 Okay, last time we were discussing the electrostatics and I said that the reason I do spend half the lectures on electrostatics is primarily because that is where most of the fundamentals are dealt with and what happens in magnetostatics which I will deal today is basically there is a correspondence that you can work out. But let us look at what did we say yesterday the day before when I met. So we said that originally when we were dealing with vacuum electrostatics we had basically 2 equations. One was the divergence of the electric field was equal to the charge density divided by epsilon 0. Epsilon 0 is the permittivity of the free space and the because of the fact that the electric field is conservative I should correct it and say that because of the fact that the electrostatic field is conservative I always have del cross of equal to 0. These were the 2 equations with which we started. Subsequent to that we said that suppose I instead of vacuum I have a dielectric medium. We talked about that there are different types of dielectric but in principle what it says is that if you take any substance the total charge of the material let us say is equal to 0 even then it may or may not happen that the positive charge centers and the negative charge centers in a molecule exactly coincide. If the coincide of course the total charge is equal to 0 even if they do not coincide when you take a macroscopic volume or if you take a small volume in which the total number of positive charges must be equal to the total number of negative charges and as a result the material as a whole is always electrically neutral. Even in the first case where we say that the positive and the negative charge centers coincide if you apply an external electric field this will lead to separation of the positive and the negative charge center and if there is a small separation between the negative and the positive charge center it will give rise to a dipole moment. So therefore our model of dielectric irrespective of what type of dielectric we are talking about is essentially a collection of dipoles. Now in the absence of the electric field what we assume is that this large number of dipoles they are randomly oriented. Now if they are randomly oriented the vector sum of the dipole moment I also told you that the direction of the dipole moment is from the negative charge to the positive charge the randomly oriented dipole moment collection will give you a net dipole moment equal to 0 the vector sum is 0. Now when you apply an external electric field these dipoles start getting partly polarized the only in the presence of a very strong electric field they will become polarized. You know it is not quite akin to the what happens in a magnetic material because the polarization or the alignment is lot more that happens there but nevertheless the principle is exactly the same. So what we said this is something which I did not derive because I said that I am trying to condense the a 40 lecture course into 4 lectures. So what we said is that inside the dielectric we said that the polarization vector is defined as the net dipole moment of the material per unit volume. In fact almost parallely we will define later the magnetization vector to be the net magnetic moment of the substance per unit volume. So this polarization vector so in a dielectric medium what happens is that there are two types of charges produced or charge density there is a volume charge density which remains inside the dielectric and this is given by the minus the divergence of the polarization vector and on the surface there are charges and the surface charge density is given by the normal component of the polarization vector. Now because of the fact these charges unlike the case in case of a metal where the charges the particularly the electrons which are the carriers they can move throughout. In this particular case what happens is that we call it the bound charges because the negative charges are tightly bound to the positive charges and as a result we have just given it a name bound charges to distinguish from free charges which happen in case of the conduction. Now with these modifications we said that the divergence of the electric field now remember I am still with electrostatics there is no time dependence nothing all that I have done is to introduce material. Now this relationship is always true that is del dot of E equal to rho by epsilon 0, epsilon 0 is still the permittivity of the free space whether you are doing the physics in free space or in a dielectric medium the divergence of the electrostatic field is still always the charge density divided by the permittivity of the free space. Now the difference that comes up now is that this charge density rho which in case of vacuum was only the free charges now gets modified and I pick up that another term which is the bound charge. So, therefore, the divergence of the electric field is the charge density divided by epsilon 0, but this charge density has two components one due to the bound charge and the other due to the free charge. So, now the reason why we like to retain this form is this. This electric field which the physicists always represented by the vector E is the actual field that an external charge will experience that is if you put a test charge in the system the force that the test charge will experience is given by the charge q the of the test charge multiplied by this electric field E and not d. The however, because of the fact that we have now two quadrature one due to free charges and another due to bound charges we mentally try to think is it possible to think of one component which arises due to the free charges and another component which arises due to the fact that there are bound charges. Now mind you there is no way in which experimentally you can actually determine. So, in that sense the vector d which we define for historical reason d and electric field are defined in different dimension. So, we define a vector d which is given by epsilon 0 times the actual real vector E plus the polarization vector. Now if you do that and substitute it back into the divergence equation we get del dot of d equal to rho free. So, basically what we have done is we have taken this equation and asked that supposing I needed an equation which is del dot of d equal to rho, but remember because this is some free charges we did not know exactly what to do with the dimensions we said all right let us put the d having the same dimensions and p. So, vector d is a mathematical construct which is defined as epsilon 0 E plus p which depends upon the properties of the medium ok. So, the divergence of this vector d is entirely determined by the free charges. The reason why I say that d is a artificial construct as there is no way in which you can actually separate them because an external charge will always experience the force due to the net electric field which we always talk as E is this clear ok. So, that was electrostatics along with the modifications that we talked about. The let us now switch over to what happens in the corresponding magnetic field problem. The first change that occurs in the magnetic field problem is that unlike electric situation where we have positive charges and negative charges and they exist separately in this world. The magnetism is a very peculiar phenomena in which the positive and the negative magnetic charges which are normally given by the name magnetics north pole and the magnetic south poles they are always found to exist together. So, there is no way in which supposing you take a magnet house over small you have a north pole and a south pole you split it into two you will get two magnets each having a north pole and a south pole and not be able to separate a north pole and a south pole. I repeat that there is no theoretical reason why a magnetic monopole as we would like to call them should not exist, but nature has sort of decided you know in nature there are many asymmetries that one finds. For example, you have heard of the matter and antimatter you have electron corresponding to their positrons, your protons correspondingly there are anti protons, but the normal matter they happen to be much more in number than the corresponding antimatter. There are theories which says when the universe was created there were equal numbers of matter and antimatter because it has to happen that way because the universe was created out of energy. And energy if it became matter must give me equal number of matter and antimatter, but however there was an asymmetry in the beginning for whatever reason and then most of the antimatter was lost. I am not talking about that this is not related, but I wanted to point out that there are many asymmetries in physics and this is one of them. Now, since the magnetic monopoles do not exist my del dot of E which was rho by epsilon 0 that has to change because the net magnetic charge that I have is always a north pole plus a south pole. Now, if it were to happen that every time that there was a positive charge there would be simultaneously a negative charge of equal magnitude. Then of course, del dot of E would not have been equal to rho by epsilon 0 it would have been 0 because the positive charge density plus negative charge density will cancel out. So, what happens in this case is that the first change is the magnetic monopoles do not exist. Now, remember the electric field was defined as the force experienced by a positive charge in the presence of the field that was the way I would. But now because of the fact that the magnetic isolated charges they do not exist. So, therefore, I cannot define the magnetic field by the force exerted on an isolated magnetic charge that is the first difference that we have. So, what do we do? Now, remember that there is a difference the what we have for testing is an electric charge not a magnetic charge. So, even if I want to test for a magnetic field my test charge is always an electric charge and what is found is this that a test charge in the presence of an electric field experiences a force which is given by Q times the electric field strength. Now, if this charge happens to be moving, if this electric charge happens to be moving then it is found that there is certain types of field in which the electric charge will experience a side wise force and the interesting thing about it is that force does not exist if the charge is not moving at all, if the charge is stationary the force is equal to this. So, if a moving electric charge experiences a side wise force where we cannot find out an obvious existence of an electric field in that direction, we would know that there is that is the region of the magnetic field and that force which is known now as the Lorentz force is given by the magnitude of the charge times the cross product of the velocity with the direction of the magnetic field. So, in other words you can work back you say that alright this is the direction in which I find this to be moving this is the direction of the original velocity. So, therefore, there is a magnetic field direction which exists like this. So, let us look at what all this. Now, magnetic monopoles do not exist. So, they could not be source of a field unlike the electric field where the source of the electric field is charged itself. Now, what happens is that the source I am now making you a few statements about how did the magnetic field arise and all that. The current actually a steady current is the source of a magnetic field wherever there is a steady current one finds and how these are experimental observation. You take a steady current put an electric charge nearby. Now, if the electric charge is not at rest you will find in the presence of a steady current the electric charge will experience the sideways. So, therefore, current just as charged is the source of an electric field. Steady currents are sources of the magnetic field. So, now the point is this that there is usually a confusion which many students have that when we say a current is the current a vector ok. The answer is no current is not a vector current density ok that is because of the way the current is defined. The current would be defined as the amount of charge that is crossing the certain area per unit time. Now, so therefore, the quantity has to be something like a q by t dimension ok and it also depends upon what is the total area because I am talking about how much charge is flowing through. So, current is not a vector, but if you make that area through which it is crossing infinitesimally small and calculate what is the rate at which the charge is passing per unit area. Then of course, since it is a infinitesimal amount and we can assign a direction in which that is moving the charges happen to be moving there that quantity has the dimension of current per unit area and that is what we call as the current density. So, current density is a vector current is not ok. So, therefore, we said that the amount of charge crossing the boundary of a surface as of a volume per unit time. Now, that is the rate of change of charge per in the volume because the that has to happen because otherwise the continuity will be. Now, the in the equation of continuity what we say essentially is that if I have a steady state supposing I am talking about a certain volume then whatever charge enters there has to leave from that volume otherwise there would be an accumulation of charges. Now, when we say something is a steady current there is the type of current that you see by connecting a wire to a battery. So, basically if you take any cross section or any small volume in it then the amount of charge that is going into that volume must be exactly equal to the amount of charge that is coming out of the volume very much like what happens in a water pipe. And so, that is the equation of continuity. Now, let us look at this equation of continuity which we have talked about earlier, but the good idea to look at it. So, this is actually the best way to visualize equation of continuity is not for electricity because people cannot actually immediately give their mind to it, but to look at what happens if you had a water pipe. So, if you had a water pipe and supposing this is the cross section of that what is the type of cross section I do not care. Now, the water or whatever the electric charges in my case would be flowing through. So, this current that I define remember I told you that the current itself is not a vector, the amount of fluid that is crossing through a pipe ok that is what we talk this is the analogous quantity for how much of fluid is getting in per unit time. Now, this is quantity. So, if the density is rho and the velocity happens to at any point happens to be v I take an infinitesimal area there. So, basically my current is rho v dot ds ok the. So, this is basically what is happen and this quantity then since it is rho v dot ds this rho times v is what I define as J by current. Now, I have given it a direction the for this because the this is flowing in. So, look at what we are talking about we said this J dot ds integrated over a closed surface is equal to the change in the total amount of charge per unit time and I write down the charge as integral rho dv rho is the charge density. You collect them and what you get is d rho by dt plus del dot J is equal to 0. Now, there would be an accumulation of charge if the amount of charge that is going in does not equal to the amount of charge which is flowing out either accumulation or there must be a source inside because after all you know more charges cannot come out unless you know from whatever it is going in ok. So, that was the equation of continuity. Let us come back to analogy with the Coulomb's law. See the electrostatic was built with a few premises. We have charges, charges are source of electric field. I have an identical premise which says steady currents are source of the magnetic field. Then we said electric field exerts a force on a positive charge in the direction of the field. We said the magnetic field exerts a side wise force on an electric charge. Now, this is the asymmetry in the magnetism. The unlike electric field where I dealt with the electric charge to test an electric field. In case of a magnetic field I still use an electric charge to test a magnetic field that is because of the nature's estimate. Now, what we do is this we say all right there is a Coulomb's law. The Coulomb's law talked about how to calculate the electric field at a particular point. Now, here the sense the source of B is the current distribution. I postulate now remember again that laws are never proved. Laws are basically certain small number of assumptions, hypothesis that you take right you never prove Newton's law. You say that this is you start with Newton's law rest of them will follow. Laws are things on which the physics rests. So, in case of a magnetic field we say that the corresponding law is what is known as the Biot-Sawar's law. Now, again the how is the law formulated? Law is formulated by the fact that I know that the magnetic field exerts a side wise force. So, therefore, I have to provide that look the magnetic field due to whatever source at a position R is given by now remember that here now I have a cross product D L cross R minus R. Now, otherwise other than dimensionality you will find the law is very similar. There is a 1 over R cube there is a 1 minus R minus R cube R in the numerator. So, as a result I still have a similarity with inverse square law. So, this is the differential thing. Now, so, this law is called Biot-Sawar's law. So, corresponding to Coulomb's law I have a Biot-Sawar's law in magnetic field. So, I said this is my the magnetic field due to a small current element D L. So, in this case the primes are the at the positions of the source and R is the observation point. So, if I have a steady current and I am looking at the net field at the position R, then obviously I need to integrate it and I told you that the quantities or the position vector corresponding to the source is given by prime here. So, therefore, all that I need is to integrate a whole volume. Now, when you do that I have a J cross R minus R prime. Now, this is R minus R prime by R minus R prime cube. So, I can actually express it as a gradient of 1 over R minus R prime with respect to the observation point. And after that I simply use some trivial algebra to convert it into this equation. I had 1 over 4 pi epsilon 0 as a constant in front of dimensional constant. In this case, this is mu 0 over 4 pi. I will come back to the relationship between them as we go along. And so, this is the magnetic field B is written as mu 0 by 4 pi the curl of this quantity. This is very interesting that I have expressed the magnetic field as a curl of some quantity. I can now forget well once more I have done some problems there is a red thing there it does not matter. But let us look at ultimately what I have said. If you forget the derivation there we have said that the magnetic field at a point R is given by mu 0 by 4 pi which is replacing 1 over 4 pi epsilon 0 curl of a current density divided by R minus R prime ok. Now, you can immediately see that I had told you earlier that the magnetic field is expressible as a curl of a vector potential. And that is what I have done and later on I will identify this factor mu 0 by 4 pi times this integral with the expression of the vector potential. Now, same divergence of a curl is identically equal to 0, then the moment I am able to write the Bayer-Sebert's law as curl of something the divergence of this field is equal to 0. Now, this is where we always say that del dot B being equal to 0 is a consequence of the fact that the magnetic monopoles do not exist. Because in the corresponding electric field problem, my del dot of B was equal to the density or related to the density of the electric charge of course, charge divided by an epsilon 0. And if you take del cross of this quantity, trivial algebra, but on the other hand you can do that if you take del cross of this quantity you with a little bit of an algebra you can show del cross of B is equal to mu 0 by mu 0 times 0. So, I have now collected the equations of the electrostatics and the equations of the magnetostatics. Del dot of B equal to rho by epsilon 0 correspondingly del dot of B equal to 0 because there is nothing like a rho there. Del cross of B is 0 because electric field is a conservative field. Magnetic field is not, magnetic field is not conservative field that you can see because its magnetic force is velocity dependent is V cross B is the Lorentz force. Now so therefore, supposing I am connecting two points by some path and I take a charge along that path from point A to point B in the presence of a magnetic field with certain velocity. I will get a value force dotted with df. Now supposing I take another path connecting the same pair of points A and B. Now I move a charge along that path the second path with a different velocity then the work that I will calculate would be obviously different. So therefore, our original definition of a conservative field that the work done is the independent of the path that need not be true because so velocity dependent forces are not conservative. So, this is one of the things that I want to say. Now you have heard of Ampere's law. Now like in the your electrostatic we had a Gauss's law and in fact, I had asked that why do you use Gauss's law if Gauss's law is equivalent to a Coulomb's law and you people immediately said that that is because under certain symmetric situation it is much easier to work with Gauss's law rather than Coulomb's law because Coulomb's law being a differential law I have to integrate. Now identical statement with Biosawart's law. You can use Biosawart's law always, but your job may be difficult. So what is done is particularly in situations which are symmetric what is found is that there is another law which goes by the name Ampere's law which is found to be very convenient. And for example, if you take a any circuit and supposing this is the plane of that the normal to that is this we have seen that our law was del cross B equal to mu 0 j. Now take the surface integral of this over on both sides. So, I know that J dot ds is nothing but my definition of current full current amount of current. So, I get del cross B dot ds, but by Stokes theorem this becomes the integral line integral over B. So, B dot dl equal to mu 0 i. So, Ampere's law is equally powerful, but Ampere's law cannot always be used. It can be used only in certain cases where there are symmetries ok. So, I just have taken one example to tell you how does it actually work that they are the same. Take the problem which you would be doing it in the class the students have done it from the schools, but it is always a good idea to explain them with reference to a problem that they have already seen. So, let us talk about a situation when I have a long straight wire. This is a simple problem. So, what we can do is that you take a small element ok. So, which is your the del is in the direction of the current. So, you take the what is the magnetic field due to this small element which is here. So, I calculate according to Bayer-Sebert's law i r minus r prime and just do some arithmetic basically. Now, if you do that arithmetic you find immediately now look the only point to work out is not the trigonometry, but the following that if you look at a small element here the magnetic field because it is a cross product is perpendicular to this direction of r minus r prime. Now, so the point actually is the following that remember I have taken this as the x direction. Now, this direction now you will have from all parts of the you can choose an origin and sets an infinite wire it is symmetrically located. So, the if I resolve it into a component along the z direction and another component along this is perpendicular to this direction another perpendicular to the plane on the blackboard let us say. Then by symmetry one component will cancel out and I will be simply left with this expression which is the same as the ampere plug ok. Likewise you might like to illustrate it with an example which is again works out very well is field of the axis of a coin. Now, this is a thing which you should always work out I am again not going to work out the details all of you know this is absolutely trivial this is to talk about that what happens to this expression for the magnetic field. Finally, if you sort of do the arithmetic that I have done there that turns out that the expression that you get there on the numerator has a current times an a square and that is nothing, but the definition of a magnetic moment. So, it turns out that the a circular coil gives you a field which is given by mu 0 by 2 pi m by z cube z is the field along the axis and this m is the magnetic moment. So, a closed current loop acts like a magnetic moment well another this there is a particular reason why I am talking about this case and this is very interesting take a solenoid once again I am not going to derive it because you have done all these things in the class all the time I will only point out certain interesting thing about this. You take a solenoid now you can use Bayer-Sawart's law to calculate the field due to a solenoid you have to assume that the solenoid is infinitely long. So, that the edge effects are neglected. Now what you find is the field outside the solenoid is 0, but field inside the solenoid is given by as you are all aware it is given by mu 0 times n times i number of turns per unit length times the current this is known to everybody. So, field outside is 0 field inside is mu 0 and i ok. Now so, this is the derivation which will be there in the notes, but that is not particularly important and similarly you always work it out with the toroid all right. So, what we are talking about is this that I will come back to this, but let me go back forward a little bit and try to explain something else the point actually is the following that later on as we go along I will calculate the magnetic vector potential due to a solenoid. Now you would find something very interesting that in case of a solenoid the magnetic field outside is 0, but the magnetic field inside is constant. Now when you calculate the magnetic vector potential what you will find is that we get an expression for the vector potential which is such that the vector potential outside does not vanish, but they will cross of that vanish. Now we will see that that has very interesting consequence the other point that you need to talk about is what is the force between two currents. Now so, you have basically the following thing that you have a current here arbitrary loops you take and you will take a small current element here with a small current element there. Now the way to understand this why two current carrying wires have force between them is this that since each one of them is carrying a current each one is a source of a magnetic field. Now suppose I look at this circuit now this circuit has a current and since there is a current in the region around it experiences it is a source of a magnetic field. Now in that magnetic field the charges which are in this current remember current is nothing but moving charges and moving charges by definition will exert forces on in a will experience forces in a magnetic field. So, this is a source of the magnetic field and the charges which are moving in the current here they experience a force due to the field created by this and so as a result the these charges will experience a force due to the field and vice versa of course. You can calculate either way you can calculate the force due to this on that or due to that on this and the point that one tries to make it that if you calculate the force due to let us say 1 on 2. Now you find an expression which is symmetric between the indices 1 and 2 and it is always a good idea to show that because of the symmetry that you find here ok. The force on one circuit due to the second circuit is equivalent opposite to the force on the second circuit due to the first circuit showing thereby that the Newton's third law is again correct ok. And these are of course the standard expressions that you know and here of course if you just let them memorize the formula they realize after all it is this is wrong should have been i 1 into i 2 after all it is i 1 into i 2. So, therefore, there must be current point ok. Let me come to something which is interesting is generally not understood, but let us spend a little bit of time on it. So, the question is we have introduced two types of fields an electric field and a magnetic field. But there is an issue that is coming up now. The issue that is coming up is we have said moving charges gives rise to magnetic field, static charges give rise to electric field. But as you are all aware that whether a charge is moving or not depends upon my frame of reference. A charge which is stationary in one frame of reference if I were to move with a velocity then in my frame of reference that charge is moving. So, therefore, depending upon the observer's frame of reference the charge could be said to be either moving or at rest. If it is at rest it gives rise to an electric field, if it is in motion it gives rise to a magnetic field. How do you understand? We are trying to essentially point out the electric and the magnetic phenomena though for convenience they are separated out. They are separated out because we say let us assume this in this laboratory everything is at rest. And if there is a steady current with respect to this laboratory the charges are moving. If there is a charge which is at rest in this laboratory then the charge is at rest. But how does one understand this? And this is what I would like to point out because they were interesting idea. Look at the following situation that I have two frames. So, this observer 1 is at rest, this observer is at rest. Now there are two current carrying conductors here ok or let us say just there are two line charges. Line charges means charges along a line which are at rest. So, according to this gentleman who is at rest there is a length line charge here and there is a line charge there. So, according to this frame which I will call as S, the line charges on this are sources of electric field because the charges are at rest. So, therefore, this gentleman calculates supposing I call it 1 call it 2 and the distance between them is d. So, you will say that the field due to this line charge remember the line charge expression it is done by always very trivially worked out using Gauss's law right. So, line charge expression is if the charge density is lambda it is lambda by 2 pi epsilon 0 d. So, the field at the location of 2 is lambda 1 supposing the charge density here is lambda 1 by 2 pi epsilon 0 d and direction is along this depending upon what charge is there the field could be this way or that way. So, this is my expression that the field at the location 2 due to the line charge 1 is like this. Now, so if I want to now calculate how much force this line charge exerts on line 2 supposing on the second line I have a charge density lambda 2. So, therefore, my field is lambda 1 by 2 pi epsilon 0 d and the force on the charge 2 because the amount of charge on that is lambda 2 times l. So, multiply the field with the amount of charge you get force is given by lambda 1 lambda 2 l divided by 2 pi epsilon 0 d. So, the second line experiences force due to first line and vice versa. Let us look at the same problem and this force I have already calculated lambda 1 lambda 2 l divided by 2 pi epsilon 0 d. Let us look at what happened how do I look at this situation from this point of view of this observer who I call as S prime who is moving with the velocity V along the line. So, the question is how does one look at it from the point of view of S prime. Now, you have just had couple of lectures on relativity. So, one of the things you have realized is that if an observer is moving along certain direction the length in that direction will appear contracted. So, what we do is this that this length which I said as l will now appear contracted by a factor gamma which is given by square root of 1 minus V square V C square. Because the conclusion is is there any assumption the value of this speed V? I have two lines which are at rest in the laboratory all right. With respect to these two lines which are at rest in the laboratory I am moving along their direction with a velocity V correct. Yeah, but what I am asking is that you talked about length contraction. Yeah. But let us say I am not moving in a velocity which is not close to C but it is very small let us say I am just walking. Yeah. Walking past that. You will not observe it. Yeah. So, then I will not happen this length contraction thing. So, what you will see as a consequence the point is let us assume there is a length contraction. Because length contraction is correct whether I observe it or not depends upon the velocity which I am moving correct. Yeah. So, my velocity if I can run with respect to those two lines with a velocity comparable to that of C you will be able to observe what I am talking about. I am not talking about an observation. I am talking about let us look at why magnetic and electric field arise and in fact it will turn out that the magnetic field effect is a relativistic effect. No, that is why my exactly my confusion is. Because you are saying that it is a relativistic effect. Now let us talk about a simple current carrying conductor. Let us say copper wire. Right. And we know that in the copper wire the electrons are really not moving in the speed of even close to C. Yeah. So, I should not get a magnetic field. But why I said the electrons are moving at all? I have just said I have a line charge. What I am asking is that in a simple conductor or let us say in a wire. Right. When electron moves I know that since this is a actually a solid state system. Right. Electron is not moving obviously in a very high velocity. Absolutely. We have this current. Right. But we still see. But we do not have a current here. See you are you are you are not misunderstood my statement. I am saying I have a line charge. In this case you have a line charge but your observer is moving. My observer is moving. Yeah. But I am talking in an opposite situation where the observer is still. Yeah. But the electron is moving. Yes. And I get still a magnetic field. Yes. That is what I get when I there is a current carrying conductor. Charge is moving I get a magnetic field. But then what I am asking is that since I know that in the real material electron is not moving. Right. Even close to a velocity c then I should not be able to observe any magnetic field out of a conductor which. I am trying to tell you that what you are calling as a magnetic field or as an electric field is a matter of your perception. Okay. There is the whole purpose of this demonstration which is the artificial demonstration is that there is something called an electric field, something called a magnetic field. But then I can always go to a frame of reference in which the electric field will appear as a magnetic field or the magnetic field will appear as an electric field. So, that is the reason why electricity and magnetism are always treated together. Okay. Now, the what you are calling as a magnetic field is because of a dominant effect that you see there. Right. So, even in a very small current that I try to pass through. Right. I see an observable effect of it. See an observable effect. But still how does it get relativistic with that that is what. The point is that you are talking about the velocity of the charges. Yes. With respect to that I am now saying that entire thing which you are calling as magnetic field I could convert it into an electric field situation. Okay. That is all. The electricity magnetism effects are effects which are perceived as electricity or magnetism. So, there is the absolutely no confusion between what you are saying what I am saying. You see if you have a steady current you will see the magnetic effect. There is no doubt about it. It has nothing to do with the velocity with which the electrons are moving. But because of the fact that there is steady magnetic field in the region around it the electron will experience a sidewise force. So, I know that there is a magnetic field in that region. I do not calculate these with respect to remember I am calculating with respect to stated laws. This is the magnetic effect. This is the magnetic effect. One is the Bayer-Sawant's law and another is the Coulomb's law. So, I am trying to point out in this example where I do not have electrons I do not have anything. I am trying to point out something else. And what I am trying to say is that your question is not related to what I am talking about now. We will come back to what you are talking about a little later. Let me complete this and then if you have a doubt you come back again. Oh, okay. Yes, sure. Please. All right. So, what I am now saying is let us look at the same situation with respect to this observer with moving. With the velocity I do not care whether he is moving with a large velocity, small velocity. It does not matter, okay. Now according to this observer then the charges there will be appear to be moving. All right. Let them move with certain velocity. Now let us then look at what is the, so firstly there is a length contraction. Now how much is the length contraction? Length contraction happens by this factor gamma which is 1 over square root of 1 minus v square by c square, okay. But now notice because the length has contracted but the amount of charge that has not contracted there is nothing like a charge contraction. So, my charge density has increased, correct. Charge density will appear to have increased. And how much since length has become L by gamma. If lambda was the charge density lambda will appear become gamma times lambda. Is it clear? Okay. Now so each one of the lengths become now gamma times lambda and length L becomes L by gamma. And the, now remember there is a distance d between them. But since the observer is moving along the direction of the length in the perpendicular direction there is no length contraction, okay. So, d remains the same. Okay. Now f d remains the same. Let us calculate the force between the two wires, okay. As calculated by my observer. Remember I am still in electrostatics. I am saying now let us look at it. The force between two charges. I am moving there are two lines there. So, I know its length has reduced. Charge density has increased. So, therefore I calculate again the force between them using the same formula as I talked about. That is 1 over 2 pi epsilon 0. The lambda 1, lambda 2, lambda 1 has become gamma lambda 1, lambda 2 has become gamma lambda 2 and the length L by gamma divided by d. Now if you do that remember there is a 1 over gamma there 1 gamma cancels out. I am left with the force as calculated by this man to be gamma times the force calculated by this man. Now if I do not know whether Professor Shri Prasad did it or not. If he also worked out what happens to a force in the presence of a boost that is in the presence of an observer who is moving with the velocity. He would have told you that the force does not transform as gamma times F. It transforms F divided by gamma. Now notice and it partly explains the question that you had raised. So, he has calculated. This observer has calculated a force between two lines to be given by gamma times F if he believed only in electrostatics all right. But a relativistic calculation of the force when an observer is moving with a velocity V will tell you that the force transforms as force by gamma. So, now this observer who incidentally is not moving with any large velocity it is not necessary that he is moving with any large velocity. He finds that he cannot reconcile the his observation with the tenets of relativity. He finds now that I seem to have lost in this game of calculating everything through a static situation. As a result of which he now said that because of reasons which I do not understand that is that is the question that we are talking about when charges are moving. When charges are moving remember we did not talk about any consideration of how fast I am moving how slow I am moving. As long as there is a motion there is a discrepancy coming in. So, he says for reasons that I do not understand I understand the part that it should be gamma F. Relativity tells me it should be F by gamma and this difference is arising purely because of my state of motion is it clear. Now if he makes that postulate that because of the fact that I am moving there is a another field which is coming in and how much is the that field that force should explain me the difference between F by gamma and gamma times. So, the field that he postulates is what he has given a name called magnetic field. Now let us see what is happening. I said that this is not consistent with relativity. Now if you now take the two differences this is the thing which he calculated gamma times F. F by gamma is what it ought to be according to Einstein. So, therefore, he says there is an additional force because of the fact that my observer is moving and that force I can calculate is given by F by gamma minus gamma F because that is the difference between them there is observation. Now you calculate this. So, this is I have taken out gamma F common it is 1 over gamma square minus 1 which is minus gamma F v square by c square and this gamma F is what he had calculated. So, I substitute for F prime this expression that he calculated and then you immediately see that we are saying that f m prime is i 1 i 2 l prime by 2 pi epsilon 0 c square d where l is i is lambda times v. So, immediately realize that the constant mu 0 which appears in the our expression for the magnetic field you can see it from here is given by 1 over epsilon 0 c square. So, please understand that we said that origin is relativistic ok. The what is an electric field and what is a magnetic field arises because of our perception. I had not in this derivation ever talked about which field is small which field is large. The if the charges are moving with velocity there is a steady current you will see a magnetic effect ok and it has nothing to do with the charges having velocity comparable to that of light they do not. But the magnetic effect is much smaller than the corresponding electric effect. No, but the magnetic effect will definitely definitely depend on how fast that charge is moving because if I have a very large is it that. No, it is a steady current it depends only on i. Ok, it does not depend. But the magnetic effect if you see you have seen this earlier also where you do optics. Now you always talk about the electric field square right intensity of light that you talk about you never talk in terms of the magnetic field square. And that is because in such situation the magnetic field strength is much smaller than the electric field strength. But if you have a charge which is moving in a steady current the main effect is the magnetic effect. So, basically the point that I wanted to make is the what is electric field and what is magnetic field is a matter of perception. In what appears as an electric field in one frame will appear as a magnetic field in another frame ok. Let us come back to this another interesting point about vector potential. We had said that I can define we said curl of E is equal to 0. Now when curl of E was equal to 0 I know that if curl of a vector field is 0 then there exists a quantity called a scalar potential with respect to which I can define that the electric field is the gradient of the scalar potential. We physicists take a minus sign there that is not particularly important. But notice magnetic field the curl is not equal we had in fact seen that the magnetic field the curl of B is 0. So, what we say is this that let us define a magnetic vector potential by the relation that the curl is of that quantity gives me the magnetic field ok. Now, notice there is a theorem in vector calculus which says if you have got a vector field the vector field is uniquely representable as one component which is whose divergence is 0 another component whose curl is 0. One component which is called solenoidal component another component which is called irrotational component. So, what we are now saying is this that curl of A A is not physical A is not physical but curl of A is physical because magnetic field is physical. So, what we are saying is this that now curl of A has been specified. Now that tells me since A is a vector field I can choose divergence of the curl is 0 I will always get this expression to be valid del cross A is defined. I can choose del dot A whatever I like. Now this is a choice which is known as a gauge choice. So, del dot A choose at your will, but of course, choosing at my will is not going to make my life easy. So, what one does is to choose the divergence of A in such a way that it makes your equations much simpler ok. So, let us look at what it means. Take what is I know that the curl of B is given by mu 0 j. So, now curl of B, but B is curl of A. So, this is curl of curl of A there should have been a bracket there and this is a standard algebra which said this is del of del dot of A minus del square A that is equal to mu 0 j. Since I said that the divergence of A can be chosen at your will why not choose it at 0 because it will simplify my equation. So, divergence of A being equal to 0 does not follow from any physics. It follows from the fact that A itself is meaningless, but curl of A is meaningful because it is a magnetic field. So, I choose the divergence of A to be equal to 0. This is called a gauge choice I will be talking more about gauge choice as I go about. So, this gauge is given a name Coulomb gauge. Many calculations in physics you will find done in Coulomb gauge and that is because ultimately the physics that comes out is gauge independent. You could have chosen del dot of A equal to 5 and nothing will go wrong, but your life will become more and more complicated. So, this gives me an equation for del square of A. Remember del cross of B is equal to mu 0 j del dot of A I have chosen to be 0. So, that gives me del square of A equal to minus mu 0 ok. This is the Poisson's equation and I can always find a solutions of the Poisson's equation. But the question is this that we have said that I choose the gauge to be a Coulomb gauge. Then they you will ask can I always choose such a gauge? My answer is yes. Suppose suppose I found I have calculated vector potential and it turns out that the vector potential that I have calculated its divergence is not equal to 0. Now, if the divergence is not equal to 0 how can I make it 0? So, let us look at this. I know that since curl of A is the only interesting thing and curl of a gradient is always equal to 0 to any vector potential that I find calculate. I can always add the gradient of a scalar function without altering the value of B. This is clear. Curl of A I am saying that I choose A a a new A prime supposing I have calculated some A. Now, whatever A I have calculated I now add to this gradient of a scalar function psi. Both A and A prime are acceptable vector potentials because curl of a gradient is 0. So, curl of A is the same as curl of A prime both of them give me the same magnetic field. So, if I do that I say that del dot of A prime then is del dot of A plus del square of psi. Now, I want del dot of A prime. Suppose I choose psi such that del square of psi is minus del dot of A minus del dot of A is a number. So, if I choose del square psi to be minus del dot of A that would make del dot of A prime equal to 0. So, I started with a del dot of A is not 0 I said that you have a choice add to this the gradient of some psi such that del square of psi is equal to minus del dot of that A that you have calculated. If you do that the del square of the new vector potential that you calculate is 0. In other words that I start with a calculation of A go over and find out a new A prime whose divergence is 0. In other words I can always express things in terms of good image. I can define some another psi which is not satisfying this equation then del dot A prime that is not equal to 0. That is not what I said. I said what was my question? My question is I have calculated some A by some method. I mean for example next problem I am going to calculate the vector potential. Now, nobody guarantees me that the A that I have calculated its divergence is 0 right because I have simply solved some equation and found it. Now, I am saying can you convert from that A go to another A whose divergence is 0. How do I do it? So, for that I am saying whatever A you have calculated right to that you can always add gradient of a scalar function and if you choose that scalar function intelligently you can make get a new A whose del dot is 0. No, A has to be a uniquely defined. A does not have to be uniquely defined is what I am trying to tell you. A is never uniquely defined. A has no physical meaning. It is del cross of A which is a magnetic field that is physical and that has to be uniquely defined. Now, what is it? What is the physics? Physics is in magnetic field. Magnetic field is meaningful because I can put a charge there find out how much is the force. Yes. So, sir I have not completed my question that if A is not a uniquely defined then we are having different A's corresponding to a B. Yes, absolutely. That is what I am trying to tell you that you can have different A corresponding to a fixed B. But A should be uniquely defined I think so. Scalar potential is not uniquely defined. Why did you not raise the question then? A scalar potential you can add any constant to it ordinary potential is not uniquely defined. So, B for a given B we are having different A's. Yes and I am trying to tell you that you can define different A's which give you the same B but it is in our interest to choose that A which makes our life simple. See, let me give you a simple example I do not know what is your confusion. This is I am dropping this from here to here right where is the 0 of the potential? This is potential you will say is mgh. Supposing I chose the 0 of the potential to be flawed will my physics change? No. So, there is nothing like a 0 of a potential energy. You define depending upon convenience when you are doing gravitational problem you say let the potential be 0 at infinity. Yesterday I did a problem where I said that when I am having a line right I mean we said in that case because the logarithms came choose the potential to be 0 at r is equal to 1. Potentials are not uniquely defined you always choose a reference point. So, what we are trying to say is magnetic field is physical. Electric field is physical. Potential is not physical. Different potentials can give rise to the same electric field. Different vector potentials can give rise to the same magnetic field yes your question. Yeah. So, actually we have been talking about that this vector potential is not physical, but the very difficult situation. No, I am going to tell you it has physical manifestation not physical I have not said. Okay, no, no, no, but the very. I am going to give you an example the physical example of an experiment which has proved the very reality of a vector. Yeah, I mean it is probably the Arunabhaum effect. Okay, because it becomes very difficult to understand that even if there is no magnetic field how does the particle really feels that magnetic field. Okay, we will talk about what is Arunabhaum effect next it is there in our my lecture. Okay, but let me again summarize because he has a confusion. I am trying to say that when things are appear in the form of differentiation right something is derivative of something then two functions may have the same derivative is it clear. And the result of the derivative if that is the only physical thing that you have the original functions are not physically important because you are expressing a scalar potential as a gradient of 5. Now two functions may have different gradients same gradient, but the functions need not be there correct. So, both of them however when you have taken a differentiation give you the same magnetic field is physical, but when you say magnetic field is del crochet I am not saying A is not physical I am saying different A can give you the same B. All that I am saying is to a particular A add the gradient of any scalar function and by definition del cross of del is equal to 0. Now, so therefore I choose a side which is equal to minus del dot A. This is since A I have calculated del dot of A is known to me. So, this thing is nothing but a Poisson's equation and solutions of Poisson's equation can always be found. So, I can always find a side. So, what I am saying is Coulomb gauge or whatever gauge you talk about you can always force a solution to satisfy that gauge ok. So, my expression for vector potential is here because this was the magnetic field and this is del cross of this and then you compare this with the expression for the scalar potential. You find there is a again a very much of a similarity with the roll of rho being taken as j by j ok. Just to point out remember I calculated the field due to a long straight wire. Now, supposing I have a steady current along z. Now, this is very simple since the vector potential is along the proportional to j direction as the j direction is uniform along z. So, therefore, my A is given by this expression. And so, I can immediately calculate my A is given by k mu 0 i by 2 pi log r plus del psi. Logarming is a very interesting function as I told you that. So, because of that this is now let us look at a solenoid. I am coming back to this experiment that we talked about. Now, we know that for a solenoid the field inside is constant field outside is 0. So, which vector potential gives me that? Look at that. So, I use again line integral of a vector potential. So, A dot dl this is another method of calculating vector potential. See integral of A dot dl by Stokes theorem is del cross A dot dl, but del cross A is B. So, it is B dot dl. So, in other words line integral of the vector potential is the flux. And what I do is I calculate the flux and use the symmetry to calculate what is vector potential. So, look at it in the inside the solenoid if I take a circular path since the field is constant ok. Since the field is constant mu 0 ni. So, if supposing I have taken a radius s. So, the flux is mu 0 ni times pi s square. But by symmetry of the line integral it is 2 pi s A and so, therefore, I get A to be given by this ok. Now, notice if I were to calculate the vector potential outside. So, vector potential inside is easily calculated. If I were to calculate vector potential outside I will enclose this solenoid with a bigger circle let us say of radius capital R. The field the flux is still because of that portion which is within the solenoid because outside the field is 0 clear. So, outside the solenoid my field is given by this. So, vector potential you notice is linear inside the solenoid and goes as 1 over s outside the solenoid. This expression for the vector potential this expression for the vector potential gives me the magnetic field correctly both inside and outside the solenoid ok. So, here I have a very interesting situation my vector potential is not equal to 0. Now, if it is not equal to 0 how do I feel its physical effect. Now, this is a very interesting experiment I will not have occasion to tell you in detail, but let us look at what is the effect it is called Arunabhumi effect. In Arunabhumi effect what you do is basically an Young's double slit set up. Now, you all know that Young's double slit experiment usually done with light will give you interference patterns on a screen. Now, you could replace light with electron beams also and sense the as we know matter also behaves like waves and will be able to see electron diffraction pattern on a wave if we have done focused beams of electrons going and hitting a Young's double slit. So, on a screen I will see a distribution pattern of the electron density. So, I can always measure it. So, this is the type of pattern that you expect in when you have a beam of electrons illuminating a Young's double slit. Now, let us do the following. In front of the slits I put a small solenoid the dimension of the solenoid is small. So, that electron beam is physically not much obstructed by that. So, therefore, I just put a solenoid there is no no current flowing flowing through it nothing just a small solenoid I keep and take it. Now, I do not expect since it does not obstruct the electron path that much the geometry of it is so small the pattern whatever was there in the absence of the solenoid is not much effect. But switch on a magnetic field that is switch on a current. Now, when you switch on a current once again the actual magnetic field is confined within that solenoid. So, it is not that the electrons have been subjected to the magnetic field because the electrons are not passing through that. But what you find experimentally is that the pattern is shifted and you as you know the shifting of the pattern has something to do with the changes of phase. And the only explanation is now mind you just switching on a current inside a small solenoid through which the electrons are not passing. So, that there is no question of Lorentz force etcetera being coming in outside the solenoid the field is equal to 0. But nevertheless you find the pattern is affected. So, what has happened outside? There is a vector potential which is not equal to 0 outside. And one can recalculate some of the things say how the presence of a vector potential is going to lead to a phase difference. This is what is known as Arun Abha. But as I said this is not usually taught. Yeah. The mathematics can be done with this. So, it is not a very difficult thing. The difficult part is to explain I mean how really I mean what is the fundamental reason that the electron which is moving around in this or tracing this two different path even if it is not in the physical magnetic field. But still it is really the electron wave function phase. Yeah, no that I understand. As you know two things can have the same phase different phases. But still in terms of the probability density don't make any difference. No, no, yeah quantum mechanically that is the way to calculate the thing. So, is there more fundamental way to understand how exactly. Again see the point is this that you are trying to understand a pure quantum effect by classical mechanics. You think the only thing that is physical is Lorentz force. That is not true. The thing is quantum mechanical. I have a wave function. The mere presence of a has changed its phase. Yeah. It has not changed the probability density. No, yes that is true. Yeah. But still it is the phase addition which leads to the pattern. I have a vector addition of the phase. Yeah. That is the whole idea of interference. So, because the phases have changed the phase with which each beam is coming in has changed. See let us do the following thing. Supposing you talk about again the Young's double cylindrical method. You have two paths which are adding up. Now when I add up what do I do? We calculate a path difference. Right? We say that alright one has come from this slit another has come from that slit and so there is a path difference that I convert to a phase difference. And then say if the this is an integral multiple of you know pi this happens if it is not something else. So, ultimately it is the phase difference between the two paths that are important. Yeah, that is true. So, I don't exactly remember the mathematics detail. But the thing I if I remember correctly then that phase difference is actually given by the line integral of the vector potential outside. Right. So, for the vector potential is giving rise to a phase difference for your electron wave, but not to its probability density. So, if I have somehow or other managed that the light path from one of them right from the beginning has a different phase difference. See remember that when you do optic you have a let us forget about electron for the moment. You have a light coming through slit number 1 light comes through slit number 2 you add them. How do you add them? You say that there is a path difference between them. What is meant by path difference? That the in one path which is longer the light has picked off certain phases. Now, suppose right from the beginning I had a mechanism by which I could change the path difference of one of the ones. Am I making it clear or not? Then what I should add will not just be the path physical path difference that I calculate, but also the addition. The in fact this is not something which is not known to you. Would you know what would happen if the in the light path of one of them one of the slits I put a polarizer and the light path of the other one I put another polarizer so that they arrive with different polarizations from the two slits. What happens to the pattern? Anybody knows? Is the question clear? The waves they do not superimpose to produce the pattern instead they produce some symmetric figures. Why? I mean the answer is right. See the question is this that normally you do Young's double slit pattern by division of amplitude right. So, but on one path you put a polarizer so that light coming out of that and hitting the first slit is polarized in certain direction. On the other path you put a polarizer in a crossed manner. So, that the light coming out of that is polarized in the reverse way direction correct. What is the pattern that you see on the screen? You would not see circular. Perpendicularly polarized lights do not interfere. The interference pattern simply vanishes. What is it that I have done? I have changed the polarization of the vector. So, what I am saying is suppose just at one of the slits you put a quarter wave plate do not change the polarization. Supposing they are all the same polarizer, but on one of them you put a quarter wave plate. How do you calculate the fringe pattern? To you have the path difference, but in addition to the path difference you have to add that extra quarter wave am I clear? So, the fringe pattern will be shifted now. I have not put a quarter wave plate, but what I have done is to provide a solenoid there which is giving that extra phase. So, nothing has changed physically because the phase that is there is not going to make any change in the intensity or any such thing, but the pattern is going to change because of the principle of support. Am I clear? Any question?