Introduction to Faraday's Law of Induction
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[{"Name":"Introduction to Faraday\u0027s Law of Induction","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Intro and Example","Duration":"22m 44s","ChapterTopicVideoID":21363,"CourseChapterTopicPlaylistID":99480,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/21363.jpeg","UploadDate":"2020-04-06T21:31:15.8330000","DurationForVideoObject":"PT22M44S","Description":null,"MetaTitle":"16-1 Intro and Example: Video + Workbook | Proprep","MetaDescription":"Faradays Law of Induction - Introduction to Faraday\u0027s Law of Induction. Watch the video made by an expert in the field. Download the workbook and maximize your learning.","Canonical":"https://www.proprep.uk/general-modules/all/physics-2-electricity-and-magnetism/faradays-law-of-induction/introduction-to-faraday%27s-law-of-induction/vid21443","VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:05.385","Text":"Hello. In this lesson we\u0027re going to be learning about Faraday\u0027s law of induction."},{"Start":"00:05.385 ","End":"00:10.035","Text":"This is one of the basic laws in electricity."},{"Start":"00:10.035 ","End":"00:13.200","Text":"What it says is that the emf,"},{"Start":"00:13.200 ","End":"00:18.164","Text":"which is something called the electromotive force,"},{"Start":"00:18.164 ","End":"00:28.050","Text":"is equal to the negative derivative of the magnetic flux with respect to time."},{"Start":"00:28.050 ","End":"00:32.560","Text":"Let\u0027s explain this now in simpler terms."},{"Start":"00:32.890 ","End":"00:41.030","Text":"The emf is the electromotive force and it\u0027s voltage, and right now,"},{"Start":"00:41.030 ","End":"00:45.065","Text":"we can just think of it just like what we\u0027ve seen before,"},{"Start":"00:45.065 ","End":"00:50.090","Text":"the voltage going through a regular electric circuit."},{"Start":"00:50.090 ","End":"00:53.765","Text":"That\u0027s enough to know for the meantime later,"},{"Start":"00:53.765 ","End":"00:56.240","Text":"we\u0027re going to learn about the emf in"},{"Start":"00:56.240 ","End":"01:01.100","Text":"slightly more detail and we\u0027ll see that it\u0027s a little bit different but for now we"},{"Start":"01:01.100 ","End":"01:04.610","Text":"just have to know that it\u0027s a type of voltage and it\u0027s equal to"},{"Start":"01:04.610 ","End":"01:08.915","Text":"the negative change in time of the magnetic flux."},{"Start":"01:08.915 ","End":"01:14.560","Text":"We\u0027ve already seen what electric flux is when we were speaking about Gauss\u0027s law."},{"Start":"01:14.560 ","End":"01:21.635","Text":"We always know that the flux is going to be some integral with respect to area,"},{"Start":"01:21.635 ","End":"01:24.620","Text":"so this is equal to the integral of,"},{"Start":"01:24.620 ","End":"01:33.300","Text":"so here we have the magnetic flux of the magnetic field dot ds, the area."},{"Start":"01:33.440 ","End":"01:42.939","Text":"What does this mean if we have some electric circuit which is enclosed in a circle."},{"Start":"01:42.939 ","End":"01:45.650","Text":"So we have a complete circuit and we have"},{"Start":"01:45.650 ","End":"01:49.340","Text":"this emf where there\u0027s a voltage flowing through the circuit."},{"Start":"01:49.340 ","End":"01:51.860","Text":"When we\u0027re speaking about the magnetic flux,"},{"Start":"01:51.860 ","End":"02:00.950","Text":"it\u0027s all of the magnetic flux that is passing through in this circuit."},{"Start":"02:00.950 ","End":"02:09.220","Text":"If we measure the magnetic flux through this circle and we see how it changes in time,"},{"Start":"02:09.220 ","End":"02:11.730","Text":"and then we multiply it by negative 1,"},{"Start":"02:11.730 ","End":"02:15.653","Text":"then we\u0027ll get the voltage flowing through the circuit."},{"Start":"02:15.653 ","End":"02:17.990","Text":"We\u0027ll get the emf."},{"Start":"02:18.170 ","End":"02:23.795","Text":"Right now, what we\u0027re going to do is we\u0027re going to look at a simpler question,"},{"Start":"02:23.795 ","End":"02:30.420","Text":"and this will help us to understand how to solve questions dealing with Faraday\u0027s law."},{"Start":"02:31.370 ","End":"02:39.305","Text":"Here we have a permanent conducting wire that doesn\u0027t move."},{"Start":"02:39.305 ","End":"02:44.960","Text":"Here we have a resistor of resistance R,"},{"Start":"02:44.960 ","End":"02:49.730","Text":"and then here we have another wire that doesn\u0027t move."},{"Start":"02:49.730 ","End":"02:59.540","Text":"Here we have some rod or wire that is moving in this direction with a velocity of V_0,"},{"Start":"02:59.540 ","End":"03:07.420","Text":"and the length of this or the width of this second over here is L,"},{"Start":"03:07.420 ","End":"03:12.720","Text":"and we can say that this distance over here is x,"},{"Start":"03:12.720 ","End":"03:17.765","Text":"and notice that x is always changing as the rod moves along."},{"Start":"03:17.765 ","End":"03:25.980","Text":"We can say that the initial position of the rod is at x_0."},{"Start":"03:25.980 ","End":"03:29.210","Text":"Let\u0027s say that x_0 was somewhere over here."},{"Start":"03:29.210 ","End":"03:30.860","Text":"It doesn\u0027t really matter."},{"Start":"03:30.860 ","End":"03:34.295","Text":"Now the first question that we want to answer,"},{"Start":"03:34.295 ","End":"03:38.280","Text":"is we want to calculate the emf."},{"Start":"03:38.600 ","End":"03:41.560","Text":"In order to find me emf,"},{"Start":"03:41.560 ","End":"03:46.535","Text":"we know that we have to take the negative derivative with"},{"Start":"03:46.535 ","End":"03:52.495","Text":"respect to time of the magnetic flux."},{"Start":"03:52.495 ","End":"03:55.910","Text":"Here we have with a dot over here,"},{"Start":"03:55.910 ","End":"03:59.030","Text":"which means that it\u0027s the derivative with respect to time."},{"Start":"03:59.030 ","End":"04:05.225","Text":"That means that first we have to calculate the magnetic flux, which as we saw,"},{"Start":"04:05.225 ","End":"04:12.350","Text":"is equal to the integral of the magnetic field dot ds."},{"Start":"04:12.350 ","End":"04:16.520","Text":"I forgot to say that we have a magnetic field"},{"Start":"04:16.520 ","End":"04:22.050","Text":"throughout the region which is equal to B_0."},{"Start":"04:24.560 ","End":"04:29.090","Text":"Because we have this constant magnetic field throughout the region,"},{"Start":"04:29.090 ","End":"04:30.950","Text":"we don\u0027t have to do this integral."},{"Start":"04:30.950 ","End":"04:34.580","Text":"We can just say that this is equal to the magnetic field,"},{"Start":"04:34.580 ","End":"04:36.740","Text":"which is constant throughout, B_0,"},{"Start":"04:36.740 ","End":"04:40.055","Text":"multiplied by the surface area."},{"Start":"04:40.055 ","End":"04:44.630","Text":"Where, of course, the area that we\u0027re calculating is this area over here,"},{"Start":"04:44.630 ","End":"04:49.835","Text":"which is of course always changing depending on where this rod is,"},{"Start":"04:49.835 ","End":"04:54.055","Text":"and this rod is moving with this velocity V_0."},{"Start":"04:54.055 ","End":"05:00.930","Text":"We can say that this is equal to B_0 multiplied by this width,"},{"Start":"05:00.930 ","End":"05:05.385","Text":"so L, multiplied by this length over here."},{"Start":"05:05.385 ","End":"05:09.155","Text":"We said that this length is changing."},{"Start":"05:09.155 ","End":"05:12.275","Text":"What is this length equal to?"},{"Start":"05:12.275 ","End":"05:17.960","Text":"It\u0027s equal to, we said that the rod starts off at some starting position,"},{"Start":"05:17.960 ","End":"05:20.975","Text":"so we can say that the total x,"},{"Start":"05:20.975 ","End":"05:26.640","Text":"let\u0027s call this x tag."},{"Start":"05:26.640 ","End":"05:30.665","Text":"Sorry, this is X. The total x that is"},{"Start":"05:30.665 ","End":"05:36.750","Text":"changing is equal to the initial starting point which was X_0,"},{"Start":"05:37.550 ","End":"05:44.090","Text":"plus the velocity multiplied by time V naught time."},{"Start":"05:44.090 ","End":"05:49.399","Text":"Because velocity is equal to distance over time,"},{"Start":"05:49.399 ","End":"05:52.480","Text":"where distance over here is x divided by t,"},{"Start":"05:52.480 ","End":"05:54.620","Text":"so if we isolate out the x,"},{"Start":"05:54.620 ","End":"05:59.085","Text":"we get vt, so that\u0027s where this comes from."},{"Start":"05:59.085 ","End":"06:01.829","Text":"Now we can add this n over here,"},{"Start":"06:01.829 ","End":"06:04.595","Text":"so we\u0027re multiplying by this distance x,"},{"Start":"06:04.595 ","End":"06:10.220","Text":"which is equal to X_0 plus V_0 multiplied by"},{"Start":"06:10.220 ","End":"06:18.065","Text":"t. Then we want to take the derivative of this with respect to time."},{"Start":"06:18.065 ","End":"06:21.334","Text":"If we calculate that,"},{"Start":"06:21.334 ","End":"06:24.990","Text":"we\u0027re going to be left with B_0 L X_0,"},{"Start":"06:24.990 ","End":"06:27.860","Text":"so when we take the derivative of that with respect to time,"},{"Start":"06:27.860 ","End":"06:29.195","Text":"it will cancel out,"},{"Start":"06:29.195 ","End":"06:32.900","Text":"and then we have B_0 L V_0t,"},{"Start":"06:32.900 ","End":"06:38.490","Text":"and the derivative of that with respect to time is simply"},{"Start":"06:38.490 ","End":"06:45.330","Text":"equal to B_0 L V_0t."},{"Start":"06:45.330 ","End":"06:51.650","Text":"Therefore, we get that the emf is equal to the negative of this time"},{"Start":"06:51.650 ","End":"06:58.630","Text":"derivative so it\u0027s equal to negative B_0 L V_0t."},{"Start":"06:59.300 ","End":"07:05.315","Text":"Of course, there\u0027s no t because we took the derivative with respect to time."},{"Start":"07:05.315 ","End":"07:10.260","Text":"Just remember, there\u0027s no t. Now,"},{"Start":"07:10.260 ","End":"07:12.565","Text":"question number 2,"},{"Start":"07:12.565 ","End":"07:19.956","Text":"is we want to find the current flowing through the circuit."},{"Start":"07:19.956 ","End":"07:22.690","Text":"As we know this EMF,"},{"Start":"07:22.690 ","End":"07:27.280","Text":"the electromotive force is voltage."},{"Start":"07:27.280 ","End":"07:30.970","Text":"Therefore, what we can do is we can use Ohm\u0027s law,"},{"Start":"07:30.970 ","End":"07:37.915","Text":"which says that the current is equal to the voltage divided by the resistance,"},{"Start":"07:37.915 ","End":"07:42.400","Text":"which is equal to the EMF divided by the resistance R."},{"Start":"07:42.400 ","End":"07:45.730","Text":"What we\u0027re going to do is we\u0027re going to work out"},{"Start":"07:45.730 ","End":"07:49.840","Text":"first the size of the current and then the direction."},{"Start":"07:49.840 ","End":"07:52.150","Text":"In order to do that,"},{"Start":"07:52.150 ","End":"07:57.205","Text":"we\u0027re going to calculate this using the absolute value of the EMF."},{"Start":"07:57.205 ","End":"08:07.660","Text":"What we have is the absolute value of negative B naught L. V naught divided by R,"},{"Start":"08:07.660 ","End":"08:13.585","Text":"which is going to give us just B naught V naught L"},{"Start":"08:13.585 ","End":"08:20.350","Text":"divided by R. Now we have the size of the current,"},{"Start":"08:20.350 ","End":"08:24.230","Text":"but what we wanna do is we want to calculate the direction."},{"Start":"08:24.480 ","End":"08:27.685","Text":"In order to calculate the direction,"},{"Start":"08:27.685 ","End":"08:30.160","Text":"we\u0027re going to use Lenz\u0027s law."},{"Start":"08:30.160 ","End":"08:32.140","Text":"What does Lenz\u0027s law say?"},{"Start":"08:32.140 ","End":"08:36.790","Text":"It says that the current that is created is going to act in"},{"Start":"08:36.790 ","End":"08:43.648","Text":"the direction such as to oppose the change in flux."},{"Start":"08:43.648 ","End":"08:44.830","Text":"So in other words,"},{"Start":"08:44.830 ","End":"08:49.030","Text":"it\u0027s just stating in words this equation over here."},{"Start":"08:49.030 ","End":"08:54.070","Text":"We want the direction of the current to be in"},{"Start":"08:54.070 ","End":"08:58.690","Text":"the opposite direction to the change in flux,"},{"Start":"08:58.690 ","End":"09:01.430","Text":"so we see over here,"},{"Start":"09:01.440 ","End":"09:10.130","Text":"VB dot is the change in flux and we want a current in the negative direction."},{"Start":"09:10.650 ","End":"09:14.455","Text":"Here is Lenz\u0027s law written out,"},{"Start":"09:14.455 ","End":"09:18.040","Text":"so what we want to do is we want to see"},{"Start":"09:18.040 ","End":"09:22.090","Text":"which direction the current is flowing in this circuit."},{"Start":"09:22.090 ","End":"09:27.895","Text":"The first question that we\u0027re going to ask ourselves is whether the flux is"},{"Start":"09:27.895 ","End":"09:34.730","Text":"increasing or decreasing as this rod moves in this direction."},{"Start":"09:34.980 ","End":"09:38.470","Text":"In this, you ask yourself for each question,"},{"Start":"09:38.470 ","End":"09:41.740","Text":"if the flux is increasing or decreasing."},{"Start":"09:41.740 ","End":"09:46.210","Text":"Over here, as the rod moves in this rightwards direction,"},{"Start":"09:46.210 ","End":"09:54.520","Text":"we can see that the surface area of the circuit increases because x is increasing."},{"Start":"09:54.520 ","End":"09:57.145","Text":"If the surface area is increasing,"},{"Start":"09:57.145 ","End":"10:01.600","Text":"that means more magnetic flux can pass through."},{"Start":"10:01.600 ","End":"10:04.045","Text":"In that case, the flux,"},{"Start":"10:04.045 ","End":"10:07.790","Text":"in this case is increasing."},{"Start":"10:07.890 ","End":"10:11.935","Text":"Then what we want to do according to Lenz\u0027s law,"},{"Start":"10:11.935 ","End":"10:15.070","Text":"is if the flux is increasing,"},{"Start":"10:15.070 ","End":"10:17.395","Text":"we want to decrease it."},{"Start":"10:17.395 ","End":"10:22.750","Text":"If the rod was traveling and the leftwards direction where the flux was decreasing,"},{"Start":"10:22.750 ","End":"10:25.720","Text":"then according to Lenz\u0027s law want to oppose that,"},{"Start":"10:25.720 ","End":"10:28.945","Text":"so we\u0027d want to increase the flux."},{"Start":"10:28.945 ","End":"10:34.520","Text":"Therefore, we want to decrease the flux."},{"Start":"10:34.800 ","End":"10:38.245","Text":"If we want to decrease the flux,"},{"Start":"10:38.245 ","End":"10:42.460","Text":"what we\u0027re going to do is we\u0027re going to imagine that we have"},{"Start":"10:42.460 ","End":"10:47.110","Text":"an electric field in the opposite direction."},{"Start":"10:47.110 ","End":"10:48.700","Text":"Let\u0027s join them gray,"},{"Start":"10:48.700 ","End":"10:51.995","Text":"so we\u0027re pretending like we have a magnetic field in the opposite direction."},{"Start":"10:51.995 ","End":"10:56.590","Text":"Here the magnetic field is going into the page."},{"Start":"10:56.590 ","End":"11:00.625","Text":"What we\u0027re going to want to do is we\u0027re going to form a magnetic field,"},{"Start":"11:00.625 ","End":"11:04.400","Text":"which comes out of the page."},{"Start":"11:04.830 ","End":"11:07.720","Text":"This is what we just did."},{"Start":"11:07.720 ","End":"11:11.590","Text":"We created a magnetic fields in the opposite direction,"},{"Start":"11:11.590 ","End":"11:13.600","Text":"so coming out of the page."},{"Start":"11:13.600 ","End":"11:18.340","Text":"Then what we\u0027re going to do is we\u0027re going to find"},{"Start":"11:18.340 ","End":"11:24.770","Text":"the direction of the current that would create this magnetic fields."},{"Start":"11:25.290 ","End":"11:30.430","Text":"We\u0027re finding the direction of the current which would create this opposite field."},{"Start":"11:30.430 ","End":"11:34.540","Text":"Now we can use the Right hand rule."},{"Start":"11:34.540 ","End":"11:37.510","Text":"We point our thumb in the direction of"},{"Start":"11:37.510 ","End":"11:42.010","Text":"this magnetic fields which we\u0027ve created in the opposite direction."},{"Start":"11:42.010 ","End":"11:44.650","Text":"In this case, our thumb will point out of"},{"Start":"11:44.650 ","End":"11:48.070","Text":"the page or out of the screen and then our fingers of"},{"Start":"11:48.070 ","End":"11:54.160","Text":"our right hand are going to curl in the direction of the current."},{"Start":"11:54.160 ","End":"11:57.190","Text":"In this case, our thumb is pointing out of the page,"},{"Start":"11:57.190 ","End":"12:06.170","Text":"which means that our current is going to flow in this direction anticlockwise."},{"Start":"12:07.260 ","End":"12:13.030","Text":"This is the magnitude of the current and this is the direction."},{"Start":"12:13.030 ","End":"12:18.520","Text":"Now, let\u0027s just imagine that we had a decreasing flux."},{"Start":"12:18.520 ","End":"12:20.688","Text":"An example with a decreasing flux,"},{"Start":"12:20.688 ","End":"12:22.390","Text":"what would we do?"},{"Start":"12:22.390 ","End":"12:24.523","Text":"In that case,"},{"Start":"12:24.523 ","End":"12:26.775","Text":"just like over here,"},{"Start":"12:26.775 ","End":"12:32.590","Text":"we\u0027d want to increase because we\u0027re doing the opposite,"},{"Start":"12:32.590 ","End":"12:37.825","Text":"so we\u0027re doing increase the flux."},{"Start":"12:37.825 ","End":"12:39.820","Text":"Then in that case,"},{"Start":"12:39.820 ","End":"12:42.610","Text":"we\u0027re going to make a B field,"},{"Start":"12:42.610 ","End":"12:44.950","Text":"but instead of in the opposite direction,"},{"Start":"12:44.950 ","End":"12:49.300","Text":"it will be in the same direction."},{"Start":"12:49.300 ","End":"12:53.185","Text":"Then we\u0027re going to find the direction of current,"},{"Start":"12:53.185 ","End":"13:01.885","Text":"which would create this magnetic field"},{"Start":"13:01.885 ","End":"13:04.760","Text":"in the same direction."},{"Start":"13:06.480 ","End":"13:11.380","Text":"What\u0027s important to remember here is that when you\u0027re"},{"Start":"13:11.380 ","End":"13:15.490","Text":"using Faraday\u0027s law to calculate the current in the second,"},{"Start":"13:15.490 ","End":"13:19.315","Text":"to also use Lenz\u0027s law."},{"Start":"13:19.315 ","End":"13:22.825","Text":"Now let\u0027s go on to question number 3."},{"Start":"13:22.825 ","End":"13:24.824","Text":"Question number 3 is,"},{"Start":"13:24.824 ","End":"13:31.990","Text":"what external force is needed in order to pull the rod at a constant velocity?"},{"Start":"13:31.990 ","End":"13:34.105","Text":"Why is there a force?"},{"Start":"13:34.105 ","End":"13:37.765","Text":"What we have is we have wires that have a"},{"Start":"13:37.765 ","End":"13:41.680","Text":"current flowing through them and this whole system"},{"Start":"13:41.680 ","End":"13:45.835","Text":"is located inside an external magnetic field"},{"Start":"13:45.835 ","End":"13:49.555","Text":"and so we know that there\u0027s going to be a force."},{"Start":"13:49.555 ","End":"13:56.695","Text":"The equation for this force is given as dF,"},{"Start":"13:56.695 ","End":"14:01.900","Text":"which is equal to I,"},{"Start":"14:01.900 ","End":"14:04.525","Text":"the current, dl,"},{"Start":"14:04.525 ","End":"14:08.380","Text":"the length of the wire cross B,"},{"Start":"14:08.380 ","End":"14:09.730","Text":"and of course,"},{"Start":"14:09.730 ","End":"14:12.370","Text":"B is the magnetic field."},{"Start":"14:12.370 ","End":"14:17.200","Text":"If our current, the angle between the current and"},{"Start":"14:17.200 ","End":"14:21.880","Text":"the magnetic field and the magnetic field itself are all constants,"},{"Start":"14:21.880 ","End":"14:26.140","Text":"then we can say that the force is equal to"},{"Start":"14:26.140 ","End":"14:34.865","Text":"BIL multiplied by sine of the angle between the 2."},{"Start":"14:34.865 ","End":"14:37.995","Text":"As we already saw,"},{"Start":"14:37.995 ","End":"14:44.880","Text":"we have this rod which is moving outwards at a velocity of V naught,"},{"Start":"14:44.880 ","End":"14:50.640","Text":"and we want to ensure that this velocity V naught is constant."},{"Start":"14:50.640 ","End":"14:57.600","Text":"What we\u0027re going to have is we\u0027re going to have a magnetic force which is acting in"},{"Start":"14:57.600 ","End":"15:06.015","Text":"the direction so as to oppose this movement or to oppose the increase of magnetic flux."},{"Start":"15:06.015 ","End":"15:09.705","Text":"Let\u0027s call this FB."},{"Start":"15:09.705 ","End":"15:15.030","Text":"Then of course, because we have equal and opposite forces."},{"Start":"15:15.030 ","End":"15:20.550","Text":"We\u0027re going to have this opposite force which has to be equal and opposite."},{"Start":"15:20.550 ","End":"15:23.430","Text":"It\u0027s an external force F external."},{"Start":"15:23.430 ","End":"15:28.005","Text":"This is the force that we\u0027re trying to calculate which is going to ensure"},{"Start":"15:28.005 ","End":"15:34.480","Text":"that this rod over here moves at a constant velocity V naught."},{"Start":"15:34.490 ","End":"15:38.220","Text":"Again, what we have is this rod which is moving in"},{"Start":"15:38.220 ","End":"15:43.950","Text":"this direction in a way such as to increase the magnetic flux."},{"Start":"15:43.950 ","End":"15:46.170","Text":"The force is going to be applied in"},{"Start":"15:46.170 ","End":"15:51.270","Text":"the opposite direction in order to stop this increase in flux."},{"Start":"15:51.270 ","End":"15:58.680","Text":"That\u0027s this FB but then in order for this to move at a constant velocity,"},{"Start":"15:58.680 ","End":"16:04.395","Text":"we have the force F external which is equal and opposite."},{"Start":"16:04.395 ","End":"16:08.490","Text":"Now let\u0027s calculate this force."},{"Start":"16:08.490 ","End":"16:13.980","Text":"We have that FB is equal to what we saw over here."},{"Start":"16:13.980 ","End":"16:16.455","Text":"We have a constant magnetic field."},{"Start":"16:16.455 ","End":"16:18.420","Text":"That\u0027s why we can use this equation."},{"Start":"16:18.420 ","End":"16:21.675","Text":"We have a constant magnetic field of B naught,"},{"Start":"16:21.675 ","End":"16:25.335","Text":"a constant current which we already calculated."},{"Start":"16:25.335 ","End":"16:28.605","Text":"We\u0027ll just put an I. We won\u0027t write all of this."},{"Start":"16:28.605 ","End":"16:32.235","Text":"L is of course this length over here."},{"Start":"16:32.235 ","End":"16:34.665","Text":"Then sine Alpha."},{"Start":"16:34.665 ","End":"16:41.580","Text":"The angle between the 2 is 90 degrees. What are we talking about?"},{"Start":"16:41.580 ","End":"16:44.460","Text":"The angle between the current,"},{"Start":"16:44.460 ","End":"16:47.340","Text":"the direction that it\u0027s traveling and the force."},{"Start":"16:47.340 ","End":"16:50.385","Text":"We can see it\u0027s this over here, it\u0027s 90 degrees."},{"Start":"16:50.385 ","End":"16:53.620","Text":"Sine of 90 is just 1."},{"Start":"16:54.140 ","End":"16:59.700","Text":"This is FB which is of course equal and opposite to F external."},{"Start":"16:59.700 ","End":"17:04.990","Text":"This is equal to negative F external."},{"Start":"17:06.080 ","End":"17:12.060","Text":"Now what we want is we want to know what F external is equal to."},{"Start":"17:12.060 ","End":"17:13.875","Text":"We can just substitute in."},{"Start":"17:13.875 ","End":"17:17.435","Text":"We have B naught multiplied by I,"},{"Start":"17:17.435 ","End":"17:26.030","Text":"where I is also B naught c squared v naught L divided by R. All of this is multiplied"},{"Start":"17:26.030 ","End":"17:35.130","Text":"by L. We can just square this and we can define this as the positive x-direction."},{"Start":"17:35.130 ","End":"17:39.630","Text":"Of course, F external is pointing in the positive x-direction,"},{"Start":"17:39.630 ","End":"17:45.010","Text":"so we don\u0027t have to add in this minus over here."},{"Start":"17:45.260 ","End":"17:48.735","Text":"This is the answer to Question 3."},{"Start":"17:48.735 ","End":"17:54.405","Text":"This is the external force needed in order to pull the rod at a constant velocity."},{"Start":"17:54.405 ","End":"17:57.660","Text":"Now let\u0027s answer Question 4,"},{"Start":"17:57.660 ","End":"18:04.420","Text":"where Question 4 is asking us what the power is equal to."},{"Start":"18:04.820 ","End":"18:08.130","Text":"We want to also find the external power."},{"Start":"18:08.130 ","End":"18:13.260","Text":"We\u0027re talking about the power due to this external force."},{"Start":"18:13.260 ","End":"18:18.045","Text":"The equation for power when we\u0027re dealing with a force."},{"Start":"18:18.045 ","End":"18:22.500","Text":"We can either see it as the change in work divided"},{"Start":"18:22.500 ","End":"18:26.820","Text":"by the change in time but when we have a force,"},{"Start":"18:26.820 ","End":"18:32.940","Text":"this is equal to the force dot the velocity."},{"Start":"18:32.940 ","End":"18:35.775","Text":"Here we have the force."},{"Start":"18:35.775 ","End":"18:40.920","Text":"We have B naught squared V naught L"},{"Start":"18:40.920 ","End":"18:47.535","Text":"squared divided by R dot product with the velocity which is V naught."},{"Start":"18:47.535 ","End":"18:49.890","Text":"We can add in a square root over here."},{"Start":"18:49.890 ","End":"18:56.565","Text":"Then both the external force and the velocity are in the positive x-direction."},{"Start":"18:56.565 ","End":"19:00.640","Text":"We\u0027re just left with this."},{"Start":"19:02.120 ","End":"19:04.920","Text":"This is the answer to Question 4."},{"Start":"19:04.920 ","End":"19:09.585","Text":"What is the power of the external force?"},{"Start":"19:09.585 ","End":"19:13.905","Text":"Now, let\u0027s answer question Number 5."},{"Start":"19:13.905 ","End":"19:21.880","Text":"Where question Number 5 is asking us what is the power of the resistor?"},{"Start":"19:24.350 ","End":"19:29.070","Text":"When we\u0027re dealing with the power of the resistor,"},{"Start":"19:29.070 ","End":"19:34.320","Text":"what we want to do is we want to use the voltage and the current."},{"Start":"19:34.320 ","End":"19:39.315","Text":"We know that the equation for power is equal to"},{"Start":"19:39.315 ","End":"19:46.665","Text":"the voltage across the resistor multiplied by the current flowing through it."},{"Start":"19:46.665 ","End":"19:54.270","Text":"If we substitute in here that the voltage is equal to I multiplied by r,"},{"Start":"19:54.270 ","End":"19:56.460","Text":"we get that another equation,"},{"Start":"19:56.460 ","End":"20:01.170","Text":"parallel equation for power is I squared multiplied by"},{"Start":"20:01.170 ","End":"20:08.795","Text":"r. If we do a few more substitutions of various ways of writing out this equation,"},{"Start":"20:08.795 ","End":"20:12.320","Text":"we can also get another equation for power,"},{"Start":"20:12.320 ","End":"20:15.740","Text":"which is V squared divided by"},{"Start":"20:15.740 ","End":"20:21.690","Text":"R. What we want to do is we want to calculate now the power of the resistor."},{"Start":"20:21.690 ","End":"20:24.855","Text":"As we can see, we have these 3 parallel equations"},{"Start":"20:24.855 ","End":"20:27.345","Text":"where of course the voltage is the EMF,"},{"Start":"20:27.345 ","End":"20:30.390","Text":"electromotive force that we spoke about before."},{"Start":"20:30.390 ","End":"20:36.030","Text":"However, the easiest equation that we can calculate over here,"},{"Start":"20:36.030 ","End":"20:40.635","Text":"we will have to do less canceling out is this equation over here."},{"Start":"20:40.635 ","End":"20:42.240","Text":"We take I squared,"},{"Start":"20:42.240 ","End":"20:46.710","Text":"so we have B naught squared multiplied by"},{"Start":"20:46.710 ","End":"20:51.270","Text":"V naught squared multiplied by L squared divided by I"},{"Start":"20:51.270 ","End":"20:55.095","Text":"squared so this was I squared multiplied by"},{"Start":"20:55.095 ","End":"21:00.825","Text":"R. Then this R over here cancels out with the squared over here."},{"Start":"21:00.825 ","End":"21:05.280","Text":"What we\u0027re left with is that the power is equal to B naught squared,"},{"Start":"21:05.280 ","End":"21:06.959","Text":"V naught squared,"},{"Start":"21:06.959 ","End":"21:13.515","Text":"L squared divided by R. We can see that the power"},{"Start":"21:13.515 ","End":"21:21.465","Text":"in the resistor is the exact same power that we got for the external force."},{"Start":"21:21.465 ","End":"21:25.515","Text":"In this case they are equal,"},{"Start":"21:25.515 ","End":"21:28.305","Text":"but not always will they be equal."},{"Start":"21:28.305 ","End":"21:33.000","Text":"The power in the resistor is how much of the energy in"},{"Start":"21:33.000 ","End":"21:39.060","Text":"the resistor is lost to external things such as heat."},{"Start":"21:39.060 ","End":"21:43.290","Text":"That is of course per unit of time, so per second."},{"Start":"21:43.290 ","End":"21:51.270","Text":"The power of the external force is the power which is entering into the circuit."},{"Start":"21:51.270 ","End":"21:55.980","Text":"In this case it\u0027s entering the circuit from this single external force."},{"Start":"21:55.980 ","End":"22:04.050","Text":"In this case, they\u0027re both equal because I only have the resistor and the external force."},{"Start":"22:04.050 ","End":"22:08.723","Text":"However, if this circuit consisted of another electrical component,"},{"Start":"22:08.723 ","End":"22:10.800","Text":"so something aside from the resistor,"},{"Start":"22:10.800 ","End":"22:13.592","Text":"let\u0027s say over here we also had a capacitor,"},{"Start":"22:13.592 ","End":"22:20.475","Text":"then this power would be divided between the resistor and the capacitor."},{"Start":"22:20.475 ","End":"22:23.910","Text":"Then of course, the power in the resistor would"},{"Start":"22:23.910 ","End":"22:27.165","Text":"not be equal to the power of the external force."},{"Start":"22:27.165 ","End":"22:31.860","Text":"The reason that they are equal in this question is because all the power,"},{"Start":"22:31.860 ","End":"22:38.175","Text":"all the energy that is lost in the resistor goes in to the external force."},{"Start":"22:38.175 ","End":"22:41.700","Text":"Only in this specific case are they equal."},{"Start":"22:41.700 ","End":"22:44.740","Text":"That\u0027s the end of this lesson."}],"ID":21443},{"Watched":false,"Name":"Exercise 1","Duration":"20m 37s","ChapterTopicVideoID":21364,"CourseChapterTopicPlaylistID":99480,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.770","Text":"Hello. In this lesson,"},{"Start":"00:01.770 ","End":"00:04.380","Text":"we\u0027re going to be answering the following question."},{"Start":"00:04.380 ","End":"00:11.340","Text":"An infinite wire has a current i-naught flowing through it and a square frame of"},{"Start":"00:11.340 ","End":"00:19.515","Text":"side length a and resistance r is located a distance of x-Naught away from the wire."},{"Start":"00:19.515 ","End":"00:26.820","Text":"At t=0 the frame begins to move with a velocity of v-naught in the x-direction."},{"Start":"00:26.820 ","End":"00:35.400","Text":"There is a magnetic field B and question number 1 is to calculate the EMF."},{"Start":"00:35.400 ","End":"00:41.210","Text":"First of all, we know that the EMF is a type of voltage and it is equal to"},{"Start":"00:41.210 ","End":"00:47.430","Text":"the negative time derivative of the magnetic flux."},{"Start":"00:47.430 ","End":"00:49.420","Text":"In order to calculate the EMF,"},{"Start":"00:49.420 ","End":"00:52.820","Text":"we have to calculate the magnetic Flux."},{"Start":"00:52.820 ","End":"00:56.090","Text":"However, in order to calculate the magnetic flux,"},{"Start":"00:56.090 ","End":"01:00.025","Text":"we have to calculate the magnetic field."},{"Start":"01:00.025 ","End":"01:03.285","Text":"Let\u0027s see what the magnetic field is."},{"Start":"01:03.285 ","End":"01:06.290","Text":"According to Ampere\u0027s law if we have"},{"Start":"01:06.290 ","End":"01:13.055","Text":"a current carrying wire that means that we\u0027re going to have a magnetic field."},{"Start":"01:13.055 ","End":"01:16.520","Text":"From Ampere\u0027s law which you can look if you can\u0027t"},{"Start":"01:16.520 ","End":"01:20.240","Text":"remember how to calculate it in the chapter dealing with Ampere\u0027s law,"},{"Start":"01:20.240 ","End":"01:23.870","Text":"the magnetic field due to a current carrying wire is equal"},{"Start":"01:23.870 ","End":"01:29.270","Text":"to Mu naught multiplied by the current flowing through it."},{"Start":"01:29.270 ","End":"01:38.580","Text":"Here it is i-naught divided by 2 Pi r where r is the distance away from the wire"},{"Start":"01:38.580 ","End":"01:43.280","Text":"and its direction is in the Theta direction so"},{"Start":"01:43.280 ","End":"01:49.880","Text":"the magnetic field goes around the wire so we can get this from the right-hand rule."},{"Start":"01:49.880 ","End":"01:53.630","Text":"We can say that this is the y-direction so"},{"Start":"01:53.630 ","End":"01:57.980","Text":"our thumb points in the direction of the current and then our fingers curl"},{"Start":"01:57.980 ","End":"02:06.970","Text":"in the direction of the magnetic field which here will be going like so."},{"Start":"02:06.970 ","End":"02:09.105","Text":"This is the magnetic field,"},{"Start":"02:09.105 ","End":"02:13.265","Text":"so this is the Theta direction and so if we look on this side"},{"Start":"02:13.265 ","End":"02:17.660","Text":"of the wire because this is where our frame is,"},{"Start":"02:17.660 ","End":"02:21.995","Text":"we can see that on this side of the y if you use the right-hand rule,"},{"Start":"02:21.995 ","End":"02:28.770","Text":"the magnetic field over here is going inside the page."},{"Start":"02:29.620 ","End":"02:36.410","Text":"As we can see we have a magnetic field going inside the page and we can"},{"Start":"02:36.410 ","End":"02:39.470","Text":"see that it isn\u0027t a constant magnetic field because it is"},{"Start":"02:39.470 ","End":"02:42.710","Text":"dependent on r which is the distance away,"},{"Start":"02:42.710 ","End":"02:46.109","Text":"from the wire itself."},{"Start":"02:46.730 ","End":"02:49.405","Text":"In the previous videos,"},{"Start":"02:49.405 ","End":"02:52.970","Text":"we\u0027ve seen that in order to calculate"},{"Start":"02:52.970 ","End":"03:00.950","Text":"the magnetic flux we can just use the area of the square,"},{"Start":"03:00.950 ","End":"03:04.760","Text":"and see the magnetic flux through this area."},{"Start":"03:04.760 ","End":"03:08.420","Text":"However, here as we move across the square in"},{"Start":"03:08.420 ","End":"03:13.320","Text":"this direction we can see that the magnetic flux is going to be"},{"Start":"03:13.320 ","End":"03:17.165","Text":"decreasing as r increases so we don\u0027t have"},{"Start":"03:17.165 ","End":"03:23.160","Text":"a constant magnetic field so we\u0027re going to have to integrate."},{"Start":"03:24.490 ","End":"03:29.765","Text":"Of course, aside from inside the frame,"},{"Start":"03:29.765 ","End":"03:33.085","Text":"we have this different magnetic field,"},{"Start":"03:33.085 ","End":"03:37.645","Text":"also we forgot that our frame is also moving so as it moves"},{"Start":"03:37.645 ","End":"03:43.035","Text":"the magnetic field is also going to change within the frame and so,"},{"Start":"03:43.035 ","End":"03:46.350","Text":"therefore, the flux is also changing."},{"Start":"03:46.350 ","End":"03:49.845","Text":"This is great. We can see that the magnetic flux is"},{"Start":"03:49.845 ","End":"03:54.639","Text":"changing as a function of time which means that we have this EMF."},{"Start":"03:54.639 ","End":"04:00.395","Text":"If it wasn\u0027t as a function of time this derivative would be equal to 0."},{"Start":"04:00.395 ","End":"04:02.925","Text":"Let\u0027s work out the magnetic flux."},{"Start":"04:02.925 ","End":"04:09.190","Text":"The magnetic flux is the integral of b dot ds."},{"Start":"04:10.490 ","End":"04:15.975","Text":"As we can see our ds which is our area goes along"},{"Start":"04:15.975 ","End":"04:21.410","Text":"the x-axis which is in this direction and the y-axis so we have"},{"Start":"04:21.410 ","End":"04:29.205","Text":"this double integral of b so we have Mu naught i-naught divided by"},{"Start":"04:29.205 ","End":"04:38.560","Text":"2 Pi r. Then ds is in the xy direction so dx, dy."},{"Start":"04:38.780 ","End":"04:42.760","Text":"Now what we want to see is we have this r but we"},{"Start":"04:42.760 ","End":"04:46.270","Text":"want r in Cartesian coordinates so according to x and"},{"Start":"04:46.270 ","End":"04:53.030","Text":"y and as we saw i is our distance away from the wire itself."},{"Start":"04:53.030 ","End":"04:56.420","Text":"As we can see it\u0027s always going to be in the x-direction no"},{"Start":"04:56.420 ","End":"05:00.055","Text":"matter which point we choose because the wire is infinitely long."},{"Start":"05:00.055 ","End":"05:04.790","Text":"If we choose any point is just going to be this distance away from"},{"Start":"05:04.790 ","End":"05:10.005","Text":"the wire so instead of i we can replace it over here with x."},{"Start":"05:10.005 ","End":"05:14.850","Text":"Now what we want to do is we want to substitute in our bounds."},{"Start":"05:15.500 ","End":"05:19.880","Text":"Along the y-axis we can see we\u0027re going from this point over"},{"Start":"05:19.880 ","End":"05:24.665","Text":"here so we can say that that\u0027s at y is equal to 0"},{"Start":"05:24.665 ","End":"05:29.660","Text":"and up until the top of the frame over here which is at y is equal"},{"Start":"05:29.660 ","End":"05:35.885","Text":"to a and then with x we have something a bit more complicated."},{"Start":"05:35.885 ","End":"05:40.410","Text":"At t is equal to 0 our frame is located at"},{"Start":"05:40.410 ","End":"05:47.850","Text":"x-Naught and then from then on it carries on moving with this constant velocity v-naught."},{"Start":"05:47.850 ","End":"05:51.110","Text":"If I substitute in overhear from"},{"Start":"05:51.110 ","End":"06:00.680","Text":"x-naught until this over here which is just x-naught plus a,"},{"Start":"06:00.680 ","End":"06:07.340","Text":"we\u0027re going to get the magnetic flux just at the initial time t is equal to 0 and as we"},{"Start":"06:07.340 ","End":"06:10.730","Text":"can see our magnetic flux is independent of"},{"Start":"06:10.730 ","End":"06:15.160","Text":"time at this moment which means that we will have a 0 EMF."},{"Start":"06:15.160 ","End":"06:19.130","Text":"What we want is we want to know the general equation for"},{"Start":"06:19.130 ","End":"06:24.370","Text":"the magnetic flux over a period of time not just at t is equal to 0."},{"Start":"06:24.370 ","End":"06:28.760","Text":"What we can see is if we place our frame"},{"Start":"06:28.760 ","End":"06:33.635","Text":"a few moments later where it\u0027s at some random distance x,"},{"Start":"06:33.635 ","End":"06:37.350","Text":"and then the frame is over here."},{"Start":"06:38.180 ","End":"06:40.995","Text":"What will x be?"},{"Start":"06:40.995 ","End":"06:49.020","Text":"x is going to be equal to the initial position which was x-naught plus,"},{"Start":"06:49.180 ","End":"06:52.055","Text":"this distance over here."},{"Start":"06:52.055 ","End":"06:53.479","Text":"What is this distance?"},{"Start":"06:53.479 ","End":"06:55.145","Text":"It\u0027s the velocity,"},{"Start":"06:55.145 ","End":"07:01.520","Text":"v-naught multiplied by time t. What we\u0027re going to do"},{"Start":"07:01.520 ","End":"07:04.670","Text":"is we\u0027re going to plug this in here so instead of writing"},{"Start":"07:04.670 ","End":"07:08.585","Text":"x-naught we\u0027re just going to write x."},{"Start":"07:08.585 ","End":"07:13.380","Text":"Then x is just going to be,"},{"Start":"07:13.580 ","End":"07:21.140","Text":"this over here where of course x is this side of the frame and then we always have to"},{"Start":"07:21.140 ","End":"07:28.799","Text":"add in this a to get the change in flux throughout the frame itself."},{"Start":"07:29.570 ","End":"07:37.310","Text":"Now all that\u0027s left is to just do the integral so we can take out all the constants and"},{"Start":"07:37.310 ","End":"07:46.050","Text":"integrate with respect to y first so we will have Mu naught i-naught a divided by 2 Pi."},{"Start":"07:46.210 ","End":"07:55.475","Text":"Then what we\u0027re going to have is we\u0027re going to multiply this by 1 divided by x, dx,"},{"Start":"07:55.475 ","End":"07:59.450","Text":"and when we plug in our bounds we\u0027re just going to have"},{"Start":"07:59.450 ","End":"08:07.270","Text":"ln(x) plus a minus ln(x)."},{"Start":"08:08.960 ","End":"08:15.810","Text":"This is the integral and now what we want to do we have our magnetic flux,"},{"Start":"08:15.810 ","End":"08:17.475","Text":"this is the magnetic flux."},{"Start":"08:17.475 ","End":"08:22.530","Text":"We want to get the EMF which is the negative time derivative."},{"Start":"08:23.540 ","End":"08:26.810","Text":"As we can see we have a function within a function,"},{"Start":"08:26.810 ","End":"08:32.790","Text":"we have ln of something where our something is as a function of"},{"Start":"08:32.790 ","End":"08:37.100","Text":"t. What we want to do in order to take"},{"Start":"08:37.100 ","End":"08:42.170","Text":"the derivative of this with respect to time is to use the chain rule."},{"Start":"08:42.170 ","End":"08:51.920","Text":"Our EMF is equal to the negative time derivative of the magnetic flux"},{"Start":"08:51.920 ","End":"08:54.260","Text":"which says we have a function within"},{"Start":"08:54.260 ","End":"08:58.685","Text":"a function it\u0027s according to the chain rule going to be equal to"},{"Start":"08:58.685 ","End":"09:08.680","Text":"the negative derivative of the magnetic flux with respect to its variable which is x,"},{"Start":"09:08.710 ","End":"09:15.755","Text":"multiplied by the derivative of"},{"Start":"09:15.755 ","End":"09:25.260","Text":"the variable x with respect to its variable t so dx by dt."},{"Start":"09:27.400 ","End":"09:34.449","Text":"We have negative and then the derivative of this so we have our constants"},{"Start":"09:34.449 ","End":"09:42.140","Text":"Mu naught i-naught a divided by 2 Pi."},{"Start":"09:43.410 ","End":"09:52.695","Text":"Then the derivatives of these with respect to x is just going to be 1 divided by x plus a"},{"Start":"09:52.695 ","End":"10:02.700","Text":"minus 1 divided by x and then this is multiplied by dx by dt where this is x,"},{"Start":"10:02.700 ","End":"10:10.720","Text":"remember x-naught is a constant which is independent of time so when we take"},{"Start":"10:10.720 ","End":"10:14.930","Text":"the time derivative of that it\u0027s equal to 0 plus the time derivative"},{"Start":"10:14.930 ","End":"10:19.695","Text":"of v-naught t is just going to leave us with v-naught,"},{"Start":"10:19.695 ","End":"10:22.990","Text":"so this is the EMF."},{"Start":"10:22.990 ","End":"10:26.720","Text":"This has the answer to question number 1."},{"Start":"10:26.720 ","End":"10:32.280","Text":"Now, let\u0027s answer question number 2."},{"Start":"10:32.600 ","End":"10:36.770","Text":"Question number 2 is to calculate the current"},{"Start":"10:36.770 ","End":"10:41.130","Text":"and what is meant by this is the current in the frame."},{"Start":"10:41.480 ","End":"10:45.515","Text":"We were told in the question that"},{"Start":"10:45.515 ","End":"10:48.770","Text":"the frame has a resistance r so let\u0027s imagine that there\u0027s"},{"Start":"10:48.770 ","End":"10:56.210","Text":"a little resistor over here with the resistance r. In order to calculate the current,"},{"Start":"10:56.210 ","End":"11:01.295","Text":"we know that current is equal to voltage divided by"},{"Start":"11:01.295 ","End":"11:07.070","Text":"resistance where we know that the EMF is a voltage,"},{"Start":"11:07.070 ","End":"11:13.669","Text":"so this is a voltage so it\u0027s just equal to the EMF divided by the resistance"},{"Start":"11:13.669 ","End":"11:18.920","Text":"r. I\u0027m not going to substitute this in but all you have"},{"Start":"11:18.920 ","End":"11:24.542","Text":"to do is just take this and plug it in over here."},{"Start":"11:24.542 ","End":"11:31.030","Text":"Remember of course x is also a substitution of this."},{"Start":"11:31.030 ","End":"11:35.950","Text":"Now, we know the current has both size and direction."},{"Start":"11:35.950 ","End":"11:39.535","Text":"This is the size and now we want to know the direction."},{"Start":"11:39.535 ","End":"11:45.955","Text":"What we do is we use Lenz\u0027s law in order to calculate the direction."},{"Start":"11:45.955 ","End":"11:51.160","Text":"Lenz\u0027s law states that the direction of the current is going to"},{"Start":"11:51.160 ","End":"11:56.350","Text":"be such as to oppose the change in magnetic flux."},{"Start":"11:56.350 ","End":"12:01.465","Text":"First of all, we want to see that the magnetic flux is changing,"},{"Start":"12:01.465 ","End":"12:03.655","Text":"which it is, we saw over here,"},{"Start":"12:03.655 ","End":"12:06.025","Text":"it\u0027s dependent on time."},{"Start":"12:06.025 ","End":"12:13.030","Text":"Also, the fact that we just have an EMF means that the magnetic flux is changing."},{"Start":"12:13.030 ","End":"12:17.230","Text":"Now, what we want to do is we want to oppose its change."},{"Start":"12:17.230 ","End":"12:22.045","Text":"We want to see if the magnetic flux is increasing or decreasing,"},{"Start":"12:22.045 ","End":"12:25.810","Text":"and what we can see is from the magnetic field,"},{"Start":"12:25.810 ","End":"12:30.580","Text":"as the frame moves away from the current carrying wire,"},{"Start":"12:30.580 ","End":"12:34.820","Text":"the magnetic field is decreasing."},{"Start":"12:35.190 ","End":"12:38.890","Text":"If the magnetic field is decreasing,"},{"Start":"12:38.890 ","End":"12:43.360","Text":"then that means that the magnetic flux is also decreasing."},{"Start":"12:43.360 ","End":"12:45.145","Text":"According to Lenz\u0027s law,"},{"Start":"12:45.145 ","End":"12:52.405","Text":"we\u0027re going to want to have a current that is opposing this decrease in magnetic flux."},{"Start":"12:52.405 ","End":"12:57.640","Text":"What we\u0027re going to do is we\u0027re going to add another magnetic field,"},{"Start":"12:57.640 ","End":"13:04.195","Text":"which is in the same direction as this original magnetic field."},{"Start":"13:04.195 ","End":"13:06.340","Text":"It\u0027s also going into the page,"},{"Start":"13:06.340 ","End":"13:13.150","Text":"and then we want to see which current is inducing this magnetic field."},{"Start":"13:13.150 ","End":"13:15.805","Text":"According to the right-hand rule,"},{"Start":"13:15.805 ","End":"13:22.220","Text":"if our thumb is pointing in the direction of the magnetic field,"},{"Start":"13:22.320 ","End":"13:26.590","Text":"then what we will get is that our fingers will curl around in"},{"Start":"13:26.590 ","End":"13:31.105","Text":"the direction of the current which is in this direction,"},{"Start":"13:31.105 ","End":"13:36.290","Text":"so our current is traveling in the clockwise direction."},{"Start":"13:37.500 ","End":"13:43.375","Text":"Now let\u0027s answer question number 3."},{"Start":"13:43.375 ","End":"13:50.030","Text":"What external force is required in order for the frame to move at a constant velocity?"},{"Start":"13:50.190 ","End":"14:02.050","Text":"As we saw, our force is given by the equation dF=Idl cross B."},{"Start":"14:02.050 ","End":"14:07.720","Text":"Assuming that this magnetic field over here that we induced is constant,"},{"Start":"14:07.720 ","End":"14:11.350","Text":"so we can say that therefore the force is equal"},{"Start":"14:11.350 ","End":"14:16.855","Text":"to B multiplied by I multiplied by the side length,"},{"Start":"14:16.855 ","End":"14:19.660","Text":"multiplied by sine of the angle"},{"Start":"14:19.660 ","End":"14:23.890","Text":"between the direction of the current and the magnetic field."},{"Start":"14:23.890 ","End":"14:29.530","Text":"As we can see, the current is always on the xy plane,"},{"Start":"14:29.530 ","End":"14:34.675","Text":"and the magnetic field is always in the z-direction."},{"Start":"14:34.675 ","End":"14:40.120","Text":"We can see that the angle between the current and the magnetic field is 90 degrees,"},{"Start":"14:40.120 ","End":"14:44.950","Text":"so we can say that this over here sine of 90 degrees,"},{"Start":"14:44.950 ","End":"14:46.390","Text":"which is equal to 1,"},{"Start":"14:46.390 ","End":"14:51.890","Text":"so therefore what we get is that the force is equal to BIL,"},{"Start":"14:52.230 ","End":"14:58.820","Text":"where L over here is the side length, which is a."},{"Start":"15:00.930 ","End":"15:04.690","Text":"Now, what we want to do is we want to find out"},{"Start":"15:04.690 ","End":"15:08.335","Text":"what magnetic force is being applied to this frame,"},{"Start":"15:08.335 ","End":"15:14.750","Text":"and then we will see that the external force is in the opposite direction."},{"Start":"15:15.660 ","End":"15:19.570","Text":"First of all, on these 2 sides,"},{"Start":"15:19.570 ","End":"15:21.100","Text":"the top and the bottom,"},{"Start":"15:21.100 ","End":"15:25.600","Text":"we can see that they\u0027re mirror images of one another."},{"Start":"15:25.600 ","End":"15:29.170","Text":"That\u0027s the exact same magnetic field acting on them,"},{"Start":"15:29.170 ","End":"15:35.470","Text":"so we\u0027re going to have this force up in this direction,"},{"Start":"15:35.470 ","End":"15:38.620","Text":"from this side according to the right-hand rule,"},{"Start":"15:38.620 ","End":"15:44.260","Text":"because the current is traveling on this side in the right direction."},{"Start":"15:44.260 ","End":"15:47.080","Text":"Remember according to this equation,"},{"Start":"15:47.080 ","End":"15:50.095","Text":"the direction of the current is your thumb,"},{"Start":"15:50.095 ","End":"15:53.380","Text":"the direction of the magnetic field is your forefinger,"},{"Start":"15:53.380 ","End":"15:58.195","Text":"and then the middle finger represents the direction of the force,"},{"Start":"15:58.195 ","End":"16:02.815","Text":"and we get that the force is pointing in this direction,"},{"Start":"16:02.815 ","End":"16:06.445","Text":"and on the bottom side for the exact same reason,"},{"Start":"16:06.445 ","End":"16:10.150","Text":"because the current is now in the leftward direction,"},{"Start":"16:10.150 ","End":"16:13.345","Text":"so the force is pointing downwards,"},{"Start":"16:13.345 ","End":"16:15.670","Text":"and because both of these forces as we\u0027ve"},{"Start":"16:15.670 ","End":"16:18.130","Text":"already said are exactly equal and opposite because"},{"Start":"16:18.130 ","End":"16:23.260","Text":"both of these sides have the exact same magnetic field applied to them,"},{"Start":"16:23.260 ","End":"16:25.835","Text":"so they\u0027re just going to cancel out."},{"Start":"16:25.835 ","End":"16:34.600","Text":"What we\u0027re left with is some force on this side and some force on this side."},{"Start":"16:35.580 ","End":"16:41.530","Text":"What\u0027s important to note is that these 2 forces are in the opposite direction,"},{"Start":"16:41.530 ","End":"16:43.435","Text":"however, they\u0027re not equal."},{"Start":"16:43.435 ","End":"16:47.095","Text":"This is F_1 and this is F_2."},{"Start":"16:47.095 ","End":"16:48.895","Text":"Why are they not equal?"},{"Start":"16:48.895 ","End":"16:50.560","Text":"According to this equation,"},{"Start":"16:50.560 ","End":"16:55.930","Text":"we can see that the force is dependent on the magnetic field,"},{"Start":"16:55.930 ","End":"17:01.030","Text":"which as we saw is changing as we get further away from the wire."},{"Start":"17:01.030 ","End":"17:05.499","Text":"We can see that the force on this edge is going to be less"},{"Start":"17:05.499 ","End":"17:11.440","Text":"than the force on this edge because the magnetic field at this point is weaker."},{"Start":"17:11.440 ","End":"17:14.800","Text":"Let\u0027s calculate F_1."},{"Start":"17:14.800 ","End":"17:21.685","Text":"F_1 is equal to the magnetic field B that we calculated,"},{"Start":"17:21.685 ","End":"17:29.290","Text":"so that is equal to Mu_0 I_0, divided by 2Pi."},{"Start":"17:29.290 ","End":"17:33.220","Text":"Then our F_1 is at a general position"},{"Start":"17:33.220 ","End":"17:39.880","Text":"of some x which will take us to this left side,"},{"Start":"17:39.880 ","End":"17:45.985","Text":"plus a, which will take us to the right side of this frame."},{"Start":"17:45.985 ","End":"17:51.260","Text":"R over here is some general x plus a."},{"Start":"17:52.860 ","End":"17:56.110","Text":"This is our B multiplied by I."},{"Start":"17:56.110 ","End":"18:00.980","Text":"Now, I over here is the current in the frame."},{"Start":"18:01.050 ","End":"18:05.710","Text":"The current in the frame is this I that we calculated,"},{"Start":"18:05.710 ","End":"18:08.125","Text":"so I\u0027m just going to leave it as I,"},{"Start":"18:08.125 ","End":"18:10.074","Text":"the current in the frame,"},{"Start":"18:10.074 ","End":"18:13.360","Text":"which is our emf divided by R,"},{"Start":"18:13.360 ","End":"18:16.420","Text":"and then we multiply it by the side length,"},{"Start":"18:16.420 ","End":"18:19.405","Text":"which we\u0027ve already said is equal to a."},{"Start":"18:19.405 ","End":"18:21.805","Text":"This is our F_1,"},{"Start":"18:21.805 ","End":"18:28.300","Text":"and it is in this rightwards direction which we said was the positive x direction,"},{"Start":"18:28.300 ","End":"18:33.940","Text":"and then F_2 is equal to again the magnetic field,"},{"Start":"18:33.940 ","End":"18:38.410","Text":"but this time it\u0027s the magnetic field on this side over here,"},{"Start":"18:38.410 ","End":"18:41.800","Text":"so we\u0027re just at a general position x,"},{"Start":"18:41.800 ","End":"18:45.670","Text":"so we have the magnetic field at the general position x,"},{"Start":"18:45.670 ","End":"18:47.960","Text":"not x plus a."},{"Start":"18:48.150 ","End":"18:51.145","Text":"That is going to be equal to"},{"Start":"18:51.145 ","End":"19:01.330","Text":"Mu_0 I_0 divided by 2Pi x multiplied by the current,"},{"Start":"19:01.330 ","End":"19:03.115","Text":"again, the current in the frame,"},{"Start":"19:03.115 ","End":"19:07.255","Text":"not the current in the wire which is forming the magnetic field,"},{"Start":"19:07.255 ","End":"19:08.635","Text":"multiplied by I,"},{"Start":"19:08.635 ","End":"19:10.735","Text":"we already said what I is equal to,"},{"Start":"19:10.735 ","End":"19:17.860","Text":"and again multiplied by L which is the side length which again is a."},{"Start":"19:17.860 ","End":"19:21.160","Text":"This is in this leftwards direction,"},{"Start":"19:21.160 ","End":"19:25.340","Text":"which is the negative x direction."},{"Start":"19:25.950 ","End":"19:29.650","Text":"Now, in order to find the external force required"},{"Start":"19:29.650 ","End":"19:32.830","Text":"in order for the frame to move at a constant velocity,"},{"Start":"19:32.830 ","End":"19:38.270","Text":"what we do is we\u0027re going to find just the magnitude first,"},{"Start":"19:38.370 ","End":"19:42.730","Text":"and it\u0027s equal to the sum of these 2 forces,"},{"Start":"19:42.730 ","End":"19:50.395","Text":"which is just taking the magnitude of F_1 minus because it\u0027s in the opposite direction,"},{"Start":"19:50.395 ","End":"19:54.295","Text":"the magnitude of F_2."},{"Start":"19:54.295 ","End":"19:56.335","Text":"You can just plug that in."},{"Start":"19:56.335 ","End":"20:03.670","Text":"We just take the absolute value of F_1 minus the absolute value of F_2,"},{"Start":"20:03.670 ","End":"20:06.880","Text":"and this will give us the magnitude of the external force,"},{"Start":"20:06.880 ","End":"20:08.650","Text":"and its direction, of course,"},{"Start":"20:08.650 ","End":"20:11.765","Text":"is going to be in the direction of the velocity,"},{"Start":"20:11.765 ","End":"20:16.710","Text":"because the whole point in this question is to find what external force"},{"Start":"20:16.710 ","End":"20:21.060","Text":"is required in order for the frame to move at this constant velocity,"},{"Start":"20:21.060 ","End":"20:27.355","Text":"which means that the force is going to be in the same direction as the velocity,"},{"Start":"20:27.355 ","End":"20:31.000","Text":"which of course means it\u0027s in the x direction."},{"Start":"20:31.000 ","End":"20:34.315","Text":"This is the answer to question number 3,"},{"Start":"20:34.315 ","End":"20:37.460","Text":"and that is the end of this lesson."}],"ID":21444},{"Watched":false,"Name":"Exercise 2","Duration":"12m 3s","ChapterTopicVideoID":21365,"CourseChapterTopicPlaylistID":99480,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.995","Text":"Hello. In this lesson,"},{"Start":"00:01.995 ","End":"00:04.650","Text":"we\u0027re going to be answering the following question."},{"Start":"00:04.650 ","End":"00:13.395","Text":"A conducting ring of radius A and resistance I is attached to 2 non-conducting rods."},{"Start":"00:13.395 ","End":"00:18.105","Text":"These rods rotate the ring with an angular velocity of Omega."},{"Start":"00:18.105 ","End":"00:23.145","Text":"You can imagine that the ring is just going around these rods."},{"Start":"00:23.145 ","End":"00:30.590","Text":"Like if you had some barbecue spit and the meat is going around,"},{"Start":"00:30.590 ","End":"00:33.125","Text":"the rod is spinning it around."},{"Start":"00:33.125 ","End":"00:36.470","Text":"The angular velocity is Omega."},{"Start":"00:36.470 ","End":"00:41.450","Text":"There is a magnetic field of B_naught throughout in this direction"},{"Start":"00:41.450 ","End":"00:47.865","Text":"as so and Question number 1 is to calculate the EMF."},{"Start":"00:47.865 ","End":"00:49.505","Text":"First of all, as we know,"},{"Start":"00:49.505 ","End":"00:56.615","Text":"the EMF is equal to the negative time derivative of the magnetic flux."},{"Start":"00:56.615 ","End":"01:02.430","Text":"The first thing that we have to do is we have to calculate the magnetic flux."},{"Start":"01:02.510 ","End":"01:09.155","Text":"What we can see is that because this ring is spinning around,"},{"Start":"01:09.155 ","End":"01:15.015","Text":"so it\u0027s not always going to be exactly parallel to the magnetic field."},{"Start":"01:15.015 ","End":"01:21.640","Text":"First of all, let\u0027s say that the direction of the magnetic field is in the z-direction."},{"Start":"01:21.640 ","End":"01:24.225","Text":"Now, to make this drawing clearer,"},{"Start":"01:24.225 ","End":"01:31.130","Text":"let\u0027s draw this diagram when we\u0027re looking at it from the side."},{"Start":"01:31.130 ","End":"01:35.000","Text":"Then what we\u0027re going to have is we\u0027re going to have"},{"Start":"01:35.000 ","End":"01:41.280","Text":"the ring that\u0027s at some angle like so."},{"Start":"01:41.280 ","End":"01:44.750","Text":"It will look something like this and just to make it even simpler,"},{"Start":"01:44.750 ","End":"01:47.375","Text":"we\u0027ll just draw it as a line."},{"Start":"01:47.375 ","End":"01:49.940","Text":"We\u0027re looking at the ring from this side,"},{"Start":"01:49.940 ","End":"01:53.070","Text":"and we can just see the line."},{"Start":"01:55.160 ","End":"02:01.290","Text":"This is of course the z-axis,"},{"Start":"02:01.290 ","End":"02:05.415","Text":"the z-direction, coming from over here."},{"Start":"02:05.415 ","End":"02:16.205","Text":"As we know, the magnetic flux is given as the integral of B.ds vector."},{"Start":"02:16.205 ","End":"02:18.305","Text":"Our ds vector, as we know,"},{"Start":"02:18.305 ","End":"02:23.385","Text":"always has to be perpendicular to the plane."},{"Start":"02:23.385 ","End":"02:32.720","Text":"This is our ds vector and this is our z-axis over here."},{"Start":"02:32.720 ","End":"02:41.070","Text":"Now what we can do is we can label this angle between the 2 as Theta."},{"Start":"02:41.920 ","End":"02:50.795","Text":"Then when we do the dot product between the B vector or the magnetic field,"},{"Start":"02:50.795 ","End":"02:52.685","Text":"which as we know,"},{"Start":"02:52.685 ","End":"02:55.840","Text":"is also in the z-direction."},{"Start":"02:55.840 ","End":"02:59.680","Text":"Then when we do B.ds,"},{"Start":"02:59.680 ","End":"03:03.590","Text":"we can say that this is equal to the integral of"},{"Start":"03:03.590 ","End":"03:12.160","Text":"Bds multiplied by the cosine of the angle between the 2."},{"Start":"03:12.530 ","End":"03:19.400","Text":"In that case, we can say that this is equal to the magnetic field,"},{"Start":"03:19.400 ","End":"03:24.560","Text":"which is B_naught multiplied by cosine of Theta,"},{"Start":"03:24.560 ","End":"03:27.770","Text":"and then we just have to integrate along ds,"},{"Start":"03:27.770 ","End":"03:31.430","Text":"where ds is the area of the ring."},{"Start":"03:31.430 ","End":"03:39.214","Text":"What we get is B_naught cosine of Theta multiplied by Pi,"},{"Start":"03:39.214 ","End":"03:43.420","Text":"the radius^2, so A^2."},{"Start":"03:43.420 ","End":"03:50.495","Text":"Here we see how also the area is dependent on this angle Theta."},{"Start":"03:50.495 ","End":"03:56.600","Text":"The angle that the ring is located at and of course,"},{"Start":"03:56.600 ","End":"03:57.935","Text":"because we\u0027re given Omega,"},{"Start":"03:57.935 ","End":"04:04.535","Text":"we know that Theta is dependent on time and on angular velocity."},{"Start":"04:04.535 ","End":"04:08.645","Text":"Theta is equal to Omega t. In that case,"},{"Start":"04:08.645 ","End":"04:16.970","Text":"we get that our magnetic flux is equal to B_naught cosine of Theta,"},{"Start":"04:16.970 ","End":"04:23.170","Text":"which is Omega t multiplied by Pia^2."},{"Start":"04:23.350 ","End":"04:29.720","Text":"Now, the EMF is just the negative time derivative."},{"Start":"04:29.720 ","End":"04:32.180","Text":"It\u0027s equal to negative,"},{"Start":"04:32.180 ","End":"04:35.990","Text":"and then the time derivative of this."},{"Start":"04:36.830 ","End":"04:41.205","Text":"The inner derivative will be Omega."},{"Start":"04:41.205 ","End":"04:45.765","Text":"We have negative multiplied by Omega,"},{"Start":"04:45.765 ","End":"04:48.930","Text":"multiplied by B_naught,"},{"Start":"04:48.930 ","End":"04:56.170","Text":"and then the derivative of cosine or the outer derivative of cosine Omega t is equal to"},{"Start":"04:56.170 ","End":"05:06.775","Text":"negative sine of Omega t. Then this is all multiplied by Pia^2."},{"Start":"05:06.775 ","End":"05:10.180","Text":"Then this minus and this minus can cancel out."},{"Start":"05:10.180 ","End":"05:13.030","Text":"What we get is that the EMF is equal to"},{"Start":"05:13.030 ","End":"05:23.155","Text":"Omega B_naught sine of Omega t multiplied by Pia^2."},{"Start":"05:23.155 ","End":"05:26.160","Text":"That\u0027s the answer to Question number 1."},{"Start":"05:26.160 ","End":"05:29.835","Text":"Now let\u0027s answer Question number 2, calculate the current."},{"Start":"05:29.835 ","End":"05:34.715","Text":"As we know, our current is equal to voltage divided by resistance."},{"Start":"05:34.715 ","End":"05:43.410","Text":"Our voltage is Omega B_naught sine of Omega t multiplied by"},{"Start":"05:43.410 ","End":"05:51.900","Text":"Pia^2 and our resistance is R, so that\u0027s it."},{"Start":"05:51.900 ","End":"05:54.425","Text":"This is our current and as we can see,"},{"Start":"05:54.425 ","End":"05:56.659","Text":"our current is changing."},{"Start":"05:56.659 ","End":"06:00.850","Text":"It\u0027s constantly changing as a function of Omega."},{"Start":"06:00.850 ","End":"06:05.300","Text":"This is how we get AC, an alternating current,"},{"Start":"06:05.300 ","End":"06:12.420","Text":"and this is also what they do in the various power plants."},{"Start":"06:12.420 ","End":"06:19.580","Text":"They take a ring or lots of rings and that they have a current flowing through them,"},{"Start":"06:19.580 ","End":"06:23.795","Text":"and they just rotate them or that they can conduct current,"},{"Start":"06:23.795 ","End":"06:27.530","Text":"and they rotate them inside an external magnetic field,"},{"Start":"06:27.530 ","End":"06:29.975","Text":"and then you get this AC current,"},{"Start":"06:29.975 ","End":"06:32.590","Text":"and you also get this voltage."},{"Start":"06:32.590 ","End":"06:35.655","Text":"That\u0027s the answer to Question number 2."},{"Start":"06:35.655 ","End":"06:39.090","Text":"Now let\u0027s answer Question number 3."},{"Start":"06:39.090 ","End":"06:44.465","Text":"Question number 3 is to calculate the EMF again,"},{"Start":"06:44.465 ","End":"06:48.500","Text":"where now the magnetic field is changing in time."},{"Start":"06:48.500 ","End":"06:51.305","Text":"So it\u0027s in the same direction,"},{"Start":"06:51.305 ","End":"06:55.245","Text":"but it\u0027s dependent on time."},{"Start":"06:55.245 ","End":"07:01.650","Text":"It\u0027s equal to B_naught cosine of Omega t. This is our new magnetic fields."},{"Start":"07:01.650 ","End":"07:06.290","Text":"Again, the first thing we have to do is just like before,"},{"Start":"07:06.290 ","End":"07:09.785","Text":"we have to calculate the magnetic flux."},{"Start":"07:09.785 ","End":"07:16.520","Text":"The magnetic flux is again the integral of Bds or Bds cosine of Theta."},{"Start":"07:16.520 ","End":"07:20.760","Text":"We have the integral of the magnetic fields."},{"Start":"07:20.760 ","End":"07:28.280","Text":"B_naught cosine of Omega t multiplied by,"},{"Start":"07:28.280 ","End":"07:32.525","Text":"so that was B multiplied by cosine of Theta."},{"Start":"07:32.525 ","End":"07:37.670","Text":"Of course, cosine of Theta where Theta is equal to Omega t. Again,"},{"Start":"07:37.670 ","End":"07:45.495","Text":"we have Omega t. That\u0027s Theta and then ds, finally."},{"Start":"07:45.495 ","End":"07:49.485","Text":"We have B cosine Theta ds."},{"Start":"07:49.485 ","End":"07:53.480","Text":"Notice that the integral that we\u0027re doing here to"},{"Start":"07:53.480 ","End":"07:57.095","Text":"find the magnetic flux is dependent on ds."},{"Start":"07:57.095 ","End":"08:03.580","Text":"We\u0027re integrating with respect to the surface area of the ring."},{"Start":"08:03.580 ","End":"08:07.775","Text":"We\u0027re not integrating with respect to the magnetic field."},{"Start":"08:07.775 ","End":"08:13.145","Text":"That means that we can take all of this as a constant because this,"},{"Start":"08:13.145 ","End":"08:16.470","Text":"although the magnetic field is changing,"},{"Start":"08:17.140 ","End":"08:22.640","Text":"we\u0027re integrating this integral is with respect to the area,"},{"Start":"08:22.640 ","End":"08:24.920","Text":"not with respect to the magnetic fields,"},{"Start":"08:24.920 ","End":"08:26.570","Text":"so it\u0027s considered a constant."},{"Start":"08:26.570 ","End":"08:30.355","Text":"You can say that this is equal to B_naught."},{"Start":"08:30.355 ","End":"08:41.120","Text":"Then we have cosine^2 of Omega t. Then we have the integral on ds,"},{"Start":"08:41.120 ","End":"08:43.805","Text":"which is of course equal to Pia^2."},{"Start":"08:43.805 ","End":"08:53.220","Text":"We have B_naught cosine^2 of Omega t multiplied by Pia^2."},{"Start":"08:53.660 ","End":"08:58.600","Text":"Now, with the EMF, we want to take the negative time derivative."},{"Start":"08:58.600 ","End":"09:01.085","Text":"Because we have cosine over here,"},{"Start":"09:01.085 ","End":"09:04.030","Text":"we know that one of the derivatives,"},{"Start":"09:04.030 ","End":"09:06.180","Text":"so we have cosine^2 Omega t. We have"},{"Start":"09:06.180 ","End":"09:11.795","Text":"the middle derivative of cosine and the outer derivative of 2."},{"Start":"09:11.795 ","End":"09:14.510","Text":"The middle derivative is going to be sine,"},{"Start":"09:14.510 ","End":"09:16.160","Text":"which will give us a minus and then the"},{"Start":"09:16.160 ","End":"09:21.550","Text":"minus for calculating the EMF with a negative time derivatives."},{"Start":"09:21.550 ","End":"09:22.895","Text":"This minus over here,"},{"Start":"09:22.895 ","End":"09:29.595","Text":"together with a minus for the derivative of cosine, will cancel out."},{"Start":"09:29.595 ","End":"09:32.440","Text":"We can just leave out the minuses."},{"Start":"09:32.630 ","End":"09:37.080","Text":"What we\u0027re left with is B_naught."},{"Start":"09:37.080 ","End":"09:40.590","Text":"Then we have 2,"},{"Start":"09:40.590 ","End":"09:47.130","Text":"and then we have cosine of Omega t. Then we"},{"Start":"09:47.130 ","End":"09:56.840","Text":"have sine of Omega t. Then we have from the anti-derivative Omega."},{"Start":"09:56.840 ","End":"10:01.220","Text":"All of this is of course multiplied by Pia^2."},{"Start":"10:01.220 ","End":"10:06.875","Text":"What we have here is 2 cosine Omega t sine Omega t,"},{"Start":"10:06.875 ","End":"10:16.960","Text":"which is equal to simply sine of 2 Omega t. We can just rewrite this"},{"Start":"10:16.960 ","End":"10:26.180","Text":"as B_naught Omega Pia^2 multiplied"},{"Start":"10:26.180 ","End":"10:35.390","Text":"by sine of 2 Omega t. We can see that we have exactly the same onset,"},{"Start":"10:35.390 ","End":"10:37.565","Text":"just written in slightly different order."},{"Start":"10:37.565 ","End":"10:44.605","Text":"The only difference is that when we had the constant magnetic field of B_naught,"},{"Start":"10:44.605 ","End":"10:52.359","Text":"we\u0027re multiplying by sine of Omega t. When we have this changing magnetic field,"},{"Start":"10:52.359 ","End":"10:59.380","Text":"then we\u0027re multiplying by sine of 2 Omega t. We can see that the EMF or"},{"Start":"10:59.380 ","End":"11:05.620","Text":"the voltage is also changing,"},{"Start":"11:05.620 ","End":"11:09.770","Text":"but this time with a frequency of times 2."},{"Start":"11:10.340 ","End":"11:13.555","Text":"If the ring is rotating,"},{"Start":"11:13.555 ","End":"11:17.785","Text":"we get this alternating voltage over here,"},{"Start":"11:17.785 ","End":"11:20.725","Text":"which leads to an alternating current."},{"Start":"11:20.725 ","End":"11:26.450","Text":"But if our magnetic field is also alternating with time,"},{"Start":"11:26.450 ","End":"11:31.025","Text":"then the frequency of the voltage is going to be twice as much."},{"Start":"11:31.025 ","End":"11:34.805","Text":"You can think about it as we have 2 things alternating."},{"Start":"11:34.805 ","End":"11:37.640","Text":"Also, the ring is moving around,"},{"Start":"11:37.640 ","End":"11:39.050","Text":"and it\u0027s alternating with time,"},{"Start":"11:39.050 ","End":"11:42.505","Text":"and also the magnetic field is alternating with time."},{"Start":"11:42.505 ","End":"11:45.300","Text":"We get double the frequency."},{"Start":"11:45.300 ","End":"11:47.870","Text":"Of course, to calculate the current,"},{"Start":"11:47.870 ","End":"11:51.680","Text":"if we were asked, we would use the exact same equation."},{"Start":"11:51.680 ","End":"11:58.035","Text":"Again, the current would have also this double frequency."},{"Start":"11:58.035 ","End":"12:00.770","Text":"This is the answer to Question number 3,"},{"Start":"12:00.770 ","End":"12:03.930","Text":"and that is the end of this lesson."}],"ID":21445},{"Watched":false,"Name":"EMF in Conducting Rod","Duration":"14m 37s","ChapterTopicVideoID":21366,"CourseChapterTopicPlaylistID":99480,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.545","Text":"Hello. In this lesson,"},{"Start":"00:01.545 ","End":"00:04.920","Text":"we\u0027re going to be speaking about the EMF of a rod."},{"Start":"00:04.920 ","End":"00:08.580","Text":"Here we have a rod of length L,"},{"Start":"00:08.580 ","End":"00:15.300","Text":"which is traveling through a magnetic field with a velocity of v_0."},{"Start":"00:15.300 ","End":"00:19.380","Text":"Now, the rod here has been drawn with a bit of a width,"},{"Start":"00:19.380 ","End":"00:22.080","Text":"and in questions, we\u0027re not going to focus on the width,"},{"Start":"00:22.080 ","End":"00:24.075","Text":"but it\u0027s just so that here,"},{"Start":"00:24.075 ","End":"00:26.010","Text":"now during the explanation,"},{"Start":"00:26.010 ","End":"00:28.800","Text":"we can draw it and explain it properly."},{"Start":"00:28.800 ","End":"00:36.580","Text":"What\u0027s important to remember in this rod is that the rod is a conductor."},{"Start":"00:36.680 ","End":"00:42.410","Text":"Inside the rod, we have lots of different charges."},{"Start":"00:42.410 ","End":"00:44.659","Text":"We have positive charges,"},{"Start":"00:44.659 ","End":"00:51.675","Text":"negative charges, distributed throughout like so."},{"Start":"00:51.675 ","End":"00:55.100","Text":"As the rod moves,"},{"Start":"00:55.100 ","End":"00:59.110","Text":"the charges are also moving, of course."},{"Start":"00:59.110 ","End":"01:02.965","Text":"These charges can move around because the rod is a conductor."},{"Start":"01:02.965 ","End":"01:05.645","Text":"Seeing as these charges are moving,"},{"Start":"01:05.645 ","End":"01:08.570","Text":"that means that we have Lorentz\u0027s force."},{"Start":"01:08.570 ","End":"01:13.350","Text":"A force is being applied to them and this force is called Lorentz\u0027s force."},{"Start":"01:13.730 ","End":"01:22.370","Text":"Lorentz\u0027s force, so we have F=q multiplied by v,"},{"Start":"01:22.370 ","End":"01:26.480","Text":"the velocity, cross product with the magnetic field."},{"Start":"01:26.480 ","End":"01:30.140","Text":"If the velocity and the magnetic field are constant,"},{"Start":"01:30.140 ","End":"01:33.335","Text":"which is, let\u0027s say the situation that we have over here,"},{"Start":"01:33.335 ","End":"01:40.670","Text":"then we can say that the magnitude of the force is equal to q multiplied by the velocity,"},{"Start":"01:40.670 ","End":"01:45.110","Text":"which here is v_0, multiplied by the magnetic field."},{"Start":"01:45.110 ","End":"01:50.390","Text":"Here, we multiply it by sine of the angle between the two,"},{"Start":"01:50.390 ","End":"01:56.390","Text":"but because the velocity is perpendicular to the magnetic field,"},{"Start":"01:56.390 ","End":"02:00.740","Text":"sine of 90 is just equal to 1,"},{"Start":"02:00.740 ","End":"02:03.900","Text":"so it\u0027s just left like so."},{"Start":"02:04.220 ","End":"02:08.825","Text":"This would give us the magnitude and if we want to know the direction,"},{"Start":"02:08.825 ","End":"02:12.140","Text":"we just have to use the right-hand rule."},{"Start":"02:12.140 ","End":"02:20.764","Text":"The positive charges according to the right-hand rule will be pointing in this direction."},{"Start":"02:20.764 ","End":"02:23.180","Text":"I\u0027ll do F_B plus."},{"Start":"02:23.180 ","End":"02:25.070","Text":"According to the right-hand rule,"},{"Start":"02:25.070 ","End":"02:29.502","Text":"the negative charges will be pointing in the opposite direction,"},{"Start":"02:29.502 ","End":"02:32.270","Text":"so F_B minus, and this is because"},{"Start":"02:32.270 ","End":"02:35.345","Text":"we\u0027ll get the exact same directions with positive charges."},{"Start":"02:35.345 ","End":"02:41.910","Text":"But then over here, we\u0027ll have a minus sign symbolizing the negative charge."},{"Start":"02:42.440 ","End":"02:46.050","Text":"What happens to these forces?"},{"Start":"02:46.050 ","End":"02:53.615","Text":"What we\u0027re going to get is an accumulation of charges on either side of the rod."},{"Start":"02:53.615 ","End":"02:56.150","Text":"Over here, because the force on"},{"Start":"02:56.150 ","End":"03:00.375","Text":"the positive charges is pushing the positive charges up here,"},{"Start":"03:00.375 ","End":"03:03.635","Text":"so here we\u0027ll get an accumulation of positive charges"},{"Start":"03:03.635 ","End":"03:08.225","Text":"and the force on the negative charges is pushing them down here."},{"Start":"03:08.225 ","End":"03:12.440","Text":"At the bottom, we\u0027ll have this accumulation of negative charges."},{"Start":"03:12.440 ","End":"03:18.235","Text":"Then what we can see is that we have this polarization."},{"Start":"03:18.235 ","End":"03:23.015","Text":"Here we have one edge of the rod which is positive,"},{"Start":"03:23.015 ","End":"03:26.329","Text":"and here we have the opposite side which is negative."},{"Start":"03:26.329 ","End":"03:31.355","Text":"What we can see is that this is very similar to a battery where we have"},{"Start":"03:31.355 ","End":"03:38.220","Text":"the positive exit of the battery and we have the negative side."},{"Start":"03:38.530 ","End":"03:47.645","Text":"Then this polarization is going to form a magnetic field inside the rod,"},{"Start":"03:47.645 ","End":"03:49.715","Text":"which is going to go as we know,"},{"Start":"03:49.715 ","End":"03:52.205","Text":"from positive to negative."},{"Start":"03:52.205 ","End":"03:58.209","Text":"We have this electric field E going from positive to negative."},{"Start":"03:58.209 ","End":"04:02.150","Text":"Then of course, if we have an electric field,"},{"Start":"04:02.150 ","End":"04:07.700","Text":"then that means that we\u0027re going to have an electric force,"},{"Start":"04:07.700 ","End":"04:10.759","Text":"which is given by q,"},{"Start":"04:10.759 ","End":"04:16.110","Text":"the charge, multiplied by E, the electric field."},{"Start":"04:16.400 ","End":"04:24.370","Text":"What we get also is this electric force over here."},{"Start":"04:24.370 ","End":"04:30.755","Text":"At the beginning, we just started with a magnetic field and saw the magnetic force,"},{"Start":"04:30.755 ","End":"04:32.270","Text":"which we already saw,"},{"Start":"04:32.270 ","End":"04:34.100","Text":"but then over time,"},{"Start":"04:34.100 ","End":"04:36.245","Text":"due to the magnetic force,"},{"Start":"04:36.245 ","End":"04:38.720","Text":"we get this polarization."},{"Start":"04:38.720 ","End":"04:46.755","Text":"This polarization causes an electric field which in turn gives us an electric force."},{"Start":"04:46.755 ","End":"04:52.980","Text":"What we want to do is we want to see what happens as time goes by."},{"Start":"04:53.060 ","End":"04:57.785","Text":"Over time, as this polarization increases,"},{"Start":"04:57.785 ","End":"05:02.460","Text":"we get that the electric force"},{"Start":"05:06.310 ","End":"05:10.220","Text":"that we calculated over here,"},{"Start":"05:10.220 ","End":"05:12.140","Text":"actually I\u0027ll do it without the vector,"},{"Start":"05:12.140 ","End":"05:18.275","Text":"is going to be equal to the magnetic force that we calculated over here."},{"Start":"05:18.275 ","End":"05:20.840","Text":"When these forces are equal,"},{"Start":"05:20.840 ","End":"05:29.940","Text":"then the sum of the forces on all of these charges over here is going to be equal to 0."},{"Start":"05:30.650 ","End":"05:37.555","Text":"Let\u0027s write that down. The sum of all of the forces is going to be equal to 0."},{"Start":"05:37.555 ","End":"05:40.205","Text":"If the sum of all of the forces is equal to 0,"},{"Start":"05:40.205 ","End":"05:44.345","Text":"that means that our charges aren\u0027t going to be moving."},{"Start":"05:44.345 ","End":"05:46.100","Text":"What does that mean?"},{"Start":"05:46.100 ","End":"05:50.010","Text":"That means that we\u0027ve reached steady state."},{"Start":"05:50.690 ","End":"05:55.505","Text":"When the electric force is equal to the magnetic force,"},{"Start":"05:55.505 ","End":"05:59.515","Text":"we say that we\u0027ve reached the steady state."},{"Start":"05:59.515 ","End":"06:03.215","Text":"We can assume that this happens relatively quickly,"},{"Start":"06:03.215 ","End":"06:07.830","Text":"probably within the first few seconds."},{"Start":"06:08.870 ","End":"06:11.435","Text":"Using this steady state,"},{"Start":"06:11.435 ","End":"06:14.705","Text":"we can calculate this electric field."},{"Start":"06:14.705 ","End":"06:17.335","Text":"Our electric field,"},{"Start":"06:17.335 ","End":"06:21.545","Text":"as we saw or the equation for the electric force,"},{"Start":"06:21.545 ","End":"06:26.345","Text":"is q multiplied by E. This is of course,"},{"Start":"06:26.345 ","End":"06:28.520","Text":"equal to our magnetic force,"},{"Start":"06:28.520 ","End":"06:31.705","Text":"which is equal to qv,"},{"Start":"06:31.705 ","End":"06:35.640","Text":"or let\u0027s just leave it general qvB."},{"Start":"06:35.640 ","End":"06:40.205","Text":"This is of course just if our v and our B are perpendicular."},{"Start":"06:40.205 ","End":"06:45.425","Text":"If not, then we multiply this side by sine of the angle between the two."},{"Start":"06:45.425 ","End":"06:48.110","Text":"The only reason we\u0027re not multiplying here by"},{"Start":"06:48.110 ","End":"06:52.190","Text":"anything is because if they\u0027re perpendicular,"},{"Start":"06:52.190 ","End":"06:58.200","Text":"then the angle between the two is 90 and sine of 90 is 1."},{"Start":"06:58.760 ","End":"07:03.125","Text":"At this stage, I can divide both sides by q."},{"Start":"07:03.125 ","End":"07:06.380","Text":"What I get is that my electric field is equal to"},{"Start":"07:06.380 ","End":"07:12.027","Text":"the velocity multiplied by the magnetic field."},{"Start":"07:12.027 ","End":"07:15.075","Text":"Now if I have my electric field,"},{"Start":"07:15.075 ","End":"07:18.855","Text":"that means that I can calculate the voltage."},{"Start":"07:18.855 ","End":"07:24.390","Text":"As we know, the voltage with a V is equal to"},{"Start":"07:24.390 ","End":"07:30.370","Text":"the negative integral of E.dl or dr,"},{"Start":"07:30.370 ","End":"07:35.875","Text":"it doesn\u0027t really matter, so just the length of this route or this trajectory."},{"Start":"07:35.875 ","End":"07:38.185","Text":"The minus, we don\u0027t really have to"},{"Start":"07:38.185 ","End":"07:41.170","Text":"look at it right now because it just gives us the direction."},{"Start":"07:41.170 ","End":"07:44.410","Text":"All I want to do really is do the integral."},{"Start":"07:44.410 ","End":"07:50.935","Text":"Now, we already saw that we\u0027re in steady state in order to get to this point."},{"Start":"07:50.935 ","End":"07:52.225","Text":"If we\u0027re in steady state,"},{"Start":"07:52.225 ","End":"07:58.300","Text":"that means the sum of all of the forces on all of the charges is equal to 0,"},{"Start":"07:58.300 ","End":"08:01.270","Text":"which means that at every point,"},{"Start":"08:01.270 ","End":"08:07.640","Text":"our electric force is equal to our magnetic force."},{"Start":"08:08.070 ","End":"08:11.695","Text":"As we\u0027ve already seen, or we know,"},{"Start":"08:11.695 ","End":"08:17.965","Text":"the magnetic force on every single point on this rod is constant, it\u0027s the same."},{"Start":"08:17.965 ","End":"08:19.570","Text":"If you do v cross B,"},{"Start":"08:19.570 ","End":"08:25.090","Text":"you\u0027ll see that we have this constant magnetic force on the rod, in which case,"},{"Start":"08:25.090 ","End":"08:28.600","Text":"according to this equation and being in steady state,"},{"Start":"08:28.600 ","End":"08:34.690","Text":"that means that our electric force also has to be constant on every point on the rod."},{"Start":"08:34.690 ","End":"08:37.510","Text":"If our electric force is constant,"},{"Start":"08:37.510 ","End":"08:45.200","Text":"that means that our electric field is also constant on every point on the rod."},{"Start":"08:45.690 ","End":"08:50.200","Text":"In that case, once I\u0027ve seen that my electric field is constant,"},{"Start":"08:50.200 ","End":"08:52.015","Text":"I don\u0027t have to do this integral."},{"Start":"08:52.015 ","End":"08:53.980","Text":"I can just say that this is equal to"},{"Start":"08:53.980 ","End":"08:58.660","Text":"the electric force multiplied by the length of the rod,"},{"Start":"08:58.660 ","End":"09:01.180","Text":"which over here is of length"},{"Start":"09:01.180 ","End":"09:08.680","Text":"L. The minus just symbolizes if we\u0027re integrating from bottom to top or top to bottom."},{"Start":"09:08.680 ","End":"09:12.310","Text":"Let\u0027s just put this all in absolute value"},{"Start":"09:12.310 ","End":"09:16.699","Text":"so that we don\u0027t have to focus too much on the minus."},{"Start":"09:18.240 ","End":"09:22.270","Text":"That means that now I have the voltage,"},{"Start":"09:22.270 ","End":"09:23.815","Text":"which is E.L,"},{"Start":"09:23.815 ","End":"09:27.175","Text":"and then if I substitute L what E is equal to,"},{"Start":"09:27.175 ","End":"09:31.510","Text":"so we get vBL,"},{"Start":"09:31.510 ","End":"09:34.720","Text":"or was generally written in the equation sheets,"},{"Start":"09:34.720 ","End":"09:35.770","Text":"but it doesn\u0027t really matter."},{"Start":"09:35.770 ","End":"09:39.025","Text":"I\u0027m just rearranging the order of the letters."},{"Start":"09:39.025 ","End":"09:41.660","Text":"It\u0027s usually written as BLv,"},{"Start":"09:41.660 ","End":"09:46.250","Text":"where v over here is of course the velocity."},{"Start":"09:47.340 ","End":"09:53.320","Text":"We see that we have this voltage flowing through the rod and if I were to"},{"Start":"09:53.320 ","End":"10:01.405","Text":"connect the rod to some resistor of resistance R,"},{"Start":"10:01.405 ","End":"10:06.700","Text":"so it\u0027s going to act like a battery or a voltage source."},{"Start":"10:06.700 ","End":"10:10.930","Text":"What we\u0027re going to get is current flowing"},{"Start":"10:10.930 ","End":"10:16.645","Text":"through the rod and through these conducting wires."},{"Start":"10:16.645 ","End":"10:21.940","Text":"What we\u0027ll get is that the current is equal to what we\u0027ve seen before,"},{"Start":"10:21.940 ","End":"10:27.320","Text":"the voltage divided by the resistance of the resistor."},{"Start":"10:27.390 ","End":"10:30.790","Text":"Basically, what we can see is that if we have"},{"Start":"10:30.790 ","End":"10:36.100","Text":"a conducting rod which moves throughout a magnetic field,"},{"Start":"10:36.100 ","End":"10:40.675","Text":"what we\u0027re going to get is this electric field,"},{"Start":"10:40.675 ","End":"10:43.719","Text":"which will in turn give us a voltage."},{"Start":"10:43.719 ","End":"10:48.175","Text":"We\u0027ll have a voltage in the rods."},{"Start":"10:48.175 ","End":"10:54.829","Text":"Therefore, we\u0027re also going to get a current if we attach it to a resistor."},{"Start":"10:55.200 ","End":"11:01.599","Text":"This voltage that is in this rod is usually not referred to as voltage,"},{"Start":"11:01.599 ","End":"11:04.720","Text":"but it\u0027s rather referred to as EMF,"},{"Start":"11:04.720 ","End":"11:08.260","Text":"which is a term that we may have spoken about earlier."},{"Start":"11:08.260 ","End":"11:09.969","Text":"EMF is just voltage,"},{"Start":"11:09.969 ","End":"11:12.084","Text":"but it has a slightly different name,"},{"Start":"11:12.084 ","End":"11:18.955","Text":"and the EMF in a conducting rod in a magnetic field is equal to what we saw over here,"},{"Start":"11:18.955 ","End":"11:25.610","Text":"BLv, where of course this v is the velocity."},{"Start":"11:26.820 ","End":"11:31.490","Text":"This is an equation to write in your equation sheet."},{"Start":"11:32.640 ","End":"11:36.400","Text":"A conducting rod moving through a magnetic field,"},{"Start":"11:36.400 ","End":"11:39.760","Text":"the EMF or the voltage is equal to BLv,"},{"Start":"11:39.760 ","End":"11:42.535","Text":"where this v is the velocity,"},{"Start":"11:42.535 ","End":"11:43.750","Text":"L is the length of the rod,"},{"Start":"11:43.750 ","End":"11:45.625","Text":"and B is the magnetic field."},{"Start":"11:45.625 ","End":"11:48.850","Text":"This is, of course, given that the magnetic field and"},{"Start":"11:48.850 ","End":"11:52.030","Text":"the velocity are perpendicular to one another,"},{"Start":"11:52.030 ","End":"11:55.210","Text":"if they are not, you just take this equation and you multiply"},{"Start":"11:55.210 ","End":"11:59.450","Text":"it by sine of the angle between the two."},{"Start":"11:59.610 ","End":"12:01.810","Text":"Just a little side note,"},{"Start":"12:01.810 ","End":"12:06.565","Text":"we\u0027ve seen another question where we had some tracks."},{"Start":"12:06.565 ","End":"12:11.095","Text":"Where here we had a resistor and we had"},{"Start":"12:11.095 ","End":"12:16.315","Text":"a rod placed on the tracks which was moving with some velocity like so,"},{"Start":"12:16.315 ","End":"12:21.485","Text":"and of course, there\u0027s a magnetic field throughout."},{"Start":"12:21.485 ","End":"12:25.740","Text":"We answer this question in order to find the current flowing through,"},{"Start":"12:25.740 ","End":"12:28.125","Text":"we use the idea of magnetic flux."},{"Start":"12:28.125 ","End":"12:31.110","Text":"However, we could have just also used"},{"Start":"12:31.110 ","End":"12:35.944","Text":"this equation straight away instead of calculating the magnetic flux."},{"Start":"12:35.944 ","End":"12:38.725","Text":"What we get is that the EMF,"},{"Start":"12:38.725 ","End":"12:40.360","Text":"which is what we got over here,"},{"Start":"12:40.360 ","End":"12:48.310","Text":"is equal to the negative time derivative of the magnetic flux."},{"Start":"12:48.310 ","End":"12:52.435","Text":"If we found the magnetic flux and took its time derivative,"},{"Start":"12:52.435 ","End":"12:55.150","Text":"we would come up with this exact equation"},{"Start":"12:55.150 ","End":"12:58.880","Text":"that we got over here for the EMF, which is BLv."},{"Start":"13:00.000 ","End":"13:03.820","Text":"Of course again, we\u0027re just taking the absolute value"},{"Start":"13:03.820 ","End":"13:06.670","Text":"because we don\u0027t want to deal with the minus,"},{"Start":"13:06.670 ","End":"13:09.500","Text":"which is just giving us the direction."},{"Start":"13:09.870 ","End":"13:12.640","Text":"Either way would have given us the answer,"},{"Start":"13:12.640 ","End":"13:14.500","Text":"so you could use either way."},{"Start":"13:14.500 ","End":"13:16.780","Text":"Sometimes it\u0027s easier using"},{"Start":"13:16.780 ","End":"13:21.625","Text":"just this equation over here than working out the magnetic flux."},{"Start":"13:21.625 ","End":"13:24.280","Text":"Sometimes it\u0027s easier to work that out."},{"Start":"13:24.280 ","End":"13:29.260","Text":"Just remember that a conducting rod moving throughout"},{"Start":"13:29.260 ","End":"13:34.998","Text":"a magnetic field acts like a battery and it has this voltage BLv,"},{"Start":"13:34.998 ","End":"13:40.600","Text":"and then you can use that if you want to find the current or you"},{"Start":"13:40.600 ","End":"13:46.750","Text":"can just also use the magnetic flux."},{"Start":"13:46.750 ","End":"13:50.065","Text":"Sorry, I put a dot here by accident."},{"Start":"13:50.065 ","End":"14:00.445","Text":"The time derivative of the magnetic flux is the same as just writing magnetic flux dot."},{"Start":"14:00.445 ","End":"14:02.845","Text":"This dot represents the time derivative,"},{"Start":"14:02.845 ","End":"14:05.245","Text":"sub x and root it out over here."},{"Start":"14:05.245 ","End":"14:08.030","Text":"Just ignore that dot."},{"Start":"14:08.820 ","End":"14:14.335","Text":"Of course, if you have a question where you don\u0027t have this closed circuit,"},{"Start":"14:14.335 ","End":"14:17.845","Text":"so let\u0027s say you had something like so,"},{"Start":"14:17.845 ","End":"14:21.280","Text":"just a rod moving and we don\u0027t have a closed circuit,"},{"Start":"14:21.280 ","End":"14:23.375","Text":"then of course, if there\u0027s no closed circuit,"},{"Start":"14:23.375 ","End":"14:26.810","Text":"you can\u0027t calculate the magnetic flux"},{"Start":"14:26.810 ","End":"14:29.600","Text":"because you need a close circuit in order to do that."},{"Start":"14:29.600 ","End":"14:31.010","Text":"In that case,"},{"Start":"14:31.010 ","End":"14:35.025","Text":"you have to use this equation, of course."},{"Start":"14:35.025 ","End":"14:38.300","Text":"That\u0027s the end of this lesson."}],"ID":21446},{"Watched":false,"Name":"Exercise 3","Duration":"12m 17s","ChapterTopicVideoID":21367,"CourseChapterTopicPlaylistID":99480,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.685","Text":"Hello, in this lesson we\u0027re going to be answering the following question"},{"Start":"00:04.685 ","End":"00:11.310","Text":"2 conducting tracks are placed at an angle of 2 Theta to 1 another."},{"Start":"00:11.310 ","End":"00:14.220","Text":"A conducting rod is placed on top of them,"},{"Start":"00:14.220 ","End":"00:17.980","Text":"creating an equilateral triangle."},{"Start":"00:18.020 ","End":"00:23.940","Text":"At t is equal to 0 the rod is located at the vertex."},{"Start":"00:23.940 ","End":"00:29.700","Text":"The rod then moves across the tracks at a velocity of v. There is"},{"Start":"00:29.700 ","End":"00:37.120","Text":"a constant magnetic field coming out of the page that is equal to b."},{"Start":"00:37.120 ","End":"00:41.210","Text":"Question number 1 is to calculate the EMF."},{"Start":"00:41.210 ","End":"00:49.679","Text":"As we know, the EMF is equal to the negative time derivative of the magnetic flux."},{"Start":"00:49.990 ","End":"00:57.860","Text":"But another equation for a rod moving in a constant field is given as so."},{"Start":"00:57.860 ","End":"01:01.385","Text":"The EMF is equal to B,"},{"Start":"01:01.385 ","End":"01:03.875","Text":"the magnetic field L,"},{"Start":"01:03.875 ","End":"01:05.270","Text":"the length of the rods,"},{"Start":"01:05.270 ","End":"01:07.520","Text":"and multiplied by V,"},{"Start":"01:07.520 ","End":"01:10.175","Text":"the velocity at which the rod is moving."},{"Start":"01:10.175 ","End":"01:15.635","Text":"This is only if the rod is located in a constant magnetic field,"},{"Start":"01:15.635 ","End":"01:21.840","Text":"which it is, and is moving at a constant velocity V, which it is."},{"Start":"01:21.950 ","End":"01:25.520","Text":"Let\u0027s calculate the EMF both ways,"},{"Start":"01:25.520 ","End":"01:27.350","Text":"so fast, let\u0027s use this way."},{"Start":"01:27.350 ","End":"01:29.870","Text":"We already know B and V,"},{"Start":"01:29.870 ","End":"01:31.415","Text":"we are given those in the question,"},{"Start":"01:31.415 ","End":"01:33.875","Text":"but we don\u0027t know what L is."},{"Start":"01:33.875 ","End":"01:40.100","Text":"Let\u0027s calculate L. What we can do is we can"},{"Start":"01:40.100 ","End":"01:47.550","Text":"say that the distance between the vertex to the rod is a distance x."},{"Start":"01:47.630 ","End":"01:53.940","Text":"This distance over here is x."},{"Start":"01:53.940 ","End":"01:57.845","Text":"Then of course, this angle over here is Theta."},{"Start":"01:57.845 ","End":"02:02.695","Text":"Of course this will cut the rod at the mid section,"},{"Start":"02:02.695 ","End":"02:06.800","Text":"in its middle and then this distance over here,"},{"Start":"02:06.800 ","End":"02:08.705","Text":"let\u0027s call it Y."},{"Start":"02:08.705 ","End":"02:15.555","Text":"As we can see, the total length of the rod at any given point is going to be equal to 2y."},{"Start":"02:15.555 ","End":"02:19.640","Text":"L is equal to 2y."},{"Start":"02:19.640 ","End":"02:23.445","Text":"Now what we want to know is what is y?"},{"Start":"02:23.445 ","End":"02:29.060","Text":"We can see, because what we have here is a right angle triangle."},{"Start":"02:29.060 ","End":"02:34.640","Text":"We can see that y is simply"},{"Start":"02:34.640 ","End":"02:40.115","Text":"equal to x distance is over here,"},{"Start":"02:40.115 ","End":"02:44.225","Text":"multiplied by tan of Theta,"},{"Start":"02:44.225 ","End":"02:47.380","Text":"which of course we\u0027re also given in the question."},{"Start":"02:47.380 ","End":"02:49.385","Text":"Then as we know,"},{"Start":"02:49.385 ","End":"02:56.825","Text":"x is this distance traveled and we\u0027re told the velocity or we\u0027re given the velocity."},{"Start":"02:56.825 ","End":"02:59.900","Text":"The velocity is distance over time."},{"Start":"02:59.900 ","End":"03:07.445","Text":"Therefore x is equal to the velocity multiplied by the time."},{"Start":"03:07.445 ","End":"03:10.265","Text":"Then we can say that therefore,"},{"Start":"03:10.265 ","End":"03:12.950","Text":"L is equal to twice y,"},{"Start":"03:12.950 ","End":"03:15.530","Text":"where y is equal to x,"},{"Start":"03:15.530 ","End":"03:24.585","Text":"where x is equal to vt and then over here again, tan of Theta."},{"Start":"03:24.585 ","End":"03:27.065","Text":"Then we can plug this in,"},{"Start":"03:27.065 ","End":"03:30.500","Text":"therefore to our equation for EMF."},{"Start":"03:30.500 ","End":"03:33.845","Text":"This is equal to B multiplied by L,"},{"Start":"03:33.845 ","End":"03:38.990","Text":"so that\u0027s 2vt multiplied by"},{"Start":"03:38.990 ","End":"03:47.370","Text":"tan of Theta multiplied by v. We can add a v here or we can just square this v over here."},{"Start":"03:47.810 ","End":"03:50.955","Text":"Now let\u0027s calculate this way."},{"Start":"03:50.955 ","End":"03:54.950","Text":"Let\u0027s first calculate the magnetic flux."},{"Start":"03:54.950 ","End":"04:00.330","Text":"As we know, it\u0027s the integral of B dot ds"},{"Start":"04:00.330 ","End":"04:06.090","Text":"and seeing as our magnetic field is constant,"},{"Start":"04:06.090 ","End":"04:11.600","Text":"so we can just write this as B.S,"},{"Start":"04:11.600 ","End":"04:14.940","Text":"where S is, of course, the surface area."},{"Start":"04:15.220 ","End":"04:19.265","Text":"Here we\u0027re dealing with these surface area of"},{"Start":"04:19.265 ","End":"04:23.540","Text":"an equilateral triangle or other triangle where as we know,"},{"Start":"04:23.540 ","End":"04:30.350","Text":"the surface area is equal to 1/2 base times height."},{"Start":"04:30.350 ","End":"04:34.880","Text":"We have 1/2 multiplied by the base,"},{"Start":"04:34.880 ","End":"04:39.815","Text":"so that\u0027s 2y multiplied by the height,"},{"Start":"04:39.815 ","End":"04:43.030","Text":"which is x,"},{"Start":"04:43.100 ","End":"04:46.290","Text":"and then multiplied by B."},{"Start":"04:46.290 ","End":"04:48.170","Text":"The 1/2 and the 2 cancel out."},{"Start":"04:48.170 ","End":"04:57.680","Text":"Y is equal to x tan of Theta and then we have x and other x,"},{"Start":"04:57.680 ","End":"05:02.390","Text":"so x^2 multiplied by b and then we know what x is."},{"Start":"05:02.390 ","End":"05:05.600","Text":"X is equal to vt. We have x^2,"},{"Start":"05:05.600 ","End":"05:11.330","Text":"so we have v^2 t^2 multiplied by b,"},{"Start":"05:11.330 ","End":"05:16.140","Text":"multiplied by tan of Theta."},{"Start":"05:16.670 ","End":"05:24.680","Text":"This is our magnetic flux and then our EMF is equal to the negative time derivative."},{"Start":"05:24.680 ","End":"05:28.370","Text":"But let\u0027s just work out the absolute value."},{"Start":"05:28.370 ","End":"05:30.545","Text":"We take the time derivatives."},{"Start":"05:30.545 ","End":"05:40.005","Text":"We have 2v^2 t multiplied by B tan of Theta."},{"Start":"05:40.005 ","End":"05:46.340","Text":"We have 2Bv^2 T tan of Theta."},{"Start":"05:46.340 ","End":"05:50.340","Text":"We got the same answer using both methods."},{"Start":"05:50.360 ","End":"05:54.275","Text":"We show 2 ways to answer a Question number 1."},{"Start":"05:54.275 ","End":"05:57.095","Text":"Now let\u0027s answer Question number 2."},{"Start":"05:57.095 ","End":"06:03.290","Text":"The resistance of the rod per unit length is i1 and the tracks have no resistance."},{"Start":"06:03.290 ","End":"06:06.295","Text":"Calculate the current."},{"Start":"06:06.295 ","End":"06:09.530","Text":"If the tracks have no resistance,"},{"Start":"06:09.530 ","End":"06:16.280","Text":"that means that the current can only flow along the rod itself. Let\u0027s begin."},{"Start":"06:16.280 ","End":"06:23.990","Text":"We know that the current is equal to the voltage divided by the resistance"},{"Start":"06:23.990 ","End":"06:27.530","Text":"where we\u0027re trying to find the total resistance of the rods because we\u0027re"},{"Start":"06:27.530 ","End":"06:32.195","Text":"given the resistance per unit length."},{"Start":"06:32.195 ","End":"06:34.950","Text":"What is the resistance?"},{"Start":"06:34.950 ","End":"06:43.585","Text":"It\u0027s equal to the resistance per unit length multiplied by the total length, which is 2y."},{"Start":"06:43.585 ","End":"06:46.320","Text":"We\u0027ve already said what y is."},{"Start":"06:46.320 ","End":"06:49.020","Text":"We have R_1 multiplied by 2."},{"Start":"06:49.020 ","End":"06:51.495","Text":"Then y is x tan of Theta,"},{"Start":"06:51.495 ","End":"06:59.595","Text":"where x is vt. We have vt multiplied by tan of Theta."},{"Start":"06:59.595 ","End":"07:02.355","Text":"This is the total resistance."},{"Start":"07:02.355 ","End":"07:09.710","Text":"In that case, we can say that our current is equal to the voltage, which is the EMF."},{"Start":"07:09.710 ","End":"07:19.460","Text":"2v^2 tB multiplied by tan of Theta divided by the total resistance,"},{"Start":"07:19.460 ","End":"07:30.020","Text":"which is 2vt multiplied by r1 multiplied by tan of Theta."},{"Start":"07:30.020 ","End":"07:32.855","Text":"Then the tan of Theta cancels out."},{"Start":"07:32.855 ","End":"07:34.685","Text":"The t cancels out."},{"Start":"07:34.685 ","End":"07:37.385","Text":"1 of the v\u0027s from the numerator cancels out,"},{"Start":"07:37.385 ","End":"07:39.920","Text":"and the 2s cancel out."},{"Start":"07:39.920 ","End":"07:46.430","Text":"Therefore, what we\u0027re left with is that the current is equal"},{"Start":"07:46.430 ","End":"07:53.140","Text":"to vB divided by R_1."},{"Start":"07:53.510 ","End":"07:56.510","Text":"This is the magnitude of the current."},{"Start":"07:56.510 ","End":"07:59.780","Text":"As we know, current has both size and direction."},{"Start":"07:59.780 ","End":"08:03.365","Text":"We want to calculate the direction."},{"Start":"08:03.365 ","End":"08:06.650","Text":"In order to do that, we\u0027re going to use Lorentz\u0027s law."},{"Start":"08:06.650 ","End":"08:13.730","Text":"The first thing we want to see is if the magnetic flux is increasing or decreasing."},{"Start":"08:13.730 ","End":"08:18.920","Text":"As we can see, the rod is moving away like so."},{"Start":"08:18.920 ","End":"08:23.855","Text":"Meaning that the surface area is increasing."},{"Start":"08:23.855 ","End":"08:26.525","Text":"If the surface area is increasing,"},{"Start":"08:26.525 ","End":"08:30.655","Text":"then the magnetic flux is also increasing."},{"Start":"08:30.655 ","End":"08:35.285","Text":"We have an increasing magnetic flux and it causing to Lorentz\u0027s law,"},{"Start":"08:35.285 ","End":"08:38.550","Text":"we want the current to oppose this change."},{"Start":"08:38.550 ","End":"08:43.910","Text":"We want to find the current that will decrease this magnetic flux."},{"Start":"08:43.910 ","End":"08:47.060","Text":"What we do is we form a magnetic field in"},{"Start":"08:47.060 ","End":"08:50.960","Text":"the opposite direction to the external magnetic field."},{"Start":"08:50.960 ","End":"08:53.930","Text":"Here the magnetic field is coming out of the page."},{"Start":"08:53.930 ","End":"08:57.005","Text":"Now we\u0027ll have a magnetic field going"},{"Start":"08:57.005 ","End":"09:00.900","Text":"into the page and then according to the right-hand rule,"},{"Start":"09:00.900 ","End":"09:03.890","Text":"our thumb points in the direction of the magnetic field,"},{"Start":"09:03.890 ","End":"09:08.720","Text":"and our fingers curl in the direction of the current."},{"Start":"09:08.720 ","End":"09:13.735","Text":"The current is going to be flowing in the clockwise direction."},{"Start":"09:13.735 ","End":"09:18.755","Text":"As we just calculated the foresaw,"},{"Start":"09:18.755 ","End":"09:25.280","Text":"the current is only flowing along the rod because only the rod has resistance."},{"Start":"09:25.280 ","End":"09:28.970","Text":"So the current here is going in this downwards direction."},{"Start":"09:28.970 ","End":"09:34.175","Text":"But we could also say that this is the clockwise direction in this question."},{"Start":"09:34.175 ","End":"09:36.600","Text":"It\u0027s the same thing."},{"Start":"09:38.000 ","End":"09:42.790","Text":"Now let\u0027s answer Question number 3,"},{"Start":"09:42.790 ","End":"09:48.815","Text":"which is to calculate the power transferred to the system to produce current."},{"Start":"09:48.815 ","End":"09:51.965","Text":"The power comes from this movements."},{"Start":"09:51.965 ","End":"09:56.030","Text":"We\u0027re moving, which means that there\u0027s going to be some power."},{"Start":"09:56.030 ","End":"10:05.660","Text":"Power is defined as the rate of work. Dw by dt."},{"Start":"10:05.770 ","End":"10:15.860","Text":"Here, the power is lost and enters at the exact same rate because the power is lost to"},{"Start":"10:15.860 ","End":"10:20.210","Text":"pulling the rod or the power is given by pulling"},{"Start":"10:20.210 ","End":"10:25.820","Text":"the rod and the power is lost because this rod is also a resistor."},{"Start":"10:25.820 ","End":"10:30.635","Text":"We\u0027ve also seen this in the previous questions that we\u0027ve solved."},{"Start":"10:30.635 ","End":"10:38.145","Text":"Power is put in the form of pulling the rod and is lost by the resistor in the rod."},{"Start":"10:38.145 ","End":"10:42.230","Text":"In that case, we can say that power is also equal"},{"Start":"10:42.230 ","End":"10:46.505","Text":"to the power given to the rod in order to push it,"},{"Start":"10:46.505 ","End":"10:52.615","Text":"which is given by the force dot product with the velocity."},{"Start":"10:52.615 ","End":"10:57.244","Text":"It\u0027s also equal to the power lost by the resistor,"},{"Start":"10:57.244 ","End":"11:00.740","Text":"which we have a lot of different equations for."},{"Start":"11:00.740 ","End":"11:07.090","Text":"It could be either the EMF or the voltage multiplied by the current."},{"Start":"11:07.090 ","End":"11:13.310","Text":"It can also be equal to the current squared multiplied by the resistance,"},{"Start":"11:13.310 ","End":"11:14.750","Text":"the total resistance,"},{"Start":"11:14.750 ","End":"11:21.075","Text":"and it could also be the EMF or the voltage divided by the total resistance."},{"Start":"11:21.075 ","End":"11:23.580","Text":"All of these equations are the same."},{"Start":"11:23.580 ","End":"11:27.645","Text":"We can choose whichever 1 we want. Let\u0027s use this."},{"Start":"11:27.645 ","End":"11:33.630","Text":"We have that the power is equal to Epsilon I."},{"Start":"11:34.430 ","End":"11:38.850","Text":"In that case, Epsilon or EMF is equal to"},{"Start":"11:38.850 ","End":"11:46.335","Text":"2v^2 tB multiplied by tan of Theta,"},{"Start":"11:46.335 ","End":"11:55.185","Text":"and our current is equal to vB divided by R_1."},{"Start":"11:55.185 ","End":"12:03.380","Text":"Then we can see that the final answer rearranged is like so,"},{"Start":"12:03.380 ","End":"12:11.440","Text":"2 times v^3 b^2 t tan of Theta divided by R_1,"},{"Start":"12:11.440 ","End":"12:14.585","Text":"which is the resistance per unit length."},{"Start":"12:14.585 ","End":"12:18.150","Text":"That is the end of this lesson."}],"ID":21447},{"Watched":false,"Name":"Exercise 4","Duration":"10m 43s","ChapterTopicVideoID":21368,"CourseChapterTopicPlaylistID":99480,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.055","Text":"Hello. In this lesson,"},{"Start":"00:02.055 ","End":"00:05.099","Text":"we\u0027re going to be answering the following question."},{"Start":"00:05.099 ","End":"00:10.935","Text":"We have here 3 resistors and each of them has the following resistance."},{"Start":"00:10.935 ","End":"00:14.715","Text":"Resistor R_1 has a resistance of 1 ohm,"},{"Start":"00:14.715 ","End":"00:17.625","Text":"R_2 has a resistance of 2 ohms,"},{"Start":"00:17.625 ","End":"00:20.760","Text":"and R_3 has a resistance of 3 ohms."},{"Start":"00:20.760 ","End":"00:28.545","Text":"There is a magnetic field of B coming into the page and it is throughout."},{"Start":"00:28.545 ","End":"00:30.750","Text":"What is our magnetic field?"},{"Start":"00:30.750 ","End":"00:36.330","Text":"It is 2 Teslas per second multiplied by t,"},{"Start":"00:36.330 ","End":"00:37.755","Text":"where t is time."},{"Start":"00:37.755 ","End":"00:44.875","Text":"The units of the magnetic field are going to be in Tesla."},{"Start":"00:44.875 ","End":"00:48.890","Text":"The height of the circuit is 15 centimeters and"},{"Start":"00:48.890 ","End":"00:54.140","Text":"the width of each sub-circuit is 20 centimeters."},{"Start":"00:54.140 ","End":"01:00.865","Text":"What they want us to do is to calculate the current through each resistor."},{"Start":"01:00.865 ","End":"01:05.375","Text":"First of all, we can see that we have a magnetic field which is"},{"Start":"01:05.375 ","End":"01:09.545","Text":"changing with respect to time or it\u0027s just changing,"},{"Start":"01:09.545 ","End":"01:14.030","Text":"which means that we\u0027re going to have magnetic flux,"},{"Start":"01:14.030 ","End":"01:18.715","Text":"which is important because if we have magnetic flux then we have EMF or voltage,"},{"Start":"01:18.715 ","End":"01:22.055","Text":"and if we have that then we can calculate the current."},{"Start":"01:22.055 ","End":"01:25.230","Text":"As we can see we have 2 circuits."},{"Start":"01:25.230 ","End":"01:29.365","Text":"Let\u0027s call this circuit number 1 and this circuit number 2."},{"Start":"01:29.365 ","End":"01:34.480","Text":"But each circuit has the exact same surface area."},{"Start":"01:34.480 ","End":"01:37.385","Text":"We can say that S,"},{"Start":"01:37.385 ","End":"01:40.865","Text":"the surface area is equal to"},{"Start":"01:40.865 ","End":"01:49.600","Text":"0.15 multiplied by 0.2."},{"Start":"01:50.330 ","End":"01:59.340","Text":"This is going to give us 0.03 meters squared."},{"Start":"01:59.340 ","End":"02:03.180","Text":"This is the surface area of each one of these."},{"Start":"02:04.310 ","End":"02:11.820","Text":"Now let\u0027s work out the magnetic flux for circuit number 1."},{"Start":"02:11.960 ","End":"02:19.595","Text":"Our magnetic flux is equal to the magnetic field."},{"Start":"02:19.595 ","End":"02:21.650","Text":"Or rather let\u0027s explain it like this."},{"Start":"02:21.650 ","End":"02:25.985","Text":"We have an integral of B.ds."},{"Start":"02:25.985 ","End":"02:27.650","Text":"This is the magnetic flux."},{"Start":"02:27.650 ","End":"02:32.450","Text":"We\u0027re integrating along ds and as we can see, ds is constant."},{"Start":"02:32.450 ","End":"02:41.290","Text":"We can just write that as B multiplied by S. B is equal to 2t."},{"Start":"02:42.170 ","End":"02:45.315","Text":"I\u0027m going to leave out the units."},{"Start":"02:45.315 ","End":"02:48.395","Text":"This is multiplied by the surface area,"},{"Start":"02:48.395 ","End":"02:51.620","Text":"which is 0.03 meters squared."},{"Start":"02:51.620 ","End":"02:54.235","Text":"Again, I\u0027ll leave out the units."},{"Start":"02:54.235 ","End":"03:00.280","Text":"What we have is that this is equal to 0.06t."},{"Start":"03:00.280 ","End":"03:02.832","Text":"This is the magnetic flux,"},{"Start":"03:02.832 ","End":"03:05.010","Text":"in that case, the EMF."},{"Start":"03:05.010 ","End":"03:08.630","Text":"We work out the magnitude of the EMF,"},{"Start":"03:08.630 ","End":"03:14.285","Text":"but the EMF is equal to the negative time derivative of the magnetic flux,"},{"Start":"03:14.285 ","End":"03:17.120","Text":"but we\u0027re working out the magnitude."},{"Start":"03:17.120 ","End":"03:24.980","Text":"The time derivative of 0.06t is just going to leave us with 0.06."},{"Start":"03:24.980 ","End":"03:28.960","Text":"Of course, the units are volts."},{"Start":"03:30.890 ","End":"03:35.030","Text":"Now, what we want to do is we want to find first of"},{"Start":"03:35.030 ","End":"03:40.070","Text":"all the direction of the current that will be flowing through this circuit,"},{"Start":"03:40.070 ","End":"03:42.905","Text":"so just the direction,"},{"Start":"03:42.905 ","End":"03:45.515","Text":"not the magnitude just yet."},{"Start":"03:45.515 ","End":"03:47.225","Text":"We\u0027re going to use Lenz\u0027s law,"},{"Start":"03:47.225 ","End":"03:52.040","Text":"which says that the direction of the current is"},{"Start":"03:52.040 ","End":"03:59.160","Text":"going to be in a way as to oppose the change in magnetic flux."},{"Start":"03:59.300 ","End":"04:03.410","Text":"The first thing we\u0027re looking at is whether the magnetic flux"},{"Start":"04:03.410 ","End":"04:07.040","Text":"over here is increasing or decreasing."},{"Start":"04:07.040 ","End":"04:10.700","Text":"You can see that the magnetic flux is as a function of time."},{"Start":"04:10.700 ","End":"04:14.120","Text":"As time increases the flux increases."},{"Start":"04:14.120 ","End":"04:16.655","Text":"We want to oppose this change."},{"Start":"04:16.655 ","End":"04:18.965","Text":"We want to decrease the flux."},{"Start":"04:18.965 ","End":"04:22.310","Text":"That means that we\u0027re going to create or"},{"Start":"04:22.310 ","End":"04:26.390","Text":"induce a magnetic field that is in the opposite direction,"},{"Start":"04:26.390 ","End":"04:29.020","Text":"so it\u0027s coming out of the page."},{"Start":"04:29.020 ","End":"04:33.796","Text":"Then according to the right-hand rule,"},{"Start":"04:33.796 ","End":"04:38.180","Text":"our thumb points in the direction of the magnetic field,"},{"Start":"04:38.180 ","End":"04:41.585","Text":"and our fingers curl in the direction of the current."},{"Start":"04:41.585 ","End":"04:48.220","Text":"Over here we can see that the current is going to be traveling in this direction,"},{"Start":"04:48.220 ","End":"04:50.545","Text":"in the anticlockwise direction."},{"Start":"04:50.545 ","End":"04:56.170","Text":"It\u0027s as if we have a battery over here like so with a voltage,"},{"Start":"04:56.170 ","End":"05:00.170","Text":"let\u0027s call this EMF_1,"},{"Start":"05:00.170 ","End":"05:03.670","Text":"so with a voltage of Epsilon_1 or EMF_1,"},{"Start":"05:03.670 ","End":"05:07.610","Text":"and the current is flowing like so."},{"Start":"05:08.170 ","End":"05:13.750","Text":"Now, let\u0027s work out the EMF for circuit number 2."},{"Start":"05:13.750 ","End":"05:18.325","Text":"We see that we\u0027ll get the exact same result."},{"Start":"05:18.325 ","End":"05:21.010","Text":"This is equal to Epsilon_1,"},{"Start":"05:21.010 ","End":"05:24.010","Text":"which is also equal to Epsilon_2."},{"Start":"05:24.010 ","End":"05:26.365","Text":"Everything is in the same direction."},{"Start":"05:26.365 ","End":"05:28.099","Text":"Also over here,"},{"Start":"05:28.099 ","End":"05:31.340","Text":"the current will be traveling in"},{"Start":"05:31.340 ","End":"05:37.415","Text":"the same direction and the battery will also be like this with Epsilon_2."},{"Start":"05:37.415 ","End":"05:43.125","Text":"Then what we can see is that from this circuit the current is moving up on this branch,"},{"Start":"05:43.125 ","End":"05:46.430","Text":"whereas from this circuit the current is moving down in this branch."},{"Start":"05:46.430 ","End":"05:48.275","Text":"We\u0027re soon going to get to that."},{"Start":"05:48.275 ","End":"05:50.720","Text":"But as we can see, in general,"},{"Start":"05:50.720 ","End":"05:58.270","Text":"the current for both of these circuits are flowing in the anti-clockwise direction."},{"Start":"05:59.330 ","End":"06:03.200","Text":"The last stage of this question is to actually"},{"Start":"06:03.200 ","End":"06:08.030","Text":"calculate the current that is flowing through each resistor."},{"Start":"06:08.030 ","End":"06:10.370","Text":"We can use a few methods."},{"Start":"06:10.370 ","End":"06:16.414","Text":"We can use Kirchhoff\u0027s law or we can use the method that we\u0027ve seen before"},{"Start":"06:16.414 ","End":"06:19.970","Text":"where we say that over here the current is"},{"Start":"06:19.970 ","End":"06:24.560","Text":"flowing in the anti-clockwise direction and also over here."},{"Start":"06:24.560 ","End":"06:34.145","Text":"Then we just calculate for each circuit separately the currents through each resistor."},{"Start":"06:34.145 ","End":"06:36.620","Text":"Over here in this circuit,"},{"Start":"06:36.620 ","End":"06:41.660","Text":"we have 2 resistors going in"},{"Start":"06:41.660 ","End":"06:43.940","Text":"this positive direction or"},{"Start":"06:43.940 ","End":"06:48.935","Text":"the positive direction of the currents so going in the same direction as the current."},{"Start":"06:48.935 ","End":"06:52.505","Text":"We have the resistance of R_1 and R_2,"},{"Start":"06:52.505 ","End":"06:54.470","Text":"which is 1 and 2 ohms."},{"Start":"06:54.470 ","End":"06:59.785","Text":"That\u0027s 3 ohms multiplied by I_1,"},{"Start":"06:59.785 ","End":"07:03.350","Text":"the current flowing through circuit 1 minus"},{"Start":"07:03.350 ","End":"07:07.700","Text":"the current flowing through this resistor from this circuit."},{"Start":"07:07.700 ","End":"07:12.010","Text":"As we said, from circuit 1,"},{"Start":"07:12.010 ","End":"07:14.240","Text":"we have a current flowing in this direction,"},{"Start":"07:14.240 ","End":"07:18.485","Text":"but from circuit 2 we have a current flowing in this direction, the opposite."},{"Start":"07:18.485 ","End":"07:22.450","Text":"Minus R_1, which is 1 ohm,"},{"Start":"07:22.450 ","End":"07:29.435","Text":"multiplied by the current flowing through circuit 2 which is I_2."},{"Start":"07:29.435 ","End":"07:34.405","Text":"All of this is equal to the voltage source of circuit 1,"},{"Start":"07:34.405 ","End":"07:40.185","Text":"which is as we saw equal to 0.06 volts."},{"Start":"07:40.185 ","End":"07:42.480","Text":"Then for this second we do the same thing,"},{"Start":"07:42.480 ","End":"07:44.040","Text":"we have I_1 and I_3,"},{"Start":"07:44.040 ","End":"07:52.790","Text":"so 1 plus 3 so we have 4I_2 minus the resistance,"},{"Start":"07:52.790 ","End":"07:56.965","Text":"but the current flowing through from circuit number 1."},{"Start":"07:56.965 ","End":"08:04.230","Text":"That\u0027s minus 1 because that\u0027s the resistance of R_1 multiplied by I_1."},{"Start":"08:04.230 ","End":"08:07.475","Text":"This is also equal to the voltage source of the second,"},{"Start":"08:07.475 ","End":"08:12.390","Text":"which is also equal to 0.06 volts."},{"Start":"08:13.250 ","End":"08:17.535","Text":"Now we have 2 equations with 2 unknowns."},{"Start":"08:17.535 ","End":"08:20.315","Text":"All we have to do is isolate them out."},{"Start":"08:20.315 ","End":"08:23.675","Text":"I\u0027m not going to deal with the algebra now but basically,"},{"Start":"08:23.675 ","End":"08:27.290","Text":"all you have to do is let\u0027s say over here isolate out"},{"Start":"08:27.290 ","End":"08:33.260","Text":"your I_1 and then plug in that into this equation,"},{"Start":"08:33.260 ","End":"08:35.835","Text":"and then you\u0027ll get I_2."},{"Start":"08:35.835 ","End":"08:39.605","Text":"Then once you know I_2 you can plug it into one of these equations,"},{"Start":"08:39.605 ","End":"08:41.585","Text":"either the first one or the second one,"},{"Start":"08:41.585 ","End":"08:44.615","Text":"and then you\u0027ll get the answer to I_1."},{"Start":"08:44.615 ","End":"08:46.550","Text":"Feel free to do that."},{"Start":"08:46.550 ","End":"08:48.475","Text":"I\u0027ll just write out the answers."},{"Start":"08:48.475 ","End":"08:55.020","Text":"I_1, remember this is the current in circuit 1 not in resistor 1."},{"Start":"08:55.020 ","End":"09:01.365","Text":"The current in circuit 1 is equal to 3 divided by 110 amps."},{"Start":"09:01.365 ","End":"09:08.010","Text":"The current I_2 so the current in circuit 2 is equal"},{"Start":"09:08.010 ","End":"09:15.155","Text":"to 2.4 divided by 110 amps."},{"Start":"09:15.155 ","End":"09:18.800","Text":"Remember to work this out on your own and check the answer."},{"Start":"09:18.800 ","End":"09:23.840","Text":"Now what we want to do is we want to see the current through each resistor."},{"Start":"09:23.840 ","End":"09:25.925","Text":"The current at each resistor,"},{"Start":"09:25.925 ","End":"09:28.655","Text":"we can see the current in I_2."},{"Start":"09:28.655 ","End":"09:31.760","Text":"Let\u0027s write I_R_2,"},{"Start":"09:31.760 ","End":"09:37.875","Text":"the current in I_2 is simply going to be this I_1 over here."},{"Start":"09:37.875 ","End":"09:41.550","Text":"Then similarly, the current in"},{"Start":"09:41.550 ","End":"09:50.250","Text":"the R_3 resistor is going to be this current over here from circuit number 2."},{"Start":"09:50.250 ","End":"09:52.090","Text":"That\u0027s going to be I_2."},{"Start":"09:52.090 ","End":"09:56.310","Text":"But then what about the current in resistor R_1?"},{"Start":"09:56.420 ","End":"10:02.420","Text":"In resistor, R_1 remember that we have the current from circuit number 1 and we also"},{"Start":"10:02.420 ","End":"10:07.720","Text":"have the current from circuit number 2 and they\u0027re flowing in opposite directions."},{"Start":"10:07.720 ","End":"10:14.450","Text":"The total current flowing through resistor R_1 is going to be the total current."},{"Start":"10:14.450 ","End":"10:23.340","Text":"That means that we\u0027re going to have I_1 minus I_2."},{"Start":"10:23.340 ","End":"10:24.840","Text":"I_1 we have written here,"},{"Start":"10:24.840 ","End":"10:26.678","Text":"I_2 we have written here,"},{"Start":"10:26.678 ","End":"10:35.490","Text":"and I_1 minus I_2 is simply 0.6 divided by 110 amps."},{"Start":"10:35.490 ","End":"10:37.520","Text":"This is the answer to the question."},{"Start":"10:37.520 ","End":"10:40.520","Text":"These are the currents through each of these resistors."},{"Start":"10:40.520 ","End":"10:43.890","Text":"That is the end of this lesson."}],"ID":21448},{"Watched":false,"Name":"Exercise 5","Duration":"10m 3s","ChapterTopicVideoID":21369,"CourseChapterTopicPlaylistID":99480,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.800","Text":"Hello. In this lesson,"},{"Start":"00:01.800 ","End":"00:04.710","Text":"we\u0027re going to be answering the following question."},{"Start":"00:04.710 ","End":"00:09.090","Text":"A conducting rod of length L moves"},{"Start":"00:09.090 ","End":"00:14.400","Text":"along the sides of a circuit as so with this velocity v_naught,"},{"Start":"00:14.400 ","End":"00:17.040","Text":"which is also constant."},{"Start":"00:17.040 ","End":"00:24.705","Text":"Inside the circuit, we have a uniform magnetic field into the page."},{"Start":"00:24.705 ","End":"00:28.035","Text":"We are given the values for the magnetic field,"},{"Start":"00:28.035 ","End":"00:30.975","Text":"the resistances of each of the resistors,"},{"Start":"00:30.975 ","End":"00:35.100","Text":"this velocity, and the length of the rod."},{"Start":"00:35.100 ","End":"00:38.900","Text":"We\u0027re asked to calculate the current when the rod"},{"Start":"00:38.900 ","End":"00:42.680","Text":"is located between resistors R_1 and R_2."},{"Start":"00:42.680 ","End":"00:45.320","Text":"Over here, located in this block,"},{"Start":"00:45.320 ","End":"00:51.475","Text":"and the second time when the rod is located between the resistors I_2 and I_3."},{"Start":"00:51.475 ","End":"00:55.605","Text":"When the rod is in the second block, this block."},{"Start":"00:55.605 ","End":"01:01.610","Text":"Of course, what\u0027s important to remember is that there\u0027s 2 directions for the current."},{"Start":"01:01.610 ","End":"01:04.610","Text":"We have the current coming out over here,"},{"Start":"01:04.610 ","End":"01:06.274","Text":"traveling in this direction,"},{"Start":"01:06.274 ","End":"01:10.010","Text":"and the current coming out and traveling in this direction."},{"Start":"01:10.010 ","End":"01:12.415","Text":"We\u0027re going to call this current,"},{"Start":"01:12.415 ","End":"01:14.885","Text":"I left the current on the left side,"},{"Start":"01:14.885 ","End":"01:19.710","Text":"and this current on the right side, I_R."},{"Start":"01:21.140 ","End":"01:24.770","Text":"For the meantime, let\u0027s forget about I right,"},{"Start":"01:24.770 ","End":"01:27.755","Text":"and I left, we\u0027ll get back to that later."},{"Start":"01:27.755 ","End":"01:32.410","Text":"But in the meantime, we want to calculate the EMF, the voltage."},{"Start":"01:32.410 ","End":"01:39.265","Text":"As we already know, we can either take calculate the EMF first."},{"Start":"01:39.265 ","End":"01:45.360","Text":"We can either work out the magnetic flux in each case,"},{"Start":"01:45.360 ","End":"01:47.685","Text":"and then we know the EMF."},{"Start":"01:47.685 ","End":"01:50.990","Text":"Take the time derivative of the magnetic flux,"},{"Start":"01:50.990 ","End":"01:55.040","Text":"then we have the EMF, and then we can calculate the current from that."},{"Start":"01:55.040 ","End":"01:59.315","Text":"Or another way that we can use is by using"},{"Start":"01:59.315 ","End":"02:05.450","Text":"our equation for the EMF of a conducting rod moving throughout a magnetic field,"},{"Start":"02:05.450 ","End":"02:07.670","Text":"which as we saw in a previous lesson,"},{"Start":"02:07.670 ","End":"02:11.190","Text":"is equal to BLv."},{"Start":"02:12.320 ","End":"02:16.530","Text":"Over here, our B we\u0027re given its own magnetic field,"},{"Start":"02:16.530 ","End":"02:19.520","Text":"L we\u0027re also given it\u0027s the length of the rod and"},{"Start":"02:19.520 ","End":"02:23.225","Text":"the velocity we\u0027re also given it\u0027s v_naught."},{"Start":"02:23.225 ","End":"02:27.770","Text":"This way, we straight away have the EMF of the rods,"},{"Start":"02:27.770 ","End":"02:32.195","Text":"and we don\u0027t have to play around with the magnetic flux."},{"Start":"02:32.195 ","End":"02:36.320","Text":"As we saw, a conducting rod which is moving throughout"},{"Start":"02:36.320 ","End":"02:40.315","Text":"a magnetic field is going to have this EMF,"},{"Start":"02:40.315 ","End":"02:43.640","Text":"and it\u0027s going to act like a battery."},{"Start":"02:44.160 ","End":"02:47.470","Text":"Now we want to know which side of the rod is"},{"Start":"02:47.470 ","End":"02:52.225","Text":"the positive end of the battery and which is the negative end of the battery."},{"Start":"02:52.225 ","End":"02:55.990","Text":"What we\u0027re going to do is we\u0027re going to use the right-hand rule to"},{"Start":"02:55.990 ","End":"03:00.610","Text":"see in which direction the forces are being applied."},{"Start":"03:00.770 ","End":"03:04.000","Text":"According to the right-hand rule,"},{"Start":"03:04.000 ","End":"03:07.470","Text":"our v is in the right direction."},{"Start":"03:07.470 ","End":"03:10.239","Text":"Our fourth finger points in the right direction."},{"Start":"03:10.239 ","End":"03:15.040","Text":"Our middle finger points in the direction of the magnetic field,"},{"Start":"03:15.040 ","End":"03:22.690","Text":"and that means that the force is pointing in the upwards direction,"},{"Start":"03:22.690 ","End":"03:24.865","Text":"we can draw it like so,"},{"Start":"03:24.865 ","End":"03:27.770","Text":"this is our magnetic force."},{"Start":"03:27.770 ","End":"03:32.890","Text":"You could also do the right-hand rule with slightly different fingers."},{"Start":"03:32.890 ","End":"03:34.450","Text":"Again with your right hand,"},{"Start":"03:34.450 ","End":"03:38.830","Text":"and this time your thumb points in the direction of the velocity,"},{"Start":"03:38.830 ","End":"03:40.210","Text":"so it points rightwards."},{"Start":"03:40.210 ","End":"03:43.390","Text":"Your forefinger points in the direction of the magnetic field,"},{"Start":"03:43.390 ","End":"03:46.360","Text":"so it points into the page and then your middle finger,"},{"Start":"03:46.360 ","End":"03:49.615","Text":"which is actually how we\u0027ve been doing it up until now."},{"Start":"03:49.615 ","End":"03:53.695","Text":"Your middle finger is going into point in the direction of the force,"},{"Start":"03:53.695 ","End":"03:55.300","Text":"and it will point upwards."},{"Start":"03:55.300 ","End":"03:57.190","Text":"This is how we\u0027ve been doing up until now,"},{"Start":"03:57.190 ","End":"04:01.365","Text":"with your thumb representing the first elements."},{"Start":"04:01.365 ","End":"04:03.445","Text":"Thumb in the direction of velocity,"},{"Start":"04:03.445 ","End":"04:05.650","Text":"forefinger in the direction of magnetic field."},{"Start":"04:05.650 ","End":"04:09.730","Text":"Middle finger is in the direction of the answer."},{"Start":"04:09.730 ","End":"04:12.410","Text":"Just use that version."},{"Start":"04:12.950 ","End":"04:21.260","Text":"From that, we know that the positive charges going to be"},{"Start":"04:21.260 ","End":"04:31.270","Text":"accumulating up here and the negative charges will be accumulating down here."},{"Start":"04:32.090 ","End":"04:40.130","Text":"Now we can ignore the whole issue of this being a rod that is moving in a magnetic field,"},{"Start":"04:40.130 ","End":"04:46.565","Text":"as we saw in the lesson dealing with the EMF in a conducting rod."},{"Start":"04:46.565 ","End":"04:50.765","Text":"We can just look at this rod rarely as a battery."},{"Start":"04:50.765 ","End":"04:53.690","Text":"Let\u0027s redraw this."},{"Start":"04:53.690 ","End":"04:57.790","Text":"Here we have our resistor R_1."},{"Start":"04:57.790 ","End":"05:00.260","Text":"Then here we have"},{"Start":"05:00.260 ","End":"05:08.690","Text":"our resistor R_2 and here we have our resistor R_3."},{"Start":"05:08.690 ","End":"05:11.045","Text":"Of course, it\u0027s not drawn to scale."},{"Start":"05:11.045 ","End":"05:16.805","Text":"Then we can pretend like here as if here we have a battery."},{"Start":"05:16.805 ","End":"05:19.010","Text":"Well, this is the positive end."},{"Start":"05:19.010 ","End":"05:20.735","Text":"This is the negative end,"},{"Start":"05:20.735 ","End":"05:22.280","Text":"just like we had here."},{"Start":"05:22.280 ","End":"05:28.375","Text":"It has an EMF or voltage of this Epsilon over here BLv."},{"Start":"05:28.375 ","End":"05:32.959","Text":"What we can notice is that each of these resistors is"},{"Start":"05:32.959 ","End":"05:37.805","Text":"connected in parallel with our voltage source."},{"Start":"05:37.805 ","End":"05:45.665","Text":"We have the exact same voltage drop between these 2 points and the same over here,"},{"Start":"05:45.665 ","End":"05:47.975","Text":"and the same over here."},{"Start":"05:47.975 ","End":"05:51.005","Text":"They\u0027re all connected in parallel."},{"Start":"05:51.005 ","End":"05:55.640","Text":"In that case we know that the current I_1,"},{"Start":"05:55.640 ","End":"05:58.025","Text":"the current over here,"},{"Start":"05:58.025 ","End":"06:00.170","Text":"let\u0027s draw this I_1,"},{"Start":"06:00.170 ","End":"06:04.715","Text":"is going to be equal to the voltage divided by its resistance,"},{"Start":"06:04.715 ","End":"06:11.745","Text":"I_1, which is simply equal to BLv_naught divided by R_1."},{"Start":"06:11.745 ","End":"06:15.525","Text":"Then also the current I_2."},{"Start":"06:15.525 ","End":"06:19.725","Text":"That is going to be like so."},{"Start":"06:19.725 ","End":"06:25.125","Text":"The current flowing through I_2 is going to be BLv_naught"},{"Start":"06:25.125 ","End":"06:32.440","Text":"divided by R_2 and the current flowing through I_3."},{"Start":"06:32.600 ","End":"06:43.390","Text":"We have I_3 is also equal to BLv_naught divided by R_3."},{"Start":"06:43.850 ","End":"06:48.740","Text":"That\u0027s great, and now what we want to do is what we were speaking about before,"},{"Start":"06:48.740 ","End":"06:51.515","Text":"where the current travels up here,"},{"Start":"06:51.515 ","End":"06:58.155","Text":"and then it splits into our I left over here,"},{"Start":"06:58.155 ","End":"07:01.425","Text":"and I right over here."},{"Start":"07:01.425 ","End":"07:08.830","Text":"It\u0027s easy to see that I left only goes through the first resistor, resistor R_1."},{"Start":"07:08.830 ","End":"07:12.560","Text":"I left is equal to the current flowing through this resistor,"},{"Start":"07:12.560 ","End":"07:15.010","Text":"which is just I_1."},{"Start":"07:15.010 ","End":"07:20.040","Text":"However, our I right is going over here,"},{"Start":"07:20.040 ","End":"07:26.750","Text":"and then when it reaches this vertex over here,"},{"Start":"07:26.750 ","End":"07:29.315","Text":"we can see that the current is going to split."},{"Start":"07:29.315 ","End":"07:32.570","Text":"We\u0027re going to have our I_2 going down here,"},{"Start":"07:32.570 ","End":"07:35.675","Text":"and I_3 going down here."},{"Start":"07:35.675 ","End":"07:40.805","Text":"We have current going through resistor I_2 and current going through resistor I_3."},{"Start":"07:40.805 ","End":"07:45.140","Text":"In that case, I right is going to be equal"},{"Start":"07:45.140 ","End":"07:51.825","Text":"to I_2 plus I_3."},{"Start":"07:51.825 ","End":"07:56.890","Text":"This comes from Kirchhoff\u0027s law when the current splits."},{"Start":"07:56.890 ","End":"08:00.740","Text":"This is the answer to Question number 1,"},{"Start":"08:00.740 ","End":"08:02.585","Text":"and now what we want to do,"},{"Start":"08:02.585 ","End":"08:03.830","Text":"Question number 2,"},{"Start":"08:03.830 ","End":"08:09.900","Text":"where the rod is located this time between resistors I_2 and I_3."},{"Start":"08:10.400 ","End":"08:13.230","Text":"This is what we\u0027re looking at."},{"Start":"08:13.230 ","End":"08:19.520","Text":"Now, our EMF is going to remain the same because it\u0027s dependent on BLv."},{"Start":"08:19.520 ","End":"08:23.150","Text":"The magnetic field, the length of the rod,"},{"Start":"08:23.150 ","End":"08:25.310","Text":"and the velocity, which hasn\u0027t changed."},{"Start":"08:25.310 ","End":"08:28.470","Text":"Our EMF is the same."},{"Start":"08:28.470 ","End":"08:32.330","Text":"Of course, because all our resistors are still"},{"Start":"08:32.330 ","End":"08:37.905","Text":"connected in parallel with respect to our voltage source."},{"Start":"08:37.905 ","End":"08:40.380","Text":"The current I_1, I_2,"},{"Start":"08:40.380 ","End":"08:43.230","Text":"and I_3 across the resistors R_1, R_2,"},{"Start":"08:43.230 ","End":"08:48.160","Text":"and R_3 are also going to stay the same."},{"Start":"08:48.320 ","End":"08:50.525","Text":"Everything is the same."},{"Start":"08:50.525 ","End":"08:57.140","Text":"The only difference is now we still have this I left, and I right."},{"Start":"08:57.140 ","End":"08:59.150","Text":"That\u0027s the only thing that\u0027s going to change,"},{"Start":"08:59.150 ","End":"09:03.000","Text":"so let\u0027s do I left first, like we did before."},{"Start":"09:03.000 ","End":"09:04.980","Text":"Our current goes up."},{"Start":"09:04.980 ","End":"09:06.990","Text":"We have I left, and again,"},{"Start":"09:06.990 ","End":"09:10.230","Text":"it reaches this vertex."},{"Start":"09:10.230 ","End":"09:12.975","Text":"The current again splits."},{"Start":"09:12.975 ","End":"09:15.885","Text":"We get the currents blessing into I_1,"},{"Start":"09:15.885 ","End":"09:17.970","Text":"which goes through resistor R_1,"},{"Start":"09:17.970 ","End":"09:21.360","Text":"and it splits into or so I_2,"},{"Start":"09:21.360 ","End":"09:25.695","Text":"which is the current going through resistor R_2."},{"Start":"09:25.695 ","End":"09:33.735","Text":"I left is simply equal to I_1 plus I_2."},{"Start":"09:33.735 ","End":"09:38.030","Text":"I right simply goes up and goes down here."},{"Start":"09:38.030 ","End":"09:43.190","Text":"The current I right only goes through a resistor I_3."},{"Start":"09:43.190 ","End":"09:48.690","Text":"That means that I write is equal to I_3."},{"Start":"09:49.760 ","End":"09:55.280","Text":"That is the only difference between the battery source or"},{"Start":"09:55.280 ","End":"10:00.830","Text":"the rod being located here in this side or here in this side."},{"Start":"10:00.830 ","End":"10:04.170","Text":"That is the end of this lesson."}],"ID":21449},{"Watched":false,"Name":"Exercise 6","Duration":"30m 2s","ChapterTopicVideoID":21370,"CourseChapterTopicPlaylistID":99480,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.965","Text":"Hello. In this lesson,"},{"Start":"00:01.965 ","End":"00:04.365","Text":"we\u0027re going to be answering the following question."},{"Start":"00:04.365 ","End":"00:08.070","Text":"Given is a rectangular frame of length,"},{"Start":"00:08.070 ","End":"00:09.615","Text":"d, and width,"},{"Start":"00:09.615 ","End":"00:12.945","Text":"L. It moves with a constant velocity of v_0"},{"Start":"00:12.945 ","End":"00:18.075","Text":"in the direction of a constant magnetic field, B."},{"Start":"00:18.075 ","End":"00:22.200","Text":"The length of the region is 1.5d,"},{"Start":"00:22.200 ","End":"00:24.615","Text":"and its width is infinite."},{"Start":"00:24.615 ","End":"00:30.210","Text":"The frame has a total resistance of R. At t is equal to 0,"},{"Start":"00:30.210 ","End":"00:35.780","Text":"the right side of the frame enters the region of the magnetic field."},{"Start":"00:35.780 ","End":"00:38.885","Text":"Question Number 1 is to calculate the emf,"},{"Start":"00:38.885 ","End":"00:42.635","Text":"or the electromotive force, of the frame."},{"Start":"00:42.635 ","End":"00:44.855","Text":"As we saw in the previous lesson,"},{"Start":"00:44.855 ","End":"00:47.195","Text":"in order to calculate the emf,"},{"Start":"00:47.195 ","End":"00:49.630","Text":"which is denoted by Epsilon,"},{"Start":"00:49.630 ","End":"00:51.800","Text":"we have to calculate"},{"Start":"00:51.800 ","End":"01:00.375","Text":"the negative change in magnetic flux with respect to time."},{"Start":"01:00.375 ","End":"01:02.870","Text":"The first thing that we\u0027re going to do is we\u0027re going to"},{"Start":"01:02.870 ","End":"01:06.055","Text":"calculate the magnetic flux which,"},{"Start":"01:06.055 ","End":"01:08.015","Text":"as we saw also in the previous lesson,"},{"Start":"01:08.015 ","End":"01:12.935","Text":"is equal to the integral of B. ds,"},{"Start":"01:12.935 ","End":"01:18.650","Text":"where we said that this is just going to be equal to b multiplied by"},{"Start":"01:18.650 ","End":"01:26.248","Text":"the area if both the magnetic field and the region are constant, which it is."},{"Start":"01:26.248 ","End":"01:27.920","Text":"In this question,"},{"Start":"01:27.920 ","End":"01:32.600","Text":"we\u0027re told we have a magnetic field which is constant."},{"Start":"01:32.600 ","End":"01:37.879","Text":"So if our frame enters the region,"},{"Start":"01:37.879 ","End":"01:42.915","Text":"then it will look something like so,"},{"Start":"01:42.915 ","End":"01:46.500","Text":"where here we have our resistor."},{"Start":"01:46.500 ","End":"01:54.450","Text":"Here we have some area which is inside the magnetic region,"},{"Start":"01:54.450 ","End":"01:59.220","Text":"and we can call this length over here, x."},{"Start":"01:59.220 ","End":"02:04.505","Text":"In that case, we have B multiplied by this area,"},{"Start":"02:04.505 ","End":"02:09.795","Text":"which would be L, multiplied by x."},{"Start":"02:09.795 ","End":"02:11.970","Text":"What is x?"},{"Start":"02:11.970 ","End":"02:14.547","Text":"We\u0027re told that this is moving."},{"Start":"02:14.547 ","End":"02:20.160","Text":"The frame is moving in with a velocity of v_0, which is constant."},{"Start":"02:20.160 ","End":"02:24.815","Text":"We know that velocity is equal to distance over time,"},{"Start":"02:24.815 ","End":"02:27.275","Text":"in which case distance, x,"},{"Start":"02:27.275 ","End":"02:30.990","Text":"is equal to velocity here, v_0,"},{"Start":"02:30.990 ","End":"02:33.465","Text":"multiplied by time,"},{"Start":"02:33.465 ","End":"02:35.600","Text":"so that we can substitute in."},{"Start":"02:35.600 ","End":"02:38.320","Text":"Now, we just want to calculate the emf,"},{"Start":"02:38.320 ","End":"02:42.500","Text":"so let\u0027s just calculate the absolute value of it."},{"Start":"02:42.500 ","End":"02:45.365","Text":"So it\u0027s just going to be simply equal to"},{"Start":"02:45.365 ","End":"02:49.340","Text":"the change in magnetic flux with respect to time,"},{"Start":"02:49.340 ","End":"02:52.700","Text":"or the time derivative of the magnetic flux."},{"Start":"02:52.700 ","End":"02:54.260","Text":"This will be equal to,"},{"Start":"02:54.260 ","End":"02:56.315","Text":"once we substitute in x,"},{"Start":"02:56.315 ","End":"03:00.480","Text":"it will be equal to BLv_0t,"},{"Start":"03:00.480 ","End":"03:03.455","Text":"and if we take the time derivative of this,"},{"Start":"03:03.455 ","End":"03:07.890","Text":"we\u0027re just left with BLv_0."},{"Start":"03:07.970 ","End":"03:10.505","Text":"This is, of course,"},{"Start":"03:10.505 ","End":"03:14.180","Text":"the emf in this region over here,"},{"Start":"03:14.180 ","End":"03:17.345","Text":"where we have part of the frame out."},{"Start":"03:17.345 ","End":"03:22.340","Text":"So we can say that this is the emf in the region where x is"},{"Start":"03:22.340 ","End":"03:27.935","Text":"between 0 and d. Once it\u0027s greater than d,"},{"Start":"03:27.935 ","End":"03:31.415","Text":"we\u0027ll have a slightly different emf,"},{"Start":"03:31.415 ","End":"03:34.325","Text":"and if we want to know this with respect to time,"},{"Start":"03:34.325 ","End":"03:39.410","Text":"this is between t is equal to 0 and t,"},{"Start":"03:39.410 ","End":"03:41.480","Text":"which is equal to this over here,"},{"Start":"03:41.480 ","End":"03:44.929","Text":"which is d, the distance,"},{"Start":"03:44.929 ","End":"03:50.185","Text":"the total distance over here, divided by v_0."},{"Start":"03:50.185 ","End":"03:53.865","Text":"So this is when the whole frame is inside."},{"Start":"03:53.865 ","End":"03:59.150","Text":"What we\u0027re going to want to do is we\u0027re going to calculate the emf for each part,"},{"Start":"03:59.150 ","End":"04:04.700","Text":"so as the frame is moving in to the region of constant magnetic field,"},{"Start":"04:04.700 ","End":"04:06.725","Text":"when the whole frame is inside,"},{"Start":"04:06.725 ","End":"04:09.909","Text":"and as the frame leaves,"},{"Start":"04:09.909 ","End":"04:12.830","Text":"and then also each of these other questions we\u0027re"},{"Start":"04:12.830 ","End":"04:16.550","Text":"also going to have to calculate for each region."},{"Start":"04:16.550 ","End":"04:21.470","Text":"So what we\u0027re going to do is we\u0027re going to just stick with this region right now,"},{"Start":"04:21.470 ","End":"04:26.060","Text":"answer the rest of these points, these questions,"},{"Start":"04:26.060 ","End":"04:28.520","Text":"and then we\u0027ll move on to the second region,"},{"Start":"04:28.520 ","End":"04:31.710","Text":"and we\u0027ll do the same, and so on."},{"Start":"04:32.000 ","End":"04:39.500","Text":"Question 2 for this section is to calculate the current in the frame."},{"Start":"04:39.500 ","End":"04:43.940","Text":"As we know, current has both size and direction."},{"Start":"04:43.940 ","End":"04:47.510","Text":"The size, or the magnitude of the current,"},{"Start":"04:47.510 ","End":"04:51.725","Text":"is equal to the absolute value of the emf,"},{"Start":"04:51.725 ","End":"04:55.535","Text":"of the electromotive force, divided by resistance."},{"Start":"04:55.535 ","End":"04:57.815","Text":"This just comes from V,"},{"Start":"04:57.815 ","End":"05:00.410","Text":"the voltage, is equal to IR."},{"Start":"05:00.410 ","End":"05:03.110","Text":"I\u0027m just rearranging this for I."},{"Start":"05:03.110 ","End":"05:08.000","Text":"So we have our absolute value for the voltage for the emf,"},{"Start":"05:08.000 ","End":"05:14.285","Text":"which is equal to BLv_0 and divided by R,"},{"Start":"05:14.285 ","End":"05:17.315","Text":"which we\u0027re also given in the question."},{"Start":"05:17.315 ","End":"05:21.900","Text":"Now we want the direction of the current."},{"Start":"05:22.120 ","End":"05:24.790","Text":"Just like in the previous lesson,"},{"Start":"05:24.790 ","End":"05:26.640","Text":"we have to use Lenz\u0027s Law."},{"Start":"05:26.640 ","End":"05:30.790","Text":"Lenz\u0027s Law states that the current is going to be in"},{"Start":"05:30.790 ","End":"05:35.770","Text":"the direction such as to oppose the change in flux."},{"Start":"05:35.770 ","End":"05:38.890","Text":"The first step that we need to think about is,"},{"Start":"05:38.890 ","End":"05:42.430","Text":"whether as this frame moves into this region,"},{"Start":"05:42.430 ","End":"05:47.195","Text":"whether the flux is increasing or decreasing."},{"Start":"05:47.195 ","End":"05:50.604","Text":"As the frame moves further into this region,"},{"Start":"05:50.604 ","End":"05:57.205","Text":"the area that the magnetic field passes through in the frame is increasing,"},{"Start":"05:57.205 ","End":"06:02.320","Text":"which means that the magnetic flux is also increasing."},{"Start":"06:02.320 ","End":"06:07.610","Text":"So what we want to do is we want to find the direction of"},{"Start":"06:07.610 ","End":"06:14.105","Text":"the current such that it will oppose this increase in magnetic flux."},{"Start":"06:14.105 ","End":"06:17.135","Text":"What we do, as we saw in the previous lesson,"},{"Start":"06:17.135 ","End":"06:23.856","Text":"we create a magnetic field which is in the opposite direction,"},{"Start":"06:23.856 ","End":"06:27.385","Text":"so in this case, the opposite direction is coming out of the page."},{"Start":"06:27.385 ","End":"06:30.230","Text":"Then using the right-hand rule,"},{"Start":"06:30.230 ","End":"06:33.950","Text":"we point our thumb in the direction of the magnetic field,"},{"Start":"06:33.950 ","End":"06:38.630","Text":"and our fingers curl in the direction of the current."},{"Start":"06:38.630 ","End":"06:39.770","Text":"In this case,"},{"Start":"06:39.770 ","End":"06:47.305","Text":"we can see that the current is going to be flowing in this direction anticlockwise."},{"Start":"06:47.305 ","End":"06:51.735","Text":"That\u0027s the answer to Question 2 for this region."},{"Start":"06:51.735 ","End":"06:55.530","Text":"Now, let\u0027s answer Question 3 for this region."},{"Start":"06:55.530 ","End":"07:01.505","Text":"Calculate the force required in order for the frame to move at a constant velocity."},{"Start":"07:01.505 ","End":"07:08.015","Text":"Because we have these current carrying tiles that are in a magnetic field,"},{"Start":"07:08.015 ","End":"07:10.370","Text":"we know that there\u0027s going to be a force,"},{"Start":"07:10.370 ","End":"07:12.785","Text":"and this comes from Lorentz\u0027s law."},{"Start":"07:12.785 ","End":"07:15.140","Text":"We know that according to Lorentz\u0027s law,"},{"Start":"07:15.140 ","End":"07:18.485","Text":"the force on this wire is going to be,"},{"Start":"07:18.485 ","End":"07:23.910","Text":"so dF is equal to IdL,"},{"Start":"07:23.910 ","End":"07:25.200","Text":"where I is the current,"},{"Start":"07:25.200 ","End":"07:29.915","Text":"and dL is the wire across with B,"},{"Start":"07:29.915 ","End":"07:32.045","Text":"where B is the magnetic field."},{"Start":"07:32.045 ","End":"07:38.030","Text":"We already saw that if the current and the magnetic field are constant,"},{"Start":"07:38.030 ","End":"07:46.655","Text":"then our force is given as BIL multiplied by sine of Alpha,"},{"Start":"07:46.655 ","End":"07:51.300","Text":"where Alpha is the angle between dL and B."},{"Start":"07:51.620 ","End":"07:56.630","Text":"So over here, the current is constant and so is the magnetic field."},{"Start":"07:56.630 ","End":"07:59.510","Text":"In that case, we can say that the force is equal"},{"Start":"07:59.510 ","End":"08:03.635","Text":"to the magnetic field multiplied by the current,"},{"Start":"08:03.635 ","End":"08:08.840","Text":"where the current is BLv_0 divided by"},{"Start":"08:08.840 ","End":"08:15.380","Text":"R multiplied by L and multiplied by sine of the angle between the 2,"},{"Start":"08:15.380 ","End":"08:19.325","Text":"and we can see that the angle between the 2 is 90 degrees."},{"Start":"08:19.325 ","End":"08:22.450","Text":"So we\u0027re looking at this side right now."},{"Start":"08:22.450 ","End":"08:26.285","Text":"This side, later, we\u0027ll deal with these 2 sides."},{"Start":"08:26.285 ","End":"08:29.720","Text":"Sine of 90 degrees is equal to 1,"},{"Start":"08:29.720 ","End":"08:33.305","Text":"so we get that the force is equal to"},{"Start":"08:33.305 ","End":"08:40.480","Text":"B^2L^2v_0 divided by R. Now,"},{"Start":"08:40.480 ","End":"08:43.070","Text":"if we want to know the direction,"},{"Start":"08:43.070 ","End":"08:47.330","Text":"all we have to do is use the right-hand rule with this equation over here."},{"Start":"08:47.330 ","End":"08:49.790","Text":"We\u0027re looking at this side right now,"},{"Start":"08:49.790 ","End":"08:51.575","Text":"So as we can see,"},{"Start":"08:51.575 ","End":"08:54.980","Text":"our current is in this direction."},{"Start":"08:54.980 ","End":"08:59.690","Text":"That means that our thumb represents this IdL."},{"Start":"08:59.690 ","End":"09:04.160","Text":"Our thumb is pointing like so,"},{"Start":"09:04.160 ","End":"09:07.665","Text":"and this represents the direction of the current dL."},{"Start":"09:07.665 ","End":"09:12.275","Text":"Then our forefinger represents B,"},{"Start":"09:12.275 ","End":"09:18.630","Text":"so our B is coming into the page."},{"Start":"09:18.630 ","End":"09:21.960","Text":"Let\u0027s say that this is our forefinger,"},{"Start":"09:21.960 ","End":"09:25.620","Text":"so it\u0027s going into the page."},{"Start":"09:25.620 ","End":"09:27.515","Text":"Then if we have that,"},{"Start":"09:27.515 ","End":"09:30.125","Text":"then what we get is that our middle finger,"},{"Start":"09:30.125 ","End":"09:32.525","Text":"which represents the direction of the force,"},{"Start":"09:32.525 ","End":"09:38.657","Text":"is pointing in the leftward direction."},{"Start":"09:38.657 ","End":"09:41.245","Text":"Feel free to give this a try."},{"Start":"09:41.245 ","End":"09:44.080","Text":"What we get is that our force due to"},{"Start":"09:44.080 ","End":"09:49.486","Text":"the magnetic field is pointing in this leftward direction,"},{"Start":"09:49.486 ","End":"09:51.715","Text":"so this is F_B."},{"Start":"09:51.715 ","End":"09:57.145","Text":"What we can do is we can say that this is the positive x direction."},{"Start":"09:57.145 ","End":"10:04.180","Text":"In that case, we can see that this is in the negative x direction,"},{"Start":"10:04.180 ","End":"10:06.025","Text":"and this is F_B."},{"Start":"10:06.025 ","End":"10:10.120","Text":"So what we want is we want our F_external,"},{"Start":"10:10.120 ","End":"10:13.165","Text":"which is equal and opposite,"},{"Start":"10:13.165 ","End":"10:19.490","Text":"and that\u0027s how we get this constant velocity because the forces are equal and opposite."},{"Start":"10:19.530 ","End":"10:30.190","Text":"In that case, so we know that F_B is going to be equal and opposite to F_external."},{"Start":"10:30.190 ","End":"10:37.030","Text":"In that case, we get that F_external is equal to"},{"Start":"10:37.030 ","End":"10:46.820","Text":"positive B^2L^2v_0 divided by R in the x direction."},{"Start":"10:48.030 ","End":"10:50.725","Text":"This is this F_external."},{"Start":"10:50.725 ","End":"10:53.965","Text":"Now what about these wires over here?"},{"Start":"10:53.965 ","End":"10:57.370","Text":"What we can see is that one is going to have"},{"Start":"10:57.370 ","End":"11:02.634","Text":"a magnetic field force going in this direction,"},{"Start":"11:02.634 ","End":"11:04.945","Text":"and the other is going to have"},{"Start":"11:04.945 ","End":"11:09.525","Text":"the magnetic fields force going into the equal and opposite direction."},{"Start":"11:09.525 ","End":"11:12.055","Text":"So due to the symmetry,"},{"Start":"11:12.055 ","End":"11:14.980","Text":"these forces will cancel out,"},{"Start":"11:14.980 ","End":"11:17.935","Text":"and so the external forces also cancel out."},{"Start":"11:17.935 ","End":"11:21.715","Text":"This is the only force that is being applied."},{"Start":"11:21.715 ","End":"11:25.060","Text":"That is the answer to Question Number 3."},{"Start":"11:25.060 ","End":"11:28.660","Text":"Let\u0027s go on to Question Number 4."},{"Start":"11:28.660 ","End":"11:31.780","Text":"Question Number 4 is asking us what is"},{"Start":"11:31.780 ","End":"11:40.300","Text":"the power of the force"},{"Start":"11:40.300 ","End":"11:42.781","Text":"and the power turned to heat in the resistor."},{"Start":"11:42.781 ","End":"11:45.640","Text":"So let\u0027s first find the power of the force,"},{"Start":"11:45.640 ","End":"11:49.000","Text":"so P_external, and as we know,"},{"Start":"11:49.000 ","End":"11:52.810","Text":"power is equal to dw by dt,"},{"Start":"11:52.810 ","End":"11:57.160","Text":"so the rate of change in the work done,"},{"Start":"11:57.160 ","End":"12:03.115","Text":"or it\u0027s also equal to F.v."},{"Start":"12:03.115 ","End":"12:05.500","Text":"So here we have our force,"},{"Start":"12:05.500 ","End":"12:09.100","Text":"so let\u0027s use this version of the equation."},{"Start":"12:09.100 ","End":"12:11.075","Text":"We have our force which is"},{"Start":"12:11.075 ","End":"12:21.149","Text":"B^2L^2v_0 divided by R in the x direction dot the velocity,"},{"Start":"12:21.149 ","End":"12:25.405","Text":"where we\u0027re told the velocity is v_0,"},{"Start":"12:25.405 ","End":"12:30.280","Text":"and it is also in the x direction."},{"Start":"12:30.280 ","End":"12:34.090","Text":"Then what we\u0027ll get is that this is simply equal to"},{"Start":"12:34.090 ","End":"12:42.760","Text":"B^2L^2v_0^2 divided by R."},{"Start":"12:42.760 ","End":"12:46.060","Text":"So this is the power of this external force,"},{"Start":"12:46.060 ","End":"12:51.269","Text":"and now let\u0027s see what the power is in the resistor, so PR."},{"Start":"12:51.269 ","End":"12:56.920","Text":"As we know, the power in a resistor is equal to either v_I,"},{"Start":"12:56.920 ","End":"12:59.935","Text":"so the voltage, or here the emf,"},{"Start":"12:59.935 ","End":"13:02.710","Text":"multiplied by the current,"},{"Start":"13:02.710 ","End":"13:09.745","Text":"or another version is I^2 multiplied by the resistance,"},{"Start":"13:09.745 ","End":"13:12.880","Text":"R. Here we can do I^2,"},{"Start":"13:12.880 ","End":"13:17.440","Text":"so we have B^2L^2v_0^2"},{"Start":"13:17.440 ","End":"13:25.495","Text":"divided by R^2 multiplied by the resistance,"},{"Start":"13:25.495 ","End":"13:28.000","Text":"R. This and this will cancel out,"},{"Start":"13:28.000 ","End":"13:34.480","Text":"and what we\u0027re left with is B^2L^2v_0^2 divided"},{"Start":"13:34.480 ","End":"13:37.330","Text":"by R. We can see that the power of"},{"Start":"13:37.330 ","End":"13:42.670","Text":"the external force is equal to the power of the resistor."},{"Start":"13:42.670 ","End":"13:45.070","Text":"Specifically, in this example,"},{"Start":"13:45.070 ","End":"13:46.465","Text":"they turned out equal,"},{"Start":"13:46.465 ","End":"13:49.480","Text":"but notice that they won\u0027t always be equal."},{"Start":"13:49.480 ","End":"13:51.055","Text":"Why are they equal?"},{"Start":"13:51.055 ","End":"13:54.580","Text":"Because the energy or the power lost"},{"Start":"13:54.580 ","End":"13:58.885","Text":"in the resistor is what goes into the external force."},{"Start":"13:58.885 ","End":"14:01.405","Text":"However, if we had, let\u0027s say,"},{"Start":"14:01.405 ","End":"14:05.545","Text":"a capacitor over here or any other electrical component,"},{"Start":"14:05.545 ","End":"14:07.524","Text":"but for instance, a capacitor,"},{"Start":"14:07.524 ","End":"14:11.770","Text":"then the power that we now find in the resistor would be"},{"Start":"14:11.770 ","End":"14:17.110","Text":"divided somehow between both the resistor and the capacitor,"},{"Start":"14:17.110 ","End":"14:24.445","Text":"and then the power in the resistor would be not equal to the power of the external force."},{"Start":"14:24.445 ","End":"14:26.110","Text":"But here, specifically,"},{"Start":"14:26.110 ","End":"14:32.090","Text":"because we only have 1 component and 1 external force, they are equal."},{"Start":"14:33.120 ","End":"14:39.160","Text":"Now we\u0027ve answered all of the questions for this region over here."},{"Start":"14:39.160 ","End":"14:44.185","Text":"Now what we\u0027re going to do is we\u0027re going to answer all of these questions again."},{"Start":"14:44.185 ","End":"14:52.760","Text":"But for the region when this entire frame is located inside this magnetic field."},{"Start":"14:53.060 ","End":"14:59.760","Text":"This region starts as soon as this side of the frame"},{"Start":"14:59.760 ","End":"15:07.675","Text":"is reaching this point over here in the entrance into this region of the magnetic field."},{"Start":"15:07.675 ","End":"15:15.535","Text":"As in from when x is greater or equal to d,"},{"Start":"15:15.535 ","End":"15:21.250","Text":"so when this d over here is at this point over here,"},{"Start":"15:21.250 ","End":"15:29.275","Text":"until this side of the frame over here hits this region over here,"},{"Start":"15:29.275 ","End":"15:32.710","Text":"symbolizing that the frame is now exiting from"},{"Start":"15:32.710 ","End":"15:36.820","Text":"the other side of the region with the magnetic field,"},{"Start":"15:36.820 ","End":"15:42.470","Text":"so until this 1.5d,"},{"Start":"15:45.270 ","End":"15:51.340","Text":"or in other words, we\u0027re speaking of this region where the entire frame is"},{"Start":"15:51.340 ","End":"15:54.820","Text":"completely enclosed within this region"},{"Start":"15:54.820 ","End":"15:58.240","Text":"of the magnetic field so that means that the earliest that,"},{"Start":"15:58.240 ","End":"16:03.715","Text":"that can happen is when this side is at this point,"},{"Start":"16:03.715 ","End":"16:10.060","Text":"and the latest time that this can happen is when this side, the rightmost side,"},{"Start":"16:10.060 ","End":"16:16.193","Text":"is over here on the border for the magnetic field,"},{"Start":"16:16.193 ","End":"16:21.144","Text":"so the whole time that all of the sides of the frame are in this region,"},{"Start":"16:21.144 ","End":"16:24.880","Text":"and it is denoted by this."},{"Start":"16:24.880 ","End":"16:31.540","Text":"Calculating all of these questions for this region is super easy."},{"Start":"16:31.540 ","End":"16:34.375","Text":"Number 1, calculate the emf."},{"Start":"16:34.375 ","End":"16:38.440","Text":"We know that the emf is the negative time derivative of"},{"Start":"16:38.440 ","End":"16:42.625","Text":"the magnetic flux so the magnetic flux over here,"},{"Start":"16:42.625 ","End":"16:45.384","Text":"because we know that the magnetic field is constant,"},{"Start":"16:45.384 ","End":"16:47.980","Text":"it\u0027s simply equal to the magnetic field"},{"Start":"16:47.980 ","End":"16:51.415","Text":"multiplied by the total surface area of the frame,"},{"Start":"16:51.415 ","End":"16:53.109","Text":"which is this length,"},{"Start":"16:53.109 ","End":"16:56.260","Text":"L, multiplied by the width,"},{"Start":"16:56.260 ","End":"17:02.050","Text":"which is d. Then the emf, as we said,"},{"Start":"17:02.050 ","End":"17:09.670","Text":"is equal to the negative time derivative of the magnetic flux."},{"Start":"17:09.670 ","End":"17:14.110","Text":"As we can see, the magnetic flux is independent of time over here,"},{"Start":"17:14.110 ","End":"17:19.630","Text":"which means that the time derivative of it is equal to 0 and then that case,"},{"Start":"17:19.630 ","End":"17:22.720","Text":"if its time derivative is equal to 0,"},{"Start":"17:22.720 ","End":"17:25.630","Text":"then the answer to Question Number 2 calculate the current in"},{"Start":"17:25.630 ","End":"17:30.040","Text":"the frame so that means that the current is also equal to 0."},{"Start":"17:30.040 ","End":"17:33.610","Text":"Question Number 3, calculate the force required in order for"},{"Start":"17:33.610 ","End":"17:37.600","Text":"the frame to move at a constant velocity so the force,"},{"Start":"17:37.600 ","End":"17:38.890","Text":"if there\u0027s no current,"},{"Start":"17:38.890 ","End":"17:40.370","Text":"will also be equal to 0."},{"Start":"17:40.370 ","End":"17:42.595","Text":"Question Number 4,"},{"Start":"17:42.595 ","End":"17:48.250","Text":"the power of the force so because the power in"},{"Start":"17:48.250 ","End":"17:57.220","Text":"the resistor is equal"},{"Start":"17:57.220 ","End":"18:00.415","Text":"to the power of the force in this example over here,"},{"Start":"18:00.415 ","End":"18:02.845","Text":"because we only have 1 electrical component,"},{"Start":"18:02.845 ","End":"18:05.185","Text":"and we see that the force is 0,"},{"Start":"18:05.185 ","End":"18:15.080","Text":"and that means that the power is 0 in both the resistor and in the external force."},{"Start":"18:17.220 ","End":"18:21.550","Text":"These are all of the answers for this region over here so"},{"Start":"18:21.550 ","End":"18:25.149","Text":"when the frame is totally inside"},{"Start":"18:25.149 ","End":"18:29.320","Text":"the magnetic field region and now we\u0027re going to look at the case when"},{"Start":"18:29.320 ","End":"18:35.150","Text":"the frame begins exiting the magnetic field region."},{"Start":"18:35.400 ","End":"18:41.350","Text":"Now we\u0027re looking at this situation over here when our frame is exiting,"},{"Start":"18:41.350 ","End":"18:43.570","Text":"and it starts from the region."},{"Start":"18:43.570 ","End":"18:45.850","Text":"We have x,"},{"Start":"18:45.850 ","End":"18:48.033","Text":"where x is between 1.5d,"},{"Start":"18:48.033 ","End":"18:54.204","Text":"so as in this side of the frame has exited."},{"Start":"18:54.204 ","End":"18:56.893","Text":"So between 1.5d,"},{"Start":"18:56.893 ","End":"18:59.800","Text":"this side has left the region,"},{"Start":"18:59.800 ","End":"19:03.640","Text":"and until the entire frame has left the region,"},{"Start":"19:03.640 ","End":"19:09.627","Text":"which is just going to be 1.5d plus this whole length until we get to this side,"},{"Start":"19:09.627 ","End":"19:15.270","Text":"so it\u0027s going to be 2.5d."},{"Start":"19:15.270 ","End":"19:22.313","Text":"From here, so this is 1.5d when it\u0027s at this position until this whole length,"},{"Start":"19:22.313 ","End":"19:25.330","Text":"d, so we\u0027ve got into this side has also exited."},{"Start":"19:25.330 ","End":"19:30.025","Text":"So this whole region is between 1.5 and 2.5d."},{"Start":"19:30.025 ","End":"19:32.770","Text":"So now let\u0027s answer the questions."},{"Start":"19:32.770 ","End":"19:36.505","Text":"Question 1 is to calculate the emf of the frame."},{"Start":"19:36.505 ","End":"19:38.710","Text":"Of course, as we\u0027ve seen,"},{"Start":"19:38.710 ","End":"19:40.790","Text":"to calculate the emf, first,"},{"Start":"19:40.790 ","End":"19:45.145","Text":"we have to calculate the magnetic flux."},{"Start":"19:45.145 ","End":"19:49.330","Text":"What we want to do is we want to see how much of"},{"Start":"19:49.330 ","End":"19:55.165","Text":"the frame or what area of the frame is inside this region."},{"Start":"19:55.165 ","End":"19:57.040","Text":"As we\u0027ve already seen,"},{"Start":"19:57.040 ","End":"20:03.100","Text":"this whole length is going to be x."},{"Start":"20:03.100 ","End":"20:05.920","Text":"So from here until here,"},{"Start":"20:05.920 ","End":"20:09.235","Text":"this region is x,"},{"Start":"20:09.235 ","End":"20:14.095","Text":"and we know that this is"},{"Start":"20:14.095 ","End":"20:20.635","Text":"d. So what we want to do is we want to find what space this is over here."},{"Start":"20:20.635 ","End":"20:29.720","Text":"So it\u0027s going to be easier for us to calculate this dimension over here."},{"Start":"20:29.970 ","End":"20:35.510","Text":"This is x minus d,"},{"Start":"20:35.510 ","End":"20:42.505","Text":"so we have x minus d. Then this over here,"},{"Start":"20:42.505 ","End":"20:45.085","Text":"we\u0027re going to call z,"},{"Start":"20:45.085 ","End":"20:52.910","Text":"and then we\u0027re going to use z in order to calculate this area."},{"Start":"20:53.190 ","End":"21:03.160","Text":"What we can see over here is that x minus d plus z is going to give us 1.5d."},{"Start":"21:03.160 ","End":"21:10.165","Text":"So we can say that 1.5d is equal to x minus d,"},{"Start":"21:10.165 ","End":"21:15.350","Text":"this length, plus this length, so plus z."},{"Start":"21:15.900 ","End":"21:20.799","Text":"In that case, we get, from isolation,"},{"Start":"21:20.799 ","End":"21:29.330","Text":"that z is equal to 1.5d plus d minus x."},{"Start":"21:30.900 ","End":"21:36.520","Text":"In that case, our magnetic flux is equal to B,"},{"Start":"21:36.520 ","End":"21:39.580","Text":"because this is a constant magnetic field,"},{"Start":"21:39.580 ","End":"21:42.620","Text":"multiplied by this area over here,"},{"Start":"21:42.620 ","End":"21:44.517","Text":"so that\u0027s going to be the width,"},{"Start":"21:44.517 ","End":"21:51.235","Text":"so we have z multiplied by this length over here which, as we know,"},{"Start":"21:51.235 ","End":"21:54.990","Text":"is L. Then we can substitute in our z,"},{"Start":"21:54.990 ","End":"21:57.465","Text":"so we have BL multiplied by z,"},{"Start":"21:57.465 ","End":"22:04.065","Text":"which we saw is equal to 2.5d minus x."},{"Start":"22:04.065 ","End":"22:07.485","Text":"Of course, x is dependent on time,"},{"Start":"22:07.485 ","End":"22:12.350","Text":"so we know that speed or velocity is distance divided by time,"},{"Start":"22:12.350 ","End":"22:15.100","Text":"and this is distance."},{"Start":"22:15.100 ","End":"22:22.480","Text":"Therefore, what we get is BL multiplied by 2.5d minus,"},{"Start":"22:22.480 ","End":"22:26.635","Text":"so distance is velocity times time,"},{"Start":"22:26.635 ","End":"22:30.250","Text":"so v_0 times time."},{"Start":"22:30.250 ","End":"22:33.370","Text":"In order to find the emf,"},{"Start":"22:33.370 ","End":"22:35.455","Text":"we\u0027ll just do the absolute value,"},{"Start":"22:35.455 ","End":"22:38.470","Text":"so not taking into account negative values."},{"Start":"22:38.470 ","End":"22:42.730","Text":"When we take the time derivative of this, 2.5d is constant,"},{"Start":"22:42.730 ","End":"22:44.635","Text":"so that will cancel out,"},{"Start":"22:44.635 ","End":"22:48.745","Text":"and then we have BLv_0t."},{"Start":"22:48.745 ","End":"22:50.710","Text":"So when we take the time derivative of that,"},{"Start":"22:50.710 ","End":"22:56.240","Text":"we\u0027re just left with BLv_0."},{"Start":"22:56.850 ","End":"22:59.950","Text":"That\u0027s the answer to Question Number 1."},{"Start":"22:59.950 ","End":"23:02.215","Text":"Now, let\u0027s answer Question Number 2."},{"Start":"23:02.215 ","End":"23:04.750","Text":"Calculate the current in the frame."},{"Start":"23:04.750 ","End":"23:07.390","Text":"The current, as we know,"},{"Start":"23:07.390 ","End":"23:12.070","Text":"is simply equal to V,"},{"Start":"23:12.070 ","End":"23:14.530","Text":"the voltage, divided by R,"},{"Start":"23:14.530 ","End":"23:17.490","Text":"the resistance, where the voltage is, of course, emf."},{"Start":"23:17.490 ","End":"23:23.800","Text":"Emf is voltage, so we get BLv_0 divided by R,"},{"Start":"23:23.800 ","End":"23:25.765","Text":"voltage divided by resistance."},{"Start":"23:25.765 ","End":"23:29.620","Text":"Of course, we know that the current has both size,"},{"Start":"23:29.620 ","End":"23:31.315","Text":"which is this, and direction."},{"Start":"23:31.315 ","End":"23:35.770","Text":"So now, what we want to do is we want to calculate the direction."},{"Start":"23:35.770 ","End":"23:38.980","Text":"The first thing we ask ourselves is whether"},{"Start":"23:38.980 ","End":"23:43.600","Text":"the magnetic flux is increasing or decreasing."},{"Start":"23:43.600 ","End":"23:47.545","Text":"As our frame carries on moving in this rightwards direction,"},{"Start":"23:47.545 ","End":"23:50.995","Text":"we can see that z is decreasing,"},{"Start":"23:50.995 ","End":"23:56.035","Text":"which means that the area that the magnetic field can pass through is decreasing,"},{"Start":"23:56.035 ","End":"24:00.685","Text":"which means that the magnetic flux is also decreasing."},{"Start":"24:00.685 ","End":"24:05.080","Text":"We have a decreasing magnetic flux and in that case,"},{"Start":"24:05.080 ","End":"24:07.675","Text":"what we want to do is we want to do the opposite."},{"Start":"24:07.675 ","End":"24:11.710","Text":"We want to increase the magnetic flux."},{"Start":"24:11.710 ","End":"24:15.320","Text":"We want to oppose this change."},{"Start":"24:15.960 ","End":"24:19.405","Text":"We want to increase the magnetic flux,"},{"Start":"24:19.405 ","End":"24:21.190","Text":"so if we want to do that,"},{"Start":"24:21.190 ","End":"24:24.550","Text":"then we want to add a magnetic field,"},{"Start":"24:24.550 ","End":"24:29.905","Text":"which is in the same direction as the current magnetic field that we have."},{"Start":"24:29.905 ","End":"24:33.265","Text":"It\u0027s also going into the page."},{"Start":"24:33.265 ","End":"24:38.080","Text":"Then what we\u0027re going to do is we\u0027re going to find the direction of the current,"},{"Start":"24:38.080 ","End":"24:43.255","Text":"which induces this magnetic field over here."},{"Start":"24:43.255 ","End":"24:47.330","Text":"Now we\u0027re finding the direction of the current,"},{"Start":"24:47.330 ","End":"24:49.675","Text":"which induces this magnetic field,"},{"Start":"24:49.675 ","End":"24:52.240","Text":"which aims to oppose the change,"},{"Start":"24:52.240 ","End":"24:55.360","Text":"which means that it\u0027s aiming to increase the magnetic flux."},{"Start":"24:55.360 ","End":"24:59.710","Text":"So we point our thumb into the page and what we get is that our fingers"},{"Start":"24:59.710 ","End":"25:04.090","Text":"curl in the clockwise direction,"},{"Start":"25:04.090 ","End":"25:06.519","Text":"so the current is flowing,"},{"Start":"25:06.519 ","End":"25:10.010","Text":"therefore, in the clockwise direction."},{"Start":"25:10.470 ","End":"25:14.253","Text":"Now we want to answer Question Number 3."},{"Start":"25:14.253 ","End":"25:19.465","Text":"Calculate the force required in order for the frame to move at a constant velocity."},{"Start":"25:19.465 ","End":"25:23.875","Text":"Again, because we have this constant magnetic field,"},{"Start":"25:23.875 ","End":"25:27.040","Text":"so our force is equal to BIL, so B-I-L,"},{"Start":"25:27.040 ","End":"25:34.180","Text":"so that is equal to B multiplied by the current, which is B,"},{"Start":"25:34.180 ","End":"25:41.920","Text":"so B^2Lv_0 divided by R multiplied by L again."},{"Start":"25:41.920 ","End":"25:45.040","Text":"So this is L^2."},{"Start":"25:45.040 ","End":"25:48.595","Text":"Now what we want to do is we want to know the direction,"},{"Start":"25:48.595 ","End":"25:56.800","Text":"so F_external is going to simply be equal to the force itself,"},{"Start":"25:56.800 ","End":"26:01.180","Text":"so B^2L^2v_0 divided by"},{"Start":"26:01.180 ","End":"26:06.340","Text":"R. Now what we want to do is we want to know the direction of the force."},{"Start":"26:06.340 ","End":"26:11.920","Text":"The velocity is still in the same direction,"},{"Start":"26:11.920 ","End":"26:15.820","Text":"so that can give us the clue that the external force is going to be in"},{"Start":"26:15.820 ","End":"26:20.890","Text":"the same direction as what we saw before in our first region over here,"},{"Start":"26:20.890 ","End":"26:24.310","Text":"but also we can just use the right-hand rule."},{"Start":"26:24.310 ","End":"26:28.810","Text":"Our thumb, if we remember the rule,"},{"Start":"26:28.810 ","End":"26:32.770","Text":"so df is equal to,"},{"Start":"26:32.770 ","End":"26:38.990","Text":"so we have I dl cross B,"},{"Start":"26:39.180 ","End":"26:44.146","Text":"which means that our thumb points in the direction of the current,"},{"Start":"26:44.146 ","End":"26:46.465","Text":"which if we look at this side over here,"},{"Start":"26:46.465 ","End":"26:49.510","Text":"which is the side that\u0027s longest inside,"},{"Start":"26:49.510 ","End":"26:55.390","Text":"so this side or that is always in the region during this,"},{"Start":"26:55.390 ","End":"26:58.060","Text":"where x adheres to these bounds."},{"Start":"26:58.060 ","End":"27:00.820","Text":"This side is always in that region."},{"Start":"27:00.820 ","End":"27:02.980","Text":"Here the current is going up,"},{"Start":"27:02.980 ","End":"27:04.795","Text":"so your thumb points upwards,"},{"Start":"27:04.795 ","End":"27:08.380","Text":"your forefinger points at the direction of the magnetic field,"},{"Start":"27:08.380 ","End":"27:10.150","Text":"so points into the page,"},{"Start":"27:10.150 ","End":"27:15.340","Text":"and then your middle finger shows the direction of the force."},{"Start":"27:15.340 ","End":"27:20.170","Text":"What we get is that the force is going to"},{"Start":"27:20.170 ","End":"27:25.390","Text":"be in the negative over here, negative x direction."},{"Start":"27:25.390 ","End":"27:29.755","Text":"As we know, F_external is equal and opposite to this,"},{"Start":"27:29.755 ","End":"27:34.520","Text":"so it\u0027s going to be in the positive x direction."},{"Start":"27:35.790 ","End":"27:39.250","Text":"You can either work it out like that or just know that,"},{"Start":"27:39.250 ","End":"27:40.540","Text":"as we already said,"},{"Start":"27:40.540 ","End":"27:43.990","Text":"the force is always in the direction of the velocity,"},{"Start":"27:43.990 ","End":"27:46.530","Text":"and the velocity is in the positive x direction,"},{"Start":"27:46.530 ","End":"27:52.870","Text":"so we could have known that the external force will also be in the positive x direction."},{"Start":"27:52.870 ","End":"27:55.945","Text":"Of course, this is the magnetic force."},{"Start":"27:55.945 ","End":"28:01.075","Text":"So what we got is that our F_external is in this direction,"},{"Start":"28:01.075 ","End":"28:05.930","Text":"and our F_B is in this direction."},{"Start":"28:07.080 ","End":"28:10.765","Text":"Here, the force is due to the magnetic field,"},{"Start":"28:10.765 ","End":"28:15.144","Text":"and here we have this equal and opposite force in order to have constant velocity,"},{"Start":"28:15.144 ","End":"28:17.154","Text":"which is in the opposite direction,"},{"Start":"28:17.154 ","End":"28:19.570","Text":"but also the force is in"},{"Start":"28:19.570 ","End":"28:22.720","Text":"the direction of the velocity because that\u0027s what we\u0027re trying to find,"},{"Start":"28:22.720 ","End":"28:28.495","Text":"the force needed in order for this velocity to remain constant."},{"Start":"28:28.495 ","End":"28:31.240","Text":"Question Number 4, what is the power of"},{"Start":"28:31.240 ","End":"28:34.660","Text":"the force and the power turned to heat in the resistor?"},{"Start":"28:34.660 ","End":"28:38.305","Text":"P of the force, so P_external,"},{"Start":"28:38.305 ","End":"28:44.430","Text":"is equal simply to the force dot product with the velocity,"},{"Start":"28:44.430 ","End":"28:51.450","Text":"so it will have B^2L^2v_0^2 divided by R,"},{"Start":"28:51.450 ","End":"28:55.743","Text":"which we also saw in the first region,"},{"Start":"28:55.743 ","End":"29:02.515","Text":"and the power in the resistor is equal to I^2R,"},{"Start":"29:02.515 ","End":"29:06.130","Text":"which is going to be equal to the exact same thing,"},{"Start":"29:06.130 ","End":"29:10.660","Text":"B^2L^2v_0^2 divided by R,"},{"Start":"29:10.660 ","End":"29:14.230","Text":"so we can see that these 2 are also equal,"},{"Start":"29:14.230 ","End":"29:17.750","Text":"just like we saw in the first region."},{"Start":"29:18.030 ","End":"29:21.340","Text":"These answers remain the same."},{"Start":"29:21.340 ","End":"29:24.025","Text":"So even though the current has flipped direction,"},{"Start":"29:24.025 ","End":"29:28.765","Text":"we also have just a mirroring of the same problem."},{"Start":"29:28.765 ","End":"29:34.855","Text":"That\u0027s why the force is the same also because the velocity doesn\u0027t change direction,"},{"Start":"29:34.855 ","End":"29:38.935","Text":"and the power will also be the same, again,"},{"Start":"29:38.935 ","End":"29:41.395","Text":"because the velocity is in the same direction,"},{"Start":"29:41.395 ","End":"29:43.045","Text":"and the same magnitude,"},{"Start":"29:43.045 ","End":"29:46.675","Text":"and the force is also the same."},{"Start":"29:46.675 ","End":"29:49.000","Text":"Similarly, the current is still going to"},{"Start":"29:49.000 ","End":"29:51.820","Text":"be the exact same, the current doesn\u0027t change,"},{"Start":"29:51.820 ","End":"29:53.515","Text":"just its direction changes,"},{"Start":"29:53.515 ","End":"29:57.680","Text":"and of course the resistance is constant."},{"Start":"29:58.670 ","End":"30:02.470","Text":"That\u0027s the end of this lesson."}],"ID":21450},{"Watched":false,"Name":"Exercise 7","Duration":"13m 56s","ChapterTopicVideoID":21371,"CourseChapterTopicPlaylistID":99480,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.380","Text":"Hello. In this lesson we\u0027re going to be answering the following question."},{"Start":"00:04.380 ","End":"00:13.770","Text":"A rod of length L rotates about one of its edges at a constant angular velocity of Omega."},{"Start":"00:13.770 ","End":"00:17.115","Text":"The rods is in a uniform magnetic field B,"},{"Start":"00:17.115 ","End":"00:18.600","Text":"which is into the page,"},{"Start":"00:18.600 ","End":"00:22.110","Text":"and it\u0027s perpendicular to the plane of rotation."},{"Start":"00:22.110 ","End":"00:26.925","Text":"Question number 1 is to calculate the voltage between the 2 edges of"},{"Start":"00:26.925 ","End":"00:32.590","Text":"the rod by using integration using Lorentz\u0027s law."},{"Start":"00:33.380 ","End":"00:36.825","Text":"Let\u0027s answer this question."},{"Start":"00:36.825 ","End":"00:41.270","Text":"We know that the voltage is equal to the negative"},{"Start":"00:41.270 ","End":"00:47.400","Text":"integral of E.dr or dL,"},{"Start":"00:47.400 ","End":"00:49.986","Text":"it doesn\u0027t matter, it\u0027s the roots, it\u0027s the trajectory,"},{"Start":"00:49.986 ","End":"00:54.005","Text":"and we know that Lorentz\u0027s law is as such,"},{"Start":"00:54.005 ","End":"00:56.555","Text":"force is equal to"},{"Start":"00:56.555 ","End":"01:04.910","Text":"q_v cross B,"},{"Start":"01:04.910 ","End":"01:10.070","Text":"plus q multiplied by E. What I can do is I"},{"Start":"01:10.070 ","End":"01:15.625","Text":"can take the q out of the brackets and then I have v cross,"},{"Start":"01:15.625 ","End":"01:22.490","Text":"B plus E. If"},{"Start":"01:22.490 ","End":"01:26.720","Text":"I look at a point along the rod,"},{"Start":"01:26.720 ","End":"01:32.879","Text":"I can see that it\u0027s going to be moving in a circle at this constant velocity,"},{"Start":"01:32.879 ","End":"01:36.660","Text":"this constant angular velocity."},{"Start":"01:38.450 ","End":"01:41.600","Text":"Because everything over here is constant,"},{"Start":"01:41.600 ","End":"01:44.825","Text":"we can assume that we have arrived at the steady-state,"},{"Start":"01:44.825 ","End":"01:48.109","Text":"which as we saw in the previous lesson, if we\u0027re in steady-state,"},{"Start":"01:48.109 ","End":"01:51.670","Text":"we can say that the sum of all of the forces is equal to 0,"},{"Start":"01:51.670 ","End":"01:59.650","Text":"and that means that this over here is equal to 0,"},{"Start":"02:00.410 ","End":"02:05.090","Text":"because of this, and therefore what we get is that what is"},{"Start":"02:05.090 ","End":"02:09.110","Text":"inside the brackets is equal to 0. if we rearrange,"},{"Start":"02:09.110 ","End":"02:13.460","Text":"we get that V cross B is equal to"},{"Start":"02:13.460 ","End":"02:20.880","Text":"negative E. V cross B is something that we can know,"},{"Start":"02:20.880 ","End":"02:24.270","Text":"we can replace it over here in this integral."},{"Start":"02:24.270 ","End":"02:31.480","Text":"Our integral becomes the integral and then we can of course put the minus inside."},{"Start":"02:31.480 ","End":"02:33.419","Text":"It makes no difference,"},{"Start":"02:33.419 ","End":"02:36.380","Text":"we can just move it here, It\u0027s the same thing."},{"Start":"02:36.380 ","End":"02:45.210","Text":"Then we can say that the integral is of V cross B.dr."},{"Start":"02:46.610 ","End":"02:50.235","Text":"This is of course the velocity,"},{"Start":"02:50.235 ","End":"02:52.560","Text":"just in case it wasn\u0027t clear."},{"Start":"02:52.560 ","End":"02:54.525","Text":"This is the velocity."},{"Start":"02:54.525 ","End":"03:00.050","Text":"Let\u0027s just make it clear that this is the voltage by writing this like so with a capital"},{"Start":"03:00.050 ","End":"03:03.560","Text":"V. What we need here is"},{"Start":"03:03.560 ","End":"03:09.080","Text":"the linear velocity and what we\u0027re given in this question is the angular velocity."},{"Start":"03:09.080 ","End":"03:13.115","Text":"As we know, the linear velocity is equal to"},{"Start":"03:13.115 ","End":"03:20.545","Text":"the angular velocity multiplied by r_i."},{"Start":"03:20.545 ","End":"03:22.845","Text":"Now what is r_i?"},{"Start":"03:22.845 ","End":"03:24.720","Text":"If we look over here,"},{"Start":"03:24.720 ","End":"03:26.355","Text":"this is the origin,"},{"Start":"03:26.355 ","End":"03:30.065","Text":"and we take a little piece over here and we look"},{"Start":"03:30.065 ","End":"03:34.880","Text":"at the distance or the trajectory that it covers."},{"Start":"03:34.880 ","End":"03:41.141","Text":"The distance from the origin until this point is r_i,"},{"Start":"03:41.141 ","End":"03:45.455","Text":"and of course, we can make this into a vector,"},{"Start":"03:45.455 ","End":"03:47.840","Text":"as we can see, it\u0027s going around in a circle,"},{"Start":"03:47.840 ","End":"03:51.630","Text":"so it\u0027s in the Theta direction."},{"Start":"03:52.760 ","End":"04:02.005","Text":"Now we can just say that this is the positive Theta direction anticlockwise,"},{"Start":"04:02.005 ","End":"04:04.450","Text":"we can say that and in that case,"},{"Start":"04:04.450 ","End":"04:05.680","Text":"using the right-hand rule,"},{"Start":"04:05.680 ","End":"04:11.200","Text":"we can see that the positive z direction is towards us."},{"Start":"04:11.200 ","End":"04:15.415","Text":"We can see that the magnetic field is in the negative z-hat direction."},{"Start":"04:15.415 ","End":"04:17.440","Text":"This matters when we plug it in,"},{"Start":"04:17.440 ","End":"04:20.710","Text":"we have this integral of v,"},{"Start":"04:20.710 ","End":"04:24.715","Text":"which is Omega r_i."},{"Start":"04:24.715 ","End":"04:28.960","Text":"We can just leave it as i in the Theta direction,"},{"Start":"04:28.960 ","End":"04:31.210","Text":"cross with our magnetic fields,"},{"Start":"04:31.210 ","End":"04:33.730","Text":"which we saw is in the negative z direction,"},{"Start":"04:33.730 ","End":"04:35.980","Text":"and its of magnitude B."},{"Start":"04:35.980 ","End":"04:41.385","Text":"Cross negative B in the z direction,"},{"Start":"04:41.385 ","End":"04:42.840","Text":"and all of this is"},{"Start":"04:42.840 ","End":"04:48.810","Text":"dot dr. Let\u0027s just"},{"Start":"04:48.810 ","End":"04:55.500","Text":"see what our dr is equal to, our dr vector."},{"Start":"04:55.500 ","End":"04:58.310","Text":"What we can see as we\u0027re looking at this rod,"},{"Start":"04:58.310 ","End":"05:00.787","Text":"and we see that this is the I direction,"},{"Start":"05:00.787 ","End":"05:05.705","Text":"and this little piece that we said before that we look at its trajectory."},{"Start":"05:05.705 ","End":"05:11.885","Text":"This piece is dr. Each little increments in the rod is dr,"},{"Start":"05:11.885 ","End":"05:15.200","Text":"that means that dr vector is just dr,"},{"Start":"05:15.200 ","End":"05:22.410","Text":"a little increments in the rod in the radial direction."},{"Start":"05:22.760 ","End":"05:26.710","Text":"That\u0027s all that dr. is, therefore,"},{"Start":"05:26.710 ","End":"05:31.670","Text":"we can say that our voltage from here is equal to,"},{"Start":"05:31.670 ","End":"05:34.940","Text":"we can take out the Omega and negative B out of"},{"Start":"05:34.940 ","End":"05:38.660","Text":"the integration because we\u0027re not integrating with respect to it,"},{"Start":"05:38.660 ","End":"05:45.890","Text":"we\u0027re integrating with respect to r. What we have is negative Omega B,"},{"Start":"05:45.890 ","End":"05:50.885","Text":"and then we have our integral of r,"},{"Start":"05:50.885 ","End":"05:59.025","Text":"and then we have Theta hat cross z-hat,"},{"Start":"05:59.025 ","End":"06:07.650","Text":"and all of this is dot-product with dr-hat."},{"Start":"06:07.790 ","End":"06:10.650","Text":"Now if we do here,"},{"Start":"06:10.650 ","End":"06:13.655","Text":"we have Theta hat cross product,"},{"Start":"06:13.655 ","End":"06:16.115","Text":"or a cross multiplied with a z-hat,"},{"Start":"06:16.115 ","End":"06:19.250","Text":"which is going to give us our hats as we know."},{"Start":"06:19.250 ","End":"06:27.735","Text":"What we have is negative Omega B and then the integral of r, r-hat dot with."},{"Start":"06:27.735 ","End":"06:31.670","Text":"Now we\u0027ve done this whole bracket we can leave the brackets"},{"Start":"06:31.670 ","End":"06:36.955","Text":"now, dot-product with drr-hat."},{"Start":"06:36.955 ","End":"06:40.760","Text":"First we solve what was in the brackets and now we"},{"Start":"06:40.760 ","End":"06:44.540","Text":"can do the dot-product between what\u0027s outside."},{"Start":"06:44.540 ","End":"06:46.430","Text":"When we have r-hat."},{"Start":"06:46.430 ","End":"06:49.225","Text":"r-hat, we get 1,"},{"Start":"06:49.225 ","End":"06:53.990","Text":"what we\u0027re left with is rdr."},{"Start":"06:53.990 ","End":"06:57.810","Text":"Rdr multiplied by 1,"},{"Start":"06:57.810 ","End":"07:03.200","Text":"which is just this, and of course I have to write out the whole integrals."},{"Start":"07:03.200 ","End":"07:09.155","Text":"We have negative Omega B and then we have the integral of rdr,"},{"Start":"07:09.155 ","End":"07:12.470","Text":"which we could also set in the balance."},{"Start":"07:12.470 ","End":"07:14.075","Text":"We\u0027re going from here,"},{"Start":"07:14.075 ","End":"07:16.535","Text":"from 0 from the origin,"},{"Start":"07:16.535 ","End":"07:19.295","Text":"until the top of the rod,"},{"Start":"07:19.295 ","End":"07:21.440","Text":"where we\u0027re told the rod is of length L,"},{"Start":"07:21.440 ","End":"07:23.105","Text":"so until L,"},{"Start":"07:23.105 ","End":"07:26.190","Text":"then we have negative Omega B,"},{"Start":"07:26.190 ","End":"07:29.960","Text":"and then this is multiplied by r^2 divided by 2,"},{"Start":"07:29.960 ","End":"07:31.850","Text":"where we plug in the balance,"},{"Start":"07:31.850 ","End":"07:37.265","Text":"we have multiplied by L^2 divided by 2 minus,"},{"Start":"07:37.265 ","End":"07:41.480","Text":"where we plug in 0, we get of course 0."},{"Start":"07:41.480 ","End":"07:43.950","Text":"This is the answer."},{"Start":"07:44.260 ","End":"07:47.780","Text":"Now another way we could have solved this question is by"},{"Start":"07:47.780 ","End":"07:52.670","Text":"using the equation that we\u0027ve already seen for the EMF in a rod."},{"Start":"07:52.670 ","End":"07:54.425","Text":"As we\u0027ve already seen,"},{"Start":"07:54.425 ","End":"07:59.010","Text":"each points in the rod has a different linear velocity."},{"Start":"08:00.110 ","End":"08:03.060","Text":"What we would do just as a reminder,"},{"Start":"08:03.060 ","End":"08:05.750","Text":"the EMF or the voltage in"},{"Start":"08:05.750 ","End":"08:12.455","Text":"a conducting rod that is moving through a magnetic field is equal to BLv,"},{"Start":"08:12.455 ","End":"08:16.445","Text":"where this is of course the linear velocity."},{"Start":"08:16.445 ","End":"08:21.740","Text":"Over here, because each point has the same angular velocity,"},{"Start":"08:21.740 ","End":"08:24.520","Text":"but a different linear velocity,"},{"Start":"08:24.520 ","End":"08:30.410","Text":"what we would do is we would split up this rod into lots of tiny different rods,"},{"Start":"08:30.410 ","End":"08:32.795","Text":"that means that instead of L,"},{"Start":"08:32.795 ","End":"08:35.480","Text":"we\u0027d have the length of each tiny little rod,"},{"Start":"08:35.480 ","End":"08:36.834","Text":"which as we said was dr,"},{"Start":"08:36.834 ","End":"08:39.635","Text":"and instead of V,"},{"Start":"08:39.635 ","End":"08:46.370","Text":"we\u0027d substitute in what Omega is in terms of V. The linear velocity is equal to Omega r,"},{"Start":"08:46.370 ","End":"08:52.445","Text":"and then what we\u0027d have here is the d epsilon."},{"Start":"08:52.445 ","End":"08:55.355","Text":"In order to get the total EMF,"},{"Start":"08:55.355 ","End":"08:56.840","Text":"we\u0027d have to integrate on that,"},{"Start":"08:56.840 ","End":"09:00.560","Text":"and we\u0027d set in the same bounds from 0 to L,"},{"Start":"09:00.560 ","End":"09:06.020","Text":"and we would get the exact same answer."},{"Start":"09:06.020 ","End":"09:11.970","Text":"Then this of course would just give us the magnitude."},{"Start":"09:12.880 ","End":"09:15.695","Text":"We get the magnitude,"},{"Start":"09:15.695 ","End":"09:20.525","Text":"which will just work out the same and using the right-hand rule,"},{"Start":"09:20.525 ","End":"09:23.075","Text":"we can get the minus."},{"Start":"09:23.075 ","End":"09:26.960","Text":"What we\u0027ll see is that using the right-hand rule,"},{"Start":"09:26.960 ","End":"09:31.100","Text":"the magnetic force will be pointing into the circle."},{"Start":"09:31.100 ","End":"09:32.600","Text":"We\u0027ll get all of"},{"Start":"09:32.600 ","End":"09:38.120","Text":"the positive charges here and all of the negative charges accumulating on"},{"Start":"09:38.120 ","End":"09:45.320","Text":"this edge and that would mean that the arrow will be pointing into the circle,"},{"Start":"09:45.320 ","End":"09:48.600","Text":"which will give us this minus."},{"Start":"09:49.150 ","End":"09:54.670","Text":"Now let\u0027s answer Question number 2."},{"Start":"09:54.670 ","End":"10:02.040","Text":"Question number 2 is asking us to calculate the voltage in the rod using Faraday\u0027s law."},{"Start":"10:02.360 ","End":"10:08.105","Text":"Faraday\u0027s law says that the EMF for the voltage is equal to"},{"Start":"10:08.105 ","End":"10:14.965","Text":"the negative time derivative of the magnetic flux."},{"Start":"10:14.965 ","End":"10:17.285","Text":"The first thing that I need,"},{"Start":"10:17.285 ","End":"10:18.955","Text":"we can rub all of this out."},{"Start":"10:18.955 ","End":"10:24.960","Text":"Is I need to have a closed circuit in order to calculate the magnetic flux."},{"Start":"10:24.960 ","End":"10:28.115","Text":"what I say is I assume that I\u0027m,"},{"Start":"10:28.115 ","End":"10:34.415","Text":"or I pretend that I\u0027m connecting the 2 edges of the rod via some wire,"},{"Start":"10:34.415 ","End":"10:39.435","Text":"and then I form this closed circuit where this angle over here,"},{"Start":"10:39.435 ","End":"10:43.780","Text":"like so is equal to Theta."},{"Start":"10:44.360 ","End":"10:50.180","Text":"Now what we want to do is we want to calculate this magnetic flux."},{"Start":"10:50.180 ","End":"10:52.955","Text":"We say that the magnetic flux, as we know,"},{"Start":"10:52.955 ","End":"10:56.930","Text":"is equal to the integral of B.ds,"},{"Start":"10:56.930 ","End":"11:02.765","Text":"where ds is this surface area over here and"},{"Start":"11:02.765 ","End":"11:09.330","Text":"because we can see that our magnetic field is constant or is uniform,"},{"Start":"11:09.330 ","End":"11:13.290","Text":"we can say that this is simply equal to B.S."},{"Start":"11:14.770 ","End":"11:19.415","Text":"Now if we want to see what S is the surface area."},{"Start":"11:19.415 ","End":"11:25.160","Text":"The surface area of a part of a circle is simply"},{"Start":"11:25.160 ","End":"11:30.650","Text":"equal to the double integral from 0 until the radius of the circle,"},{"Start":"11:30.650 ","End":"11:37.584","Text":"and from 0 until Theta of rdrd Theta"},{"Start":"11:37.584 ","End":"11:45.810","Text":"and what we get from this integral is Theta r^2 divided by 2."},{"Start":"11:45.810 ","End":"11:49.545","Text":"We can just multiply that over here,"},{"Start":"11:49.545 ","End":"11:55.780","Text":"we get Theta r^2 divided by 2."},{"Start":"11:56.240 ","End":"12:00.565","Text":"Now of course this r is the radius of the circle,"},{"Start":"12:00.565 ","End":"12:04.060","Text":"where here the radius of the circle is simply the length of the rod,"},{"Start":"12:04.060 ","End":"12:07.089","Text":"which is L. Over here specifically,"},{"Start":"12:07.089 ","End":"12:11.740","Text":"I can replace this by L. Instead of I^2, L^2,"},{"Start":"12:11.740 ","End":"12:17.680","Text":"and now we have to take the negative time derivative of the magnetic flux."},{"Start":"12:17.680 ","End":"12:22.450","Text":"What over here is dependent on time, that is Theta."},{"Start":"12:22.450 ","End":"12:25.300","Text":"As we may know,"},{"Start":"12:25.300 ","End":"12:28.345","Text":"the time derivative of Theta."},{"Start":"12:28.345 ","End":"12:29.815","Text":"If I do a dot on top,"},{"Start":"12:29.815 ","End":"12:31.270","Text":"it\u0027s just the time derivative."},{"Start":"12:31.270 ","End":"12:37.910","Text":"It\u0027s the same as doing d Theta by dt is equal to Omega, the angular velocity."},{"Start":"12:38.600 ","End":"12:42.950","Text":"If we take the negative time derivative of this,"},{"Start":"12:42.950 ","End":"12:47.545","Text":"what we will get there for is if we look at,"},{"Start":"12:47.545 ","End":"12:55.850","Text":"we can just write EMF is equal to negative time derivative of the magnetic flux,"},{"Start":"12:55.850 ","End":"12:59.795","Text":"which is equal to Theta dot."},{"Start":"12:59.795 ","End":"13:01.340","Text":"We\u0027ve taken the time derivative,"},{"Start":"13:01.340 ","End":"13:05.360","Text":"added a negative and multiplied by L^2,"},{"Start":"13:05.360 ","End":"13:10.110","Text":"divided by 2, and of course there\u0027s a B here as well."},{"Start":"13:10.670 ","End":"13:14.010","Text":"We multiply the area,"},{"Start":"13:14.010 ","End":"13:16.335","Text":"this is just the area by B."},{"Start":"13:16.335 ","End":"13:22.280","Text":"Then we can see that this is just equal to negative Theta dot,"},{"Start":"13:22.280 ","End":"13:27.330","Text":"which we saw as Omega L^2 divided by 2B."},{"Start":"13:28.420 ","End":"13:30.440","Text":"As we can see,"},{"Start":"13:30.440 ","End":"13:36.950","Text":"it\u0027s the exact same answer that we got for Question number 1. we can see that this EMF is"},{"Start":"13:36.950 ","End":"13:44.465","Text":"basically a voltage and we can calculate it in various ways using Lorentz\u0027s law,"},{"Start":"13:44.465 ","End":"13:50.285","Text":"using the equation that we saw in one of the previous lessons for the EMF in a rod,"},{"Start":"13:50.285 ","End":"13:53.185","Text":"and by using Faraday\u0027s law."},{"Start":"13:53.185 ","End":"13:56.500","Text":"That is the end of this lesson."}],"ID":21451}],"Thumbnail":null,"ID":99480}]

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[21443,21444,21445,21446,21447,21448,21449,21450,21451];

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