 Good morning to all of you again. So let me take up the questions which came up through the chat. Some of them are fairly simple. So first question that I take up is a very clarification question from BDT college centre number 1176. The question was that when I talk about force exerted by Q1 at Q2. If Q1 is at position R1, Q2 at R2 do I take R12 or R21? This is really a fairly straightforward question. Supposing you have a charge Q1 located at a position vector R1 and you want to find out what is the force on another charge which is located at a position vector Q2. So this is R2. So by vector addition you notice that this is R2 minus R1. So the point is that when we write down the force between Q1, Q2 we will write down the force as 1 over 4 pi epsilon 0 Q1, Q2 divided by R2 minus R1 square and R2 vector minus R1 vector. If Q1, Q2 is they have opposite sign then of course automatically the direction will change. Now the point is that what do you want to write it? It is a question of a shorthand notation. Usually this is denoted by R12 but if you want to call it by R21 anyway you have to make a statement that this is what I mean. It is your choice. The shorthand notations never have any standard rule. So this is just Coulomb's law explanation sorry this is cube. After number 1313 the question is why do you eliminate del dot a? You remember that the question was based on del dot of a equal to 0 which I said is a there is a gauge choice. I said this is Coulomb gauge and I mentioned that if you have chosen a particular a then to that a I can add another gradient. So I could define something called a plus grad psi. So then we said that look we have this this is called a gauge choice and I explained why because you see the physically the only thing that is important is del cross a and del cross a is your magnetic field b. Now you cannot eliminate it. You cannot choose del cross a to be something because this is the physical quantity which you are interested in calculating. So the choice is what is the del psi that you choose and that I said I even proved yesterday that even if initially you have chosen a gauge for which del dot a is not equal to 0. This is a choice you do not want it do not take it. Even if initially you have done that later on by solving a Poisson's equation it is possible to define an a prime such that del dot of a is equal to 0. So this is always possible and so this is what makes sense physically. This is a physical field. Now in this connection I would also quickly try to answer because this is not really within the scope of my lecture. However yesterday I was giving an example of how one can establish the reality of a and the example that I gave in that connection is to say that suppose you are doing a Young's double slit experiment but this time with let us say beam of electrons. Now what I do is this that supposing I have a solenoid and I sort of make the two beams one go around it like this and another go around in front of it. And the solenoid is small enough so that the physical path etc. does not matter. So when you look on a screen there would be interference pattern on this ignore the solenoids current there is no current in the solenoid. Then I made a statement that look if you put on the current in the solenoid then you find that the fringe pattern is shifted. As we know that shift of fringe pattern is connected with the presence of an additional phase. Now there is nothing that has changed in this problem other than the fact that this solenoid has been given a current. As I said this is your Aron of Bomb effect. Now the question that has been asked by somebody I have forgotten the name now but the is what gives you this change of phase. Now that is not a question which unfortunately I can answer within classical electromagnetism. The problem is actually the following that what you try to do is this that this is a usually a quantum mechanical problem and you try to solve a quantum mechanical problem which take this as a free electron take this as another free electron beam. As you know the free electron wave functions are given by ik.x this has been talked about in your quantum mechanics lecture. So what we do is this that we rewrite the Hamiltonian of the problem instead of p square by 2 m in the presence of a magnetic field in it is known that the kinetic energy term becomes 1 over 2 m p minus E a by c where a is actually the vector potential. This is of course very well known to anybody who has done even a course in classical mechanics. So basically you have to solve the Schrodinger equation in the presence of a vector potential. I will little while later point out how to calculate how to write down a vector potential for a constant magnetic field. Now this is a easy problem to solve and then what you find is that the wave function that you find there the solution of the Schrodinger equation that you find there it picks up an additional phase because of the presence of it. Now and it is that which is responsible for giving me the fringe shape. The point that I was trying to make is that since it is a solenoid the solenoid has a magnetic field inside B is not equal to 0 inside you know that this is given by mu 0 n i. But B is equal to 0 outside the solenoid but when you calculated we found out an expression for a both inside and outside the solenoid but a is such that del cross of a is equal to mu 0 n i inside and is equal to 0 outside. Remember if a del cross of a quantity is 0 the quantity does not have to be 0. So a has a value even outside and these two beams one on that side one on that side they are not going inside the solenoid. So they are not being exposed to a magnetic field but they are being exposed to a vector potential. And the vector potential as I told you if you solve this problem in quantum mechanics picks up an additional phase and that is the reason why we get a fringe shift. As I said that this question cannot be answered completely classically but since you have done several lectures on quantum mechanics I thought it is not too inappropriate to mention it. Next question came from Netaji Subhash Chandra Bosch Institute, Guaraya 1 to 6 6 centers. This question is on magnetic monopoles. Remember I mentioned to you that monopoles have not been found to exist. Now the statement is monopoles have not been found to exist I am not saying monopoles do not exist because I do not have that much of knowledge. So the point is that the originally the you will realize that the electromagnetic Maxwell's equations are asymmetric with respect to magnetism and electricity. And for the simple reason that electric single charges exist but magnetism north poles and south poles must coexist together. This is what has been found nobody has really found out the reason why. Now ever since Paul Dirac Dirac of course you must have heard the name he had suggested that there is no reason why monopoles do not exist physicists have naturally been searching for monopoles. But so far the only problem is what has been found is that the monopoles if they exist this was born out even by Dirac's calculations they will be massive by that what I mean is their energy will be huge. Now in fact the energy will be so huge that even the fastest accelerators may not be able to detect it. So therefore what one tries to do is that is there any monopoles which or are where there any monopoles which existed in the early universe. Now as you know that the physicists have been trying to reconstruct what happened in the early universe by many types of experiments and particularly experiments connected with LHC which takes care of very high energy particles. But so the searches are going on and the physicists at this moment really speaking there is no great reason to believe that they will be found but nevertheless the there is no theoretical reason why the monopoles do not exist. So therefore the answer to that question is let me end it with a question mark. The next question again is a quick question 1256 Hyderabad the question that they asked is that we discussed the method of images and in both cases I used gave two examples in both cases I gave the example of a single charge located in front of or in the presence of a conductor. Actually it is not necessary that it is conductor method of images is even applicable to dielectrics but let us not worry about that now because I did not have time to go through it. But in any case the method of images was illustrated with the presence of a charge the question that our friend from Hyderabad wanted to ask is supposing instead of a single charge I had a charge cloud electron cloud for example how would these things change. See the point is that technically there is no reason why the method of images which simply use two principles one principle it used is that there is a uniqueness theorem. So uniqueness theorem simply said that if Laplace's equation has a solution and I have found a solution that must be the only solution there are no two solutions of Laplace's equation correspond to a given boundary condition. So therefore there is no reason theoretically why when you put a electron cloud you will not be in a position to use the method of images. But however the problem is that you see if you put an electron cloud the image will also be a cloud. So therefore you are going to lose the simple principle that we had that there was one ray coming from the image charge and another from the charge which I call as the object charge and these two the forces they cancelled because now there will be multiple forces between different members of that charge and one does not quite know how to handle it mathematically. So method of images is a technique which is unlikely to work for this case though there is no reason why theoretically it is infeasible. The next question is from centre number 1266 which is again I think from Calcutta again the question is a little tricky question. The question is why are there no dipole moments in dielectrics without an electric field or why is there no dipole moment in the absence of E. So let me first make one statement this statement is not even true but on the other hand why did I say it is an interesting question. Why the statement is not true is this the molecules do have permanent dipole moment in fact classic example is water molecule. So if you look at the structure of water H2O so you the molecule is not a symmetric molecule. So you have this oxygen here and the two hydrogens they are at an angle of 105 degrees to each other. There is a hydrogen here there is a hydrogen there. An asymmetric molecule by in principle has an dipole moment and in fact water dipole moment has been measured and it is 6.2 10 to the power minus 30 coulomb meter. So this is actually the permanent dipole moment of the water molecule. If you look at molecules which have mirror symmetry for example symmetric molecules like oxygen, nitrogen, carbon dioxide, carbon tetrachloride these are all mirror symmetric molecules. These are no dipole moment and the reason is very simple that if the molecule is symmetrical then of course the charge centers will be at the center itself. I still think that a statement I should make on why did this question come up. So why no dipole moment this question came up primarily because of certain popular writings that have been coming up but on the other hand because you some of us may not have in a background to understand in what connection they have been written. So the point is there has been a question of why the fundamental particles by that I mean the you know the constituents of matter the why do not they have dipole moment. So a distributed system like molecules of course have a dipole moment but on the other hand the dipole moment has not been found in things which are fundamental the reason is a little tricky. In fact once again there is no reason why it should not be there but they it has not been found. The for reasons that I cannot go to right now it turns out that they of course really it is sort of intuitive that having a permanent dipole moment would violate what is known as parity situation and they actually it turns out you probably have all heard of that parity can actually be violated but on the other hand people have not been able to find the situation of having a dipole moment in systems like this. Another quick question so I wanted to point out that the question on dipole moment was interesting but the way it has been phrased that why are there no dipole moment in molecules is incorrect because molecules do have permanent dipole moment in the absence of electric field but the write ups that you occasionally read are connected with that there are no dipole moments in the fundamental particles of the nature. The next question again is a very straight forward query centre number 1313 it is asking that why do you put a negative sign before the scalar potential. Why electric field should be minus gradient of five quick reason no particular reason in fact the mathematicians use the definition of scalar potential as just E equal to del five but you realize the mathematicians are not dealing with physical fields they are dealing with that if there is a vector field if there is a vector field whose curl is equal to 0 then that vector field can be expressed as a gradient of a scalar function scalar field that is the way the mathematicians look at that is not quite the way the physicists should look at it the physicists would say they put a minus sign for the simple reason that look at say I have a gravity now I know the gravitational force f is minus mg z z is unit vector z it is downward now this I cannot change its direction is not now if I now define a potential without this minus sign then suppose suppose suppose we say that my gravitational force is written as instead of minus del phi I write it as plus del phi the expression for my phi will be minus mg z that is a height or h plus a constant. Now there is no problem there accepting this would mean that if the height decreases the system has more potential unless you want to change the whole thing this type of changes if you want would also need to be to correct for example the torque expression which is e cross p right p cross e it should now become e cross p. Now so therefore what I am trying to say this is just a convention this is just a convention this is to keep be consistent with the fact that we know that the potential energy is defined such that higher the height more is the potential energy etcetera. Another quick question was the from center well I am not very sure whether this was actually asked from 1313 or not but probably 1299 but there was one of them asked a question that the you know is the decomposition of a vector the same as resolution of a vector the answer is yes it is just a English language that we use ok there is a question on dielectric by 1 0 2 0 the question was that we talked about rho b bound charges volume charges and sigma b now the question is this that since there are no free charges inside a dielectric why take these 2 into account because the total charge enclosed is 0 that statement is not true see the point is what we said is that the total polarization charges 0 that the polarization charges 2 components the one of the component is the volume component the other component is the surface component now it is the volume component plus the surface component total charge that would cancel out in a neutral medium but on the other hand since I know for example when I have a uniformly charged sphere non conducting sphere of course I calculate that how much a charge is contained inside at a distance r that is a separate question but you see if I do this take such a Gaussian volume then it does not enclose any charges on the surface if you take the total surface area of course both of them have to be taken care of and there are no polarization charges inside but because there are volume charges and surface charges so you have to put it in the integral expressions there ok I think another quick question before I take up the last question which will take a bit of a time is from center 1 2 2 5 which asked why did I use the word virtual charge the reason for using the statement virtual charge phrase virtual charge was primary because I have been using the language of optics you know that when you stand in front of a mirror you have an image and that image is virtual the image is virtual because you will not be able to put a screen at an equal distance behind this mirror and catch it you know that real images can be focused on a screen the virtual images cannot. So in that sense that an image by a mirror particularly plane mirror because in case of spherical mirrors we may even have situation where we have real image these are virtual. So in the method of images I have been saying imagine a charge q prime put for example in case of a plane behind the conductor now we know that behind the conductor behind the conductor I have material body of the conductor there are no real charges floating around there. So this is an imagined charge that if it were there then I will be able to do it. So that is a quick question next question was the lots of questions but one of the centers I have is 1 1 5 8 the question asked is can you discuss the boundary conditions that are applicable at the interface between two dielectric media the reason why I am taking this question in a bit of a detail is that this is a very relevant and a practical situation. See as you know light rays are nothing but electromagnetic waves I know for example when you have a ray of light falling on let us say a glass medium interface between glass and air let us say then what happens is that here is a ray coming in and you have of course a reflected ray and you have a transmitted ray. Now we will see later that if you want to prove this is called well one is law of reflection angle of incidence is equal to angle of reflection the second one is called the Snell's law which simply says that the velocity in the second medium divided by the velocity in the first medium or rather the other way round velocity in air divided by the velocity in this is the refractive index of the medium and that is equal to sign of the incident angle divided by the sign of the refracted angle. So this is your theta i this is your theta transmitted. Now we have of course in school learnt that this is given but this is not really given one can derive this from electromagnetic theory and the way one derives them from electromagnetic theory is to find out what is the ray of light which is falling is an electromagnetic wave. So therefore I have to worry about electric field. Now this electric field when it falls from one dielectric medium this in this case it is air to the second dielectric medium in this case it is glass then I have to satisfy the condition at the surface certain amount of continuity conditions have to be applied on the surface. Remember if this was an interface between air and a conductor then we would have said that if there are free charges on the surface of the conductor then the normal component of the electric field is discontinuous but the tangential component should be continuous. It is the tangential components continuity that will lead us to the law of reflection refraction etcetera. Now so therefore it is important to understand what are the boundary conditions at the interface between the dielectrics. So though it is something which I took it up previous day but because it takes a bit of a time I did not take it up yesterday. So let us look at supposing these are two I have two dielectric medium remember that is no longer conductor. So let me take one of the dielectric medium to be air and the other one glass or whatever you have. Now again if you recall the way we found out the continuity condition in case of a conductor we said that take a Gaussian pillbox part of it outside the conductor and part of it inside the conductor. Now we had a great advantage there because as we know inside a conductor there were no electric field. So now you remember what we do in case of a Gaussian pillbox we ultimately reduce the size of this to 0 negligible. So that there is no contribution to the flux electric flux from the sides of the cylinder and there are no contribution to the flux in case of a conductor from the part of the base which is inside the conductor. Now so I have only one side. Now this is not really true when I talk about the situation that happens in case of a dielectric. So let us let us proceed with that again. So let me draw that picture once more. So I have this Gaussian pillbox part of it is inside part of it outside. The way this derivation differs from what we did in case of a conductor is I again as I take this height to be negligible there is no contribution to the surface integral from the two sides. But now I have both the faces because previously the electric field was 0 inside the conductor but now I have polarization charges. Now I have polarization charges inside the dielectric. Let us assume for the moment that there are no free charges on the surface. There could be free charges on the surface. Now if no free charges on the surface then I deal only with the polarization charge. Now so what would be my situation notice this that the normal on this face is n normal on this face is opposite direction. So n prime is equal to minus n. So what we said is this that look what is my del dot e volume integral. So we said that this is equal to 1 over epsilon 0 rho b this is nothing but my Maxwell's equation del dot e is rho divided by epsilon 0 I have integrated both side but rho b now is not say remember my Maxwell's equation was del dot e equal to rho by epsilon 0. But now I have only bound charges there are no free charges because I am talking about dielectrics and I have said that there are no free charges. So if there are no free charges del dot of p is simply equal to rho bound divided by epsilon 0 I have integrated both sides. Now I also showed yesterday that this is equal to 1 over epsilon 0 rho b is minus del dot of p d q 1. If you compare these two expressions now what I am going to get the left hand side becomes by divergence theorem is e dot d s and the right hand side has now picked up a minus sign. So I have got minus 1 over epsilon 0 p dot d s now but look at this quantity what is this quantity this is minus 1 over epsilon 0 p dot d s is nothing but the surface charge density that we have in the problem. So that is delta b times the of course the surface area delta s. Now if you now realize what is e dot d s by taking the fact that there is this surface and that surface take the surface to be small enough. So you get e outward minus e inward normal component of that because surface integral is equal to sigma b divided by epsilon 0 this is what you get. So this you notice this expression is essentially the same as that of a conductor. The expression is the same as that of a conductor accepting that instead of the surface charge density being free charge density I now have a bound charge density. Now supposing I had both supposing my surface had some free charges then this would be replaced by sigma free plus sigma bound both the charges are there in the contribution and the reason for this minus sign is remember sigma b outside the dielectric is 0 there is sigma b only from this side. So it picks up that additional minus sign there. So this is this is the way the normal component of the electric field will have a discontinuity. This is the same as that of the conductor and this is the correction due to the bound charges. Now this is very easy to write in terms of the d vector that I introduced we said the vector d is epsilon 0 e plus b. Now substitute these there in my definition and remember that outside the medium sigma b is equal to 0 a little trivial arithmetic tells you d outside normal component minus d inside normal component this is simply given by sigma f that is free sigma b cancels out because of this definition which means which is a very simple fact that if there is no free charge on the surface normal component of d is continuous. Because if sigma f is equal to 0 this quantity on the right becomes equal to 0 tangential component the same old rule is applicable e out tangential is still continuous about e these two are the boundary conditions to be satisfied at the interface between two dielectrics in our case may be air may be and glass. Now that essentially brings me to the end of the chat sessions that we had yesterday and if some questions have not been answered it means it has been answered by other people's question the other possibility is that the question does not quite fall into my whatever I am doing. Having disposed of the chats that came yesterday let me take off from where I started. So, I was telling you that this is the actually more in the sense of a problem that very frequently you use uniform magnetic field. And so therefore I need to have a ready made answer for what is the vector potential corresponding to a uniform or a constant magnetic field let us take the magnetic field to be in the z direction. Now my claim is a of r the vector potential at the position r is given by the cross product of b with r divided by 2 if you take this remember this is obviously not unique it is not unique because air is never unique to this I can add the gradient of any scalar function. So, look at what is del dot of a so essentially I am doing what is del dot of b cross r. Now if you look up any calculus book it will tell you that the del dot of b cross r it is more like a chain rule differentiation accepting that because there are cross product you have some minus signs coming it will turn out to be r dot del cross b minus b dot del cross r. Now it is a it takes one minute to calculate the del cross r that is the curl of the position vector and show that this is equal to 0. So, I am left with del dot of a equal to half r dot del cross b, but the if you look at the so this is 0 because b is constant. So, del cross of a constant field has to be 0 remember I am trying to find out the vector potential of a uniform magnetic field a constant magnetic field. So, I can differentiate a constant and get a non-zero value. So, this term is 0 this term is 0 because del cross of r is equal to 0. So, this choice of b cross r by 2 satisfies that 2 is unimportant, but we choose it to give me the correct magnetic field this gives me the right gain choice namely del dot of a is equal to 0. So, I have done that and del cross you can immediately calculate the del cross and show that this indeed is equal to b. Now so there are 2 choices which are very frequently done one is to say that x component of the vector potential is minus b y by 2 and y component is b into x by 2 this is one choice and this if you you can calculate del cross of a and you will find that this is equal to b instead you need not have these factors to there and make the second component 0 write this as minus b y 0 0 and again you will find that the del cross of a is equal to b. Now if you work both of these out you will find that the difference between these 2 form of a is that these 2 differ by a del of b times x y by 2 you can check it immediately because this is del of b times x y by 2 is just the difference that you are having. So, the reason I am pointing it out is very important to almost remember what is the expression that you take for the vector potential of a uniform magnetic field I mean either this or this this is more commonly used. Since we have been talking about boundary conditions we have finished about conductor we have talked about your dialectrics and things like that supposing I have a current carrying surface in this example. So, this is this the I have a sheet of current essentially the currents direction is coming out. So, please understand that this brown color thing that you are seeing actually represents the intersection of the surface with the paper or with the screen. So, therefore, the surface is perpendicular to the screen and so therefore, the direction of the current is out of the paper this is the direction I have taken the current is coming out of the paper. Now, I want to find out the boundary conditions that must be applicable on such surfaces. Remember again our statement that if I have a charged surface then my normal component for the electrostatics then my normal component of the electric field has a discontinuity, but the tangential component is continuous. In case of the corresponding magnetic situation the situation is reverse as you can see it and this is because of the Gauss's law. So, I take again a Gaussian pale box now remember my b dot d s must be equal to 0 always. So, therefore, if these heights go to 0 negligible then I have my b dot d s which is simply s times b 2 n minus b 1 n because the normal directions are opposite right, but now previously my e dot d s was equal to charge enclosed, but I do not have a charge enclosed now. So, when I had electrostatics we said the integral of e dot d s is 1 over epsilon 0 times charge enclosed, but now integral of b dot d s is 0 there is no charge enclosed. So, therefore, as b 2 n minus b 1 n the previous case we have e 2 n minus e 1 n is equal to the whatever is the charge density that you had, but now this side is 0. So, therefore, I have b 2 n is equal to b m b 1 n right. So, this simply means that the normal component of b is continuous. Supposing you are working in a coulomb gauge I know I have del dot of a is equal to 0 and you can immediately show that it means the normal component of the magnetic vector potential is also continuous. What about the tangential component? Remember the tangential component in case of the electric field was continuous the normal component was discontinuous. Now, here again I have the same situation, but my problem is now reverse my normal component of the magnetic field is continuous, but it turns out that if there are surface currents if there are surface currents the situation is parallel to what happened in case of surfaces having charges. Now, we are saying if I have surface currents then the tangential component of b will be discontinuous. Now, one can immediately see it why? See remember I told you the direction of the current direction of the current on the surface is coming out. So, let us call it the our direction is that of let us say in the S direction. Now, the normal to the surface is on this this is the surface normal and I define a tangential direction also. So, I such that I define a system of axis such that this S which is the direction in which is coming out the normal to the surface which is the intersection is in the plane of the paper and another tangential direction which is perpendicular to is this is the tangential direction. These are a right handed system namely t is equal to n cross s n is equal to s cross t etcetera. Now, what I do is this I do exactly what I did in case of electrostatic I take an amperian loop with small heights and I say integral of b dot dl equal to mu 0 i. Now, remember that in the other case it is important to understand why different boundary conditions are arising for electrostatics and for magnetostatics. See in electrostatics the normal component was discontinuous because e dot ds was not equal to 0, but b dot ds was equal to 0. But now because of the fact the electrostatic field is a conservative field curl of e equal to 0. In other words integral of e dot dl is equal to 0, but in this case I do not have that magnetic field is not a conservative field integral b dot dl instead of being 0 by ampere's law is mu 0 i. So, therefore if I take the amperian loop and write this down if the surface charge density is written by j. So, I have mu 0 j integral j dot ds and of course I have put in the area there. So, you take this integral out you know I mean contribution only from here and there that is b 2 t minus b 1 t t meaning tangential is given by this. So, there is a discontinuity there is a discontinuity in the tangential component of the magnetic field at the surface and this is simply rewriting it in the nature of a vector. Having done that I spent a lot of time in discussing dielectric. Now because the techniques are similar we do not need to spend a lot of time in studying magnetostatics also we have seen that every time I am teaching magnetostatics I am pointing out remember what happened in case of electrostatics because the mathematics everything is the same. Let us do that as to what is happening here. Now I have magnetic material of course you are all very familiar with the magnetic material. The magnetic material that we have are because of the following this. You see ultimately my matter consists of atoms and in a very crude level the atoms have electrons. These electrons are moving around now moving electrons are moving charges. So, they are responsible for atomic currents. So, basically as you have done the picture of an atom is around a positive charge the electrons or some charges is moving around. So, this part is understandable classically this is the orbital motion. There is a second motion there is a second contribution to the atomic current and that is due to the spin motion of the electrons. Now though in many school books you will find a comparison that orbital motion is like the earth moving around the sun the spin motion is like the earth moving about itself. Please understand this second comparison is a meaningless comparison. Second comparison because electron is not moving around itself and that is not the reason for giving me spin. Spin unfortunately is a concept which cannot be explained on the basis of classical physics. Spin is purely quantum. So, spin of an electron or other fundamental particles are intrinsic in fact you know that neutron has a spin magnetic moment though it is neutrally. So, we cannot really discuss spin in any understandable way when we are discussing classical physics, but let us look at what happens. Let us just talk about the orbital motion. Now you see so I have electrons moving around in a circuit enclosing obviously a certain area. Now when I put in a magnetic field there is an attempt there is an attempt to increase the flux through the atomic the whatever circuit which the moving electrons are occupied. Now this will mean this will mean that there is an attempt to change the flux through the area which is being spanned by the atomic movements. Now we will learn in our next lecture that there are you already know it anyway that Faraday's law will tell you that this means that the currents must change or adjust themselves to oppose such a. Now this is the reason for diamagnetism. Diamagnetism is there in all materials because the origin of diamagnetism is the Faraday's law. We will have more to talk about Faraday's law in the next lecture. The other thing is paramagnetism this is primarily I am talking about the spin motion of the electrons or the spin of the electrons. So, again if you want to look at it in a very crude way the picture that I gave for the dielectric is still applicable here that is what we say is that a random piece of iron is not magnetic and that is because a random piece of iron inside has presence of small domains each domain has a magnetic moment. But at an arbitrary temperature in the absence of the magnetic field these domains have their magnetic moments which are aligned in arbitrary random directions and as a result the total magnetic moment adds up to 0. Now that is a paramagnet that is a paramagnet. Now when you put in a magnetic field these domains the magnetic moment of these domains they try to align in the direction of the magnetic field which leads to a net magnetic moment of this sample. And there are of course other things that if you make the magnetic field strong enough once all the domains have their magnetic moments aligned there is nothing more to align. So, the magnetization saturates and that is what we are talking about is a ferromagnetic situation. But in any case there are two basic processes one is diamagnetism arising out of the orbital motion of the electron and because the Faraday's law will say that if you are trying to change the flux through the atomic circuits then the internal currents that are there they will try to oppose it. So all materials are diamagnetism to a greater or lesser degree. The paramagnetism as I told you arises because at a given temperature at an arbitrary temperature in the absence of a magnetic field the spin magnetic moments are aligned arbitrarily and it is only when you put in a magnetic field they would try to align themselves. So, this is what we talked about in the preceding couple of minutes. So, therefore, my total current my total current consists of two parts I have of course the conduction current which is due to the bulk transports that is the macroscopic charge transports. And I have the currents inside currents inside which are atomic currents, but since I cannot deal with them individually I deal with them in an average way. Once again I want to point out please understand what we talked about in case of electrostatics. We said there are charges there are free charges in a metal for example and in a dielectric there are these the charges which are bound. So basically I am talking about the same thing in a different language. We said there are these free electrons or free charges which are moving and that gives me what you can call as the conduction current. And then there are the internal currents which in some sense are bound to the atoms which are there. See there are also currents, but their currents localized within an atom or a molecule. Now I define the magnetization remember I define polarization as net electric dipole moment per unit volume. I define magnetization vector m as net magnetic moment vector per unit volume this is a v and as of course delta v going to 0. So once again a very parallel definition. So having done that let us calculate now the vector potential of a magnetic material. So what do you do there? So what I do is I split my material magnetic material into small domains let us take a infinitesimally small domain having a magnetic moment and the origin is here the position of the magnetic moment is here and this is the point at which I want to calculate the vector potential. Now remember my vector potential expressions I had given you earlier there was in terms of that it was in the direction of the current. So therefore the corresponding expression happens to be mu 0 by 4 pi m which is the magnetic moment cross r by r cube. So this is basically like Bayes-Watts law that the vector potential due to the magnetic moment remember the magnetic moment is defined in terms of the your the loop area times the current itself. So the vector potential which we had shown it to be in the direction of current. So this is the expression that you get. So if I now sum it over all the materials then this expression becomes then mu 0 by 4 pi magnetization at r prime cross r minus r prime by r minus r prime cube d cube r. What has happened? There was a conduction current earlier there that role has been taken over by magnetization current the expression the mathematics and everything remains the same. So this is my a of r. So this is what I did now what I do is I am not going to be redoing this algebra again because that is the reason why we spent a lot of time in working it out in electrostatics. Remember one of the ways in which this is done is write it as m r prime cross del prime of 1 over r minus r prime because this is r minus r prime by r minus r prime cube is nothing but minus gradient of 1 over r minus r prime. But that gradient is taken with respect to r here I am taking a gradient with respect to r prime. So therefore, mu minus r prime. So this is what it is now you you notice that this is something like a m prime cross gradient of something. Now you can use your vector algebra and say that del cross of scalar time a vector is given by this and I had essentially this expression. So it is this minus this and that is what I have done here. I have said this quantity is del cross of this into this and a compensating term from here. So that is the way it is. If you it would be instructive for you to compare this expression with what we did in case of electrostatics. We said there that there are two expressions there of course the cross did not come. But important point was that we split it exactly into two parts. One of the parts I kept it as a volume integral. The other part I converted into a surface integral and this is precisely what I am going to do now. So this is my expression since this is del cross acting on dq bar. This becomes a surface integral immediately the other one stays like that. The algebra the method everything is identical. So if you look at it now and come back to the comparison which we made we notice that there is a surface component here and there is a volume component and then you compare it with that what happens then. What you find is that if you calculate my del cross B I have consisting of two terms. One is the pure magnetization term. So this is my jm del cross m and there is a surface term. The surface term is what I showed you here. There is a surface term and there is a volume term and of course add to that my the usual conduction current. So this is the expression that we said. We said all right you now calculate the del cross of that quantity. Now you do the do a bit of an algebra and you find del cross of B happens to be equal to mu 0 del cross m and this quantity del cross m is my has the dimension of current density and is equal to jm volume. So the comparison is one to one comparison is one to one. I have the Ampere's law earlier which was del cross B equal to mu 0 j. But now I am saying in addition to the conduction current which is treated exactly the same way as we did earlier. Now there is a contribution to the volume current due to the atomic things because they are there. Now so let us summarize the Maxwell's equation that we have got so far. So my integral B dot dl which was earlier mu 0 i has two parts. One is mu 0 times conduction current plus im and im I have said is del cross m dot dl and that is nothing but m dot dl by again by applying Stokes law. So like we defined a displacement vector d or d vector d I can define a different quantity because this B dot dl and here im is m dot dl so but there is a mu 0 there. So take that mu 0 on the other side. So I define B by mu 0 minus m remember we have defined epsilon 0 e plus p but this is B by mu 0 minus m this is defined as a h vector this is defined as an h vector and the corresponding Ampere's law for h will be h dot dl equal to ic where ic is the conduction current. Now once again notice the similarity when we talked about the electrostatics our d vector integral of the d vector was determined by entirely by real charges true charges free charges. Now we are saying the h vector is determined in terms of true conduction currents only the free currents not by bound currents. So similarity the electric vector e determined by the actual net charge which has both the polarization charge and the free charge correspondingly the vector B is determined by total currents the corresponding expression for B dot dl is mu 0 times i this is the there is a full this is here both the conduction current and the internal currents but h dot dl like d is determined by only the conduction part remember that d was determined only by my free charges. Similarly before I close this not I do not close the session but I am closing this discussion of magnetostatics we just end it with some definitions that very often we have linear magnetic material where the magnetization is proportional to h you write magnetization that proportionality constant is usually called the susceptibility if it is a linear material. So therefore B can be written as in this fashion and this is then called permeability if you remember mu 0 was called permeability of the vacuum now 1 plus chi that is the susceptibility is the way the permeability factor is modified in a linear material not in general in general it could be different. So this is this is the way that we complete the discussion of electrostatics. So by finishing electrostatic electrostatics and magnetostatics we have been able to get into all the time independent component of the Maxwell's equation in next few minutes that I have and in the second lecture I will now bring in time dependence phenomena. If there are queries connected with what we are discussing right now okay I will take it up and of course as usual if I do not have time I will take it up and 1 3 3 0 which is Mothilal Nehru yeah go ahead please. What is the difference between insulator and dielectric how can we explain this relation? Are you asking what is the difference between an insulator and a dielectric? Yes sir yes. There is none see the point is this that the when you are discussing the material properties of a substance in relations to whether it is a conductor or not okay the materials are classified as conducting if there are free charges and if there are no free charges they are considered to be dielectric the word insulator okay as you know that in terms of conductivity properties the substances are classified differently they are talked about as conductors the insulators I mean and also there are semiconductors but you remember the semiconductors is closer to an insulator than to a conductor because accepting that the band gap there is substantially less than that of an insulator so basically an insulator is a perfect dielectric if you like but on the other hand if I have materials which do not have free charges I call it a dielectric the in study of electromagnetism accepting when we are talking about the conductivity property the word insulator is not used but but they are an insulator and dielectric are similar is that Meghna Saha Institute yes from somewhere in West Bengal yes go ahead. We talked about displacement vector and that is given by epsilon 0 e plus p yes where p is the polarization and we find polarization as a total number of dipole per unit volume yes right it is not number of dipoles it is net dipole moment added vectorially and then divided by the volume yes yes so dimensionally it is fine that it represents a dimension of field no problem physically how could we I mean realize that this polarization is some sort of field well you see the point is that what you know that we go back to the first class that we had first hour what is the field okay the field was there are we talked about scalar field we talked about vector field and things like that okay the vector field is a field which is a point function defined at every point but has a direction okay so that was a vector field now the polarization as I have defined remember I said that per unit volume yes but we also said that it is defined in a point way in other words what you actually do is to take a small enough volume find out what is the net direction of the dipole moment in that divide it by that volume so therefore now that the volume thing has cancelled out from both numerator and denominator what you have got is polarization per unit you see this is similar to the way the density mass density for example is a point function because what you do even though you say that density is mass per unit volume it doesn't mean that you take 1 meter cube of the mass and then find out how much is that mass what you actually do is to take any small volume let that volume go to 0 okay and divide the mass of contained in that divided by the volume and in that sense it is called mass per unit volume in that sense this is called polarization dipole moment per unit volume so it is a vector point vector it is defined from point to point and so therefore it is a field and the fact that it has a at the it gives rise to a measurable electric property that is if you put up a charge there it will experience a force is of course automatic because that is the way we defined remember it was defined electric field on the other hand e field the e field is something which a test charge would actually feel so this is something which I can experimentally determine the polarization quantity I determine by the conceptually I can determine of course the magnetic moment per unit volume yes we can realize the origin of electric field right and then we talked about polarization that is p and then we added with e to have the total displacement vector I can realize conceptually that e will have the dimension of electric field no problem but on what basis I mean physically on what basis we are adding it with electric field to find out the net I will tell you what see suppose I don't have a charged object okay but I just have a dielectric now if I have a dielectric now remember though this thing is electrically neutral but microscopically that macroscopic body has an electric structure you know that is the reason why I spent a bit of a time in doing a multiple expansion and then we found that to the our leading term is not just a matter of charge but it's also the fact that the charge separation takes place so I basically if I have a dipole in fact the question is very properly answered see we said a dipole is basically two opposite charges separated by a small distance now if this distance becomes zero then of course there is no charge at all in that but if you do it in a limiting sense the dipole has a moment and it has an electric field because when I take any point outside the distance from the positive charge and the negative charge though my newtley different but they are different so therefore a dipole technically is also a source of electric field you know we calculate this all the time and given it in terms of the thing so therefore and now I have a collection of dipoles that's all thank you okay one more question yeah we talked about this multiple expansion and then we concentrated about two terms one the monopole term I ignore the dipole term we had the potential and then we had two terms one is minus at the divergence V the term was P dot n cap right then we associated them yes yes and the second one with the surface charge bound charge both of them the physical significance of these volume bound charge because as I can understand when we are talking about dipoles they are they come in pairs right and if they carry equal and opposite charges so if I consider a small volume there I would find net charge zero and how do you define this volume charge bound charge to be non-zero and then you this is this you are you have now come back to in such a situation there is no contribution to the first term but you see now I am looking at the moment term okay the you are tied right my total charge is zero but the dipole moment is not zero okay so you are absolutely right that if I am only looking at that then I don't have a term which consists of total charge in neutral material I don't take that term at all but that doesn't mean that the dipole or the quadrupole term is zero the quadrupole of course is too small for us to look at but of course as you know nuclear phase is worry about quadrupole there was a question on quadrupole but that's a male question I will answer it later but so so please don't be under the impression see the total charge is zero you are absolutely correct total charge is zero for a dipole also even in the limiting sense the total charge of a dipole is zero but but the dipole moment is not zero that's the way you define it and so because of that I have a contribution thank you yes go ahead which places it this is Samaya College local Bombay yeah go ahead when we talk about the electric and magnetic potential yes electric potential had some practical implication like potential difference of our voltage in what situation the magnetic potential becomes relevant you know good question you asked but if you recall I spent a bit of a time in the morning in explaining that electric potential you see it's a question of your familiarity electric processes or things now for example gravitational potential you don't use that word but people who work with gravity they use it all the time because they are exactly the same they are they accepting that the gravitational force is always attractive but there is no difference between the nature of the forces the the problem of the magnetic potential is that they it is it is not unique but you will say so is the scalar potential not unique but the difference is the following a vector potential has a gauge choice associated with them so corresponding to the same situation I can choose different vector potentials as long as they differ only by the gradient of a scalar function now the question was that what is the use now because of this reason that it's a choice of a gauge and all that the effect of this is very difficult to demonstrate the mathematically it's a very useful thing because just the way you say electric field is minus gradient of 5 I say magnetic field is del cross a but remember it is not you know I mean not as elegant as talking about a scalar because scalars are much easier to deal with than vectors so I have not really gained anything by defined a vector potential because my magnetic field was a vector I am expressing it as a vector of another del cross of another quantity which is a vector so I am not gained much in case of electrostatics the scalar potential gave me a great advantage because the num they being scalars I could simply add them up but vector potential adding is going to be very difficult because they are vectors so but on the other hand there that is the reason why I talked about a very elegant experiment in the morning called the Arunov bomb effect see the what I told you there is the vector potential has a physical effect which you can see okay see while the scalar potential is simply a mathematical device which made your calculations easy because they were scalars now my vector potential is doesn't make my calculation that easy okay but on the other hand this Arunov bomb effect showed that you see what was Arunov bomb effect we said inside a solenoid field is non-zero outside it is 0 magnetic vector potential the way I have chosen is non-zero both inside and outside so now this problem I couldn't handle here but it is very easy to work out with quantum mechanics and see that that leads to a change in phase I would since all of you are teachers I would request you to look up okay and read this beautiful thing about what is Arunov bomb effect okay so there is a lot of interest there Marathavada Institute of Technology yes right okay good morning sir my question is you said a steady line charge produces magnetic field for an observer who is moving okay so my first question is is it experimentally verified and my second question is suppose the observer is steady and charge starts moving then it becomes the case of time varying field okay let me explain see what I said is this you have a steady line charge okay the line charge is static but you realize that it is static provided you are sitting down with respect to that line it's in your lab there is a line charge or any charge and you are static then only that charge is static but as we know the velocities are always relative so if the observer started moving the I have given a line charge just to provide it a long enough supposing it's a closed circuit but then if I start moving I think that with respect to me those charges are moving now if the charges are moving that's equivalent to a current a current should give me a magnetic field now what I said is the following let me let me take couple of minutes this will be the last question I'll take up in this session what we said is this that supposing I have an observer static inside a lab the he sees only a line charge he sees a line charge and he can calculate what is the field due to that line charge now he can not only calculate but he can put a test charge somewhere find the force on that test mu 0 electric field is given by lambda by 2 pi 2 pi epsilon 0 r and you can calculate how much is the electric field statement number one statement number two is now let a person move walk in a direction or move in a direction run in a direction parallel to the direction of that charge parallel to a direction of that line so with respect to him the charges which were there are now moving in the reverse direction now but he has now learned relativity which you will do it little later so he says that look if I have a length and I am moving along the line of length the length gets contracted but there is no way that he will delete the charges since the length has got contracted by a factor gamma the charges have remained the same it means the charge density has gone up but that is not a problem your charges your lambda you know how to calculate if the charge density is lambda times gamma you calculate that then you find out what is the force between these two current carrying the charges if you like now when you calculate that force you find an expression which you gave yesterday which is not the same as the actual force that the relativity would predict that what happens if something is moving in the line along that so there is a shortage and this shortage he has to explain so notice with respect to the moving person he has both charges which are giving rise to electric field and current steady currents which are steady because he is moving in a constant velocity steady currents which are giving rise to magnetic field so he has to invoke something that because in my reference frame the charges are moving there is an additional field which is the magnetic field so in that sense I said magnetic field and electric field are two manifestations of the same thing next question was can you experimentally verify the answer is both yes and no what do I mean by that see the point is this you notice that when you do electromagnetic waves you always talk about electric field you never talk about magnetic field you do not talk about magnetic field though it could be a equally good way of talking about it and the reason is that the strength of the magnetic field in such situation is 1 over c times the strength of the electric field so the magnetic field strength in an electromagnetic wave is not a prominent thing so you need okay the relativistic effects in order to detect it so the answer is that if your person could move with speeds comparable to that of light which of course is idiotic to talk about then this would be possible but on the other hand experimental verification of the type of thing that you are talking about is certainly not okay thank you