 In the first part of this video, we saw how an emf is induced in a moving rod and we derived the magnitude of emf that is blv by figuring out the magnitude of electric field strength and multiplying it with the length of the conducting rod. In this video, we will see how there is one component of Lorentz force that does positive work and there is one more component that does negative work. So the overall work done by the Lorentz force still comes out to be a zero. And in understanding that process also see how magnetic field acts as a link between us and the electrons transferring our power spent to the electrons and thus lighting up the lamp. Now it turns out what we saw was not the full picture. The electrons over here they have two velocities. So let me bring in the rod and let's say we are looking at the rod when there is a steady current that is flowing through the rod and the circuit. Let me just bring in the lamp and the wires as well and the lamp will be glowing. Now this rod is moving to the right with some velocity V. So at different instants in time the rod will it could kind of look like this. And you could imagine that the wires are flexible so they move along with the rod. Now the rod is moving to the right and there is a current that is flowing through the rod. So that means that the electrons must be moving with some drift velocity along the length of the rod. And because the current is in upward direction the electrons must be moving downwards. So if we try and see the position of the electron. So to begin with let's say the electron is over here and then after some time the electron has moved forward but it has also moved vertically down. And then after some more time the electron has moved further forward and further down. And so on and so forth the electron is right over here. So the resultant velocity of the electron is at a certain angle and this is how the resultant velocity looks like. If we try and look at it closely there is a velocity V to the right and that is the velocity with which the rod is moving. And there is a velocity U. So this velocity is there because there is a current flowing in the rod and there will be some drift velocity that the electrons will have. Now there will be a Lorentz force due to both of these velocities. And we know that a Lorentz force is perpendicular to its velocity. So the Lorentz force due to the velocity V will be acting vertically down. This will be the direction of that force. And the magnitude of this force would be QVB. Now Q over here is just the charge of an electron. So E into V the velocity the horizontal velocity into the magnetic field strength B. And there will be one more Lorentz force which will act horizontally on this electron. And that Lorentz force will be due to the vertical velocity because force always has to be perpendicular to velocity. So the direction of that force will be just like this. And the magnitude of the force would be E into U the velocity U into B. Velocity U because it is generated due to this velocity U. Now because there are two forces both of these forces will do some work. So let's try and figure out what that work would be. So for that we would need to know the displacement of this electron for some time interval. So let's take the time interval to be as dt and try to figure out the displacement both horizontal and the vertical displacement of the electron in this small time interval. We know that displacement is given by velocity into time. So that is V dot dt. So for horizontal displacement that will be the velocity V into the time interval dt. So if we write the horizontal displacement that would be V into the time interval dt. And the vertical displacement will be due to the velocity U. So that will be U into the time interval dt. Now work turn is given by the dot product between the force and the displacement. So the force F dot the displacement which is D. If we expand this, if we expand this, this would be F D cos theta. This would be F D cos theta. Theta being the angle between the force and the displacement vector. So now what is a work turn by the vertical force E V B. So that would be the magnitude of the force that is E into V into B. Multiply it by the displacement. Now this force and the displacement, the vertical displacement, they are in the same direction. So the angle between them would be 0 degrees and cos 0 is just 1. So we can simply write the magnitude of work turn by the vertical force would be the force magnitude into the displacement. And displacement, the vertical displacement is U dt. And now the work turn by the horizontal force. That would be the magnitude of the force that is E U B. And we multiply that with the horizontal displacement. That is V dt. But notice that over here, the displacement and the force, they are in opposite directions. So the angle between them is 180 degrees. So cos 180 would be minus 1. So we write, we write, we add a minus sign over here. And of course we write V dt. Now notice what has happened. These two values are in magnitude are exactly the same. This is a work turn by the vertical force. This is a work turn by the horizontal force. So net work turn comes out to be as 0. And that is how magnetic field is still doing zero network. And this vertical component of the force, if we try and think about the work turn by this vertical component along the entire length, that would be E into V into B dot L. That is the entire length of the rod. So let's just write L with this color, A V B into L. This is a work turn by the vertical component along the entire length. And if we are interested in EMF, we know that EMF is work turn per unit charge. So we divide this, we divide this with E and we get, and we get P V L. And that is the EMF induced. That is the emotional EMF that is induced. Now in the NCRT textbook, it is written that the work turn by the Lorentz force is B V L. But they have not clearly specified that only the vertical component is doing a positive work. But there is also a horizontal component that is doing equal amount of negative work. But one question still stands, that is who is then really lighting up this lamp? How is this lamp getting any sort of power if the magnetic field is not doing any work? Now for that we need to look at this left component closely. And the left component, what this one will do, this one will slow down the wire. But the wire is moving at a constant velocity. So that means there must be a force to the right. And let's say I am applying that force. I am applying this force on the rod to the right to balance the force on the left. So that the rod continues to move with a constant velocity. And my force must exactly be equal to the magnitude of the horizontal component. And now if I am applying some force, it means that I must be spending some power. So to think about that, to think about the power that I am exerting on the rod, we will take the force that I am exerting and multiply that with the horizontal velocity V. Because that is the velocity in the direction of the force. So this dot V. Now this will become the force that is exerted by me. It's exactly equal to EUB. So this will be E into U to B. And that is multiplied with V. Now if we look at the power that is being delivered to the electrons by the vertical component. So that would be, let me just write that over here. That would be F vertical into U. Because that is the velocity along the force. So this will be dot U. So this will come out to be as E. The magnitude of the vertical force which is EVB. So this will be EVB into U. Now notice what has happened. These two powers are exactly the same. That means that the vertical component of the magnetic force is taking my macroscopic mechanical power and turning it into a microscopic power for the electrons. And then the electrons leave the lower end, they flow through the wire, they pass the lamp and then they deliver that part to the lamp and the lamp close. Now we can see that the bulb is also glowing. So there must be some power dissipated in the lamp in the bulb. Let's see if the power that I am spending is the same power that is being delivered to the lamp. So we know that the power that is being delivered to the lamp can be given by EMF multiplied by the current. There is an EMF E induced and the current I that is flowing in the circuit. And we know that the EMF is, that's the emotional EMF that is BVL. And we know that current is how fast the charge is flowing. So charge per unit time and that will be E divided by T. Electrons are flowing over here so the charge is just the charge of an electron. And we know that the electrons are moving with a drift velocity of U. So we can write this as U equals, U equals the entire length of this conductor that is L, that is L divided by T divided by time. Now time can be written as L divided by U. Now if you place these values of time in this equation right here. So when we do that we will get this is equal to BVL multiplied by E divided by L by U. And now we can see that L just gets cancelled right away. And what you get finally is EVPU and that is the same power that I am spending on moving the rod. That is the same power that the magnetic field, the vertical component of the Lorentz force is delivering to the electrons. And then the electrons are moving through the wire and delivering the same power to the lamp. So that is how magnetic field is acting as a link. It is transferring the power that we spend, that I spend to the lamp. It is just like, it is just like as if let's say, let's say we had, let's say we had this wedge and there are these bunch of, this does not look like a wedge. Let's say if we had this wedge and there are these bunch of balls and there is a person who is pushing, who is pushing this wedge to the right. And as this person pushes the balls will get displaced and lifted. And in this situation all the work is done by this person. All the power is spent by this person. The wedge is not really doing any work. So the wedge over here acts like the magnetic field. It is just transferring the power spent by this person to the balls which act as electrons. Similar to what magnetic field is doing over here, overall the magnetic force is not doing any work. Only the vertical component is transferring our power to the electrons.