Hall Effect
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[{"Name":"Hall Effect","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Explanation and Equation","Duration":"20m 36s","ChapterTopicVideoID":21346,"CourseChapterTopicPlaylistID":246912,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/21346.jpeg","UploadDate":"2020-04-06T21:43:59.0270000","DurationForVideoObject":"PT20M36S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.875","Text":"Hello. In this lesson,"},{"Start":"00:01.875 ","End":"00:04.860","Text":"we\u0027re going to be learning about the Hall effect."},{"Start":"00:04.860 ","End":"00:07.035","Text":"We\u0027re going to see what it is,"},{"Start":"00:07.035 ","End":"00:10.185","Text":"its equation and an example."},{"Start":"00:10.185 ","End":"00:15.584","Text":"Let\u0027s imagine that we have a wire which is generally"},{"Start":"00:15.584 ","End":"00:21.990","Text":"just a very thin cylinder and this cylinder is very long,"},{"Start":"00:21.990 ","End":"00:23.790","Text":"it carries on like so,"},{"Start":"00:23.790 ","End":"00:25.860","Text":"but we\u0027re just drawing a certain section of it."},{"Start":"00:25.860 ","End":"00:30.360","Text":"Through the cylinder, a current I flows through,"},{"Start":"00:30.360 ","End":"00:35.520","Text":"and of course, if we have a current we know we are also going to have a magnetic field."},{"Start":"00:35.520 ","End":"00:37.845","Text":"Let\u0027s imagine that at this stage,"},{"Start":"00:37.845 ","End":"00:42.615","Text":"we have a magnetic field going in to the page."},{"Start":"00:42.615 ","End":"00:46.130","Text":"That means that at least at this stage,"},{"Start":"00:46.130 ","End":"00:53.700","Text":"we have a magnetic field which is exactly perpendicular to the direction of the current."},{"Start":"00:54.040 ","End":"01:01.250","Text":"Up until now, it wasn\u0027t important for us to know if the current was formed by"},{"Start":"01:01.250 ","End":"01:08.750","Text":"positive charges moving forwards or negative charges moving backwards or both."},{"Start":"01:08.750 ","End":"01:12.500","Text":"All we knew was that the current was moving in a certain direction,"},{"Start":"01:12.500 ","End":"01:17.000","Text":"and we didn\u0027t know if it was from positive or negative charges or both."},{"Start":"01:17.000 ","End":"01:20.180","Text":"The Hall effects comes to deal with that."},{"Start":"01:20.180 ","End":"01:22.070","Text":"Via the Hall effect,"},{"Start":"01:22.070 ","End":"01:27.559","Text":"we can see that current is only formed by negative charges moving."},{"Start":"01:27.559 ","End":"01:30.140","Text":"Only the negative charges move,"},{"Start":"01:30.140 ","End":"01:33.609","Text":"and that is what gives us current."},{"Start":"01:33.609 ","End":"01:37.670","Text":"You might have known this before or not,"},{"Start":"01:37.670 ","End":"01:40.670","Text":"but up until now in all of our calculations,"},{"Start":"01:40.670 ","End":"01:44.660","Text":"this fact hasn\u0027t been important and hasn\u0027t changed"},{"Start":"01:44.660 ","End":"01:48.890","Text":"how we calculate or how we work out or solve our question."},{"Start":"01:48.890 ","End":"01:51.425","Text":"Whereas now when dealing with the Hall effect,"},{"Start":"01:51.425 ","End":"01:53.765","Text":"it really is important that it\u0027s"},{"Start":"01:53.765 ","End":"01:59.430","Text":"only the negative charges that are moving and therefore causing current."},{"Start":"02:00.230 ","End":"02:03.195","Text":"Let\u0027s give an example."},{"Start":"02:03.195 ","End":"02:09.030","Text":"Let\u0027s imagine that we have these positive charges, like so."},{"Start":"02:09.030 ","End":"02:12.620","Text":"As we know, because I\u0027ve already mentioned it,"},{"Start":"02:12.620 ","End":"02:15.680","Text":"the positive charges aren\u0027t moving,"},{"Start":"02:15.680 ","End":"02:18.395","Text":"they\u0027re stuck at this exact position."},{"Start":"02:18.395 ","End":"02:24.030","Text":"Then we have our negative charges like so located around,"},{"Start":"02:24.030 ","End":"02:31.320","Text":"and we know that our negative charges are moving and that is what is causing the current."},{"Start":"02:32.030 ","End":"02:36.080","Text":"Seeing as we know that our current is moving upwards,"},{"Start":"02:36.080 ","End":"02:41.105","Text":"that tells us that a negative charges are moving downwards."},{"Start":"02:41.105 ","End":"02:45.530","Text":"Remember, the motion or the direction of travel"},{"Start":"02:45.530 ","End":"02:50.645","Text":"of the negative charges is opposite to the direction of travel of the current."},{"Start":"02:50.645 ","End":"02:55.670","Text":"Around here, we\u0027ve already seen that we have this magnetic field."},{"Start":"02:55.670 ","End":"03:04.625","Text":"Via Lorentz\u0027s law, we know that the force on the charges"},{"Start":"03:04.625 ","End":"03:14.390","Text":"is equal to q multiplied by the velocity cross product with the magnetic field,"},{"Start":"03:14.390 ","End":"03:19.070","Text":"or if we want to know what the magnitude of the force,"},{"Start":"03:19.070 ","End":"03:26.180","Text":"this is equal to the size of the charge multiplied by the magnitude of the velocity,"},{"Start":"03:26.180 ","End":"03:31.940","Text":"and multiplied by the perpendicular component of the magnetic fields."},{"Start":"03:31.940 ","End":"03:36.230","Text":"Of course, we\u0027re speaking about the component that is perpendicular"},{"Start":"03:36.230 ","End":"03:43.019","Text":"to the motion of the charges or to the direction of travel of the charges."},{"Start":"03:43.310 ","End":"03:47.090","Text":"That\u0027s the size and now let\u0027s speak about the direction."},{"Start":"03:47.090 ","End":"03:48.740","Text":"We\u0027re using the right-hand rule,"},{"Start":"03:48.740 ","End":"03:52.640","Text":"so our thumb points in the direction of the velocity,"},{"Start":"03:52.640 ","End":"03:54.245","Text":"which means points down."},{"Start":"03:54.245 ","End":"03:57.890","Text":"Our fourth finger points in the direction of the magnetic field,"},{"Start":"03:57.890 ","End":"04:01.520","Text":"which means into the page."},{"Start":"04:01.520 ","End":"04:07.250","Text":"Then what we get is our middle finger is pointing in the direction of the force,"},{"Start":"04:07.250 ","End":"04:10.955","Text":"which is pointing in the rightwards direction."},{"Start":"04:10.955 ","End":"04:15.425","Text":"Note that we know that the charges that we\u0027re looking at,"},{"Start":"04:15.425 ","End":"04:20.659","Text":"the charges causing the current is the charge of the electron."},{"Start":"04:20.659 ","End":"04:24.365","Text":"We said it\u0027s the negative charges which are causing the current,"},{"Start":"04:24.365 ","End":"04:28.529","Text":"and the negative charges are in this case the electron."},{"Start":"04:28.529 ","End":"04:34.340","Text":"What we in fact have is that our force is equal to"},{"Start":"04:34.340 ","End":"04:43.360","Text":"negative the charge of the electron multiplied by V cross B."},{"Start":"04:43.360 ","End":"04:46.125","Text":"This is the force acting."},{"Start":"04:46.125 ","End":"04:49.880","Text":"Because we have a negative number or"},{"Start":"04:49.880 ","End":"04:54.687","Text":"a negative coefficient multiplying this cross-product,"},{"Start":"04:54.687 ","End":"04:56.630","Text":"when we used the right-hand rule,"},{"Start":"04:56.630 ","End":"04:58.040","Text":"we got that the force was in"},{"Start":"04:58.040 ","End":"05:01.430","Text":"the rightwards direction because we have this negative over here,"},{"Start":"05:01.430 ","End":"05:03.640","Text":"so it\u0027s like we\u0027re multiplying by negative 1."},{"Start":"05:03.640 ","End":"05:08.420","Text":"The direction of the force that we got from the right-hand rule flips over."},{"Start":"05:08.420 ","End":"05:13.290","Text":"Now, the force is in the leftwards direction."},{"Start":"05:13.790 ","End":"05:18.545","Text":"That means that the force is acting in this direction,"},{"Start":"05:18.545 ","End":"05:22.770","Text":"I\u0027ll call this FL, Lorentz\u0027s force."},{"Start":"05:23.870 ","End":"05:27.620","Text":"Now we\u0027ve seen that we have this leftwards force"},{"Start":"05:27.620 ","End":"05:30.950","Text":"acting on the negative charges and the negative charges,"},{"Start":"05:30.950 ","End":"05:33.545","Text":"of course, are free to move."},{"Start":"05:33.545 ","End":"05:35.540","Text":"What is going to happen?"},{"Start":"05:35.540 ","End":"05:44.940","Text":"That means that our negative charges are going to be pushed to the left side of the wire."},{"Start":"05:45.890 ","End":"05:51.170","Text":"We get this charge separation over here,"},{"Start":"05:51.170 ","End":"05:54.995","Text":"where the positive charge stays stationary,"},{"Start":"05:54.995 ","End":"05:59.345","Text":"whereas the negative charges are pushed to one side of the wire,"},{"Start":"05:59.345 ","End":"06:02.975","Text":"in this case over here to the left side of the wire."},{"Start":"06:02.975 ","End":"06:08.900","Text":"Then we get this charge separation where the left side has an overall negative charge,"},{"Start":"06:08.900 ","End":"06:13.205","Text":"and the right-hand side has an overall positive charge."},{"Start":"06:13.205 ","End":"06:16.640","Text":"The direction of force acting on the electrons,"},{"Start":"06:16.640 ","End":"06:17.990","Text":"the negative charge,"},{"Start":"06:17.990 ","End":"06:19.865","Text":"is in the leftwards direction."},{"Start":"06:19.865 ","End":"06:23.915","Text":"However, the direction of force on the positive charge,"},{"Start":"06:23.915 ","End":"06:25.655","Text":"there is no force."},{"Start":"06:25.655 ","End":"06:28.235","Text":"No force because they are stationary,"},{"Start":"06:28.235 ","End":"06:31.830","Text":"they aren\u0027t moving, no force is acting on them."},{"Start":"06:33.020 ","End":"06:39.905","Text":"From the forces only acting on the negative charges and not on the positive charges,"},{"Start":"06:39.905 ","End":"06:43.520","Text":"we get this charge separation."},{"Start":"06:43.520 ","End":"06:48.830","Text":"If we have a charge separation where we have some polarization,"},{"Start":"06:48.830 ","End":"06:51.950","Text":"so here the negative side and here the positive side."},{"Start":"06:51.950 ","End":"06:55.339","Text":"That means that an electric field is formed,"},{"Start":"06:55.339 ","End":"06:58.340","Text":"and the direction of the electric field is of course"},{"Start":"06:58.340 ","End":"07:02.130","Text":"from the positive charge to the negative charge,"},{"Start":"07:02.130 ","End":"07:03.950","Text":"so these pink lines,"},{"Start":"07:03.950 ","End":"07:07.235","Text":"so it\u0027s going from the positive side to"},{"Start":"07:07.235 ","End":"07:13.710","Text":"the negative side like so and this is the E field."},{"Start":"07:14.720 ","End":"07:22.370","Text":"This electric field produces a force that acts on the electrons."},{"Start":"07:22.370 ","End":"07:23.990","Text":"If there\u0027s an electric field,"},{"Start":"07:23.990 ","End":"07:27.110","Text":"that means there\u0027s an electric force."},{"Start":"07:27.110 ","End":"07:29.285","Text":"What is the equation?"},{"Start":"07:29.285 ","End":"07:35.495","Text":"We know that the force is equal to the charge multiplied by the electric field,"},{"Start":"07:35.495 ","End":"07:39.140","Text":"where we said that E field exerts a force on the electrons."},{"Start":"07:39.140 ","End":"07:41.540","Text":"Our charge is a negative charge,"},{"Start":"07:41.540 ","End":"07:48.275","Text":"so we take the negative charge of the electron and multiply it by the electric field."},{"Start":"07:48.275 ","End":"07:52.490","Text":"What we can see is that the direction of the electric force"},{"Start":"07:52.490 ","End":"07:56.810","Text":"is in the opposite direction to the electric field."},{"Start":"07:56.810 ","End":"07:59.030","Text":"In other words, in our example,"},{"Start":"07:59.030 ","End":"08:02.360","Text":"if the electric field is in the leftwards direction,"},{"Start":"08:02.360 ","End":"08:08.405","Text":"then the electric force will be in the rightwards direction."},{"Start":"08:08.405 ","End":"08:13.440","Text":"This is our force due to the electric field."},{"Start":"08:14.450 ","End":"08:17.655","Text":"This is something to always know,"},{"Start":"08:17.655 ","End":"08:21.310","Text":"the electric force is always"},{"Start":"08:21.390 ","End":"08:29.215","Text":"opposite to the direction of the Lorentz\u0027s force or the magnetic force."},{"Start":"08:29.215 ","End":"08:37.108","Text":"The electric force is always opposite to the magnetic force."},{"Start":"08:37.108 ","End":"08:42.930","Text":"What we can see is that we have this magnetic force"},{"Start":"08:42.930 ","End":"08:52.080","Text":"going in one direction and we have in the opposite direction this electric force."},{"Start":"08:52.080 ","End":"08:55.470","Text":"If we suddenly turn on the generator,"},{"Start":"08:55.470 ","End":"08:58.095","Text":"current is flowing through the wire."},{"Start":"08:58.095 ","End":"08:59.617","Text":"At the beginning,"},{"Start":"08:59.617 ","End":"09:01.725","Text":"due to the magnetic force,"},{"Start":"09:01.725 ","End":"09:05.175","Text":"all of the electrons are going to move to one side,"},{"Start":"09:05.175 ","End":"09:09.840","Text":"which means that we\u0027re going to have the magnetic force overpowering over"},{"Start":"09:09.840 ","End":"09:15.765","Text":"here and as more and more electrons are concentrated to the one side,"},{"Start":"09:15.765 ","End":"09:19.170","Text":"the electric force is going to grow"},{"Start":"09:19.170 ","End":"09:24.000","Text":"and at some stage we\u0027re going to reach the steady-state."},{"Start":"09:24.000 ","End":"09:26.850","Text":"What is the steady state?"},{"Start":"09:26.850 ","End":"09:32.190","Text":"The steady state is where the sum of all of the forces is equal to 0,"},{"Start":"09:32.190 ","End":"09:33.840","Text":"or in other words,"},{"Start":"09:33.840 ","End":"09:39.285","Text":"that our electric force is equal to our magnetic force."},{"Start":"09:39.285 ","End":"09:43.718","Text":"There\u0027s going to be a battle between the two forces who is going to be stronger,"},{"Start":"09:43.718 ","End":"09:47.265","Text":"and eventually they\u0027re going to be equal to one another."},{"Start":"09:47.265 ","End":"09:50.640","Text":"The sum of all of the forces will be equal to 0,"},{"Start":"09:50.640 ","End":"09:54.195","Text":"which means that the magnitude of"},{"Start":"09:54.195 ","End":"10:01.420","Text":"each force is going to be equal to the magnitude of the other."},{"Start":"10:02.960 ","End":"10:05.895","Text":"That\u0027s what we have written here."},{"Start":"10:05.895 ","End":"10:09.360","Text":"I just rubbed out another line because it was written twice,"},{"Start":"10:09.360 ","End":"10:11.355","Text":"so you\u0027re not missing anything."},{"Start":"10:11.355 ","End":"10:14.670","Text":"When that happens, I can write out the magnitude."},{"Start":"10:14.670 ","End":"10:19.590","Text":"The magnitude for the electric force is equal to"},{"Start":"10:19.590 ","End":"10:26.459","Text":"the charge multiplied by the electric field."},{"Start":"10:26.459 ","End":"10:31.980","Text":"I\u0027m just going to leave it like this without the vector arrow on top showing that this is"},{"Start":"10:31.980 ","End":"10:34.860","Text":"just the magnitude of the electric field and this is equal"},{"Start":"10:34.860 ","End":"10:37.935","Text":"to the magnitude of the magnetic force,"},{"Start":"10:37.935 ","End":"10:42.900","Text":"which is equal to the charge multiplied by the velocity, again,"},{"Start":"10:42.900 ","End":"10:44.670","Text":"just the magnitude of the velocity,"},{"Start":"10:44.670 ","End":"10:51.300","Text":"and multiplied by the perpendicular component of the magnetic field."},{"Start":"10:51.300 ","End":"10:54.000","Text":"Then we can divide both sides by Q."},{"Start":"10:54.000 ","End":"11:00.975","Text":"What we get is that our electric field is equal to the velocity."},{"Start":"11:00.975 ","End":"11:03.630","Text":"The magnitude of the velocity multiplied"},{"Start":"11:03.630 ","End":"11:06.795","Text":"by the perpendicular component of the magnetic field."},{"Start":"11:06.795 ","End":"11:13.485","Text":"Of course, if you need to calculate the perpendicular component of the magnetic field,"},{"Start":"11:13.485 ","End":"11:17.580","Text":"all you have to do is take the magnetic field and multiply it by sine of"},{"Start":"11:17.580 ","End":"11:23.130","Text":"the angle between the magnetic field and the direction of the velocity."},{"Start":"11:23.130 ","End":"11:26.700","Text":"It\u0027s either the perpendicular component or"},{"Start":"11:26.700 ","End":"11:30.150","Text":"B multiplied by sine of the angle between the two."},{"Start":"11:30.150 ","End":"11:33.040","Text":"They both just mean the same thing."},{"Start":"11:33.770 ","End":"11:38.805","Text":"We know that if we have a current flowing in this direction,"},{"Start":"11:38.805 ","End":"11:49.185","Text":"that means that we\u0027ve probably attached some battery or some generator like so"},{"Start":"11:49.185 ","End":"11:52.860","Text":"too with the wire which is causing this current will be"},{"Start":"11:52.860 ","End":"11:57.000","Text":"flowing in this direction so that we know and then"},{"Start":"11:57.000 ","End":"12:00.795","Text":"we can check what the voltage drop is across the battery or"},{"Start":"12:00.795 ","End":"12:05.025","Text":"across the wire from the bottom section until the top section."},{"Start":"12:05.025 ","End":"12:12.510","Text":"However, now, because we have this split in the charges,"},{"Start":"12:12.510 ","End":"12:18.870","Text":"so we have overall negative charge on this side and overall positive charge on this side."},{"Start":"12:18.870 ","End":"12:23.535","Text":"That means that between this edge over here,"},{"Start":"12:23.535 ","End":"12:30.705","Text":"so between this edge over here and this edge over here,"},{"Start":"12:30.705 ","End":"12:34.200","Text":"we\u0027re also going to have a voltage drop."},{"Start":"12:34.200 ","End":"12:36.450","Text":"Now, my question is,"},{"Start":"12:36.450 ","End":"12:38.310","Text":"what is the voltage drop?"},{"Start":"12:38.310 ","End":"12:40.185","Text":"There are two options."},{"Start":"12:40.185 ","End":"12:45.030","Text":"Either I can attach a volt meter over here,"},{"Start":"12:45.030 ","End":"12:49.530","Text":"like so to measure the voltage drop between the two edges of"},{"Start":"12:49.530 ","End":"12:55.665","Text":"the wire or we can use this equation."},{"Start":"12:55.665 ","End":"12:58.380","Text":"What equation is this going to be?"},{"Start":"12:58.380 ","End":"13:03.010","Text":"So V, the voltage,"},{"Start":"13:03.820 ","End":"13:07.085","Text":"is going to be equal to"},{"Start":"13:07.085 ","End":"13:12.695","Text":"the electric field multiplied by the distance between these two edges."},{"Start":"13:12.695 ","End":"13:16.460","Text":"Let\u0027s say that the radius of this wire is"},{"Start":"13:16.460 ","End":"13:23.760","Text":"R. The distance between this edge to this edge is just twice the radius,"},{"Start":"13:23.760 ","End":"13:28.860","Text":"so multiplied by 2R or in other words,"},{"Start":"13:28.860 ","End":"13:32.400","Text":"this is also equal to the velocity"},{"Start":"13:32.400 ","End":"13:36.720","Text":"multiplied by the perpendicular component of the magnetic field."},{"Start":"13:36.720 ","End":"13:39.790","Text":"Then multiplied by 2R."},{"Start":"13:41.840 ","End":"13:52.200","Text":"Now what we want to know is what is the velocity of the electrons in the conductor."},{"Start":"13:52.200 ","End":"14:00.090","Text":"The velocity of the electrons in the conductor are called V_drift."},{"Start":"14:00.090 ","End":"14:04.860","Text":"Velocity of electrons, then conductor is also called V_drift."},{"Start":"14:04.860 ","End":"14:06.945","Text":"Now, let\u0027s see exactly what this is."},{"Start":"14:06.945 ","End":"14:10.230","Text":"V_drift, we can get from Jordaz model,"},{"Start":"14:10.230 ","End":"14:11.940","Text":"but we\u0027re not going to go into that."},{"Start":"14:11.940 ","End":"14:14.970","Text":"You can look it up online if you haven\u0027t heard that term yet,"},{"Start":"14:14.970 ","End":"14:17.580","Text":"but we\u0027re not going to go into that right now."},{"Start":"14:17.580 ","End":"14:19.905","Text":"Let\u0027s see what this is."},{"Start":"14:19.905 ","End":"14:23.160","Text":"We know that our current density, J,"},{"Start":"14:23.160 ","End":"14:28.215","Text":"is equal to the number of charges multiplied by their charge,"},{"Start":"14:28.215 ","End":"14:35.295","Text":"multiplied by the velocity and of course all of this is per unit volume."},{"Start":"14:35.295 ","End":"14:40.200","Text":"N is the number of charges per unit volume,"},{"Start":"14:40.200 ","End":"14:43.200","Text":"q is the charge and this V,"},{"Start":"14:43.200 ","End":"14:46.125","Text":"the velocity of the electrons is of course,"},{"Start":"14:46.125 ","End":"14:48.370","Text":"now as we know what V_drift."},{"Start":"14:48.370 ","End":"14:51.780","Text":"This is one equation that we have seen,"},{"Start":"14:51.780 ","End":"14:54.585","Text":"I think, previously in one of the chapters for J."},{"Start":"14:54.585 ","End":"14:57.840","Text":"Another equation for j that we\u0027ve seen,"},{"Start":"14:57.840 ","End":"15:02.040","Text":"we\u0027ve seen that J is simply equal"},{"Start":"15:02.040 ","End":"15:08.745","Text":"to our current per unit area."},{"Start":"15:08.745 ","End":"15:15.310","Text":"The amount of current that we have per unit area is also equal to J."},{"Start":"15:15.500 ","End":"15:19.110","Text":"Now if I equate these two equations,"},{"Start":"15:19.110 ","End":"15:21.855","Text":"I can isolate out my V_drift."},{"Start":"15:21.855 ","End":"15:26.415","Text":"I get that nqV_drift"},{"Start":"15:26.415 ","End":"15:35.855","Text":"is equal to the current divided by the surface area."},{"Start":"15:35.855 ","End":"15:39.315","Text":"If we\u0027re dealing over here with this cylinder,"},{"Start":"15:39.315 ","End":"15:44.265","Text":"I get that V_drift is equal to"},{"Start":"15:44.265 ","End":"15:53.055","Text":"the current divided by nq multiplied by the surface area over here."},{"Start":"15:53.055 ","End":"15:57.750","Text":"Here it\u0027s the surface area at the top over here, which is a circle."},{"Start":"15:57.750 ","End":"16:03.460","Text":"That\u0027s of course Pi multiplied by the radius squared."},{"Start":"16:03.920 ","End":"16:10.290","Text":"Now, I can take this equation for V_drift and plug it in over here."},{"Start":"16:10.290 ","End":"16:12.990","Text":"This is also equal to,"},{"Start":"16:12.990 ","End":"16:14.295","Text":"so my V_drift,"},{"Start":"16:14.295 ","End":"16:20.175","Text":"which is I divided by nqPir^2."},{"Start":"16:20.175 ","End":"16:22.095","Text":"Of course, this is just for a cylinder,"},{"Start":"16:22.095 ","End":"16:31.095","Text":"multiplied by the perpendicular component of my B field and then multiply it by 2R."},{"Start":"16:31.095 ","End":"16:38.385","Text":"I can of course cancel out one of these Is and let me just rewrite this."},{"Start":"16:38.385 ","End":"16:42.030","Text":"Then we get this equation over here,"},{"Start":"16:42.030 ","End":"16:47.055","Text":"which is of course the equation for the voltage between these two points."},{"Start":"16:47.055 ","End":"16:51.600","Text":"This is the voltage across the wire"},{"Start":"16:51.600 ","End":"16:56.925","Text":"when we\u0027re looking at this horizontal direction across."},{"Start":"16:56.925 ","End":"17:03.250","Text":"This is the voltage that a volt meter collected at these two points will measure."},{"Start":"17:04.370 ","End":"17:07.455","Text":"This again is our equation."},{"Start":"17:07.455 ","End":"17:10.575","Text":"Let\u0027s just go over each part of this equation."},{"Start":"17:10.575 ","End":"17:15.645","Text":"The voltage across the wire between these two points,"},{"Start":"17:15.645 ","End":"17:20.480","Text":"across from one side or one edge of the wire to the other is equal to"},{"Start":"17:20.480 ","End":"17:26.300","Text":"twice the current multiplied by the perpendicular component of the magnetic field."},{"Start":"17:26.300 ","End":"17:30.065","Text":"Of course, you can take the whole magnetic fields and multiply it by"},{"Start":"17:30.065 ","End":"17:35.390","Text":"sine of the angle between the magnetic field and the velocity."},{"Start":"17:35.390 ","End":"17:37.115","Text":"This is divided by n,"},{"Start":"17:37.115 ","End":"17:43.955","Text":"where n represents the charge density or the number of charges per unit volume."},{"Start":"17:43.955 ","End":"17:46.865","Text":"Q is the charge itself."},{"Start":"17:46.865 ","End":"17:52.020","Text":"Pi is Pi and R is the radius of the cylinder."},{"Start":"17:53.410 ","End":"17:59.120","Text":"Let\u0027s imagine that now we can see the sign,"},{"Start":"17:59.120 ","End":"18:02.045","Text":"the direction of the voltage."},{"Start":"18:02.045 ","End":"18:08.930","Text":"Let\u0027s say that this is the positive side of the voltage and this is the negative."},{"Start":"18:08.930 ","End":"18:12.530","Text":"That, we can see because the"},{"Start":"18:12.530 ","End":"18:16.085","Text":"positive is where the positive charges are and the negative is where"},{"Start":"18:16.085 ","End":"18:20.000","Text":"the negative charge is and then we can say that"},{"Start":"18:20.000 ","End":"18:24.200","Text":"this is the positive side of the voltmeter,"},{"Start":"18:24.200 ","End":"18:27.715","Text":"and this is the negative side of the voltmeter."},{"Start":"18:27.715 ","End":"18:29.870","Text":"If we align it like"},{"Start":"18:29.870 ","End":"18:33.560","Text":"so negative side with a negative side and the positive side or the positive side,"},{"Start":"18:33.560 ","End":"18:38.920","Text":"we\u0027re going to get a positive reading on the voltmeter."},{"Start":"18:38.920 ","End":"18:40.865","Text":"If I flip the wires around,"},{"Start":"18:40.865 ","End":"18:46.470","Text":"then my voltmeter will show a negative voltage."},{"Start":"18:46.910 ","End":"18:52.565","Text":"Something interesting that the direction of the voltage shows us,"},{"Start":"18:52.565 ","End":"18:54.560","Text":"so the sign of the voltage shows us,"},{"Start":"18:54.560 ","End":"18:59.210","Text":"is that it can show us and it proves that it\u0027s in fact"},{"Start":"18:59.210 ","End":"19:04.605","Text":"the negative charges which are moving and not the positive charges and"},{"Start":"19:04.605 ","End":"19:10.565","Text":"that is because if we had the exact same scenario over here where the current was"},{"Start":"19:10.565 ","End":"19:13.360","Text":"traveling in the same direction"},{"Start":"19:13.360 ","End":"19:17.525","Text":"upwards and we still had this magnetic field into the page."},{"Start":"19:17.525 ","End":"19:21.485","Text":"If it was in fact the positive charges, which we\u0027re moving,"},{"Start":"19:21.485 ","End":"19:25.460","Text":"then our voltmeter would show a negative voltage,"},{"Start":"19:25.460 ","End":"19:27.590","Text":"but because if we hook it up like this,"},{"Start":"19:27.590 ","End":"19:29.190","Text":"it shows a positive voltage."},{"Start":"19:29.190 ","End":"19:34.385","Text":"We can see that it was in fact the negative charges that are moving"},{"Start":"19:34.385 ","End":"19:40.040","Text":"and the positive charges are stationary so that is the whole effect."},{"Start":"19:40.040 ","End":"19:46.865","Text":"The whole effect is where due to the current interacting with the magnetic field,"},{"Start":"19:46.865 ","End":"19:49.910","Text":"we get that the negative charges are"},{"Start":"19:49.910 ","End":"19:54.815","Text":"the electrons get pushed to one side of the wire, therefore,"},{"Start":"19:54.815 ","End":"19:57.500","Text":"forming some charge separation,"},{"Start":"19:57.500 ","End":"20:03.935","Text":"meaning that we have an electric field between the sides of the wire."},{"Start":"20:03.935 ","End":"20:09.185","Text":"This electric field in turn produces an electric force and then"},{"Start":"20:09.185 ","End":"20:16.070","Text":"via getting into a steady-state where the magnetic force is equal to the electric force."},{"Start":"20:16.070 ","End":"20:21.650","Text":"We can calculate the voltage between the two sides of the wire,"},{"Start":"20:21.650 ","End":"20:24.710","Text":"which is equal to this and this equation,"},{"Start":"20:24.710 ","End":"20:27.740","Text":"we have course worked out by using the idea of"},{"Start":"20:27.740 ","End":"20:33.305","Text":"the drift velocity and calculating the drift velocity of the electrons."},{"Start":"20:33.305 ","End":"20:37.140","Text":"That\u0027s it. That is the end of the lesson."}],"ID":21420},{"Watched":false,"Name":"Exercise 1","Duration":"9m 51s","ChapterTopicVideoID":21535,"CourseChapterTopicPlaylistID":246912,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/21535.jpeg","UploadDate":"2020-04-21T16:17:05.3970000","DurationForVideoObject":"PT9M51S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.905","Text":"Hello. In this lesson,"},{"Start":"00:01.905 ","End":"00:04.440","Text":"we\u0027re going to be answering the following question."},{"Start":"00:04.440 ","End":"00:11.220","Text":"A current I flows along the length of a rectangular conductor, in the x-direction."},{"Start":"00:11.220 ","End":"00:17.740","Text":"The width of the conductor is W and that\u0027s parallel to the y direction."},{"Start":"00:17.740 ","End":"00:20.985","Text":"The depth of the conductor is t,"},{"Start":"00:20.985 ","End":"00:27.520","Text":"and that is parallel to the z-direction which is coming out of the page."},{"Start":"00:27.560 ","End":"00:32.240","Text":"We have a uniform magnetic field B,"},{"Start":"00:32.240 ","End":"00:36.470","Text":"which is present, and it too is in the z-direction."},{"Start":"00:36.470 ","End":"00:40.220","Text":"The magnetic field is also coming out of the pitch."},{"Start":"00:40.220 ","End":"00:45.230","Text":"Calculate the size and direction of the voltage at the ends of the conductor."},{"Start":"00:45.230 ","End":"00:49.590","Text":"Assume that the electron density is given."},{"Start":"00:49.850 ","End":"00:55.235","Text":"First of all, because we have a current and magnetic field,"},{"Start":"00:55.235 ","End":"00:58.570","Text":"we know that we\u0027re going to have some force."},{"Start":"00:58.570 ","End":"01:04.805","Text":"From Lorentz\u0027s law, we know that F is equal to q"},{"Start":"01:04.805 ","End":"01:12.190","Text":"multiplied by the velocity cross-product with the magnetic field."},{"Start":"01:12.190 ","End":"01:17.180","Text":"If we know that our current is traveling in the upwards direction,"},{"Start":"01:17.180 ","End":"01:25.700","Text":"then we know that that means that our electrons are traveling in the downwards direction."},{"Start":"01:25.700 ","End":"01:29.430","Text":"This is the direction of the velocity of the electrons,"},{"Start":"01:29.430 ","End":"01:32.545","Text":"in the opposite direction to the current."},{"Start":"01:32.545 ","End":"01:36.690","Text":"This is going to be equal to,"},{"Start":"01:36.690 ","End":"01:38.985","Text":"first of all, q,"},{"Start":"01:38.985 ","End":"01:41.690","Text":"we know that q, of course,"},{"Start":"01:41.690 ","End":"01:44.420","Text":"we\u0027re talking about the movement of the electrons,"},{"Start":"01:44.420 ","End":"01:47.645","Text":"so q is a negative charge because electrons are negative."},{"Start":"01:47.645 ","End":"01:56.490","Text":"It\u0027s negative the charge of an electron then multiplied by our V drift."},{"Start":"01:56.490 ","End":"02:00.560","Text":"We said that this is our V drift,"},{"Start":"02:00.560 ","End":"02:03.430","Text":"the drift velocity of the electrons."},{"Start":"02:03.430 ","End":"02:07.445","Text":"Add a d over here to make it clearer."},{"Start":"02:07.445 ","End":"02:12.275","Text":"Then cross-product with our magnetic field B,"},{"Start":"02:12.275 ","End":"02:16.700","Text":"which is uniform, and then the z-direction."},{"Start":"02:16.700 ","End":"02:23.975","Text":"What we can do is we can then rewrite this as this."},{"Start":"02:23.975 ","End":"02:28.145","Text":"We have V drift, V_d,"},{"Start":"02:28.145 ","End":"02:32.000","Text":"which is traveling in this downwards direction,"},{"Start":"02:32.000 ","End":"02:36.705","Text":"which we see is the negative x direction."},{"Start":"02:36.705 ","End":"02:42.920","Text":"Then cross-product with our B in the positive z direction."},{"Start":"02:42.920 ","End":"02:46.460","Text":"Here this minus and minus cancel out."},{"Start":"02:46.460 ","End":"02:52.670","Text":"Then what we have is the magnitude of the electrons charge multiplied"},{"Start":"02:52.670 ","End":"03:00.510","Text":"by the magnitude of v drift multiplied by B,"},{"Start":"03:00.510 ","End":"03:03.380","Text":"and then when we take x cross z,"},{"Start":"03:03.380 ","End":"03:08.760","Text":"what we get is in the negative y direction."},{"Start":"03:09.080 ","End":"03:13.280","Text":"This is our force, it\u0027s in the negative y direction."},{"Start":"03:13.280 ","End":"03:17.120","Text":"Another way that we could do this is through the right hand rule."},{"Start":"03:17.120 ","End":"03:20.375","Text":"Our thumb points in the direction of v,"},{"Start":"03:20.375 ","End":"03:24.170","Text":"so a thumb is pointing downwards."},{"Start":"03:24.170 ","End":"03:28.940","Text":"Then our pointing finger points in the direction of B."},{"Start":"03:28.940 ","End":"03:32.615","Text":"Our pointing finger is coming out of the page,"},{"Start":"03:32.615 ","End":"03:38.525","Text":"and then our middle finger will point in the direction of the force."},{"Start":"03:38.525 ","End":"03:42.935","Text":"Then what we\u0027ll get is that this is in the leftward direction."},{"Start":"03:42.935 ","End":"03:47.750","Text":"However, because we know that we\u0027re dealing with a negative charge,"},{"Start":"03:47.750 ","End":"03:50.995","Text":"so we flip the direction over,"},{"Start":"03:50.995 ","End":"03:55.745","Text":"and then what we get is that our middle finger is pointing here."},{"Start":"03:55.745 ","End":"03:58.925","Text":"This is our thumb in the V direction."},{"Start":"03:58.925 ","End":"04:02.940","Text":"This is in the B direction,"},{"Start":"04:03.080 ","End":"04:07.670","Text":"and this is our V cross B,"},{"Start":"04:07.670 ","End":"04:08.960","Text":"or in other words,"},{"Start":"04:08.960 ","End":"04:10.865","Text":"the direction of a force."},{"Start":"04:10.865 ","End":"04:15.030","Text":"That is the negative y-direction again."},{"Start":"04:16.310 ","End":"04:24.170","Text":"What we can see is that our force is being applied in the negative y-direction,"},{"Start":"04:24.170 ","End":"04:27.810","Text":"so in this rightwards direction."},{"Start":"04:27.810 ","End":"04:35.750","Text":"I\u0027m just going to erase this and also this over here to give us some more space."},{"Start":"04:35.750 ","End":"04:41.105","Text":"What we can see is that the force is of course acting on the particles which can move,"},{"Start":"04:41.105 ","End":"04:44.390","Text":"which is, in our case the electrons."},{"Start":"04:44.390 ","End":"04:50.000","Text":"Our force F is pushing the electrons in the rightwards direction."},{"Start":"04:50.000 ","End":"04:55.805","Text":"What we\u0027re going to get is some buildup of negative charge on this side,"},{"Start":"04:55.805 ","End":"04:57.905","Text":"and then on the opposite side,"},{"Start":"04:57.905 ","End":"05:01.535","Text":"we\u0027re going to have leftover positive charge,"},{"Start":"05:01.535 ","End":"05:04.085","Text":"or an overall net positive charge,"},{"Start":"05:04.085 ","End":"05:07.280","Text":"because the electrons which were present to balance out"},{"Start":"05:07.280 ","End":"05:11.670","Text":"this positive charge have moved over to this side."},{"Start":"05:13.130 ","End":"05:19.240","Text":"Now what we can see is that from the positive to the negative,"},{"Start":"05:19.240 ","End":"05:21.985","Text":"like so in this direction,"},{"Start":"05:21.985 ","End":"05:25.430","Text":"we have an electric field."},{"Start":"05:25.620 ","End":"05:28.420","Text":"Due to this electric field,"},{"Start":"05:28.420 ","End":"05:31.120","Text":"we have an electric force."},{"Start":"05:31.120 ","End":"05:32.965","Text":"Let\u0027s call this F_E,"},{"Start":"05:32.965 ","End":"05:38.780","Text":"which is equal to q multiplied by the electric field."},{"Start":"05:38.780 ","End":"05:44.790","Text":"As we know, q is the charge of the electrons."},{"Start":"05:44.790 ","End":"05:52.450","Text":"What we have is the negative charge of the electrons multiplied by the electric field."},{"Start":"05:52.450 ","End":"05:56.540","Text":"What we can see is that our electric force is acting"},{"Start":"05:56.540 ","End":"06:01.560","Text":"in the opposite direction to our electric field."},{"Start":"06:03.170 ","End":"06:06.755","Text":"Eventually, we saw in the previous lesson,"},{"Start":"06:06.755 ","End":"06:12.305","Text":"we\u0027ll get a balance between our electric force and our magnetic force."},{"Start":"06:12.305 ","End":"06:22.395","Text":"What we can say is that our F_E is equal to our F_B. Let\u0027s write that out."},{"Start":"06:22.395 ","End":"06:26.780","Text":"We have that our charge q,"},{"Start":"06:26.780 ","End":"06:32.590","Text":"this is q multiplied by E is equal to,"},{"Start":"06:32.590 ","End":"06:37.755","Text":"here our charge, again multiplied by"},{"Start":"06:37.755 ","End":"06:41.490","Text":"our V drift multiplied by"},{"Start":"06:41.490 ","End":"06:47.015","Text":"B. I\u0027m leaving out the direction because they\u0027re on the same plane."},{"Start":"06:47.015 ","End":"06:50.045","Text":"First of all, we can cancel out the q."},{"Start":"06:50.045 ","End":"06:56.090","Text":"Then what we get is that E is equal to V drift multiplied by B,"},{"Start":"06:56.090 ","End":"06:59.360","Text":"and what I want to know is I want to know my voltage,"},{"Start":"06:59.360 ","End":"07:01.950","Text":"so this is my voltage."},{"Start":"07:01.950 ","End":"07:03.695","Text":"My voltage of course,"},{"Start":"07:03.695 ","End":"07:07.205","Text":"is the potential difference between these two edges,"},{"Start":"07:07.205 ","End":"07:08.675","Text":"which as we know,"},{"Start":"07:08.675 ","End":"07:13.955","Text":"is simply equal to E multiplied by the distance between these two edges,"},{"Start":"07:13.955 ","End":"07:19.620","Text":"which is W. That is equal to E,"},{"Start":"07:19.620 ","End":"07:23.130","Text":"which is V drift multiplied by B,"},{"Start":"07:23.130 ","End":"07:25.305","Text":"multiplied by the width."},{"Start":"07:25.305 ","End":"07:28.415","Text":"Now all that\u0027s left to know is what is our V drift?"},{"Start":"07:28.415 ","End":"07:31.840","Text":"V drift is equal to what?"},{"Start":"07:31.840 ","End":"07:36.770","Text":"This we\u0027ve already seen comes from our J. J is equal"},{"Start":"07:36.770 ","End":"07:42.545","Text":"to our current divided by the cross-sectional area."},{"Start":"07:42.545 ","End":"07:53.070","Text":"Then we saw in another equation that J is equal to nq multiplied by V drift."},{"Start":"07:54.320 ","End":"08:01.295","Text":"What we can do is we can isolate from here that our V drift is equal to"},{"Start":"08:01.295 ","End":"08:08.675","Text":"the current divided by nq multiplied by A."},{"Start":"08:08.675 ","End":"08:11.495","Text":"Where A is of course the cross-sectional area,"},{"Start":"08:11.495 ","End":"08:13.385","Text":"which is this over here."},{"Start":"08:13.385 ","End":"08:16.160","Text":"Here we have a rectangular cross sectional area,"},{"Start":"08:16.160 ","End":"08:21.440","Text":"so this is equal to t multiplied by w. Now we can"},{"Start":"08:21.440 ","End":"08:26.810","Text":"plug this n over here so we get that the voltage is equal to V drift,"},{"Start":"08:26.810 ","End":"08:36.030","Text":"which is I divided by nqtw multiplied by B multiplied"},{"Start":"08:36.030 ","End":"08:45.550","Text":"by w. The w\u0027s cancel out and we\u0027re left with IB divided by nqt,"},{"Start":"08:45.550 ","End":"08:49.280","Text":"where q is of course the charge of the electron."},{"Start":"08:49.760 ","End":"08:55.235","Text":"This is the magnitude of our voltage or our potential difference."},{"Start":"08:55.235 ","End":"08:57.380","Text":"Then if we want to see the directions,"},{"Start":"08:57.380 ","End":"09:04.370","Text":"well here we see that this side has a higher potential difference."},{"Start":"09:04.370 ","End":"09:08.795","Text":"Then this side or this side has a higher potential than the side."},{"Start":"09:08.795 ","End":"09:16.630","Text":"That means that if I attached some volt meter over here, like so,"},{"Start":"09:16.630 ","End":"09:20.615","Text":"when this is the positive side of the voltmeter,"},{"Start":"09:20.615 ","End":"09:23.825","Text":"and this is the negative side of the voltmeter,"},{"Start":"09:23.825 ","End":"09:28.190","Text":"then my voltmeter over here will show me a positive value."},{"Start":"09:28.190 ","End":"09:30.860","Text":"If I switch the sides where the negative side of"},{"Start":"09:30.860 ","End":"09:34.520","Text":"the voltmeter is attached to the positive side over here,"},{"Start":"09:34.520 ","End":"09:36.835","Text":"or the side with the higher potential,"},{"Start":"09:36.835 ","End":"09:42.200","Text":"and the positive side of the volt meter is attached over here to the side"},{"Start":"09:42.200 ","End":"09:48.095","Text":"with a lower potential then my voltmeter will show me a negative value."},{"Start":"09:48.095 ","End":"09:51.810","Text":"That\u0027s it, that\u0027s the end of this lesson."}],"ID":22368}],"Thumbnail":null,"ID":246912}]

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