 Good afternoon, so in the morning session we had talked about the remaining part of electrostatics. What we did at that time is to talk very briefly about water dielectrics, also goes by the name insulators. So, basically what I said is the following that in dielectrics unlike in the conductors there are bound charges. You see conductors are characterized by charges which are free, they can move around if you apply an electric field, but in dielectrics basically the charges are tightly bound to the atoms of the molecules to which they belong. In some situations the positive and the negative charge centers coincide and even in the absence in the absence of the electric field and if you apply an electric field there would be a separation of charge. What we did is to define what we called as the polarization vector. So, polarization vector was defined as the essentially the dipole moment per unit volume. This is vector P that is sum over I p i if you take an assembly of a volume then divide it by the volume. That is the meaning of the dipole moment per unit volume. And what we proved by looking at the potential expansion up to the dipole terms. In fact the expansion that we did is popularly called the multiple expansion and in fact if you kept another term it would be quadrupole term, but we kept only up to dipole term. So, what we said is the following that if I have a dielectric here and if I am looking at the potential at some arbitrary points then the net potential can be expanded in terms of various moments of the charge distribution. For example the zeroth moment of charge distribution is simply rho r dr that is just the charge q that is nothing but the zeroth moment of charge distribution. The dipole moment as we said has been defined as the first moment of the charge density so which is essentially r rho r dr over a volume you have to normalize it over a volume. So, this is what we kept and we said well I mean I can expand the potential in terms like this and then the next one would be quadrupole moment and octopole moment and things like that. So, the idea is something like this that if this distance and you know the you have to keep less and less terms the further and further you go. So, for instance if you are very far away then it might be even good enough to ignore even the dipole term in which case this dielectric essentially seems like a point at the location of p that is its electric field is essentially the this p should not be confused with the polarization vector this is just a point. So, and that is obvious because if you have a block of any matter and if you are looking at it from a very large distance of course it looks like a point charge. So, that is that is fine the other thing that we did is to do that and looking at the potential expression what we did is to find that the expression for potential has two terms one which essentially gave us the surface charge. So, I call them bound because they arose from different type of charges they are not free charges as we were accustomed to talk in the conductors we said this sigma bound is given by the normal component of the polarization. So, basically we found an expression for the potential which was surface integral of p dot n divided by r minus r prime etcetera. So, that by comparing it with the typical expression for the field due to a surface charge we found that the dielectric has a surface charge density and which is equal to p dot n. The other one is that there is a volume charge density again we will call it rho bound or rho b just and that is given by negative divergence of the polarization vector. So, we will later on in the tutorial sessions we will try to work out some problems and you will find that the net charge of a neutral dielectric anyway turns out to be 0 only thing is that the charge distribution works out to be different. So, other thing that we did is to say that now that we have more or less finished electrostatics let us try to summarize the equations that we have. So, remember del dot of E was shown to be equal to rho over epsilon 0 this electric field is the actual field what do I mean by an actual field by that I mean this is the field that if you put for example, a charge q there at a point the charge q will experience a force which is given by q times e. So, that is what I meant that that is the field that you always determine experimentally. So, del dot of E equal to rho over epsilon 0 is what we said, but then if I have this rho having two parts namely the free part like in a conductor the charges which are free and I have rho bound divided by epsilon 0. So, this part came from the polarization component and what we do is this that I can rewrite say that epsilon 0 del dot E is equal to rho free and rho bound if you recall I said is minus del dot P. So, therefore, what I do is I take this term to the left and find that del dot of epsilon 0 E plus P this is given by rho free notice that there is no divided by epsilon 0 here. So, the dimension of d vector originally called a displacement vector, but today the displacement vector nomenclature is reserved for different things I will continue to call it d vector del dot of E is the d is equal to rho free. So, this is my Maxwell's equation now and the this is supplemented with a definition of d in terms of E and the polarization. So, these are normally called a constitutive relation you will later on see for it for example, the Ohm's law is a constitutive relation. So, these are not parts of Maxwell's equation, but they have to be taken along with the Maxwell's equation for a proper definition. So, this rho free if I want to calculate d vector is basically a intuitive construct and in a given situation it may not be possible to actually separate rho free from the rho bound. Now, it also then says that if I try to find out what is the surface integral of d over a closed volume closed surface closed volume. Remember that this equation gave us E dot d s that was my flux of the electric field to be given by q enclosed divided by epsilon 0. Now, this q enclosed is now both the free charge as well as the bound charge. Now, however if you would rather like to do the d dot d s which is the flux of the displacement field if you like d vector that is simply given by q free. There is no divided by epsilon 0 for the dimensional reasons that I told you about. And so del dot of is equal to this. So, this so far is my collection of and of course as long as I am in electrostatics my del cross of E is equal to 0. This we have seen that this is always valid. With this let me temporarily you know shift to the discussion of magnetic phenomena and magnetostatics to begin with because in these lectures I have to ultimately come to the electromagnetic field and things like that. So, that was what we did in the morning and let us do the following. Now, going to magnetostatics first let us look at the differences. You will realize the mathematics is very similar. So, I do not really worry about that is the reason we spent a lot of time in doing the electrostatics. The first statement to make is that free electric charges exist that I have a positive charge you have a negative charge. Now one does not know of magnetic monopoles I mean one clearly does not even know whether they actually exist or not. The supposing you take a magnet there is it always consists of a north pole and a south pole you break a magnet into two each one of them will have a north pole and a south pole and things like that. Now so remember we had defined an electric field by saying that what is the force it exerts on a charge q we said electric field E times q is the force that a charge q experiences when placed in an electric field. Now, I cannot give you the same definition here the reason why I cannot give you the same definition is there are no magnetic monopoles known to exist. Now you notice that I have not been making statements magnetic monopoles do not exist well I have written it that way but that is because there is no theoretical reason why magnetic monopoles that is an isolated magnetic north pole or a magnetic south pole cannot exist there is nothing special about it. If they did we also know how to change our Maxwell's equation there will be nothing greatly wrong with it but the point is that last several years hundreds of years literally people have searched for the existence of magnetic monopoles they have occasionally there have been reports that they have found it but on the other hand as of today there is no reliable information about anybody having established the existence of magnetic monopoles. So we will assume for the purpose of this course that magnetic monopoles do not exist now if the magnetic monopoles do not exist then it means the magnetic poles always appear in pairs. In other words in any given volume the net magnetic charge if you want to use that word is equal always equal to 0. Now remember we said that our del dot of the surface integral of the electric field was given by well q or q by epsilon 0 depending upon whether you are choosing the d field or the e field but in this case since there are no isolated magnetic monopoles the corresponding flux expression that is surface integral of b over a closed surface b dot ds must be equal to 0 because the amount of magnetic charge that is contained inside enclosed is equal to 0. So what we do is this we if since we cannot isolate a magnetic charge but we know how to isolate an electric charge we define the magnetic field later on during the lecture I will point out that the magnetic field is purely a relativistic effect as I go along. So but interestingly we define the magnetic field in terms of the force experienced by an electric charge in the field just the way we define electric field by the force that and again an electric charge experience in that field. So there is a slight difference here the electric charge experiences a side wise force in the magnetic field and the force is given by the net supposing there is a combination of electric field and a magnetic field the force is given by the Lorentz force expression which is q times e this we have been talking about all these days and q times v cross b that is if the velocity so notice very interesting thing that the magnetic field exerts a force on a charge particle a side wise because it is perpendicular to the direction of v b as well as the direction of v. So in other words if the velocity direction only has a component along the direction of the magnetic field the force is equal to 0. Once again where do we find how do we create magnetic field what is found is that the current is the source of the magnetic field the steady current is the source of the magnetic field. Now let us define first what is meant by a current the there are issues there because of various reason we do not you know we do not quite distinguish sometimes the difference between a current and a current density we you know if you ask a school student that is current a vector they will say yes current is a vector unfortunately the world current when it is used in the proper sense which is the amount of charge that is crossing a boundary of surface of a volume per unit time that is the rate of change of charge per in the volume that is not a vector that is a that is a scalar quantity. But of course the magnetic the current density which is what we will define later turns out to be a vector quantity. Now if I am having a steady state there is obviously no accumulation of charge. Now since there is no accumulation of charge what it means remember this is something which we have been talking about earlier. So let us say I just given some sort of a volume and let us say this is the bounding surface this is like a current carrying wire. Now you see the if you cut a cross section there between two wires. So the electric charges are coming in traditionally the current direction is taken to be the movement of the positive charge direction but of course we know that what moves are actually electrons. So therefore the direction of current is actually opposite to the direction in which the negative charges move but that is not very particularly important. So what we did is to say my current is charge density times the velocity dotted with ds because this is my definition of current density. And so therefore my minus of j dot ds minus sign because of the reason I told you is the rate at which my total charge is changing in that volume and remember q is integral of rho dv. So therefore what I get is that the change current which we have written down and by I convert this j dot ds using our friend divergence theorem into a divergence of j and the volume integral of that. So q is integral rho dv so therefore you just combine these two and get an expression which is d rho by dt plus del dot of j is equal to 0. So this is basically what we wanted to talk about that this is called equation of continuity and if the there is no accumulation obviously in such a situation there would not be an accumulation charge. So therefore I would not have this term d rho by dt equal to 0. So I have the definition of a steady current namely del dot of j equal to 0. Having made that statement that the source of magnetic field is steady current. Now I need to formulate a law like we did for the electrostatics the Coulomb's law. We said that if there is a charge what is the electric field expression for the electric field there. Now in this case the corresponding law is called the Biot-Savart's law. So what we are saying is this that source of the magnetic field is basically a current distribution. So here this red this is my you know some sort of a wire through which let us say the current is moving. Now how do I calculate the magnetic field due to such a current. So what I do is this I split this current flowing into small sections and I define the direction in that small section by the tangential tangent to that direction of the curve at that point. So that let us call it dL prime. O is my origin the current element as we call it is located at a vector R prime. So direction of the current now notice that I have given a vector sign there and I have said that that is because the length element has the same direction as the current density and which in this case is along the tangent to the curve. So supposing I have a dL there and I am interested in finding out what is the magnetic field at a location R. Now this is now a law the law as you know that these have been found to be true by doing experiments that this is like Coulomb's law I mean you do not question where did you get it from. So I want also question so we say that the magnetic field contribution dBr due to this length element is given by now remember we had in the electrostatics 1 over 4 pi epsilon 0 in this case it turns out a different factor mu 0 is called the permeability of the free space that was called permittivity of the free space. So mu 0 by 4 pi so that is a constant which will come out this is proportional to the current flowing and the it is basically an inverse square law again because it is proportional to 1 over R minus R prime square but it is because we said that the force is side wise the magnetic field direction is given by dL prime direction cross R minus R prime in other words it is perpendicular to the direction of dL prime and this and that is the magnetic field that direction that is you know a standard sphere it is the inverse square but its direction is given by a cross product relationship. So now this is the important all important by Summert's law this is the one which takes place of Coulomb's law there is a lot of similarity the similarity is in the fact that there is a proportionality with the charge the the charge the role of charge is taken by a current here. So there was a Q in your electric field expression I have an I there there is a 1 over R square there here also we have a 1 over R square here the difference is the difference is that if there is a current element the direction of the magnetic field is given by a cross product. So other than that this is your main law comparable to the Coulomb's law for the magnetic phenomena. So let us let us look at what do I get if I want to write it in slightly different way. So what I will do is this that since it is a bit of an algebra in spite of the fact that I have several times been told that do not waste your time on derivation but sometimes a deriving helps and particularly if you want to understand something so can we just switch over. So we said my differential value of B at the position R is given by a constant mu 0 by 4 pi this incidentally like 1 over 4 pi epsilon 0 this constant keeps on coming back this times the current strength I multiplied by I have dL cross the vector R minus R prime divided by R minus R prime q as I told you that this is also in some sense an inverse square law. So this is what I have got. So in order to get B I need to of course integrate this. So I have mu 0 by 4 pi I so that is not a problem and I do the integration of well remember that here is an I there which is nothing but my current density. Now what is current? Current is the surface integral the J dot dS is the current. So therefore what I will do is this this I times dL is essentially in the direction of J. So this I will write J of R prime cross R minus R prime divided by R minus R prime cube I am determining the field at the position R. So R is not an integrated variable the variable of integration is R prime. So this is what I have got. Now I do some manipulations which we have talked about for quite some time. So we have got mu 0 by 4 pi integral. So I will write this as del of 1 over R minus R prime cross J of R prime some explanation required for this. So notice that the gradient of 1 over R minus R prime is minus R minus R prime by R minus R prime cube. So I have not taken a minus sign but what I have done is to interchange the cross product. As you know A cross B is equal to minus B cross A. So this times dQ bar. That is what I have got. Now I do a bit of a manipulation this is the only long derivation I will be doing. So mu 0 by 4 pi. So I use again a chain rule differentiation. So I will write down this as in the following way I will say this is equal to del cross J of R prime divided by R minus R prime is actually very straight forward and trivial dQ bar prime. Why was I able to do it? That is because you notice that this del cross is taken with respect to variable R not with respect to R prime. Let me put that there and J of R prime is a function which is dependent on R prime which is not the same as R. R is a fixed coordinate. So fixed means the position of the it is the position coordinate of the point where you want to do that. So I could simply borrow this here. Now since I again do this this is mu 0 by 4 pi. Now since del is taken with respect to R variable. So it has no effect on that interact integration. So I will write this as del cross integral J R prime by R minus R prime. So this expression for B this is B at the position R. You can see why I did this. You can see immediately why I did this because I could write B as a curl of a quantity. So therefore my since I know divergence of a curl is always equal to 0 I get del dot of B is equal to 0. This is the formal way of getting that where did we get that from. So therefore summarizing again the basic principle behind our derivation is the Biot-Sawart's law. That is the only thing that I have assumed. Using Biot-Sawart's law definition of current density etcetera I have been able to show that del dot of B is equal to 0. Now as you know that I have been making this statement that in order to know a field completely I need to also point find out not only the divergence of the field but I also need to point out what is the curl of that field. So what is curl B? But before I go that there okay let me repeat that anyway. So B of R is given by mu 0 by 4 pi del cross integral J of R prime which is of course a vector divided by the scalar R minus R prime B cube R prime. There are two things I want you to notice here. First is because B can be expressed as a curl of a quantity as I pointed out that this tells you my del dot of B is equal to 0. That is the magnetic Gauss's law. The second thing that I want you to realize is that B has been written as a curl of something. Let us just call this something. So this something plays a very important role there. We will see later that I will represent this something by a quantity called a vector potential okay. We will come back towards that okay. So next question is what is del cross B? So obviously this is nothing I have B already. So I say del cross B is mu 0 by 4 pi del cross. Now whenever you write del cross del cross you be very careful because these brackets are extremely important. Notice this is del cross del cross of something and this is almost a formula del cross del cross A or whatever. This is this is something which we keep on using in electromagnetic theory. So it is probably a good idea for you to almost remember it. So this turns out to be del of del dot of A minus del square A. Let us look at what is then del cross B. So my first term is this is obviously my A along with mu 0 by 4 pi term etc. So let us look at what is this del of del dot of A. Now this can be written as I have mu 0 by 4 pi which of course comes there. I need my gradient there. Now then I need a del dot of A. So del dot of this quantity. So let me let me write it like this. Del of del dot I need del dot of j of r prime divided by r minus r prime d q r prime which I will put in a minus sign there and make this del dot as del prime dot because it is only acting on the denominator and so therefore I can take care of this. So how do I take care of such a thing? I write this as the following. I have made a minor mistake but let me try to correct it. Minus mu 0 by 4 pi let me write it down. Del of del prime dot let me write that down j r prime by r minus r prime minus 1 over r minus r prime and del dot of j r prime. This is what I have got. So this quantity then what I have is the following and dotted with d q bar of course. Let me erase this line. This is wrong. Let me keep it like this. So I do not even need that minus. Now what I do is this. I realize I am dealing with steady current. You realize remember our equation of continuity. So this told me that this term must be equal to 0. So continuity equation tells me that del dot of j is equal to 0. So I am left with now a term which is mu 0 by 4 pi and here what I have is this that this is a by divergence term theorem. I can convert this into a surface integral again. So I can write down this as del of a surface integral of j r prime over r minus r prime dot d s. Now this has to be equal to 0. This is very important, small but important point. You see when I say that I am taking it over the surface one of the techniques that we very frequently do is this. We say alright it does not matter how big your surface is. So extend the surface to infinity. Now the fields that you do take care the fields all go to 0 at large enough distances. So therefore this integral which I can expand the bounding surface to take it to infinity and so therefore this integral will turn out to be 0. Now if you do that now then what I am left with is simply the term del cross of j del cross of b is equal to mu 0 j. I am sorry for minor arithmetic errors but it will be all correctly written on the in my notes. So we have got del dot of b equal to 0 del cross of b is equal to mu 0 j these in some sense are new magnetostatic Maxwell's equation. Let us just do one or two calculation because in this session I am more interested in talking about couple of things which are you will find interesting. So one law which you have found out is called Ampere's law. Remember when we did Coulomb's law and then came to Gauss's law many of you commented that well what good is Gauss's law because it is very difficult to work out in general situation which is correct. So this is simply the point that in case you have symmetric situation then it makes much more sense to trust your intuition and try to get a solution rather than trying to you know in a simple situation having a very big technique is not a great idea. So let me consider what happens to the for example supposing I am looking at a long straight wire look I said that del cross of b is equal to mu 0 j. Now in this case I take for example now notice that this tells me that supposing I am to do a surface integral del cross of b dotted with ds and take the surface integral. So this is mu 0 j dot ds but remember this is current density so j dot ds is nothing but my current. So it is mu 0 times I now what I do is this I do a little algebra this is surface integral of a curl and I use a Stokes theorem on that and convert this into a line integral of the magnetic field itself. So I get integral b dot dl is equal to mu 0 i. You may note that since this came from Biosawar's law this also has its origin ultimately in the Biosawar's law. So this is what I call as an Ampere's law is that any less rigorous than Biosawar's law the answer is no just as Gauss's law is completely equivalent to the Coulomb's law this law is also completely equivalent to the corresponding Biosawar's law. Now let us let us see why we start sometimes using this again the problem of symmetry supposing I have a long straight wire. Now if there is a long straight wire long enough so that I do not have to worry about the edge effects. Now the direction of the magnetic field due to this current is already told to you by the Biosawar's law because I have said it is dl cross r dl in supposing this is my z direction supposing I am looking for a point here. So you take any current element going like this dl cross r has to be perpendicular to dl as well as this r since both these lines are on this plane then my magnetic field contribution due to this will be in a plane perpendicular. Now this is from the Biosawar's law is the content of what we call as the right hand rule what the right hand rule tells you is this that you assume that you are holding that current wire I am assuming that there is a plastic jacket on it so that you do not get a shock and then curl your fingers around it and let the thumb point in the direction of the current in that case the sense in which your fingers of the right hand curl gives you the direction of the magnetic field. So once again the right hand rule which you have learnt right from the school days is also coming from the Biosawar's law and notice this so therefore I will since the direction keeps on changing it is much easier to take the azimuthal obviously it is an azimuthal direction so the magnetic field direction will be the phi direction. So how do I you have done this several times that now use the symmetry that all over if I take a circle a long straight wire cannot distinguish between different points on that circle it is an extremely symmetric situation so I have a long straight wire and I am looking at a point P now draw a perpendicular and supposing this distances are what I am trying to say is all points which are at the same distance from the wire they must be identical the direction we have already talked about. So what we do is this what is B dot dl now the B dot dl in that case must be whatever is the magnitude of the magnetic field at each point integral if you take it times the length and which is nothing but 2 pi times r and this we said is mu 0 times i. So therefore the magnetic field vector is given by mu 0 times i divided by 2 pi r along the direction of phi. Now I know that you are all familiar with this but the point is that what we have tried to point out here is Ampere's law comes from Beyer-Sawart's law the right hand rule comes from the Beyer-Sawart's law like the Gauss's law was a could be used in extremely symmetric situation I can also use the Ampere's law in symmetric situation. So Ampere's law has limited validity though it is equally powerful can be used gainfully in symmetric situation. Now notice I am not going to work it out because you have all done this supposing I am looking at a circular coil in the plane of the paper carrying a current. Now can we use Ampere's law here very interesting point you realize if you are working only along the axis only along the axis then there are enough symmetry there but you will find that you will be normally required to work this problem out using a Beyer-Sawart's law as I said most of you have done such things so I will not be unnecessarily reporting it. Let me take another device which is very frequently used and all of you have been worried about it and this is what is called solenoid remember when this name solenoidal came up lots of questions. So basically a solenoid is a device like this now this is what is called a solenoid this is the picture of a solenoid and so what we do normally is to say that look what is actually the way the solenoid is happening remember that the current must enter one point and come out in each turn. So this is what I am showing here that this is actually the current centering not the magnetic field the current enters here takes a turn and comes back here and like that all the turns go. Now if you look at a length much larger than the width now you will be able to see show that the solenoid gives you a constant magnetic field. Now these are important that how does one generate constant magnetic field there are other ways which we will talk about inside the solenoid today I am talking about it for a reason inside the solenoid the field is constant and as you are all aware that it is given by mu 0 n i along the axis of the but point is that one can show because of the fact that the current is going in one case coming out in the other case the net effect is to give you a 0 magnetic field outside the solenoid. So for the solenoid the magnetic field is constant mu 0 n i inside the solenoid outside it is 0. So let us remember that for something that will be coming up with later. Now let us come to another interesting point the point is the following let me try to illustrate it here itself that supposing I have two current carrying wires loops whatever arbitrary shape loops I am trying to find out what is the force exerted by one current loop on the other. So I know that that if I know the magnetic field at a particular point then the force experienced by this length element d l 2 due to the magnetic field here right remember that the magnetic field exerts a force on a moving charge and currents are nothing but moving charge. So therefore this is basically your Lorentz force law. So if I am talking about that what is the force that is exerted on this element here due to the magnetic field here. So magnetic field by this element is d b 1 and here I have got I 2 because this is I 1 on this and I 2 on this. So my force is given by I 2 d l 2 cross d b 1 and you we have already got from bio sub art's law an expression for d b 1 we did that and then take the all the you know integration over the loop which is producing the magnetic field you get force on 2 due to 1 is given by the integral. Now there is already an integral over d l 2 because I am looking at what is the force exerted by this loop on that loop. So I have an integration over d l 2 as well and there is an integral that I require to create my magnetic field. So I have got f 2 1 is mu 0 by 4 pi I 1 I 2 this is fairly straight forward but that is not the reason I wrote it down. I wrote it down to find out that if this is f 2 1 what is f 1 2 that is what is the force exerted by this on that now clearly you will say interchange 1 and 2 I interchange 1 and 2 there is no change in this I 1 I 2 because it becomes I 2 I 1 which is the same I get this as d l 1 cross d l 2 cross r 1 minus r 2 divided by of course again it is the same because r 1 minus r 2 modulus is the same as r 2 minus r 1 modulus alright. So I have to show that f 1 2 is equal to f 2 1. So how do I get f 1 2 f 1 2 is this what is f 2 1 interchange 1 and 2 all through out. Now in order to show this you again need some smart algebra. So you notice this d l 2 cross d l 1 cross r 2 minus r 1 this is interchange. I can write this that is why I told you that a cross b cross c formula is an extremely important relation what is a cross b cross c a cross b cross c that is a cross bracket b cross c is b a dot c minus c a dot b and this is all that I have you notice already the second term is already anti symmetric that is if you interchange 1 and 2 it just because the minus sign is the first term which is creating problem. But I need an integral now when I need an integral I notice that this quantity is frequently written as minus gradient of 1 over r 1 minus r 2. So this is what I have got here the moment I have a closed loop a line integral of a gradient clearly I can use stokes theorem convert it into surface integral but then it is curl of a gradient which is equal to 0. So this first term will turn out to be good and the second term is automatically anti symmetric. So in other words the Newton's third law is valid for the forces between the currents. Let me come to an interesting point now point that I am trying to tell you is this that where did this magnetic field come from there are no magnetic monopoles electric at least we had the electric charges. So we said that alright they give rise to a field what is this magnetic field how do you understand now this is something which you try to understand because this is not what you find in a standard text book the magnetism phenomena it arises essentially due to electric forces because the only force that we know okay basic force as I told you there are only four types of forces in nature you have of course coulomb force that is the only force that we know off. So magnetism also comes because but how does it happen I will show you by a simple example not a very rigorous one but a simple example that magnetism arises because of relativistic effect. So in order to do that let me give you a interesting problem supposing I have two long straight wires okay just a little while back I had written down it is though I said I would not derive it but it is important. So suppose I have two parallel wires carrying current now the we want to find out what is the force due to two parallel wires carrying current in this case it is I can calculate again by field and things like that I can show that two parallel wires carrying current separated by a distance d exert a force on each other of course in this case an attractive force which is mu 0 times the strength of I 1 strength of I 2 divided by the distance between there are factors like 2 pi etc. So mu 0 I 1 I 2 by 2 pi d the parallel attractive forces are there okay this is this is an expression which I will require. So I have told you now that consider two line charges remember what is a line charge a long straight wire which carries charge there is no current there just charges static charge let the charge density of this be lambda 1 let the charge density of this be lambda 2. Now this person here which I call as s is observing this situation from a laboratory so what does he see he sees that there are two line charges okay now let us look at let us call this frame s in the frame s because there are two static line charges there are only electric fields and we know that the field due to the line charge number 1 okay at the location of 2 the distance is d is given by lambda 1 by 2 pi epsilon 0 d r okay now what is the force that this exerts on this now remember that these are only fixed charges so therefore the force is purely electrostatic right. So and this has a charge density lambda 2 so that means it has a charge lambda 2 l so therefore the force as we just now calculated is given by lambda 1 l I mean not calculated just now but when we did line charge for electrostatics we said it is lambda 1 lambda 2 l divided by 2 pi epsilon 0 d so this is the expression for the force between the two infinite line charges alright now let us do the following. Let us look at another observer who is moving along the length of the wire with a velocity I will call this observer as s prime now you have learnt or you will learn in some of these recent courses that some of the courses to be given by professor Prashad on relativity that when the this person looks at this length this means the length gets contracted this you have learnt in special theory of relativity in your master's program now if the length gets contracted since the total charge cannot be created this means that my charge density increases the factor by which the length contraction takes place as you all know is gamma which is equal to 1 over square root of v square minus by c square. So therefore since the amount of charge must remain the same the charge density becomes gamma times lambda and the length has become l by lambda length has got contracted charge density had got increased so that this into this still remains l times lambda as we want now remember that lengths are contracted along the direction of the velocity but this distance d is perpendicular to the direction of the velocity. So therefore there is no contraction in this direction alright we still talk about what is the force as seen by the observer in s prime now once again look at what we did this this we said well instead of lambda 1 you have gamma lambda 1 instead of lambda 2 you have gamma lambda 2 instead of l you have got l by gamma and of course d did not change it does not get contracted so if you cancel the gammas 1 gamma you are left with gamma times l. So the force between 2 wires as seen by the fixed observer is f as seen by the observer s prime is gamma times l. So what is the problem 2 observers look at things differently the problem is the following if you remember this is gamma times l if you look up your relativity books which I presume it will be done towards the end here you will see that the transverse force remember the force is transverse it is perpendicular to the direction of the length the transverse force f becomes f by gamma not f into gamma. So what has happened is my observer s prime has found this force to be f into gamma whereas theory of relativity told us that it should be a bi gamma now obviously the theory of relativity is right but this how does s prime look at such a situation. So s prime says that look I can understand gamma f part of it but I should have actually got a result f by gamma now the shortage that he has the difference between f by gamma and gamma f which is able to explain he is able to explain by electrostatics and he is not able to get this. So he postulates there is an additional force which I do not understand how much is that additional force according to s prime. So f by gamma is what he should have got gamma times f is what he is getting. So take gamma f common write it as 1 over gamma square minus 1 I have given you the expression for gamma it turns out to be v square by c square with a minus sign. So this is f prime v square by c square I already gave you the expression for f prime put it there. So this is now what you are getting because I take i equal to lambda v I have converted this into i 1 i 2 l prime by this. Now remember immediately that this expression is what we wanted right I had shown that the magnetic force between them is given by the magnetic field if you remember is mu 0 i by 2 pi r etcetera. So therefore the role of mu 0 is taken by 1 over epsilon 0 c square remember mu 0 appears in the numerator. So mu 0 equal to epsilon 0 c square is actually a rigorous result from relativity. So summarizing the observer fixed in the laboratory sees only a electrostatic forces between charges the because that observer is moving he also sees that charges are moving with relative to him. So that is equivalent to a current so in his frame of reference there is also a magnetic field. The last thing that I want to do today is the definition of a vector potential. So we had curl of b equal to 0. So in electrostatic field we had said curl of e equal to 0 this enables us to define a scalar potential e to be equal to minus gradient of 5. But magnetic field is not irrotational that is its curl is not equal to 0 magnetic field on the other hand happens to be a solenoidal field. So that del dot of b equal to 0 now if del dot of b equal to 0 we ask can we define something similar. Then I know recall that divergence of a curl is identically equal to 0. So I will say that b then must be given by curl of some field a this is what is known as a vector potential. I now recall for you the famous Helmholtz theorem which said that a vector field can be split into two parts one part a solenoidal the another part that is del dot of b equal to 0 another part which is your irrotational. But here you notice that I only have b equal to del cross i. So in order to fix b uniquely I need to define another field say this is a field whose divergence is equal to 0 basically I have del cross of a right curl of a is. So b has a physical meaning now what about a notice a curl of a gradient is also equal to 0. So what I do is I say that look if curl of a gradient is equal to 0. So I will write down del cross b that is equal to del cross del cross a and I told you this is del of del dot of a minus del square a that is equal to mu 0 j. So what we have been saying is this that you can add to a a term which is gradient of a scalar field ok you can add so a is not unique for a given b. This is very similar to what we had talked about in case of a scalar potential as well we have said that gradient of phi does not specify phi uniquely only thing is there it was a much simpler thing we said that the potential could be differ by constants and it will still give you that the same thing. So what we say is this that to a to a we can always add a term which is gradient of a scalar. So it goes like this now so suppose you have you have got like this you say that look I want to choose I want to choose del dot of a equal to 0 this is incidentally called a Coulomb gauge. So suppose I want to so this thing that to a we can always add a term which is gradient of a scalar this is what is called a gauge choice. So one of the gauge choices is for example del dot of a is equal to 0 now notice we said that a vector is completely specified by giving its curl and divergence. Now we are saying here that look you have chosen a curl of a is given to you curl of a is b which is physically meaningful I can still choose the divergence of a as I like that is what we said this gauge is known as Coulomb gauge and most of the problems people work out in Coulomb gauge the question then is that can you always find a such a choice. Now what we are saying is supposing I would like to have the Coulomb gauge all right but I happen to have chosen a gauge which does not satisfy del dot of a equal to 0 the question I am asking is can I always do that supposing I have chosen a now let me say all right choose a different a given by a prime equal to a plus some gradient of some scalar sign I have not yet said what is sign then del dot of a prime remember I have chosen an a whose del dot is not 0 because and I want to find out can I choose by some method for that problem a vector potential whose del dot that is divergence is 0. So, let us find out what is del dot of a prime it is obviously del dot of a plus del square psi is what I have got now I am saying you have already chosen an a. So, we have chosen a del dot of a is not equal to 0 but I know what is it. So, therefore, if I am to choose if we choose del square psi I want to choose a function del square psi such that it is equal to minus del dot of a this will give del dot of a prime equal to 0. So, what happens what happens is this you have chosen a gauge corresponding to your magnetic field you know what it is this function you know. So, if you plug in to this equation this function that you have chosen then all that you need to do is a solution of a Poisson's equation and the solution of Poisson's equation can always be found. So, basically what we are trying to say is this that even if you have started with a gauge for which the Coulomb gauge condition is not satisfied but you are in love with Coulomb gauge you can always by this process go over to a Coulomb gauge. So, I can always choose the gauge that I want do I have an expression for a the answer is yes again. So, maybe I can go back. So, this is what we pointed out that you can always do a choice. Now, remember that for the scalar potential we had obtained an expression like this that phi at r is 1 over 4 pi epsilon 0 rho r prime by r minus r prime and integral rho r prime by this. We had shown that Bayer-Sawart's law gave us this expression for b of r that tells me that a of r is given by an expression like this mu 0 by 4 pi j of r prime r minus r prime d cube r prime. It is important to realize that the vector on the left hand side is a the vector on the right hand side is j of course, this is being integrated but let us look at some interesting consequences there are. Take for example, a long straight wire now look at this. Now, if I have a long straight wire then the direction of j of r prime is fixed. So, suppose my current density is in the z direction then the because this direction does not change integrated the direction will not change. So, a's direction will be same as the j's direction which means it will be also in the direction of k. But let us look at whether I can do a calculation like this or not. You will notice that this is not going to give you anything. If a's in z direction j is in z direction and a's in z direction and so write down that j of r prime because the direction is fixed is your current divided by the area and so that is what we have done there. Now you do that integration you get it infinite. This is obviously not a good thing but you remember which said that choice of a can be different. So, the question is is there a way of finding out a vector potential which will give me the same magnetic field. Now there are many ways of doing it. So, this is your expression for the vector potential. Let me go back to the magnetic field. Remember it is a long straight wire. So, the magnetic field is known to be to be mu 0 i by 2 pi r along the phi direction the direction in which your fingers curl and by definition then it is del cross of a r the phi component of that. But go to cylindrical coordinate your cylindrical coordinate as you know is rho phi z. So, write down what is the phi component. Remember the phi component is dou by dou z of a r minus dou by dou r of a z but we have seen that a is along the z direction only. So, this term is not there. So, I am left with d a z by d r is mu 0 i by 2 r you integrate this and you find out it is mu 0 i by 2 r log r. So, this is the relevant thing and of course, I could choose any gradient that I like. The only other thing that I want to talk about in the last 10 minutes that I have is that let us return back to the question of a solenoid. Remember what I told you and this is a very important point which because of which I am raising this issue. Remember I proved that or we you all know that the magnetic field inside a solenoid is along the axis and this constant. So, magnetic field outside the solenoid is 0. Look at this picture and try to see let us calculate what is the magnetic vector potential corresponding to this situation. Now, first let us do inside this is almost doing like ampere slope. So, around the axis inside have a circle of radius r or l does not matter. So, I am trying to find out how much is integral a dot d l integral a dot d l by Stokes law Stokes theorem is del cross a dot d s which is equal to b dot d s which is nothing but flux. So, I notice an equivalent expression for flux is a dot d l just as it is b dot d s. Now, we said that the direction of a is the same as the direction of the current and in this case the direction of current is the phi direction. So, therefore, a is a phi phi and b is along the z direction. So, therefore, if I do a line integral of radius s I get 2 pi s times phi a phi which is equal to this is equal to your phi b what is phi b phi b is the flux. So, how much is the flux? The flux is mu 0 n i times pi s square because flux is magnetic field which is mu 0 n i times pi s square which is the area. So, the magnetic field gives me a an expression for phi which is given by a phi given by mu 0 n i by 2 into s just fine more interesting thing is supposing I did the same job outside. Now, if I take a outside loop the for s greater than r the line integral is still 2 pi s s and but the flux is only due to the part which is inside. So, this is mu 0 n i times pi capital r square where r is the radius of the this circular cross section of the solenoid. So, that gives me that the azimuthal component of phi outside the solenoid is given as mu 0 n i r square by 2 s. Now, the summary of it is the following that for a solenoid the magnetic field long enough solenoid without edge effect the magnetic field is constant along inside the solenoid and along the axis of the solenoid outside it is equal to 0, but the corresponding vector potential is given by mu 0 n i s by 2 s is the radius radial distance inside the solenoid outside the solenoid unlike the magnetic field it is not equal to 0, but it has a value which is mu 0 n i r square by 2 s. Now, the next question is that you would say so what after all the vector potential is not a physical thing. So, therefore, a vector potential outside the solenoid is not equal to 0 what have we lost? The there is some very interesting experiment and the experiment is shown here it is in this way that imagine I am doing a Young's double slit experiment, but I am doing it with electrons. Now, you know that one can do since you have been having a course in quantum mechanics you know that the electrons also have a wave character. So, therefore, whatever you did with light for the Young's double slit experiment you can do it with electrons also the corresponding deep broccoli wavelength waves would also so diffraction. Now, what I do is the forming experiment. So, here I have an electron gun which is going and in its path I have put a very small solenoid initially there is no current in the solenoid I mean in the turns of the solenoid. Now, I will get the fringe pattern because now imagine that what I am doing is that let me just talk about two paths supposing one path is going around the left of this solenoid another part is going around the right of the solenoid and I get a path difference the presence of the solenoid is unimportant I have taken it negligible whatever is the path difference based on that I will get a Young's pattern. Now, what you do is put a current in that solenoid. Now, if you put a current in the solenoid you will see so what the the electron beam is passing outside the solenoid. So, therefore, if there is no path difference which is introduced because of that. However, it turns out that you will find that the fringe patterns will now be shifted the reason why the fringe pattern will be shifted is because though the magnetic field inside the solenoid is uniform outside it is equal to 0, but the a the vector potential is not equal to 0 outside and this would add to a phase. Now, this is well known as the Aranda bomb effect that you will find the fringe patterns shifting because of the fact that there is a vector potential outside which introduces a path difference or phase difference between the two paths around this because they are not exactly at the same point. So, therefore, even though the vector potential has been taken to be a sort of a technique for getting magnetic field and all that the reality of the vector potential is very much there and it can be experimentally tested. I would like to stop here and from tomorrow the first time I will of course take up your question and then we will go over to time dependent phenomena. Thank you.