 Good morning to everybody, so we are actually coming to the last session on electromagnetism and as I have been doing over the last few days, I will pick up questions which came in the chat medium because we could not take up while the lectures are on. So let me start with a few of the questions, I will only pick up those questions which we think are of sufficient general interest. So let me first pick up this question from Center number 1276, this is Sankar Chan Patel College at Meshna, the question is, are all insulators dielectric? I am clarifying this question because this is always a confusion among many are all insulators dielectric. You see the words insulator and dielectric are used interchangeably always, but there are some differences between them. So let me first put it this way, all dielectrics are insulators but all insulators are not dielectric. So I will make a statement, all dielectrics are insulators. So what distinguishes a dielectric or those insulators which are not dielectrics from dielectrics which are insulators. See the basic idea is that dielectrics are insulators which are capable of being polarized. Not all insulators are capable of being polarized because polarization depends upon the ability that application of an electric field that there would be separation of the dipole moment to an extent that this system will get polarized. So therefore we say that dielectrics are insulators which can be polarized. So in that sense the dielectrics have the property that they would dissipate less energy, they can store charges etc. And there are insulators which cannot be used as dielectrics, this is the primarily the point. So however I must sort of alert to you that the word dielectric and insulators have been so commonly used interchangeably that a lot of confusion arises. So there is absolutely no harm accepting that not all insulators can be dielectrics, dielectrics are those insulators which can be easily polarized and they have certain properties. Next question came from Nirmah University 1085. The question is can you magnetize iron by sending a current through it? So this question needs a slight bit of a clarification. See the point is if you want to take a piece of iron and make a permanent magnet out of it the answer is no. The however what I can do is supposing I take a piece of iron and you know have some windings with insulated windings and pass a current through it. Now what happens in that case is this that when there are currents in the windings of this circuit and there is iron core, soft iron core which is inside that because of the fact that when you pass a steady current it is a source of a magnetic field. So therefore the iron that in the core is subjected to a magnetic field and as a result since we have told you that a block of iron has magnetic domains which in the absence of a magnetic field and at an arbitrary temperature have their net magnetic moments of each domain pointing out randomly. But when you do this a current passes establishing a magnetic field in the core then these domains the magnetic moments of these domains they sort of try to align or partially align and the iron gets magnetized. So the answer is yes but what is the problem? So this is the standard principle of an electromagnet. The problem is that if the power is switched off, if the current goes to 0 it the process immediately reverses. In other words the it still remains random as long as there is a magnetic field then only this would happen. But a permanent magnet you cannot prepare by simply passing through a current field. The next question came from Centre number 1164, Parashuram College Delhi. The question is that do moving charge or does a moving charge exert a force on a static charge? There is there should be absolutely no difference in understanding this question. See the point is a charge whether it is static or it is moving is a source of an electric field. The difference between a moving charge and a static charge comes by the fact that when you talk about a moving charge the it is also can be considered as an equivalent to a current because a current is nothing but a moving charge. So therefore a moving charge is also a source of a magnetic field. In fact when I was bringing out the relativistic effect of the moving charge I said supposing a charge is static in the laboratory. But I have an observer who is moving with respect to the charge in the lab. So from the point of view of the lab observer it is a static charge and the only field that he sees is the electrostatic field which will of course exert a force on another charge which is also static in his frame. The question that we are asking is supposing I have a moving charge in the observer's frame then of course it is still a source of electric field as we have said that remember which said with respect to the observer who is moving in the frame of reference of the lab then though the charge is static in the laboratory with respect to our observer the charge is moving. So this observer with respect to whom the charge happens to be moving sees that that charge is a source of both an electric field and a magnetic field. And now consider a second charge which is static with respect to my observer that is that charge is moving with the same velocity as my observer. Now since with respect to my observer the charge that is moving the first charge is a source of an electric field then that charge will obviously have a force on this charge which is static with respect to my observer. But not only that the question now is that according to my observer that first charge is also a source of a magnetic field. But my observer sees that the charge which is static in his reference frame right that is has a velocity 0 in his reference frame. So there cannot be Lorentz force on it namely v cross b force must be equal to 0. So in this case they a moving charge will not exert a magnetic force on a static charge because magnetic force is Lorentz force which is q v cross b. So if the velocity is 0 if v is equal to 0 then there is no magnetic force on it. So the answer to that question is yes a moving charge exerts a force on a static charge due to the fact it is also a source of an electrostatic field. But it does not exert a Lorentz force q v cross b because in order that q v cross b is non-zero that charge there must be moving. Next question this is a question which I have not quite noted who asked for it but the question was about Lange's law. The question was why is it not applicable for spin? This question requires a little bit of understanding of what spin is about. I pointed out yesterday that unlike the picture which many school books give about the spin we say that you consider the electrons orbital motion as earth is moving around the sun. And then many textbooks say that the spin motion of an electron is like the earth moving is about its own axis. There cannot be anything more wrong or further from the truth than this. In classical physics there is no structure of the spin. In fact in classical physics there is nothing like a spin that ever occurs. So spin is a purely quantum effect and so therefore this picture that as the earth is or as a charge is moving about its own axis it is changing the flux and things like that that is not applicable for this case. It is applicable for the orbital motion of the charge that is because typically our atomic model is that there is an electron which is let us say moving around this. So that there is an area which it is sweeping so if there is a magnetic field which is threading this area then the Lange's law will say that I must have now induced current which will create an equivalent magnetic field which will oppose the change that this force is applied. Spin does not come into our discussion at all because spin has no such picture like having a circuit or things like that. So let us not worry about that. See a question that comes up yesterday also in some session the people ask this question that what is this permittivity of the free space permeability of the medium the free space or permittivity of the free space etc. Now let me tell you that these are what can be called as a measurement system constant. Now as you know that for instance if you go over from the our standard international unit SI units to another unit then that 1 over 4 pi epsilon 0 or mu 0 by 4 pi these things vanish. So these constants are specific to the type of units that you are using and the best way to understand it is this that supposing I am talking only about the electric field then it is in some sense gives me the capability of the vacuum for example to permit electric field lines. Now you would say what is this what does the sentence means see in order to understand what does that sentence mean you have to actually first consider a medium which is not vacuum. Now you will realize that as you replace the vacuum by a different medium they let us say a dielectric medium different dielectric mediums the field lines they undergo changes and the main reason is that the Coulomb's field it depends upon the force it depends upon what is the medium through which it is propagated. Now in order that I can understand for example what is the force between two charges when they are kept in a dielectric medium. Now I need to understand that what is the force means what I am talking about some absolute thing but no what I am trying to do is to say how is this force modified how is this force modified when I change from let us say air which is our model of a vacuum to whatever dielectric we are talking about. Now so then I need a reference now this should not come as a surprise to you remember we define dielectric constant for glass we say dielectric constant of gas is approximately 1.5 now what did you mean by that what you said is that the dielectric constant of glass relative to air or vacuum is 1.5. So you take the refractive index of the vacuum to be equal to 1 now this is what is called a measurement reference point the velocity of light in vacuum velocity of light in vacuum is a universal constant velocity of light in a medium is not. So therefore you compare the velocity of light inside glass with the velocity of light in vacuum and you say that I define dielectric constant refractive index as a number which is the ratio of the velocity of light in vacuum to the velocity of light in glass which happens to be 1.5. So I have a reference point in fact it turns out that these numbers mu 0 and epsilon 0 when you multiply them it turns out to be equal to 1 over square root of 1 over c square 1 over square of the velocity of light. So these are in some sense involved with the property of vacuum for instance if you look at the strength of the magnetic field and electric field you will find again this mu 0 and epsilon 0 etcetera will come up. So these are as I said measurement system constant and there are ways of defining them there are ways of measuring them for the simple reason what you do is that make the measurement on different medium and then try to scale them to a basic event and that is the way you would talk about it. Next question which came from B D T college Devangare 1176 now the question is a little specific it asks about ferroelectric medium normally I do not take up such questions because they are specific to the enquiry of a particular person and not of general interest but in this case I just decided that we will be talking about that. Similarly unlike ferromagnet this word ferro has no meaning in this context just because the ferromagnet is a well known word so when the corresponding electric phenomena was discovered they decided to call it ferroelectric. So you remember what is a ferromagnet for example a ferromagnet is a substance which can show spontaneous magnetization you subject a ferromagnet to a magnetic field for example a piece of iron to a magnetic field and you know the way it works the way it goes is something like this that as you apply an external magnetic field the magnetization of a magnet goes like this and when all the domains have been aligned it is saturated. Now the if you now withdraw the magnetic field it will not come back this way what it will do is to come back this way so that even when the magnetic field has become 0 there is a remnant of magnetization left which is called as a spontaneous magnetization okay. So what is the relevance the relevance is this these ferroelectric are substances which have spontaneous electric polarization now so in other words even in the absence of even when electric field is equal to 0 externally applied electric field is equal to 0 the electric polarization is not equal to 0 remember compare this case even when h is equal to 0 my m is not equal to 0 and if you want to now reverse this and bring it to the 0 magnetization phase you have to apply what is known as a coercive field and this is precisely what we are talking about we are showing that this shows spontaneous polarization electric polarization which can be reversed by application of an electric field next question came from 1 to 1 5 G L Bajaj Institute Mathura the question is very simple why is why are we relating the Maxwell's equation which says dive b equal to 0 to the absence of how is this related to absence of magnetic monopoles we always make a statement that dive b equal to 0 means magnetic monopoles do not exist in order to understand it what you have to do is this that instead of looking at this equation you have to convert this into the integral form you realize the integral form of this equation is that over any closed surface b dot ds is equal to 0. So if you take any closed surface you recall the corresponding statement in the electric case we had said that the integral e dot ds is equal to the charge enclosed divided by epsilon 0 this is the electrostatic Gauss's law but in this case the right hand side of this equation is 0 which means the charge enclosed inside is 0 what is this charge it is the magnetic charge which is enclosed. So if you compare the electrostatic Maxwell's equation with the magnetostatic Maxwell's equation if the electric and the magnetic phenomena were totally symmetric I should have had this side right hand side as equal to whatever is the magnetic charge let us call it q m enclosed divided by whatever actually it will not be epsilon 0 but some other constants let me just now the reason why this side is 0 is this that irrespective of how small or how big the volume that you take inside if there is one north pole there would be a south pole there is no way of separating that. So corresponding to a positive magnetic charge there is always a negative magnetic charge which makes this right hand side is equal to 0 and that is why divergence of b equal to 0 implies that magnetic monopoles do not exist there is one long question which I will go to a little later but before that let me quickly finish these questions some application questions that came up 1158 from NIT Durgapur these are these questions are being asked that you know why do you study Maxwell's equation why do you study physics for that matter the are there any applications there let me tell you make one statement and then I will answer their question the question specifically is are there any examples where Maxwell's equations are used in communications see the thing is the let us just talk about the engineering disciplines today the basic engineering disciplines can be divided into it used to be only two categories but ever since the computer science came up it is it can be considered into three categories the there is a large part of engineering which is today based on physics and these are mechanical engineering electrical engineering aerospace engineering etc etc another part of engineering is based on chemistry these are chemical engineering part of metallurgical engineering and things like that they all develop by the basic principles that are enunciated in the respective physics the computers engineering or computer science is slightly different while the computer hardware part is based on properties of semiconductors etc which is done in physics but the software part is primarily based on mathematics so therefore physics chemistry and mathematics provide a backbone to all engineering disciplines as we know today I mean things might change a little bit with the biology becoming an engineering discipline but then also it will be based on another science so in other words science provides a pillar on which the engineering rests and no engineering technique can be developed unless we understand the science behind it the to come to the specific questions on application of Maxwell's equations in communication in fact the if you like the Maxwell's equation is primarily useful in engineering in the process field of communication and this would be for example when you talk about guided waves in a either in a metallic guides or fiber optics the principle is based on the applicability of Maxwell's equation in fact if you are doing fiber optics modes you will realize that you start with Maxwell's equation then classify the modes into transverse electric transverse magnetic etc since I will be giving one or two lectures on one lecture on fiber optics towards the end and in the optics course I will again come back to this question transmission line is another place where you use Maxwell's equation there are large amount of think as I said that without the backbone of physics the engineering disciplines will never develop there are one more question before I come to little more mathematical question I am also falling short of time a little bit so I have to accelerate the question was that there is a theorem which I talked about as the Ernst's theorem in Ernst's theorem I said that you cannot achieve a static equilibrium of a system of charges by electrostatic forces alone this question came from BMS college Bangalore 1 2 2 5 now this statement is an interesting statement the in fact this would be true of for example even the gravitational forces because it is the inverse nature of the square law forces is the same now what is meant by static equilibrium a static equilibrium means remember there are various types of equilibrium out of that the static equilibrium means that if a charge is an equilibrium at rest static and if you disturb it slightly from its position they are restoring forces act to bring it back so for example on a hill this is also an equilibrium point and this is also an equilibrium point but if you disturb this from the hill why it is an equilibrium point because it is a potential maximum so therefore the force is equal to zero this is an equilibrium point because it is a minimum and the force is equal to zero but if you disturb this it does not come back if you disturb this it will again return back so this is this equilibrium is stable or a static this is unstable now how do I understand in terms of electrostatic now in electrostatic means supposing there is a charge which is an equilibrium and I am saying that if you disturb it it should come back what brings a charge back it has to be electric forces bringing it back because I have no other force so if this charge is to come back it means that the electric field lines must have a sink here no matter which way you displace it should always return back but I know a free space divergence cannot be less than zero divergence is less divey less than zero is what we require but we cannot have because in free space I must have divergence equal to zero because there is nothing else there so this requires the presence of a sink which I do not have now the question was then what other types of forces well in order to keep charges in equilibrium you could use other types of forces for example the magnetic forces you have heard of magnetic levitation okay so the word to understand is electrostatic forces alone cannot you bring in other types of forces time dependent forces and things like that of course the equilibrium can be restored it can be had as a very specific example the last one was directly related to the what I was talking about and this question came from Don Bosco Assam very good question so I would spend a bit of a time on it the question asked is how did I write emf is equal to e.dl and over a circuit while we are aware that the emf is provided between terminals of a battery I in fact explained it yesterday you are right the emf normally as we understand is the electromotive force between the terminals of a battery I use another word I said open circuit voltage so let me suppose this is my battery and this is a terminal this is a terminal let me imagine that there is a path which is connecting them now what I said is this that if the if I consider electric field if e is a conservative field then by definition integral e.dl is 0 let me let me put c for conservative now if it is 0 it cannot drive a current so what we said is this that look there are two parts of this problem the first part of the problem is outside the circuit where I have an electric field which is conservative so this part this part is not this is the outside part is a conservative electric field now in the battery a chemical process is going on converting the chemical energy into electrical energy there my electric field is not a conservative field so I will call this e n c or non-conservative according to your question that you would understand that if the electromotive force is called supposing this point is a this point is b if I say that this is from a to b inside the battery e non-conservative.dl this is perfect now I do a little bit of mathematical jugglery with it I say all right now look at it this way outside my e non-conservative is 0 now I can always add to this integral b to a outside e non-conservative how why I can do that because this term is 0 outside it is 0 I can always add 0 to anything fine so what have I got now I got now then integral closed circuit e dot dl but nc now you notice I did not write an nc there that is very simple because I know integral e dot dl conservative part is 0 by definition because if it is a conservative field it must be 0 so I add another 0 to it I add another 0 to it but this time e is conservative so by the time I add it up I get e non-conservative plus e conservative dotted with dl which means nothing but e dot dl and that is the reason why we write it so this is not electrostatic field its main contribution is the non-conservative part of the electric field which is present inside the battery. So with that I finish the questions that were asked yesterday if your question has not been asked it simply means that either it is not of sufficiently general interest or I did not have enough time to take care of it with that let me go over to continuation of Faraday's law so yesterday when we ended I introduced to you the Faraday's law and I said that look what Faraday found is the following we said a changing magnetic flux through a circuit is the source of an emf. Now I also said that an emf should not be confused with a current emf a bad name by calling it a force but emf is basically a line integral of the electric field and as we have found what is important is the non-conservative part of it but because of mathematical reasons I could define it as line integral of e dot dl. Now this thing what we are trying to say is this that if a flux changes now remember flux has two parts the magnetic field part and the area part now I could change any of them I could change both of them the circuit does not care how did you achieve the changing magnetic flux you could have changed the magnetic field you could have changed the area you could have made a relative motion between them or you could have had a combination of both the effect happens to be the same there is one thing I want to point out whenever you see a word like a law you understand that nobody can give a better explanation of this law means these are rules which have been found to be true by experiments and always we have a set of rules which we take it for as given and here for example I told you that supposing you have a magnetic constant magnetic field you have a circuit and you stretch that circuit and we showed that the as you are stretching the circuit what you are doing is that whichever charges are here now remember I have no battery no moving charges there are of course if there is a conductor or things like that there are or anything we have electrons there so when I stretch it those electrons are moving those electrons are moving now if those electrons are moving there of course they will be in a constant magnetic field they will be subjected to a Lorentz force and using purely the idea that it is a Lorentz force I could calculate the emf and so that the emf is given by minus d5 by dt so the in this case the source of emf is the Lorentz force next again the same thing here I said there is an inhomogeneous field the other question asked do you mean the same thing as non-uniform the yes these are the standard languages you can call it say uniform field means a magnetic field which is constant in magnitude as well as direction an inhomogeneous field means that its value magnitude changes in from point to point so that is what I had taken I did not change the direction I said its strength changes from point to point once again I was able to show that by taking the source of the emf to be the Lorentz force I got v cross b dot dl and once again the Faraday's law was right of course I could directly have a change in b itself now whatever I do when I convert the Maxwell's equation emf integral e dot dl is equal to d by dt of b dot ds I convert this into a differential equation I get the differential form of Faraday's law namely del cross e plus db by dt equal to 0 so this is our third Maxwell's equation which is known as Faraday's law so let us look at the following situation in fact this question has also come up that explained displacement current but it was not a part of my last lecture but today I am doing it so suppose the argument of ampere actually speaking argument of Maxwell was that look we have seen a time varying magnetic field give rise to an electric field if nature is generally symmetric why can't we assume that a time varying electric field gives rise to a magnetic field if change in magnetic field give rise to an electric field why won't change in electric field give rise to a magnetic field in fact you should know that I have written down three equations earlier del dot of e equal to rho by epsilon 0 del dot of b equal to 0 these are electric and the magnetic Gauss's law the third equation was del cross of e plus db by dt equal to 0 that was Faraday's law so why are they called Maxwell's equation the Maxwell was able to see by completing the fourth law he is the person who put all the four laws together and said these four laws form the basis of electromagnetic so it is his ability to see that there is an integration happening there that the entire pillar see put the all of them together that is the reason why they are very appropriately called Maxwell's equation now let us look at what is that fourth law so his argument was if a time varying magnetic field could give rise to an electric field the Faraday's law why can't time varying electric field give rise to a magnetic field how do I understand this now first let us try to get a time varying electric field and the simplest experiment that he thought of was the Faraday that supposing I have a circuit with a capacitor now and let us suppose this circuit is finally connected to a battery through a switch now when you switch it on when you switch it on the battery would send charges we have seen that conventionally the direction of charge is the direction of positive charge but it is actually the electrons which are moving but so current would flow let us sign this charging this capacitor now when an electric field is established here it polarizes the medium inside this and then then I will have a oppositely charged plate here that is what I am doing is to build up or I am charging these capacitors so during the process of charging it might take a short time but during the process of charging in this region between the two plates a changing electric field is happening so here is an example where my electric field changes with time this is during the process of charging of a capacitor okay now let me come back to this equation that we had that is the Ampere's law we said integral b.dl is equal to mu 0 i enclosed over any closed surface try to understand this equation the left hand side is a an integral over a boundary a curve integral b.dl the right hand side says what is the current enclosed what do I mean by current and closed the current enclosed is the amount of current that is passing through a surface because you remember the current definition is integral j.ds the definition of enclosed current is the amount of charge that is passing through that area okay the rate of change of that that's my definition now here is an example so this is equivalent to del cross b equal to mu 0 j now let me take now first so this is my outside circuit in the outside circuit let me take a loop let me take a loop of this type which I showed in the previous one there is a loop now I want to tell you defining a loop defining a boundary does not decide which surface it is a boundary of in the first example I am taking that let it be considered a boundary of this disk itself now in that case the since this charged wire cuts that surface so there is a current which is enclosed there is a current which is enclosed in here which is simply the outside conduction current so therefore if I take the line integral of the magnetic field over this circuit and if the corresponding surface is that disk surface I get this equation integral b.dl equal to mu 0 times i n close and i enclosed is nothing but the outside conduction current now let me change this a little bit what I am now going to do is this I am saying that instead of considering the disk consider this to be a boundary of a surface which looks like a pot something like a matka now what it means is this that no part of the surface ok the a see the this is a boundary but the surface is here so this closed surface does not enclose this i this is the surface and this is my boundary so but however so what is happening here if there is no in this case remember I said what is i n close i n close is 0 that is because on this surface no current passes so does it mean del cross b equal to 0 that is the question that we are asking now there is a dilemma in other words depending upon what you decided to take it as the surface whether the surface is cutting the outside wire of the conductor or the surface is intersecting surface is intersecting a the region between a capacitor plate I am getting two different answer in the first case when I take it as a disk I am getting the normal conduction current as the integral of b dot d that is because the flux is changing now the point that Maxwell thought is this that suppose I imagine that the space in the capacitor is filled with dielectric well air or anything in this space the electric field is changing with time during the process of charging because it takes time for the capacitor plate to be charged and a constant field being established so during the process of charging the electric field between two parallel to electric capacitors is not constant but you have a changing electric field so the electric flux is given by remember I am now talking about the d dot ds because we are talking about essentially polarizing the dielectric so my d phi e by dt is d by dt of d dot s and this you can convert into a volume integral by our favorite divergence theorem which is d by dt of del dot d d q r and this we had seen is nothing but the free charges remember del dot d is free charge and this integral is the total free charge so this is d q free by dt now if it is d q free by dt this is nothing but the outside current which means that outside in the conducting where I am getting a current I inside I do not have a current in the same sense but I have another quantity which has the same dimension as the current which is arising because of changing the electric flux and the two magnitudes are exact they are same as a result I imagine there is a circuit now if I do that then it means my first equation which was established from the Ampere's law del cross d equal to mu 0 i or del cross h is equal to j free so I said that now change that Maxwell said that you change that equation your del cross of h is now equal to j free plus d d by dt you notice this that there are two parts here this was the d d by dt part and so what actually happens there is no problem see in my outside circuit there is this part in between the capacitor there is this part so I can always write del cross of h equal to this plus this now look at how it I said that this then becomes a continuous process do find out what is del dot of del cross h which by definition is 0 because divergence of a curl is always 0 so del dot of the right hand side is equal to 0 I get del dot of j free plus d by dt of del dot of d equal to 0 okay so this is nothing but that del dot of g and del dot of d as you know is rho free so therefore d rho by dt this being equal to 0 is nothing but continuity relation so all my four equations have now fall into place I have four Maxwell's equations and I have now collected them all together I have del dot of the e equal to rho by epsilon 0 Gauss's law electrostatic del dot of b equal to 0 magneto static Gauss's law del cross of e equal to minus dou b by dt this is Faraday's law del cross of h equal to j free plus dd by dt this is the ampere Maxwell's law four of these equations together and consists of what we call as the Maxwell's equation but you notice here we defined not only the electric field e b which is proper name is magnetic field of induction or magnetic flux density see today what happens is there is a language problem most physicist would use b while talking about a magnetic field but actually speaking b is magnetic flux density it is h which is normally called as the magnetic field in order to avoid the confusion the physicists are all very smart we will say magnetic field b or magnetic field h these are two different things but but we do that and and and similarly for e and d we call it electric field e and we rarely call this b as the displacement we just say the field e field e but then I need a connection between e d and b h and this is called so Maxwell's equations are these four equations but Maxwell's equation go hand in hand with what we call as a constitutive relation these were auxiliary equations which define relationships between the displacement vector d with the polarization d and the magnetic field vector h with the magnetic flux density vector b and the magnetization m. So, this slide is the complete set of Maxwell's equation and this is what we wanted to do so let me now proceed further now I want to talk about these equations purely from the point of view of potentials remember I had defined two potentials I had a scalar potential which is electric field equal to minus gradient of phi I had a vector potential whose curl was equal to b. So, how do these equations translate into equations in terms of the potentials so let us see what we talk about firstly curly which for a electrostatic field was 0 because of the Faraday's law is no longer 0. So, electric field when there is time varying phenomena electric field is no longer a conservative field and that is that is important. So, del cross of e is so b is del cross of a put put a del cross of a here. So, therefore, you get del cross e plus dou a by dt equal to 0 this tells me that though electric field is not conservative I have been able to get a combination e plus dou a by dt which is still conservative so e is not conservative, but this combination e plus dou a by dt is conservative. Now, if this is conservative corresponding to this field e plus dou a by dt I can define a potential so let us call this minus del v. So, in terms of that my electric field is no longer minus del v, but it is minus del v minus dou a by dt. So, notice time varying phenomena is first it has taken is first told e is no longer minus gradient of v or phi I have now called it v instead of phi so that there is no confusion I have said this combination can be written as a gradient or of something and that is what we have done. So, my e is written as minus gradient of v minus dou a by dt. Now, I had said that in vacuum del dot of p equal to rho by epsilon 0 remember there is nothing happened to this equation so far because e is the electrostatic field. So, take del dot of both sides so del dot of p is minus del dot of del which is nothing but del square minus d by dt of del dot of a that is equal to this. So, this is this red marked equation del square v plus dou by dou t of del dot a equal to minus rho by epsilon 0 this is my equation number 1. Now, I have two potentials here I have a vector potential a I have a scalar potential v vector potentials curl is still the magnetic field, but scalar potentials gradient is not the electrostatic field, but scalar potentials gradient is electrostatic field plus time rate of change of the magnetic vector potential. So, this is this is an equation which connects the scalar potential v and the vector potential a with the charge density that is there. So, that is I have called it equation 1 I will come back to it all right. Let us go to Maxwell's equation now. So, I get remember my equation was del cross of b was equal to this is a bit involved derivation, but because it is a rather important issue I want to talk about it in some amount of detail. So, this was my fourth of Maxwell's equation del cross b equal to mu 0 j now I have taken a linear dielectric. So, that instead of writing d d by dt I have written as the correspondingly mu 0 epsilon 0 d e by dt. So, just going back these were my Maxwell's equation. So, del cross of h is j free plus d d by dt if I take linear things where h is equal to no magnetization h is equal to mu 0 b just put that. So, h is equal to b by mu 0. So, multiply mu 0 on both sides and then d is equal to epsilon 0 e. So, I get mu 0 epsilon 0. So, this is my fourth Maxwell's equation linear magnetic non magnetic system and a linear dielectric. So, that I mean I could do carry on with my d and h, but this is much more convenient and little easier ok. Now, let us look at this equation my b is. So, this equation I want to write down in terms of the vector potential. So, b is del cross of a. So, this is del cross del cross a I told you there is a very standard identity that del cross del cross a is del of del dot of a minus del square a. So, this is what I have written down equal to mu 0 j plus mu 0 epsilon 0 d by dt I have not changed the right hand side so far. So, this equation I have rewritten now del of del dot of minus del square of a is already there mu 0 j is already there mu 0 epsilon 0. Now, this d e by dt this is what I wrote in terms of the potential formulation that I did a little while back. So, I said that look at this e is equal to minus gradient of v minus dou a by dt. So, just substitute that. Now, if you do that I get a long equation I get del square a minus 1 over c square d square a by dt square. I have written it in a particular way del of del dot of a see take this to that side combine the term which has only a. So, these are the ones with only a this also has a, but I have written it because both of them are gradient of something. So, I have written it together is equal to mu 0 j. So, these two equations these two equations are what I am looking at now. So, this equation which is just now I derived and this equation which was I had called it as equation number 1. So, both these equations have written. So, in terms of the vector and the scalar potential I have been able to write down two equations, but there is a problem these equations are not uncoupled with respect to the two potentials making the solution if not impossible very difficult. So, you notice a and v are both involved in these equations. Now, notice one thing very interesting the if I choose del dot of a plus 1 over c square dv by dt to be equal to 0. If I choose I will tell you that question is can I choose it we will see that later, but supposing I can choose this to be equal to 0. Then of course, this term vanishes I will be left with a term which has only the vector potential del square a minus 1 over c square d square a by dt square equal to mu 0 j minus mu 0 j. What about the second equation if my del dot of a plus 1 over c square dv by dt is equal to 0. Then I can replace this del dot of a which is appearing here by a term which is 1 over c square dv by dt making it 1 over c square d square v by dt square. So, if I can make this term equal to 0 then this equation will only have a in it and this equation incidentally will only have v in it and my two equations will get decoupled and these are the equations that I will get. These are field equations the equations to waves and we can always solve them that is not problem, but can I guarantee that del dot of a plus 1 over c square dv by dt equal to 0 because I said that the two equations become decoupled only if this term can be shown to be equal to 0. Now, the point is this the I cannot show that term to be equal but I can do something smarter. I have realized while discussing about vector potential a and scalar v that both of them have a choice of gauge. For example, if you change a by gradient of a scalar the del cross a which is the magnetic field it remains the same that is because curl of a gradient by definition is identically 0. Similarly, if v changes by a constant then also it gives rise to the same field because what is involved is the gradient of v and not the v itself. So, therefore suppose I want to find out whether I can find choose a and v such that this is equal to 0 satisfied. This is the question I want to prove. Now, suppose I have chosen an a and v such that it is not equal to 0. Suppose we have found an a and v such that this is equal to some function of obviously of r and t f of r t not equal to 0. Now, can I make it 0? Well let me do a little bit of hand waving. I know that if I let change that a to a prime such that a prime is a plus gradient of some psi scalar function then my magnetic field b will remain unchanged. In other words a vector potential a and another vector potential a prime a plus del psi have the same physical import. So, let me choose a psi such that a prime is equal to a plus del psi some scalar function. Now, once you have chosen this psi let v prime be the old v minus dou psi by dt. Now, if I put them back. So, this is what I had del dot of a plus 1 over c square dv by dt was not equal to 0 was equal to f r t. I said let a prime b equal to a plus del psi which means a is equal to a prime minus del psi v prime b equal to v minus dou psi by dou t. So, therefore, v is this that is equal to that. Now, write it this way combine the psi term and the other terms and this is equal to f r t. So, I my purpose was if you realize to make this term 0 I can make this term 0 provided I can make this term equal to minus f. So, that f cancels from both side. So, in other words I need to choose psi such that I need to choose psi such that del square psi minus 1 over c square d square psi by dt square equal to minus f r t. This is an equation whose solution can be always found given f r t. And if that happens then my original request was can I make this term equal to 0. If I did that the two equations will be satisfied and they will be decoupled. And this if you choose a gauge like that by solving for a psi and changing your vector potential and the scalar potential accordingly your two equations are completely decoupled and this is called a Lorentz gauge. This is called a Lorentz gauge and it is convenient to solve our equations the Maxwell's equation in such a gauge because in this case the scalar and the vector potential are decoupled and these are standard wave equations and I can take care of it. So, this is same thing written down there. I will end this discussion with a discussion on the energy density of the electromagnetic field. See the point is the electromagnetic field this has been talked to you when Professor Shiv Prasad was talking about the introduction to quantum mechanics. The fields can also be considered as regions in which there is energy stored and there is momentum stored. I will not go into momentum because I have do not have enough time but let me explain it on the basis of the energy. Now let me say that I have a collection a region in which I have got charges, I have got dielectrics, I have got currents whatever you have. So, this is a region in which I have everything that you have learnt about. To make my discussion easy I will discuss only a linear medium and a non-magnetic medium. In other words I will not be dealing with the vector d and the vector h separately I will use the fact that the vector d can be written as epsilon times e and the vector h can be written as b divided by mu. So, let us look at first what is the energy of this combination. This collection of charges and currents they have. So, electric component of it is the energy is half d cube x rho x phi x this I derived yesterday and I said rho is nothing but del dot of d and so therefore I rewrite it as half integral d cube x e dot d and I pointed out half e dot d is my energy density electric energy density. For a linear medium where d is equal to epsilon e my energy density of the electric field is half epsilon e square. So that is the amount of charge amount of energy that is stored in the electric field. Let us come back to the magnetic part. In the magnetic part what I have is this that my magnetic energy is half a dot j d cube x j is del cross h. So, just redo a dot b cross c equal to b dot c cross a etcetera and you find out as I pointed out last time that the magnetic energy is half d cube x h dot b. So, therefore the half h dot b or for a linear medium b square by 2 mu is the energy density of the this is wrong this should be b energy density of the magnetic field. So, my collection of these two energies is the total energy density that is stored in the electromagnetic field. The electric field has epsilon by 2 e and the magnetic field has this. So, my total energy in a volume is given by epsilon by 2 e plus b square by 2 mu. I have said that I am dealing with linear medium. So, that my e and d and b and h do not come, but only I deal with two things. So, now my question is this. So, supposing I have a region of space where there is an electromagnetic field, electric field magnetic field etcetera. The total energy in that volume is the integral over epsilon by 2 e plus b square by 2 mu. How does the system? How does the energy change in this? This is the question I am asking. I have a volume in which I have a electric field magnetic field electromagnetic field that is how would that system change its energy and that theorem which I will be deriving is a very important theorem in physics it is known as the pointing theorem. In order to understand pointing theorem let me tell you one thing a very intuitive thing that if I have a closed volume having some energy the energy can be lost in only two ways. First is the mechanical way in which energy can be lost that is because the work is being done against the force. So, if there are charges there they are being subjected to electromagnetic or the Lorentz force and in doing so mechanical work is being done. So, the energy is changing. How much is the mechanical work? Remember the definition of the work f dot v v is the velocity this is appearing as a new, but do not worry about it. And what is the force? Force is well for single charge q e plus v cross b, but since it is continuous it is rho e v cross b dotted with v. The second term here v cross b dot v is by definition 0 because this term is perpendicular to v and this I am doting it. So, once the second term becomes 0 I am left with rho e dot v which I have written as rho v dot e, but rho v is nothing, but the energy with the current density. So, this is e dot j d cube x. So, therefore, one part of the energy which is lost by mechanical work, joule heat or whatever you want to call it. This is given by volume integral of e dot j d cube x. Now, let us return back. So, we have done the mechanical part. Mechanical part is given by p that is the power is e dot j d cube x. This is the power dissipated. Return back to the total energy expression. The total energy this is the derivation looks bad, but it is just matter of ordinary differentiation. Do not worry about it looks bad. So, this is my total energy. This is integral of some constant e square another constant b square. So, if I do dw by dt I get epsilon by 2 there is a square missing on this e that should be a square both the places. The epsilon by 2 2 d e by dt that will cancel that 2 there 1 by 2 mu 2 b d b by dt again the 2 will be cancelled and I will be left with. So, having taken care of this is my expression. Now, I do the following. I use the fact one of the Maxwell's equation del cross h equal to j free plus d d by dt which is equal to j free plus epsilon d by dt. In other words I am saying this epsilon d e by dt you replace by del cross h minus j that is what I have done epsilon d d by dt is replaced by that this term d b by dt you replace by using Faraday's law. So, d b by dt is minus del cross h and this is what I have written down here. Now, this expression which appears here again you look at now what do I do with this there is a e dot j there you notice this. So, this term minus e dot j epsilon and epsilon cancelling out which is here. Then the other 2 terms which is e dot del cross h minus h dot del cross c I put it there by a standard algebraic identity. This quantity is nothing but del dot of e cross h this is given by del dot of e cross h this is wrong this is wrong it is not d s, but it is d cube x. So, by d w by dt is del dot e cross h d cube x minus e dot j d cube x and remember what is my w my w was integral u d cube x. So, therefore, this term both both the sides and I told you this is miss written d s by mistake this is d cube x both the all the 3 terms have volume integrals in them. As a result since this equation is the for valid for any arbitrary volume I should be able to equate the corresponding integrals. So, this will be d u by dt plus del dot of this quantity e cross h I define is as s it is called the pointing vector is equal to minus e dot j. Now, remember that using divergence theorem though I told you this is d cube x this could be made into a surface integral e cross a dot h e cross a dot the surface do not confuse this s that I am talking about now is surface and not the pointing vector. Unfortunately, pointing vector always uses the same notation s. So, this is my equation what does this equation tell me it tells me if I take that del dot s terms outside that that a region can change its energy in two ways. Remember I told you that this is the mechanical work done e dot j. So, therefore, the energy is decreasing by one term which is the mechanical work done and a second term which is essentially a surface term. So, in other words radiation because this is the amount of energy which is simply leaving its surface. So, one is mechanical work done the another is simply the radiation. So, this is the entire idea of how a system a closed system loses its energy a closed system electromagnetic field can lose energy either because of the joule work or the mechanical work that is done on charged particles of the system or by simple radiation. And it is the pointing vector which tells you that what is the amount of power that is radiated per unit area because as I told you this is a divergence term d cube r and you convert that into a surface integral term which will make it e cross h dot d surface. And so, therefore, this is the literally the amount of energy which is leaving the closed surface through its surface by radiation only. And so, therefore, this is a vector which tells me how much of energy is being radiated per unit area per unit time and that is your pointing theorem. So, I have very few time left can you see if there are quick clarifications on the yes this is a mega science should Calcutta tell me. If we talk about electromagnetic wave passing through a medium we talk about skin depth and we say that is the frequency of the wave is large then the skin depth is very small right. That is only in case of this is skin depth one talks about only when you are talking about an electromagnetic wave or an electric field getting into a conducting medium and that is because inside a conductor there cannot be an electric field. So, therefore, what one assumes is that the field penetrates the surface of a conductor by a small distance and that is your skin depth. And then we know that if we consider please understand if we are talking about a dielectric medium we do not talk about skin depth. It is only in a conducting medium that when you are trying to pass an electric field you know an electric field is incident on a conducting surface because inside the conductor there cannot be a field we talk about a skin depth. Yes sure and then if we talk about normal radiation for example, x-ray we know higher the frequency of x-ray then it penetrates deeper. So, is there any contradiction between these two? No, no there is no contradiction at all. See we are talking about in one case radiation the other case is propagation through in a conducting medium. You know this is why I spent a bit of a time in talking about what we called as the boundary conditions because at the interface there has to be certain condition satisfied depending upon whether it is a boundary of conductor dielectric dielectric dielectric or whatever. So, first both your statements are correct, but what I am trying to say is one of them is related to a boundary condition. The other one is just a question on radiation and how the radiation takes place and all that yes. And one more thing this morning you discussed a question whether moving charge will exert any force on a static charge. So, could you please summarize your answer I could follow. Okay you recall yesterday when I introduced yesterday or day before when I introduced the magnetic field. I gave you an example of that if I have a moving observer let us take a charge which is static in your lab. Let us assume that there is a charge which is static in the lab and there are two observers. Now, when we say that a charge is static you mean it always with respect to an observer. So, if that charge is static with respect to you that charge only gives rise to an electro static field nothing else. Now, the same charge if it were moving with respect to you or if you are moving with respect to that charge it will constitute essentially like a current because a moving charge is a current. So, that then becomes a seat of a magnetic field as well. Now, what we showed yesterday in that relativistic argument is this that the if you look at a charge which is moving with respect to you or you are moving with respect to charge it is immaterial so far as physics is concerned. Then that charge is a seat of both electric field and magnetic field because you see if there is a charge there has to be an electric field it has nothing to do with the whether the charge is in motion or not. Now, the question that you might immediately say then what happens in case of a steady current why do not we talk about electric field of a steady current for the simple reason. See the conductor is not charged the when the all that is happening in a steady current is that the well actually electrons though we say positive charges they are moving around all that they are doing is passing on for one region of positive charge to another region. So, as a net in a circuit when you talk about a current there is no steady charges the net charge is 0 because there are positive charges and negative charges inside the conductor. So, you do not talk about an electric field there but since your question was based on a single charge I am assuming that charge there is not getting neutralized. So, if there is a single electron the single electron certainly will give rise to an electric field whether it is moving or not, but if it is moving then it also gives rise to an electric field. So, supposing with respect to you in the lab there is a charge which is moving and there is a charge which with respect to use at rest the moving charge will exert an electric force as well as a magnetic force and we have seen the in that relativistic description that I gave unless you take these two things together because I could always go to a frame in which that charge is brought to rest then there is only electrostatic field and then we have seen it creates a problem with respect to my understanding of how forces transform under when an observer is moving. In fact, this is something which Professor Shiv Prashad in his relativity course will take it up I will also take him to tell him to talk about it, but please remind him there to come back to this question. So, if I summarize that then if a moving charge is there which is equivalent to current have a point charge in my lab then it will experience force and that force I mean the second charge if it is moving then it will have magnetic force if it is not moving then only electrostatic force will be acting upon that. Absolutely and this is both consistent the thing that I proved is you see magnitude of the force the force cannot be different only thing is that in that case when you are moving you have to transform the things accordingly then you find in one case you add the electric force and the magnetic force in the other case you have only electrostatic force both of them must be the same. Thank you very much we will stop.