 So, we are gradually coming towards the end of our lecture sessions on electricity magnetism because with this I am we will have one more only tomorrow. But before I talk about the electromagnetic induction see there will be two things that I will be actually talking about now. One is that you see so far we have been talking about electrostatics in fact substantial part of today's this lecture will also be on electrostatics and magnetostatics combined the we will then talk about electromagnetic induction and displacement current and finally going over to the electromagnetic waves and an explanation of what pointing vector and the power transfer etcetera is all about. However, though we finish the electromagnetism we will be using electromagnetism in our lectures on optics which will be four of them and in fact optics lectures will be primarily your physical optics as it is called but it will be based on electromagnetic theory. And there we will be talking about phenomena such as diffraction interference polarization and of course some fiber optic phenomena as based on electromagnetic theory. So therefore, we are not quite done with the electromagnetic theory but we will come back but there is a rather important point which I have not touched upon and this is we have been talking about collection of charges either free charges or in a dielectric which we have been calling as the bound charge. So what we want to do is to find out what is the energy of charge distribution. Now once again I will be doing it for electrostatics it turns out that the you know once you have done this the method of doing the magnetostatic charge distribution that is the magnetic case the current distribution etcetera is that turn out to be extremely identical almost in terms of mathematical content. So therefore, this is very clearly understandable that what happens when you have a collection of charge. So let us suppose that we want to do the following. We have a collection of charge which I will call as Q1, Q2 etcetera. Let us say that this is a set Qi i equal to 1 to n, n number of charges and suppose I have interested in putting a charge Qi at the location let us say Ri with respect to some chosen origin. Now what is the energy of this charge distribution is my question. So let us see how does one do it and you will find there are some very interesting points that come up here. The way I would try to build up this charge distribution is the following. Let us assume that initially all the charges are at infinity. Remember we have said that when things are at infinity we define the 0 of the potential at infinity for convenience and we assume that the potential energy of such a configuration when all the charges are infinity is 0. So this is because I have to remember energy has to be measured from somewhere. We are talking about potential energy it requires a measurement origin and our origin is that the energy goes to 0 when the charges are separated and each one of the charges at infinity. How do I build up this charge distribution? The way I build up the charge distribution is this. Let us first pick up the charge Q1 and bring it from infinity to its point bring it from infinity to the point R1 vector that is its final position. Now notice that if I do that I do not have to do any work why I do not have to do any work because this charge was at infinity and there is since there is no other charge around the this charge when you bring it from infinity to the point R1 or position vector R1 there is no force it experiences because there is no other field there. So therefore I do not do any work so let me write down no work. Now let us look at the following once I have established Q1 as at its position let me bring Q2 and try to put it to the position R2. So let us see that what I am trying to do is this supposing this my origin I have already brought a charge at the position R1 and this situation I did not need any work because there is no electric field but now when I try to bring a second charge to a position let us say R2 this is my Q2 this situation becomes different the situation is different because the initially when I brought the first charge there was no field but when I am bringing the second charge this charge Q1 already has established a field the so therefore this charge Q2 when it is being brought in it is being brought in in the electric field which has already been established by Q1. So you notice that the you also know that these are conservative forces so if I want to bring it here I can bring it along the line joining Q1 and then of course go transfers for which there is no work done but let us look at what is the potential at the point R2 let us call it what is the potential at the point R2 due to Q1 where it was being established. So this I know that is given by 1 over 4 pi epsilon 0 Q1 I mean potential is Q1 divided by R1 minus or R2 minus R1 it does not matter because it is a magnitude now this is the potential now so therefore when you are bringing in the charge Q2 from infinity where its potential energy was 0 to the point where the potential energies are 2. If you remember potential is the capability potential tells me that that is the amount of energy potential energy a unit charge would have if you are to put it there. So potential I repeat is a function of space is a point function in space and when you brought in and put Q2 I have to do some work so let me put it this way the work is let me call it W2 so W1 is 0 of course. So my W2 is simply Q2 times phi of R2 so that is simply equal to 1 over 4 pi epsilon 0 Q1 which is already there and a Q2 divided by R2 minus R1 so far so good. I have established the two charges first charge did not need any work second charge I needed 1 over 4 pi epsilon 0 Q1 Q2 divided by R2 minus R1. Now that I have got two charges in place so I have got let me draw that picture again my origin here I have a at the position R1 I have a charge Q1 at the position R2 I have a charge Q2 this is already there. Now I want to bring a third charge I want to bring a third charge and let us say I want to put it somewhere here which is my R3 how do I do it. So notice that the problem has become a little more complicated now because the potential at this point is now due to two charges Q1 and Q2 which were already there. So this potential is Q1 by R3 minus R1 modulus plus Q2 by R3 minus R2 modulus remember the power of potential that you have been talking about. If we had been talking about field I could still do this job in principle I could do the calculate the work done by taking the field but then that is not an easy job. So what I do is I will say that look what is the potential my potential due to two charges phi at the position R3 is 1 over 4 pi epsilon 0 due to Q1 it is R3 minus R1 due to Q2 it is R3 minus R2. So if I now put a charge Q3 here the work that will need to be done will be phi times Q3. So this is simply 1 over 4 pi epsilon 0 Q3 times Q1 by R3 minus R1 plus Q2 by R3 minus R2. Now I could now go to fourth charge fifth charge etc but obviously that is not going to give us any new thing. So what we do is we say that look when I am trying to bring let us say jth charge. Now when I am bringing a jth charge bring jth charge to the location rj. Now it means all those charges which are already there accepting the jth charge the they have established already certain amount of potential at the position rj and how much is that? So that potential at the position of rj is given by this is given by sum over let us say i sum over i Qi divided by rj minus ri. I have to realize that since my charges are not at the jth position yet so my summation over i should not be including j. So this is my potential at that point. So therefore the work that is to be done in bringing the jth charge this is the work done in bringing the jth charge. Now that will be equal to 1 over 4 pi epsilon 0 Qj sum over i not equal to j Qi by ri minus rj. So when I now want to find out what is my expression for when I bring in a complete assembly what is my total charge a total work done. So the total work done is nothing but you now add up the work done in bringing each one of the charges the j equal to 1, j equal to 2, j equal to 3 like this. So if you want to now write it down this becomes 1 over 4 pi epsilon 0 sum over i and sum over j because sum over j was already there for the jth charge but I am summing over all the charges. So I get sum over i sum over j but then i cannot be equal to j. So I put i not equal to j then I write Qi Qj divided by rj minus ri. Now this is the problem now you notice that if you look at it like this then there is a double counting why is there a double counting because if you take i is equal to 1 j is equal to 2 I have got Q1 Q2 but if you take i is equal to 2 j is equal to 1 I also have a Q1 which means if you write it like a sum over i and j even if i is not equal to j you have counted each term twice. So therefore in order to take care of that you divide it by a factor of 2. So this is my expression and if you look at what is the potential like then you will see immediately this is nothing but half sum over i Qi and the potential at the ith position. So this is actually the work done. Now let us ask a different question that fine this is the way I build up the discrete charge distribution the charges are being brought one to one one by one and I put them in their position I have seen how this is done everywhere everything is nice but how do I extend it to a continuous case. So let me write down this expression that I wrote down first that is the work done that is the potential energy that is stored is half sum over i Qi and the potential at that point and I had given you the expression for the potential at that point. Now in order to now extend it to the continuous charge distribution what I do is I look at a small volume and element inside the space wherever I am taking this thing and I say let the charge density here supposing this is position r the charge density here is rho of r and I have taken a volume d tau so my total charge is rho r d tau. So this tells me immediately that I can extend this to a continuous case by saying this is nothing but half the charge is nothing but this quantity so I get rho of r and of course phi at the position r and then d tau. So this is the expression for the energy for a continuous charge distribution which is a straight forward generalize and from this sum to an integral. Now let us do some algebra the algebra is this that I know del dot of E is equal to rho over epsilon 0. Alternatively this tells me that this quantity is half of epsilon 0 integral del dot of E times phi of r d tau but you remember E is nothing but minus gradient of the potential. So therefore this is minus epsilon 0 by 2 so if E is minus grad phi del dot of E is minus del square phi. So I get it as phi which I bring it to the left hand side of this you should be careful in doing such things and I will tell you why because if you did not you put it on that side somebody may assume you are doing del square of phi phi but that is not what I want I want phi multiplied with the del square of r. So this is what I got d tau. Now I need to now simplify this expression so let us do that. So energy of a continuous charge distribution. So we said we found that this is given by E energy is w we write minus epsilon 0 by 2 integral I am dropping this phi of r etc phi del square phi d tau okay. Now I do a little bit of arithmetic I know the divergence of a scalar times a vector del phi is a vector and this is scalar is this is a constant and then so this is just change rule phi del square phi plus del phi dot del phi del phi dot del phi is nothing but del phi absolute square. So this is phi del square phi plus del phi absolute square. So if I want to write down this term as this minus that so what I get is w equal to minus epsilon 0 by 2 integral del dot phi del phi minus so minus minus plus epsilon 0 by 2 integral del square del phi absolute square d tau fine. Now you immediately realize what am I trying to do I am I have got it as a divergence this is d tau I have got a divergence of something and of course our friend which is the divergence theorem will tell me I can rewrite this term as epsilon 0 by 2 integral over the surface phi del phi let us put them together dotted with ds because this is the surface integral now plus epsilon 0 by 2 integral del phi absolute square d tau. Now this is a very interesting situation question now is so one term is over the surface another term is over the volume. Now supposing this is my charge distribution what is the volume now you would say well the volume is this well I will say that look you are right you could take the volume to be this but does it matter if I extend the volume to this because I can do that because that this volume encloses a charge distribution. So this is contributing to that but if I increase that volume there is nothing here no charge here no charge. So this is by extending that volume all over you are not really done anything by saying let me add a lot of 0 to this I can do that see what I am trying to tell you is this that please try to understand this supposing I have given you that integrate a function supposing I have given you that integrate some function fx dx from let us say a to b supposing the function is like this now if I decide that I will not do that I will define a new function which says let us say capital Fx is equal to f of x if x lies between a and b sorry and is equal to 0 outside now if I do that then you notice immediately that since the function is 0 outside a to b this integral is same as the integral from minus infinity to plus infinity of capital Fx dx. So I have not done anything because I have I have simply added a lot of 0s here and this is precisely what you are doing here that if you go back to the original expression for where did it come from ok remember this is nothing but a potential and we had a charge density and things like that. So therefore the even if I make the volume much bigger as big as I like the contribution to that integral will only come from place the volume region where you are you have a charge density this is like adding a lot of 0s now if I do that it tells me that I do not have to worry about this s and this v being this I can instead say that this is over all space and this is over the corresponding surface which means if I take this integration to infinity the contribution to the volume of course will come only from here. So my original expression on the left hand side will still remain the way it should but now this is a surface integral. So I have to worry about what are the values of the potential on the outside and I know that infinity at infinity the potential value is 0. So because of that now remember what is the difference between the two this being a volume integral I cannot make that statement because the volume integral is contributed not only from infinity to this point but also from here here here here here but a surface integral only is the outside surface. So as a result the this term will become 0 by extending the surface to infinity. So what will I be left with I will be left with only this term I will be left with only this term. So this tells me that my energy is simply much simpler result del phi absolute square d tau but this time very important this integration is not over the restricted volume but all space. Then you realize E is minus del phi you can write this as epsilon 0 by 2 integral absolute E square d tau and this tells me that I have an energy density of the electrostatic field that is simply equal to epsilon 0 by 2 E square this is energy density. Now supposing I had the corresponding magnetic situation magneto statics the role this I can do an identical job I will not do it once you have understood this you can look up any textbook and you will find the corresponding expression for it. The U magnetic or UB if you like happens to be given by B 1 over 2 mu sorry this is B square by 2 mu. So that the magnetic energy Wm let us call this electrical energy is 1 over 2 mu integral B square d tau fine. So what is interesting about it one of the things you immediately notices that this is a positive quantity let us let us skip this aside for some time we are saying We is a positive quantity. So the question that would you will immediately say is that correct is the energy of the charge distribution positive because if I had one positive charge and one negative charge only I know the energy is negative. So obviously something is wrong actually there is nothing wrong there this is correct for a continuous charge distribution and the reason why in discrete charge distribution we have a problem this could be negative or positive. So there is a problem and the problem is the following see we said when we are talking about discrete charge distribution we said let us bring a charge from infinity and put it somewhere. Now you could do that accepting that you have not taken into account that in order to create a charge which has some finite dimension you need some work and that thing we assumed that the charges are God given. So I have taken them to infinity and I have brought them but I did not create those charges but in the continuous case I have my every small volume contributes to an energy no matter how small that volume is so it contributes to an energy. Now suppose we assume suppose my discrete charge which I called as Q let me take a model in which my discrete charge is nothing but a sphere of uniform density. Then you look at that let me apply this formula for how much is the energy contained in that small volume let us say spherical volume the energy then would be given by epsilon 0 by 2 but this time I know what is the electric field so I will say 1 over 4 pi epsilon 0 0 to infinity Q square over r square d tau and so we have said it is constant over this and you notice this sorry 0 to r. Now this is a very trivial integration to do 1 over 4 pi epsilon 0 0 to r yes Q square by r square you realize that this integrand has nothing to do with angle so d tau will give me 4 pi times r square times near. So this thing you can integrate it no problem because the problem is so I have got this thing that this term itself right the you are going to integrate Q by r sorry I did Q square by r square d tau and if I take this the energy that this contributes to the this thing it sort of blows up there is some minor arithmetic problem here I will correct it next time instead of wasting time because I am having a bit of time problem so the basic point what we wanted to say is there is a infinity problem that comes up in the lower this thing yeah the problem was very simple I did wrong thing I should have set Q square by r because it was my potential but in any case I will correct it tomorrow morning there is some let us not waste time on it alright so the basic point is that there is a contribution to this is called a self energy contribution and this self energy contribution is infinity the point is this that this type of things for an assembly of or a charge having an infinite self energy is very common I mean this is true of any charge assembly that you can think of and that is because the potential at the position of the charge is infinite because the potential goes as 1 over r and so therefore when r goes to 0 that is at the position of the charge the energy is of course infinite so what we should do if I am building it up entirely from such things and taking it over I must subtract out the self energy contribution which is of course impossible so this is the issue that we wanted to talk about now the minor point of addition is this that what happens if I have a dielectric now what happened if I have a dielectric is this that we not not only need to worry about the real charges I have to establish also the process of polarization that I have to establish the polarization state now in order to do that supposing I have established already a situation where there is a charge density rho okay and suppose I am bringing more true charges and I want to increase this charge density by delta rho my question then is this my initial energy because we have seen is given by integral rho d phi d cube x now the thing is this I am bringing delta phi and delta rho extra and supposing phi is the potential in the presence of whatever we had and when we add the true charges I have the following I have delta of d equal to rho so therefore my delta rho which is what I am trying to increase is delta d delta d so my extra work to be done I am bringing in true charges that is nothing but phi delta rho d cube x or d tau now notice I did not put in a factor of half there in the other case I needed a factor of half because I was bringing charge number 1 charge number 2 but then 1 2 2 1 I had but here there is no such thing I already have built up a rho and I am simply adding one little bit to it so there is no factor so my delta w this is the extra work that needs to be done is phi delta rho d tau and now you substitute whatever we said just now phi del dot of delta d okay and again do the same thing as we have been doing all the time whenever you have a term like this write it as a divergence of this times this subtract out another term convert one of them into a surface integral another into a volume integral do all that so then you take the integration from 0 charge to the full and if you do that then what you get the energy to be given by half e dot d d cube x and because d already has an epsilon factor with respect to e there is no epsilon 0 by 0 there the corresponding expression in case of a magnetic field is simply b dot h divided by 2 where h is something which I define in the morning so this is a very important concept that is how do we build up charge distribution having done this let me at least this afternoon try to just give you an introduction to the time dependent field and of course please I understand that this charge distribution energy is a little tricky thing you have any question please send it in chat because this is a you know 15 minutes less session so we will not take up any question in this session so I am now going to be talking about time varying phenomena we have already defined the magnetic flux see the picture I am giving you here is interesting I have given you what can be looked upon as a fisherman's net so fisherman's net is like this that there is a rim here and over that there is a net now the flux over the curved surface is b dot nds where b is the magnetic field on this and of course n as we have been talking about is an outward not now instead of this I could look at this surface also and that will give me same thing but now you notice this is the rim which I have taken circular so its direction is fixed but if I treat it as a closed surface that is this is this cap is closed and then there is a net then the Gauss's law tells me that the net flux out of a closed surface is equal to 0 which tells me that the flux through this surface is the same as the flux through that surface now let us ask a question which very often you might have wondered that look how do I make a current going around in a circuit I will tell you what this problem is we have said electrostatic field is a conservative field and you know in order to have a current I need a closed circuit so integral of e dot dl over any closed surface is 0 okay now if the integral of e dot dl is equal to 0 see the problem that I have is the following that let us assume something like a ohm's law is valid you do not have to but let us suppose something like a ohm's law is valid so but what is this remember I said that the electric field is the force per unit charge a unit test charge whatever it experiences so what is this integral this integral then represents force f dot dl that amount of work that is done by taking a charge around the circuit okay but then that work must be 0 but notice one thing if I assume that my unit charge is moving in a circuit in a in a current carrying conductor and let us suppose that I assume for that ohm's law is valid so it tells me that energy joule heat must be continuously generated all right and so therefore but this tells me e dot dl is equal to 0 now so such a situation cannot explain the presence of a current establish a current because a current must you know generate heat and there is nothing there it says you come down one round the work done is 0 so energy is given now that is not true for a current now what do you do so you notice the following that what we do is this that we say that in any circuit in order that there is a current there must be a source I have called it a battery it is the battery which provides a motive force so this language I have used intentionally because electro motive force is not really a force as I have said in physics history of physics there have been many cases of a misnomer and this is one of those emf is not a force now but basically what the battery does if you you know the way a battery works it converts the chemical energy that is stored in that thing to electrical energy now since an energy is being changed from one form to another right the corresponding field is a non-conservative field now so in the battery I have a non-conservative field let me call it e prime do not worry about this line again so basically what I am now trying to say is this my electric field has two parts outside the battery when it is in external circuit I have an electric field which is conservative but the moment that charge enters a battery I know that work has to be done right I know the conversion of chemical energy to electrical energy is taking place so therefore this e prime that I have talked about it is a non-conservative field so outside I have got a conservative field inside I have got a non-conservative field but wait so why have I written it like this the reason I have written it like this is very interesting say I said that the I have a non-conservative field outside the battery and that non-conservative field inside the battery and the non-conservative field is 0 outside the battery so what I can do is to add up okay the non-conservative part outside which is 0 add it to the non-conservative part inside which is not 0 and that still gives me integral of e prime dot dn e prime is the non-conservative so non-conservative field inside the battery is non-zero outside the battery is 0 I can add any number of zeros and get a complete circuit there now this is another interesting thing that I can do see once you have been convinced that this is your electromotive force whatever e dot dl represents it's called electromotive force and so you will say but we don't write electromotive force as e prime dot dl where e prime is only the non-conservative part you are absolutely right we always write it only as e dot dl but then my e dot dl is contribution from both conservative and non-conservative part but you remember this integral e dot dl which is the conservative field is 0 so to this integral I can add that integral which is the conservative part and get the total electric field dot dl as my definition of the electromotive force in other words I had one part which is within the battery e prime dot dl I added first another non-conservative part which is 0 then I added the full conservative part over the circuit which is 0 so I added 0 plus 0 plus e prime dot dl is the complete e dot dl so that is my definition of the this thing all right so let us continue with this so I said the total electric field is a sum of two parts there is a conservative part and a non-conservative part that the current density that I have got is a sum of both the parts the line integral of the conservative field outside the battery is exactly cancelled by the line integral of the conservative part within the battery that of course we know so I have already explained why this I can take this as e dot dl now comes Faraday's famous experiment what Faraday found is that if you have a circuit and I mean we have talked about a battery so battery is understandable it is converting a chemical energy to an electrical energy and it is driving a current but what Faraday found is it is possible to generate a current in a closed circuit even without having a battery and that can be done by changing the magnetic flux through a circuit as you know flux has two parts the flux consists of magnetic field as well as the area so the way I can change the flux is either to change the magnetic field supposing I take a circuit I bring a magnet towards it and as the magnet comes closer to that circuit the magnetic field strength is increasing so therefore the magnetic flux is increasing I withdraw it the magnetic flux is decreasing the other possibility is to make that area change how can I change the area there are many ways I will be talking about it before in next few minutes but what Faraday found is this circuit does not care how did you change the flux you change the flux whether by changing actually the strength of the magnetic field changing the direction of the magnetic field or changing the area or by initiating a relative movement between the circuit and the magnetic field the circuit could not care less what circuit does is to say if you are trying to change my flux I am going to oppose it I am going to oppose it in other words this circuit likes to stay in a state of no change okay so very lazy circuits so as I said that the cause of changing flux could be a relative motion between the circuit and the source of the magnetic field and if there is no relative motion but that strength of the magnetic flux is changing with time in both the cases I have a changing magnetic flux so therefore I get an emf but those of you who remember your textbook you would have seen the Faraday's law is always stated with a minus sign in front of this minus sign is a very interesting minus sign it is not really an algebraic sign this is to remind you of another law known as Lange's law and this is I will spend couple of minutes on what that Lange's law is see the idea is something like this that when Faraday's statement is made that if you want to change the flux through a circuit there will be an electromotive force that statement did not care about whether this circuit was made of a conducting material or made of wood okay there would be an emf remember emf can provide a current but emf does not provide a current unless you connect it to a conducting circuit just as an ordinary battery ordinary battery you have two terminals ordinary battery if you buy it from the market a normal battery you will see it says 1.5 volts or 9 volts or 6 volts whatever but then that is not giving you a current so what is that so this is there are two terminals there now you take the integral between the two terminal points of the battery which is the non-conservative region and this is what is known as a open circuit voltage so when you buy a battery in the market what the shopkeeper is guaranteeing you is this that there is a open circuit voltage between the two but of course you will not be drawing any current because you have not connected connected the two ends of the battery by a wire now when you do that of course a battery will conduct a current now Lenge's law tells you that suppose you don't have a wooden circuit but supposing you have a circuit made with let's say aluminum wire or copper wire or whatever anything that can conduct electricity then if you try to change magnetic flux through the circuit the circuit will respond to it by opposing the change that you want to make if you are trying to increase the flux a induced current will be generated in the circuit which will try to reduce the flux if you are trying to decrease the flux the circuit still doesn't like it it will try to increase the flux by changing the direction of the induced current now remember we have the standard right hand rule that is if you have a circuit you sort of hold it and try to point your thumb in the direction of the current and the direction in which your fingers curl is the direction of the magnetic field so you know exactly which way the current should flow in order that the flux should increase or flux should decrease the fact that circuit opposes a change is this reminder of minus sign and the fact that if you have a conducting circuit there will be induced current in such a direction that the effect of the induced current will be to create its own field in such a direction which opposes this change it goes by a different name it is called Lange's law so Faraday's law is generic Faraday's law doesn't care about whether you have a wooden circuit or a aluminum circuit but because Faraday's law only talks about electromotive force Lange's law doesn't talk about electromotive force it talks about a current so in order that Lange's law is operated I need that two terminals of the battery must be connected by a conducting wire of copper aluminum etc so the Faraday's law or sometimes people who are fair to Lange they call it Faraday Lange's law is always written as electromotive force equal to minus d5 by dt so let me now give you take you through some of the techniques of generating electromotive force so first example that I take up is the following that I have a magnetic field let me take it as a constant magnetic field but I have a circuit here size shape is not important and let us suppose I am trying to stretch that circuit that is you know it's a conducting wire so I pull it remember you will not get a shock because the amount of current that is generated by this type of electromotive force is very small so you will not get a shock you take a circuit and just try to pull it out so I am stretching the loop now let me take a small place here the length is dl along the tangent and let us suppose I am pulling it out outward by with a velocity constant velocity v and so therefore the the change that I get in area v dt will be then the length by which it stretched this times dl so so my ds because I know ds has a direction so this is the direction of v dot dt and dl is this way and I know that the direction of an area is always perpendicular to it so therefore v dt cross dl is the direction of my ds all right now let us look at what is driving this current now I know that the charges which are there because you have a magnetic field and you have made these charges move when you are stretching it the charges are moving okay then the charges are moving then there is a Lorentz force there f is equal to qv cross b so my electromotive forces integral e dot dl e is nothing but f by q and f I have seen is qv cross b so my q and q cancels out I am left with v cross b dot dl so this is a this is a situation where the emf has been generated because you have made the circuit move you have made the circuit move so this is v cross b dot dl you know a dot b cross c equal to b dot c cross a equal to c dot a cross b so I write this as b dot dl cross v and if you look at it what am I really done so dl cross v what is this so this is the rate at which the area is changing because dl cross v is actually it is v cross dl is the area so I have put in a minus sign because of that so you notice the area is ds by dt is v cross dl so dl cross v is with a minus sign ds by dt so what I have got is b dot ds b dot ds which is my flux okay integration of that d by dt of that so I get my electromotive force is minus d5 by dt notice magnetic field has not changed I have taken the magnetic field to be constant so what has happened is that I have made the I have pulled out the loop and when I pull out the loop they charges in that they start moving in the direction in which I am pulling now that gives me a Lorentz force that is the Lorentz force remember I need a force to move it okay it is the Lorentz force giving me the electromotive force now so in this case the origin of the electromotive force is the Lorentz force let me continue with that example again here incidentally these situations are known as motional emf because the emf has been generated because of a motion relative motion between the circuit and the magnetic field now let me do another case supposing I have a magnetic field here which is directed outward directed outward from the plane of your screen but this is not a uniform field I have not I have got a field which is changing in magnitude not direction over the distance as we go along x in homogeneous magnetic field and let me take this loop to be small enough then I say supposing this is this is an instantaneous position of the loop suppose this end is at the position x at a particular instant this end is at a position x plus dx at a particular instant now if you take a point anywhere on let us say I take a point here you see this is a circuit supposing I take a point here and I consider a symmetrically situated point here that is x has changed but y has not changed so if I have a inhomogeneous magnetic field the strength of the magnetic field at x plus delta x will be different from the strength of the magnetic field at x so I use the Taylor series and say b at x plus delta x is equal to b of at x plus dou b by dou x times delta x so now I am not changing y at all I am just saying consider a point here consider another point which is at the same value of y now let us suppose that this circuit of mine this circuit of mine now what does what has happened here I have the magnetic field which is inhomogeneous I have the magnetic field which is inhomogeneous so what I have done is to say that look the my electromotive force which I have just now shown to be given by v cross b dot dl now if I go over the circuit then I since the direction of the magnetic field is perpendicular to these two directions there is no contribution to the line integral from these two now but that is not true here because these are the directions at which I will here you see what is happening is this that my v cross b this direction is z okay and the charges because what I am doing is I am making this circuit move in the x direction so my charges are moving in the x direction now if my charges are moving in the x direction then my v cross b b is in the z direction if b is in the z direction charges are moving in the x direction the direction of v cross b is either along plus y or along minus y so because of that since these two lengths are along the x direction when I dot it with the direction of the v cross b which is along the y direction these give me 0 so there is no contribution there because dot product is 0 but that is not true here because v cross b along this and that they will contribute now let's see how so when I am going to this point so notice that I have I have a direction which is plus j here and a minus j there so let us look at that point so I have got v cross bx because this is at the point x minus j is the direction dot minus j that is the dl's direction times delta y and when I come to this side it is v cross b but b is not at x but b is at x plus delta x but the direction is now plus j so dotted with plus j delta y so you add it up when you add it up what you get is v cross delta b dou b by dou x dot j delta x delta y you see there is a thing there because I have subtracted this from this and this gives me a delta x so therefore the what I have got now is because I know what is this v cross b db by dx dot delta x delta y so this is nothing but your dou b by dou t dx dt because there is a velocity here this is velocity is along the x direction so I have dx by dt dx dx will go away I will be left with dou b by dou t so therefore the line integral cancels from the top and the bottom side I get a contribution from here but because the strength of the magnetic field is different here by delta b by delta x over this so when I calculate the line integral over this I get a value so now my line integral this is my emf in this situation so here I have a circuit which is in an inhomogeneous magnetic field and it is moving like this the third thing with which I will end today's lecture and part of it I will repeat tomorrow now the what we said is this that I know if a magnetic field is changing then of course the flux is changing what I did not realize is there are different ways of changing flux because flux is not just the magnetic field flux is magnetic field as well as the area so in one case I changed an area in the other case I didn't change the area but I said let me take an inhomogeneous magnetic field so that the two ends of the loop are not subject to the same magnetic field as it is moving but what Faraday says is there is no way of telling what caused this emf whether it is the change in the magnetic field which was in the second case or in because of the motion there is no way of you can find it so based on that I have proved that integral e dot dl is minus d5 I repeat again that please don't take this to be the electrostatic field because this is then would be equal to 0 this has a non-conservative part that's where the battery is there so I rewrite this d5 by dt minus sign is there phi is b dot ds d by dt of b dot ds now what I do is this is a surface integral I use Tock's theorem to convert this into a line integral so what I do is I can do that also what I told you but let me convert this line integral to a surface integral and that is done by writing down del cross e dot ds so del cross e dot ds is this which is written like this since it is true for any surface the integrand must also be equal so I get del cross e plus db by dt equal to 0 this is a complete Faraday's law this is the complete Faraday's law remember in the electrostatic case my del cross of e was equal to 0 but when I have a changing magnetic field whatever be the cause changing magnetic flux actually the corresponding statement is del cross e plus db by dt equal to 0 so this is the third of the Maxwell's equation that I have derived out of that remember del dot of b equal to 0 is definite del dot of e equal to rho by epsilon 0 is there actually normally one writes it del dot of e equal to rho by epsilon 0 as the first Maxwell's equation the second one is del dot of b equal to 0 and the third one I have just now established is del cross of e plus dou b by dou t is equal to 0 this is Faraday's law in its differential form I have one more force coming up one more Maxwell's equation coming up and that is the displacement current and because the displacement current usually many teachers find it a little tricky to explain I will take it up in my lecture tomorrow if there is a quick clarification or a comment you want to make I would be happy to clarify it please yes go ahead okay this is from Kolhapur yes please go ahead in your last session you can told about the domains yeah can we see the domains okay if we see the domains then what technique can be used to see this so let me let me make this following statement which I will ask you know there will be two lectures on magnetism only not given by me but somebody else so they will tell you how the domain structures are determined there are experimental method you you see you have to realize as a physicist when you say can I see it can you see an electron say if you say by seeing I mean can I see it with my eyes the answer is always no when physicists use the word can I see it they mean is there an experimental method by which I can certainly determine whether these structures exist or not okay so certainly that it's there and experimental methods are there unfortunately I am not an experimental physicist I am a theoretician so you must have realized this that I am mostly going on talking theory but we will have two lectures on magnetism given by people who are familiar with such things so I will sort of talk to them and they will tell you yes of course you can determine the domain structure there are many ways one of them is for example neutron diffraction methods or neutron scattering methods but domain structures can certainly be determined okay any other question a quick clarification of whatever I talked just now in a bulk form net magnetic movement is zero as a whole we don't have a magnetic movement for a material but when the same material is reduced to nano size then we can observe sunlight magnetic movement means it gains magnetism so I want to know actually what happens at nano size means at nanometer level so that we get that net magnetic movement no no no so let me explain firstly that statement that for a bulk material I don't have a magnet is not correct I mean then you will not have you your magnets are sold in the market right you can buy a magnet in the market I mean that's a that's a bulk material which has a magnetic movement made of iron it's a magnetic moment now if you are asking me that look there are many materials which don't show magnetism unless you go to nano level the reason is this the other gentleman there asked me about the domains so we explained that when you have a bulk system okay of for example not iron but something which is not magnetic let's say a bulk copper then I will not see it to be a magnetic because of this reason of course even a piece of iron I will not see it but you see when you come to what you are calling as a nano level now what I have done is to tell you that there are these atomic magnetic magnetism now so therefore if you now go on reducing the size my by the way it is not true that every nano material has a magnetic moment okay but most of the cases supposing you take a piece of iron an ordinary piece of iron lying lying on the road does not have a magnetic moment I agree with that okay but supposing you keep on cutting and you come to that level where I have domains or regions okay where it has a magnetic moment because what I said is that it is the random location of the magnetic moments added up right you have a large number of vectors which are randomly oriented now what is the addition of all that it is 0 now but on the other hand supposing I went and focused telescoped on that area where there are one or two vectors they may not add up to 0 so there is nothing you know I mean nano level simply means you have decided to bring it not if not to atomic level but to sizes which are extremely small so there the addition of vectors may not give you 0 and that is the reason but it is not true not a general statement that no bulk magnetic material will have a magnetic moment iron piece of iron which has been subjected to a magnetic field of course has a magnetic moment okay you have a piece of iron you do that probably such experiments on your labs subjected to a magnetic field until saturation magnetism is established switch off the magnetic field after that that piece becomes magnetic and it has a magnetic moment am I clear okay thank you very much