:18.950","Text":"the x-direction with a velocity V and"},{"Start":"02:18.950 ","End":"02:24.725","Text":"our S\u0027 reference frame is moving with the same velocity and in the same direction."},{"Start":"02:24.725 ","End":"02:27.905","Text":"Q will appear stationary."},{"Start":"02:27.905 ","End":"02:34.320","Text":"Therefore, we can say that the E field in S\u0027,"},{"Start":"02:34.320 ","End":"02:42.855","Text":"so E\u0027 is going to be equal to kq divided by r^2 in the r hat direction,"},{"Start":"02:42.855 ","End":"02:45.150","Text":"and that similarly,"},{"Start":"02:45.150 ","End":"02:47.640","Text":"the B field in the S\u0027 frame of reference,"},{"Start":"02:47.640 ","End":"02:50.910","Text":"so B\u0027 is going to be equal to 0."},{"Start":"02:50.910 ","End":"02:56.525","Text":"Because relative to the S\u0027 frame of reference, q is stationary,"},{"Start":"02:56.525 ","End":"03:01.400","Text":"just like in this first example where q is stationary relative to"},{"Start":"03:01.400 ","End":"03:06.590","Text":"S. Now what happens if we"},{"Start":"03:06.590 ","End":"03:12.720","Text":"want to find the electric and magnetic field relative to S?"},{"Start":"03:12.800 ","End":"03:18.993","Text":"We\u0027ve already seen doing the transformation from S to S\u0027."},{"Start":"03:18.993 ","End":"03:23.630","Text":"This is from S to S\u0027."},{"Start":"03:23.630 ","End":"03:34.590","Text":"We would get that E\u0027 is equal to E plus V cross multiplied by B."},{"Start":"03:34.930 ","End":"03:36.485","Text":"We got"},{"Start":"03:36.485 ","End":"03:46.553","Text":"that B\u0027"},{"Start":"03:46.553 ","End":"03:51.805","Text":"was equal to B minus V cross E divided by C^2."},{"Start":"03:51.805 ","End":"03:53.750","Text":"Now if we want to do the opposite,"},{"Start":"03:53.750 ","End":"04:02.310","Text":"so we want to get from S\u0027 to S. All we have to do is isolate out our E and our B."},{"Start":"04:03.020 ","End":"04:13.583","Text":"What we will get is that E is equal to"},{"Start":"04:13.583 ","End":"04:24.370","Text":"E\u0027 minus V cross multiplied by B\u0027."},{"Start":"04:24.710 ","End":"04:29.170","Text":"Here E\u0027 was E and then B."},{"Start":"04:29.170 ","End":"04:33.290","Text":"For E we have E\u0027 and B and of course we\u0027re isolating,"},{"Start":"04:33.290 ","End":"04:38.424","Text":"so we\u0027re switching the sign before the V. If here there\u0027s a plus,"},{"Start":"04:38.424 ","End":"04:40.510","Text":"so here it becomes minus and here is a minus,"},{"Start":"04:40.510 ","End":"04:42.635","Text":"so we\u0027ll see it will become a plus."},{"Start":"04:42.635 ","End":"04:45.875","Text":"Then if we want the B field,"},{"Start":"04:45.875 ","End":"04:49.860","Text":"so this is going to be equal to B\u0027."},{"Start":"04:50.110 ","End":"04:53.510","Text":"Then we\u0027re moving this over to the other side,"},{"Start":"04:53.510 ","End":"04:55.585","Text":"so this becomes a plus."},{"Start":"04:55.585 ","End":"04:59.420","Text":"Then we have V cross multiplied by,"},{"Start":"04:59.420 ","End":"05:03.500","Text":"but then this time it will be E\u0027 instead of E,"},{"Start":"05:03.500 ","End":"05:06.390","Text":"and then divide it by C^2."},{"Start":"05:07.040 ","End":"05:11.030","Text":"We can think that it makes sense that the sine of"},{"Start":"05:11.030 ","End":"05:16.360","Text":"V changes because if we\u0027re looking according to S\u0027,"},{"Start":"05:16.360 ","End":"05:19.355","Text":"V is moving in a positive direction."},{"Start":"05:19.355 ","End":"05:21.740","Text":"But if we look according to S,"},{"Start":"05:21.740 ","End":"05:26.770","Text":"it could be that V is moving in the opposite direction."},{"Start":"05:26.770 ","End":"05:33.305","Text":"The origin is moving towards the left of the particle q,"},{"Start":"05:33.305 ","End":"05:39.145","Text":"so relative to S, the velocity is V in the negative direction."},{"Start":"05:39.145 ","End":"05:42.920","Text":"Now let\u0027s write down back to this example what fields will have"},{"Start":"05:42.920 ","End":"05:47.578","Text":"in our S reference frame,"},{"Start":"05:47.578 ","End":"05:50.435","Text":"the original reference frame that we were in."},{"Start":"05:50.435 ","End":"05:53.355","Text":"Here q is in stationary,"},{"Start":"05:53.355 ","End":"05:56.050","Text":"so q is moving over here."},{"Start":"05:56.120 ","End":"06:00.100","Text":"Any S frame of reference q,"},{"Start":"06:00.100 ","End":"06:03.260","Text":"even though it is in the S frame of reference,"},{"Start":"06:03.260 ","End":"06:08.854","Text":"it\u0027s still moving relative to the origin of the S frame of reference."},{"Start":"06:08.854 ","End":"06:11.944","Text":"Therefore, using this equation,"},{"Start":"06:11.944 ","End":"06:17.065","Text":"we can say that the E field is equal to E\u0027,"},{"Start":"06:17.065 ","End":"06:23.840","Text":"which over here we saw is equal to kq divided by r^2 in"},{"Start":"06:23.840 ","End":"06:31.500","Text":"the r hat direction minus V cross multiplied with B\u0027,"},{"Start":"06:31.500 ","End":"06:34.270","Text":"where B\u0027 was 0."},{"Start":"06:34.280 ","End":"06:44.355","Text":"Then we\u0027ll get that E is equal to kq divided by r^2 in the r hat direction."},{"Start":"06:44.355 ","End":"06:47.835","Text":"Then what about B, the magnetic field?"},{"Start":"06:47.835 ","End":"06:49.290","Text":"That is equal to B\u0027,"},{"Start":"06:49.290 ","End":"06:56.520","Text":"which we saw is equal to 0 plus V cross E\u0027,"},{"Start":"06:56.520 ","End":"07:02.205","Text":"which is kq divided by r^2 in the r direction,"},{"Start":"07:02.205 ","End":"07:05.565","Text":"and all of this is divided by c^2."},{"Start":"07:05.565 ","End":"07:09.740","Text":"Therefore, this is simply equal to,"},{"Start":"07:09.740 ","End":"07:12.155","Text":"if we get all the constants to one side,"},{"Start":"07:12.155 ","End":"07:18.170","Text":"kq divided by c^2 multiplied by"},{"Start":"07:18.170 ","End":"07:26.670","Text":"V vector cross this r hat vector divided by r^2."},{"Start":"07:27.980 ","End":"07:35.130","Text":"This is Biot-Savart\u0027s law for one single point charge."},{"Start":"07:35.130 ","End":"07:42.800","Text":"Now, it\u0027s important to note that usually it isn\u0027t written in terms of k and c^2."},{"Start":"07:42.800 ","End":"07:45.260","Text":"Let\u0027s see how it\u0027s written."},{"Start":"07:45.260 ","End":"07:53.960","Text":"We say that c^2 is equal to 1 divided by mu 0, epsilon 0."},{"Start":"07:53.960 ","End":"08:01.025","Text":"This is an equation that is known where mu 0 and epsilon 0 represent quantities."},{"Start":"08:01.025 ","End":"08:11.550","Text":"As we know, k is equal to 1 divided by 4 pi epsilon 0 or epsilon naught."},{"Start":"08:11.550 ","End":"08:16.980","Text":"Therefore, when we have k divided by c^2,"},{"Start":"08:16.980 ","End":"08:19.020","Text":"like we have over here,"},{"Start":"08:19.020 ","End":"08:26.610","Text":"we\u0027ll get k is 1 divided by 4 pi epsilon naught and"},{"Start":"08:26.610 ","End":"08:34.055","Text":"then multiplied by 1 over c^2 which is simply going to be mu naught,"},{"Start":"08:34.055 ","End":"08:37.580","Text":"epsilon naught and then these two cancel out."},{"Start":"08:37.580 ","End":"08:45.430","Text":"We\u0027ll get that k divided c^2 is equal to mu naught divided by 4 pi."},{"Start":"08:45.430 ","End":"08:50.820","Text":"Therefore, let\u0027s write out the law for Biot-Savart."},{"Start":"08:50.820 ","End":"08:53.715","Text":"This is for a single-point charge."},{"Start":"08:53.715 ","End":"09:01.105","Text":"This is going to be equal to mu naught Q divided by 4 pi"},{"Start":"09:01.105 ","End":"09:10.120","Text":"multiplied by V vector cross r hat vector divided by r^2."},{"Start":"09:10.120 ","End":"09:12.775","Text":"This is what\u0027s important to know."},{"Start":"09:12.775 ","End":"09:20.920","Text":"Now, we might have seen that in a wire where we have lots of charges flowing through,"},{"Start":"09:20.920 ","End":"09:26.385","Text":"and therefore we have a current flowing through,"},{"Start":"09:26.385 ","End":"09:32.020","Text":"and that the length of the wire is of length dl."},{"Start":"09:32.020 ","End":"09:35.530","Text":"In that case, the qv over here,"},{"Start":"09:35.530 ","End":"09:38.140","Text":"so qv"},{"Start":"09:38.140 ","End":"09:45.540","Text":"becomes Idl."},{"Start":"09:45.540 ","End":"09:48.190","Text":"When we have lots of point charges flowing,"},{"Start":"09:48.190 ","End":"09:53.020","Text":"it\u0027s a current and then we measure the length of the wire."},{"Start":"09:53.020 ","End":"09:56.150","Text":"Then qv in this equation is replaced by I,"},{"Start":"09:56.150 ","End":"09:58.520","Text":"the current multiplied by dl,"},{"Start":"09:58.520 ","End":"10:00.485","Text":"the length of wire."},{"Start":"10:00.485 ","End":"10:05.870","Text":"Now we\u0027ve derived the equation for Biot-Savart."},{"Start":"10:05.870 ","End":"10:09.000","Text":"That is the end of this lesson."}],"ID":23062},{"Watched":false,"Name":"Exercise 2","Duration":"7m 35s","ChapterTopicVideoID":12098,"CourseChapterTopicPlaylistID":152630,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/12098.jpeg","UploadDate":"2018-06-28T03:25:48.8570000","DurationForVideoObject":"PT7M35S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.995","Text":"Hello, in this lesson,"},{"Start":"00:01.995 ","End":"00:04.395","Text":"we\u0027re going to be answering the following question."},{"Start":"00:04.395 ","End":"00:08.535","Text":"A sphere has a charge density per unit volume of Rho."},{"Start":"00:08.535 ","End":"00:14.070","Text":"The sphere travels along the x-axis with a velocity of v. Find"},{"Start":"00:14.070 ","End":"00:19.605","Text":"the electric and magnetic fields inside the sphere relative to the lab,"},{"Start":"00:19.605 ","End":"00:23.225","Text":"exactly when the center of the sphere"},{"Start":"00:23.225 ","End":"00:28.320","Text":"passes the origin relative to the lab\u0027s reference frame."},{"Start":"00:29.420 ","End":"00:33.335","Text":"If we have the lab\u0027s reference frame,"},{"Start":"00:33.335 ","End":"00:34.970","Text":"let\u0027s say over here,"},{"Start":"00:34.970 ","End":"00:36.320","Text":"and this is S,"},{"Start":"00:36.320 ","End":"00:41.540","Text":"so when the center of the sphere over here passes this point over here,"},{"Start":"00:41.540 ","End":"00:47.700","Text":"we want to find what the electric and magnetic fields will be within the sphere."},{"Start":"00:49.670 ","End":"00:56.195","Text":"This sphere is moving so let\u0027s make a reference frame over here."},{"Start":"00:56.195 ","End":"01:01.100","Text":"And this reference frame is called S tag and it too is"},{"Start":"01:01.100 ","End":"01:07.285","Text":"moving in the x direction with a velocity of v. Just like the sphere."},{"Start":"01:07.285 ","End":"01:12.275","Text":"Therefore we can write that in the S tag reference frame,"},{"Start":"01:12.275 ","End":"01:14.390","Text":"an observer sitting over here,"},{"Start":"01:14.390 ","End":"01:18.720","Text":"we\u0027ll see that the sphere is stationary."},{"Start":"01:19.310 ","End":"01:25.565","Text":"If the sphere appears to be stationary in this S tag reference frame,"},{"Start":"01:25.565 ","End":"01:30.950","Text":"then that means that we can just use the regular Gauss\u0027s law in order to"},{"Start":"01:30.950 ","End":"01:37.290","Text":"find the electric field within the sphere."},{"Start":"01:37.290 ","End":"01:47.205","Text":"Using Gauss, so here we have a sphere,"},{"Start":"01:47.205 ","End":"01:52.955","Text":"and then we want to find the electric field over here within."},{"Start":"01:52.955 ","End":"01:57.230","Text":"From the origin until this Gaussian layer,"},{"Start":"01:57.230 ","End":"02:01.130","Text":"we have a radius of r. Therefore,"},{"Start":"02:01.130 ","End":"02:09.230","Text":"we know that our equation is E.S is equal to Q in divided by Epsilon naught."},{"Start":"02:09.230 ","End":"02:14.160","Text":"E is what we\u0027re trying to find, the area."},{"Start":"02:14.160 ","End":"02:24.605","Text":"The surface area of this spherical shell is equal to 4Pi r^2."},{"Start":"02:24.605 ","End":"02:27.740","Text":"This is equal to Q in divided by Epsilon naught,"},{"Start":"02:27.740 ","End":"02:29.105","Text":"so Epsilon naught,"},{"Start":"02:29.105 ","End":"02:33.385","Text":"and then Q in is just the charge enclosed over here."},{"Start":"02:33.385 ","End":"02:35.955","Text":"We\u0027re given the charge density,"},{"Start":"02:35.955 ","End":"02:39.710","Text":"so it\u0027s going to be Rho multiplied by the volume,"},{"Start":"02:39.710 ","End":"02:45.840","Text":"so that\u0027s going to be 4/3Pi r^3."},{"Start":"02:46.940 ","End":"02:54.010","Text":"Then we can divide both sides by 4Pi and r^2,"},{"Start":"02:54.010 ","End":"02:56.845","Text":"and then we\u0027ll be left here with 1r."},{"Start":"02:56.845 ","End":"03:01.585","Text":"Therefore, we will get that our electric field is equal to"},{"Start":"03:01.585 ","End":"03:07.080","Text":"Rho r divided by 3 Epsilon naught,"},{"Start":"03:07.080 ","End":"03:10.315","Text":"and of course the E field is a vector field."},{"Start":"03:10.315 ","End":"03:12.865","Text":"Then we know because it\u0027s a sphere,"},{"Start":"03:12.865 ","End":"03:15.010","Text":"it will be in the radial direction."},{"Start":"03:15.010 ","End":"03:24.669","Text":"But we can also write this as Rho divided by 3 Epsilon naught,"},{"Start":"03:24.669 ","End":"03:27.085","Text":"r vector as well."},{"Start":"03:27.085 ","End":"03:29.180","Text":"These mean the same thing."},{"Start":"03:31.280 ","End":"03:38.865","Text":"This is relative to the S tag reference frame."},{"Start":"03:38.865 ","End":"03:46.200","Text":"We can see that our E field is going to be equal to this and our B field,"},{"Start":"03:46.200 ","End":"03:54.170","Text":"so let\u0027s call this E tag and our B tag is of course going to be equal to 0."},{"Start":"03:54.980 ","End":"04:00.220","Text":"Now let\u0027s see what will happen in our S reference frame,"},{"Start":"04:00.220 ","End":"04:02.320","Text":"which is our lab\u0027s reference frame."},{"Start":"04:02.320 ","End":"04:05.200","Text":"First of all, we remember these equations."},{"Start":"04:05.200 ","End":"04:13.690","Text":"E is equal to E tag minus V cross B tag."},{"Start":"04:13.860 ","End":"04:24.455","Text":"B is equal to"},{"Start":"04:24.455 ","End":"04:28.865","Text":"B tag plus V"},{"Start":"04:28.865 ","End":"04:36.015","Text":"cross E tag divided by C squared."},{"Start":"04:36.015 ","End":"04:38.880","Text":"Now let\u0027s substitute in everything."},{"Start":"04:38.880 ","End":"04:41.835","Text":"E tag, let\u0027s use this equation over here,"},{"Start":"04:41.835 ","End":"04:50.675","Text":"so that is simply Rho divided by 3 Epsilon naught in the r vector direction,"},{"Start":"04:50.675 ","End":"04:54.280","Text":"minus the velocity cross multiplied with B tag,"},{"Start":"04:54.280 ","End":"04:56.870","Text":"which is equal to 0, so that\u0027s just 0."},{"Start":"04:56.870 ","End":"05:00.885","Text":"The B field, so we have B tag, which is 0,"},{"Start":"05:00.885 ","End":"05:07.425","Text":"plus and then we have V cross multiplied with E tag,"},{"Start":"05:07.425 ","End":"05:08.850","Text":"which we\u0027re using this version,"},{"Start":"05:08.850 ","End":"05:19.540","Text":"so that\u0027s Rho divided by 3 Epsilon naught r vector and all of this is divided by C^2."},{"Start":"05:19.550 ","End":"05:22.850","Text":"If we just tidy this up a bit,"},{"Start":"05:22.850 ","End":"05:31.120","Text":"we\u0027ll get V Rho divided by 3 Epsilon naught C^2"},{"Start":"05:31.120 ","End":"05:41.190","Text":"in the direction of x hat cross multiplied with r vector."},{"Start":"05:42.350 ","End":"05:46.725","Text":"What is x hat cross r vector?"},{"Start":"05:46.725 ","End":"05:49.790","Text":"So let\u0027s just write this down."},{"Start":"05:49.790 ","End":"05:51.730","Text":"Let\u0027s scroll a little bit."},{"Start":"05:51.730 ","End":"05:57.075","Text":"We have x hat cross multiplied by r vector,"},{"Start":"05:57.075 ","End":"06:01.725","Text":"so that is the same as x hat cross multiplied by,"},{"Start":"06:01.725 ","End":"06:08.900","Text":"so r vector is simply x in the x direction plus y in the y direction,"},{"Start":"06:08.900 ","End":"06:11.885","Text":"plus z in the z direction."},{"Start":"06:11.885 ","End":"06:13.430","Text":"This is useful to remember,"},{"Start":"06:13.430 ","End":"06:15.340","Text":"this is r vector."},{"Start":"06:15.340 ","End":"06:24.590","Text":"x hat cross multiplied by x hat is equal to 0. x hat cross multiplied by y hat is z,"},{"Start":"06:24.590 ","End":"06:34.145","Text":"so we\u0027ll have y in the z direction and x cross multiplied by z is simply negative y."},{"Start":"06:34.145 ","End":"06:38.845","Text":"We\u0027ll have negative z in the y direction."},{"Start":"06:38.845 ","End":"06:43.095","Text":"Then this is going to plug in over here."},{"Start":"06:43.095 ","End":"06:50.330","Text":"The final answer that we will get is V Rho divided by"},{"Start":"06:50.330 ","End":"07:01.080","Text":"3 epsilon naught C squared in the z direction minus z in the y direction."},{"Start":"07:02.410 ","End":"07:05.510","Text":"This is the final answer,"},{"Start":"07:05.510 ","End":"07:09.830","Text":"our question was to find the electric and"},{"Start":"07:09.830 ","End":"07:14.615","Text":"the magnetic fields inside the sphere relative to the lab at this exact moment."},{"Start":"07:14.615 ","End":"07:18.260","Text":"Our E field is Rho divided by 3 epsilon naught,"},{"Start":"07:18.260 ","End":"07:23.730","Text":"in the r direction and our B field is V Rho divided by"},{"Start":"07:23.730 ","End":"07:32.340","Text":"3 epsilon naught multiplied by C^2 in the y z direction minus z y direction."},{"Start":"07:32.340 ","End":"07:35.590","Text":"That is the end of this lesson."}],"ID":23063}],"Thumbnail":null,"ID":152630}]

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