 So, write down the next topic, motionless EMF, so assume that you have a rod, rod of length L, rod of length L moving with velocity V in a magnetic field which is into the ball, this is the magnetic field. Can this rod develop an EMF or potential difference? Yes or no? Is there any change in flux over here? Is there any change in flux? Yes or no? No, there is no loop, there is no question of flux, but still it has an EMF. Will it have a current? Current needs a loop, there will not be any current, but there will be EMF. EMF do not need a loop, it is like a battery moving. You need to find out how much is the EMF as in what is the potential difference between this end, let us call it as 1 and this as 2 between 1 and 2 what is the potential difference. But before you do that, first tell me, will there be charge accumulation at one of the end? Will the electron move towards 1 or 2? Where it will move? Electrons inside the metals, they are free to move. So, if they experience a force upwards, they will move upwards. If they experience a force downward, they will move upwards. Negative charge? How? V cross B, on a positive charge downward force, on negative it will be upward. So upward there will be accumulation of electrons. Electrons will accumulate there and of course lack of electrons here, so there will be a positive charge. So it is like a charge separation, so there will be potential difference between these two points. The electrons will keep on accumulating here till when? Will they keep on accumulating till whatever time this is moving or it will stop accumulating after some time? Can it accumulate infinite amount of charge just like this? Why? Magnetic force is in which direction for the electron? That way? This is the magnetic force direction? Is there any other force on the electron? Any other force on the electron? These positive charges won't it pull it down? There will be electric force, isn't it? When magnetic force in magnitude becomes equal to electric force, charges will stop accumulating. So when magnetic force become equal to electric force, okay? Negative force is how much? Let's say E is the charge on the electron, E, velocity is perpendicular to B, V cross B is just V into BV. And suppose electric field is E, electric field is E upward direction, the force in downward direction for the negative charge will be E into, so now, so electric field will become what? Equals to B into B, fine? And EMF is what? EMF is integral of E dot T L negative of that. You remember this? So when you do that, electric field is downward direction, you are also moving down, so dot product becomes 1. So this just becomes V V into L and minus sign, minus sign because you are finding this potential minus that potential, okay? But if you just consider magnitude of EMF, it will be V L V or B V L whatever, fine? So if a rod moves in a magnetic field, fine? If a rod moves in a magnetic field, its velocity is perpendicular to magnetic field and perpendicular to length. So basically B, V and L, all 3 are mutually perpendicular to each other. If that is the case, then EMF is BVL. And if that is not the case, then suppose the rod is like this and velocity is like that, this is theta, length is L, magnetic field is B. Then what is the EMF now between these 2 ends? Then take the velocity that is perpendicular component V sin theta, fine? So EMF is B L V sin theta, simple? Fine? You have to take velocity that is perpendicular to the length. Can I doubt on this? Let us analyze a situation. Suppose you have rails, resistance is R over here, okay? This is a rail, draw this. This length is L, magnetic field is into the board. Magnetic field is B. You have suppose a rod like this, okay? Rod is moving with velocity V, okay? Rod is moving with velocity V. V is constant, fine? You have to find out what is the EMF current and force required to move this rod with constant velocity. These are conductor only. These 2 are conductor connected with a resistance like this, like a U shaped one. This rod is sliding on it. These are required to make sure the loop is completed all the time. Tell me, how do you find EMF? Will EMF be B L V? It will be L V only. B L V only, right? But this is a loop also. So you will get ideas that why can't we find using flux, fine? Let us try to see whether that if we find using flux will that come the same, okay? So let us assume this to be X. This X keeps on changing, right? As it slides forward. So area is what? Area of the loop is X into L. So flux throughout is B into X into L, right? So D5 area is what? B into L is constant. So B L, DX by DT. What is velocity? So you get B L V only. If you find using flux, same thing you will get. So EMF is B L V. How much is the current? B L V by R. Now in order for this to move with constant velocity, what should happen? Net force on this should be 0, expression should be 0, right? Now if it is trying to move, is there magnetic force on this rod amount? With direction. When I ask such question, you don't have to think twice before saying that it will be on the left-hand side. Why? Because of Lenz's law, it will oppose the cause, fine? Keep it very simple. So if it is moving like this, force has to be backward direction. The cause is change in flux. That is increasing. It should oppose that. So magnetic force will be backside. How much is this value? Magnetic force. FB is what? How do you find? There is a wire which has a current in a magnetically. So I L B, I L into B, everything is perpendicular to each other, I L and B. So this will be what? Simply V square L square V by R. This is force, okay? Now this is a magnetic force. What it will do? It will try to decelerate the velocity. Velocity will decrease. You have to make sure the velocity doesn't decrease. What you have to do? You have to apply a pull here, which is equal to this. Then only the net force becomes 0. So this is your force. So how much force it applied? Equal in magnitude, but in opposite direction. How much power you have to supply? Power is what? Power supplied by a force is force into velocity. This is force into velocity becomes what? B square L square V square by R. This is the power which you are supplying. Or this is the rate of work which you are doing, okay? Where this goes? Just check what is I square R. How much is this? Put the value of I here. You will get the same thing. V square V square V square by R. So whatever power you supply is lost as a heat in the resistance. So you can see the energy conservation is valid in this scenario. Any doubt? Anything? No? Do one numerical on emotional elements. Suppose you have a rod. This rod can rotate about this axis, fine? This rod is rotating with angular velocity omega. The length of the rod is L, fine? The magnetic field is coming out everywhere. Magnetic field is uniform B, fine? You need to find potential difference between 0.1 and 2. This is a rod moving with velocity. So it will introduce a new emotional EMF or not, right? But is the entire rod moving in the same velocity? No. Every point on the rod moves with a different velocity. So assume a DR width. I can assume this DR is moving with the same velocity. Suppose this distance is R. What is the velocity? Omega into R. I can say this DR is moving with omega into R velocity. So the EMF developed across DR is what? Let's say that is DE. It will be what? DR length moving with omega R velocity. So small EMF across DR will be DLV, right? L is what? DR. So it will be what? B, L is DR into V which is omega R, fine? This is EMF across DR. So in order to find a total EMF, I should integrate 0 to E and this goes from 0 to R. Total length. So EMF will come out to be what? Half B omega R square. You have to remember this. Let it like the usual that comes out. Find will do the last question and then we will wait for the day. Should I dictate the question now? There is this rod. Let's say this is PQ. This PQ starts from here where X equal to 0. Magnetic field is from X equal to 0 till X equal to B only. From X equal to B to X equal to 2B there is no magnetic field. So this rod moves with constant velocity V. It starts from X equal to 0, goes till X equal to 2B, then comes back and again reaches X equal to 0 with the same magnitude of speed. So it goes and comes back. It goes with the velocity V, comes back and magnitude of velocity. Now in this entire process you have to observe these things and plot the graph. This is flux and not time X. Flux with X. This is EMF with X. This is force required to move this with constant velocity with X and this is power delivered with X. Of course it will cross X equal to A. So you can draw a line like this. This line represents X equal to A. Then it will go to X equal to 2B. So this line represents X equal to 2B. After this what will happen? It will come back. So what I will do is that I will draw X equal to A further ahead. So this is X equal to A. Then what will happen? Then again X equal to 0. This line is X equal to 0. So this is X equal to A, X equal to 2A, X equal to B, X equal to B, X equal to 2B, again X equal to B and then X equal to 0. Plot these graphs. What will happen to flux? Flux will increase or not? Increase. X equal to 0. The loops area is increasing. Increase or decrease. Right? Right. Once this rod goes beyond X equal to B, will the flux increase? No. Number of field lines will not increase. It will still remain that only because manually field ceases to exist after X equal to B. So flux will become constant. So flux will rise like this. It will go like that and then it becomes constant till it comes back to X equal to B. And then what will happen? Decrease. Flux will decrease. Right? What about EMF? EMF is what? Rate of change of flux. Rate of change of flux is positive right now. So EMF will be negative using sine convention and that is constant till flux is changing with constant rate. Getting it? Other than that what will happen? Zero. Zero? The flux does not change. So EMF will be zero. It will like this and then the maximum value or this value will be equal to 1 B and Z. The maximum flux will be equal to what? This value equal to B into, this is suppose L. VL into small b. Isn't it? Now what about force? Force is B square L square V by R. When the flux is increasing you need to apply force and when flux is decreasing you need to push it. So force is changing the direction right? So it is basically in the direction of the velocity. Velocity direction is same till X equal to B. After X equal to B anyway current is not there so there is no force. You do not need to apply any force after that. So basically you need to apply force like that. Something like this and no force goes like this then you have to push it down. Push it back. This is force. You need to apply force like this or not? Yes sir. So I am plotting this force what you are applying. Negative as in when it comes back here then magnetic force will be in which direction? Magnetic force will be this direction and then you have to push like this. So your direction of force is changing. What about power? Power is forced into velocity. So power becomes this square. So now it is velocity square it will not change the sign. Power remains positive throughout. Even negative will become positive. This is how you have to do this part of the question. So still few things are left in this chapter. There is inductance that is left and you have to practice many many positions at all. Minimum 50 questions if you are serious about minutes. Minimum. That is like just eligible to write changes. Fine so come up with your doubts. I have to tell you the fees man. There is a philosophy kids. Magnetism also 50 and results of 50. You have to really slog it out man. It is not a joke. 150 chapters in 5 months and let me tell you this is the better chance for you like January 1. Because when you write in April those guys who are studying in Kota they will get 3 times more time than you. While you are busy with pre-bords and boards preparation with that they will get more time. They will get ahead from you between Jan and April. So if you do well in January which is same for everyone. The difficulty you are facing they are also facing. So January gives you a better chance to get ahead of them. Otherwise by the time April comes they will probably have more time so they can get ahead. So make sure you treat January as your final attempt for minutes.