 So, welcome to the 8th lecture of this course. So, what we saw in the previous lecture was magnetic vector potential some theory and its usefulness particularly in two dimensional magnetic field calculations. Today at the start of this lecture we will see magnetic scalar potential. Although we may not be using this scalar potential in our this course, but it is quite useful and you know many commercial softwares they make use of this scalar potential and that is why it is important for us to understand this. The first question that is being asked here is, is A always suitable for 3D computations? Because when you are doing three dimensional field computations at any point in space you have three unknowns. For example, A, if you are using A formulation, then AX, AY and AZ there will be three unknowns at any point in space. So, then the computations become you know little bit cumbersome when if you are solving a large three-dimensional problem. So, it would be useful if we have some you know scalar potential in magnetics because then scalar potential will have only one unknown at a point is it not because it is a scalar. So, it will have only magnitude at that point in pressure. There is only one unknown at a point in at a given point in space. So, that is why you know this magnetic scalar potential was coined by earlier researchers and you know so that is that is alternative to you know magnetic vector potential, but then there are some limitations what are those we will see. So, now you know like we have E is equal to minus del V in you know electric field space can be actually express H as minus del Vm. So, Vm is scalar potential right, but we know that you know del cross H and curl of H and curl of gradient if we take now H as minus gradient of something curl of H will be identically 0 right. That means wherever we define this scalar potential the condition has to be that j has to be 0 right j has to be 0 throughout the region where magnetic scalar potential is defined. Second thing is starting again with now del cross H is equal to 0 because the first condition requires that. So, del cross H is equal to 0 you apply you know Stokes theorem and then you will get integral H dot dl is equal to 0 right. Normally you will get integral H dot dl is equal to i Ampere's law, but that is for when it is del cross H is equal to j right. So, here del cross H is equal to 0 because we are defining H as gradient of something right. So, any closed path in the region should not enclose free or induced current. So, that is another condition. So, not only j should be 0 right not only j should be 0, but there should not be any loop in that region where magnetic scalar potential is defined. There should not be any loop in that region which encloses current. For example, I will take one simple example here. Again suppose you know you have some magnetic material a rectangular you know piece in two dimensions and then you know you have to you know establish flux in vertical direction like we did in case of magnetic vector potential. What did we do there? We actually you know gave on the left hand side vertical boundary we gave a 1 a 2 right hand side a 2 and then we said a 2 can be made say 0 and you can adjust the value of a 1 difference of a 1 and a 2 will establish the flux condition. Same thing we are doing here. Now here Vm to establish the flux in vertical direction on this horizontal boundary you have to define Vm magnetic scalar potential say a is the boundary condition a and this is on the top Vm b. Now the H field line go from a to b boundary a to boundary b. So it is same as you know because now exact sort of duality is there. In the sense you have in electric field case it is e here it is H. There it was e is equal to minus del v here H is equal to minus del vm. In case of electric fields e goes from higher to lower potential similarly here H will go from higher Vm to lower Vm ok. So Vm a is greater than Vm b and these are the H lines and the horizontal lines will be equi Vm lines like you know equi potential lines there in case of electric fields static electric fields here it is equi Vm lines right. The other thing is you know once again you know we just now saw that integral H dot dl close integral H dot dl is 0 but open integral H dot dl here right will be Vm a minus Vm b simply the potential difference right and that is what you know we have established. So here that is what is written vertical lines are H field lines and this blue lines are equi Vm lines. If we do all this then the advantage is divergence b is equal to 0 right and then that gives mu divergence H is equal to 0 right and then you know you substitute H as minus del vm and then you simply get Laplace equation. You know if you have a 3D problem in which if you know you want to analyze some region wherein current sources need not be there you can excite that structure by using boundary conditions of this magnetic scalar potential establish the field condition and then you can probably analyze in greater details the flux distribution in that given domain. But you have to again remember this has to be done only for magneto static case because even in this core this if this is core material you cannot even analyze eddy currents. So it is mainly it is the magneto static analysis three dimensional for which this is useful. But you know then that puts severe restriction is it not if you know there should not be current nor you know there should be any loop in the circuit in that domain for example here take any loop in this region close loop it does not enclose any current. So that second condition also is you know not violated. Then the question comes you know are these conditions not too restrictive for you know exploiting the usefulness of magnetic scalar potential. Yes they are that is why you know these magnetic scalar potential then is used in conjunction with some other potentials. In the regions of the geometry where problem domain 3D domain where currents are not there nor you know any loop in it encloses current there you define magnetic scalar potential in the rest of the zone you use some different potential. Those potential that actually will not go into details of that but if you actually see the research papers published you will find you know combined with magnetic scalar potential researchers have used other potentials. See when you use hybrid formulation like that means that you are using two potentials one is a magnetic scalar potential and some other potential then there will be some interface within your you know problem domain where in you know on one side you have one potential on other side you have another potential. So at the interface you have to establish proper you know continuity of potential by using by imposing correct boundary conditions. So you know but that is matter of more details and unless you are interested in using this magnetic scalar potential we will not get into details of that because in this course we are not going to use it but at the same time I thought it is important for us to know such you know application of magnetic scalar potential and in general hybrid. Now another for example you know geometry wherein this could be useful for example is like this suppose you have so this is the geometry for example is a you know magnetic circuit with a gap. So now here suppose current source is somewhere here so this is current source dot cos and dot. Now inside this core if you are not interested to find out the eddy currents in the core right. So there is neither you know J present in this hash portion nor there is any loop which encloses current because why that was possible because there is a cut here. If this was a completed circuit then you could always have one loop like this which will enclose the current. So here you know in this core portion you could for example use magnetic scalar potential in rest of the problem domain including this current sources you could use another potential. Now what is that other potential and all that you will have to see some research papers and you know get more details but such possibilities definitely exist right. Then we go to magnetic forces now this is one equation which we have seen probably in the one of the very first slides of this course wherein we discussed Maxwell's equations and then at that point probably I said that this is additional equation which is very useful in addition to Maxwell's equations for computing many useful parameters in electrical machines and equipment. So this is Lorentz force equation it has you know as you can see in the bracket you have two terms one is E other is U cos B. So the first you know term which is Q into E is basically it is representing the electric force which acts on a charge. Now that charge could be either stationary or moving because the equation does not say whether you know charge is stationary or is moving. So it could be you know stationary or moving and this is you know standard you know the force on any positive charge will be along the corresponding electric field there. The second you know term U cross B into Q says that B exerts force on a moving charge because it is U cross B. So U is the velocity and now this U being in direction of length and U cross B will be perpendicular to U and that is why perpendicular to that DL because cross product will be orthogonal to the involved two factors. That means this force the magnetic force will be always orthogonal to the length and f m dot DL will be 0 always. So that is why there is an important you know conclusion and which sort of puzzles you know many and magnetic forces do no work. Now there is a good amount of discussion on this aspect in this Griffiths book I you know encourage some of you who are interested to know various you know you know examples wherein such this statement can be proved you can actually see this book. What I will do is I will when we see the Faraday's law and the corresponding we see the motoring action and all that. So there I will sort of you know intuitively prove that magnetic forces do not do the work somebody else is doing the work. So just wait till that time may be next lecture or so after having seen the Faraday's law and application to rotating machines we will sort of again look at this statement. So till that time we assume that what is being said is right and of course mathematically beyond out it is proven here that indeed since f dot DL will be always you know 0 because f will be perpendicular to DL. So you know there is no work done because the charges in this case are neither accelerated or decelerated by this magnetic force because the force is in the direction which is not along the path along which the particle is moving or charge is moving. So the speed and kinetic energy remains constant right. So remember that the direction is changing because now you know you have speed means it is a velocity if you take as a vector velocity has magnitude and direction. So this magnetic force cannot actually influence the magnitude but it can influence the direction right. So whereas you know the electric force f is equal to QE that definitely has the capability to impart kinetic energy to a particle right because it is simply f is equal to QE charges will get accelerated or decelerated by the electric field. So it is always mostly the electric field which is you know doing the work going further force on a current element is you know I DL is equal to DQ U because you know I is equal to DQ by DT so that DT you know DL by DT will be U. So this can be you know I DL can be written as DQ times U right and then DFM equation can be written as this equation can be written as DFM is equal to DQ into U cross B and then using this relation you get a famous equation which is very useful in rotating machines for example force is equal to ideal cross B and if suppose I flows through a closed path then of course this becomes a closed line integral so FM f magnetic is ideal cross B. So for volume current density then this becomes ideal gets replaced by JDV and just see the difference now here it was closed but now there is nothing like closed volume or open volume volume. So we will always simply say over volume V right so volume V J cross B DV and J cross B because I have already explained you earlier ideal is equal to KDS is equal to JDV is it not. So here that is why we are replacing ideal by JDV and this J cross B is the force density very you know popularly used in calculation of forces this is the force density. Now later on you know at more or less in one of the last modules of this course we will also be doing force computations using finite element method right. So there actually we will see you know three methods one is force calculation by J cross B other one by Maxwell stress tensor and third one is by virtual work method. Now J cross B we have understood now what J cross B means but Maxwell stress tensor and virtual work method we will you know defer the discussion of this till we actually get into that. So along with the FM computation itself we will understand their Maxwell stress tensor and virtual work right. Now let us you know see one or two slides. So now we will see in this you know virtual electromagnetic laboratory there is one you know case study force on a conductor. So these are four these are isolated conductor that means that there is no it is not in the vicinity of any other field it is in its own field so that which is shown by these you know field lines. The force acts on that conductor because it is current is there B is there. So if you actually calculate J cross B you will find that the force direction is sort of into that conductor right. So but here thing to note is the net force on this conductor is 0 because all these forces will be cancelling out but there is a force at each and every point on the conductor but net force is 0 right. So this is the first thing. Now let us understand the second case. Now there is a some uniform field as shown here and now the same conductor with current carried by it is placed in that field. Now you can see the field of this conductor and this uniform field basically will get superimposed and the net field will be something like this. So always remember when you know you see this kind of you know resultant field the field you can see this is like a stretched object because on this side you have field which is more on this side field is less. So this is like you know stretched you can say rubber string which basically will is pulled. So the conductor is going to get pulled on this side which is actually will be clear from the. So this shows the third figure shows the force direction. So now you can see the force on this side is more and force here is less and the vertical components of these forces are getting cancelled and there is only a net horizontal force right. So now we go to next topic in magnetic fields which is magnetic materials. Now we will see classification of magnetic materials for that we have to get into atomic model. See in an atom you have electron spinning about their own axis as well as you know they are orbiting around nucleus. So that gives what are known as spin and orbital moments because they are representing the corresponding bound currents. We already seen you know magnetic dipole moment is given by m small m is equal to i s into an hat. An hat is the knit normal to this current loop area and this i is representing the bound current. This bound current is and in turn is representing the spin of orbital moments. So capital M is the magnetic dipole moment per unit volume because there will be millions of such you know current loops tiny current loops in any given magnetic material. So then magnetic dipole moment is per unit volume quantity and it is a vector. We have already seen this torque is B i s sin alpha in one of the earlier lectures where s is the area of the loop, sin alpha is the angle between the applied field B and the unit normal An to this current loop which is this area. So magnetic properties are basically decided by either presence or absence of paired electrons in the outermost shell of the atomic structure. So for example diamagnetic material you know most common example is copper. You know you have in the outermost sub-shell you have phi orbitals. Now you can see here 1, 2, 3, 4 and 5 orbitals and each orbital has got 2 electrons in anti-parallel fashion and they basically occupy this anti-parallel way when you know these orbitals are filled because in this condition the energy gets minimized. So they are in case of diamagnetic and copper as an example you have all the 5 orbitals in the sub-shell last outermost sub-shell fully occupied and hence the net orbital and net spin moments get cancelled because of this anti-parallel you know electrons and then the spin and orbital moments get cancelled. But when and this is in absence of you know external or applied field. Now when you apply the external field what will happen is because of Lenz's law the orbital motion of electrons changes in such a way that that motion and the corresponding effect opposes the applied field by Lenz's law and that is why the net field is lower than the applied field and effectively then we get relative permeability marginally less than 1 but for practical purposes for engineering calculation purposes it is very close to 1 and that is why we take it 1. So copper as well as aluminum which is paramagnetic material we are going to discuss both are that is why called as in general non-magnetic materials because their mirror is close to 1 very close to 1 and it is as good as or as bad we should say as air because air is non-magnetic medium. Now coming to paramagnetic materials aluminum being one of the classic examples here the outermost sub shell has only 3 orbitals and in only one of the 3 orbitals you have only one electron and there is no corresponding anti-parallel you know other electron. So there is definitely net atomic moment due to this single electron but what happens is due to the thermal agitation at operating temperatures that we generally use these materials there is a randomization of these atomic moments and net effect therefore reduces to 0 in absence of applied field right. So when external field is applied to you know magnetize the material what happens is the external field required will be very high to get sufficient magnetization and in other words mirror is just marginally greater than 1 and this happens because of as I said this you know thermal energy is predominantly active in the operating temperatures and hence the external field required is very high to set up a given flux density and that is why mirror is close to 1. Now we will come to ferromagnetic materials which are quite important for electrical machines and equipment. So in ferromagnetic materials they are you know some textbooks call them as class of paramagnetic materials. So they are similar to paramagnetic materials but the main difference between the two is they have strong net atomic moment for example if we take iron which is a ferromagnetic material Fe, Fe has again phi orbitals like in case of copper but here there are you know 4 orbitals these 4 orbitals they have only single electron and that is why you have you know net atomic moment which is quite strong and many such atomic moments form what is known as magnetic domain and these magnetic domains get formed because thermal agitation energy in ferromagnetic materials is insignificant at operating temperatures in fact that becomes very high and significant only at temperatures like 800 degree centigrade only at those temperatures like 800 degree centigrade would the randomization would happen right. So randomization on account of thermal energy would happen at those temperatures. So that is why these magnetic domains get formed but these magnetic domains are randomly oriented in absence of applied field. Now this random orientation is for the minimum for you know minimum energy state in absence of applied field right. So now when external field is applied you have you know corresponding torque as we have seen in the last slide and now that torque acts on the current loop now that current loop is representing say one whole domain which has many magnetic moments. So in effect that current is significant in case of ferromagnetic materials and you have got a torque and then you have corresponding alignment of domains along the field and that sets up a strong internal field which is m bar magnetization vector and it this strong internal field aids the external field and that is why we say that mu r is high relative permeability is high. What does it mean when you apply external field to set up a given flux density you require less you know current to be drawn from the source because internal field is aiding the external field and that is why the requirements from source reduce and this is definitely beneficial in many applications. Now if you see the typical ferromagnetic material like iron and they exhibit that material exhibits what is known as hysteresis characteristic so not only is the characteristic not linear but it exhibits what is known as hysteresis phenomena where in B now this is B versus H or it can be also called as B versus I in terms of circuit parameters. So B lags H by what is known as hysteresis angle. For example here H has already gone to when you you know traverse this hysteresis loop like this H has already become 0 here when you are you know reversing H is already 0 but B is not yet 0. To make B 0 you have to apply H in negative direction and that is called as coarsely filled H c to make B 0 because here then you will get B as 0. So these are the two parameters remnant flux density at H is equal to 0 and coarsely filled H c which gives B equal to 0. So these are the two important parameters for any you know hysteresis loop. Another thing is the B saturation this is the B saturation so that is also important. So all the electrical machines typically transformers and electrical machines they are operated below saturation level. Suppose if typically for example in case of transformers saturation flux density is 2 Tesla and typical operating flux density of transformer core is around 1.7 Tesla. So now if suppose we want to have you know expression derived for relative permeability then what we have to do is we take linear region because only in linear region you have m proportional to H. In this also hysteresis loop there are some regions for example this region is linear. So if in some cases if you can approximate this material characteristics by linear segments there we can write m as chi m into H where chi m is the magnetic susceptibility. So in linear region then you can express B is equal to mu 0 H plus m because you know these are as I already mentioned since m is aiding the external field that is why total B will be mu 0 H plus m. In absence of magnetic material this will be 0 and that is why in free space you just get B equal to mu naught times H. So now we substitute m as chi m into H here and then we represent 1 plus chi m as mu R and then finally you get B is equal to mu 0 times mu R times H and finally mu 0 mu R is the magnetic permeability. So B is equal to mu H right. So in case of you know transformers and rotating machines different grades of this ferromagnetic materials are used. So for example we will see in this slide we see both transformer as well as you know on the right hand side you have transformer. So this is transformer 4 and on this left side you have a typical rotating machine. So now if you really if you closely observe this transformer flux density in the core you have the flux density in most of the part of the core that is the horizontal you know core which is called as yoke and vertical part of the core which is called which are called as lex. The flux is in this direction here vertical direction either vertically up or vertically down and this in horizontal yokes you have either you know going towards right side and after half cycle it just reverses and goes to the left side. So in most of the part of the core you have flux direction in the same direction either you know going up or down or going right or left. That is why in case of transformers you have you know the grade that is used is grain oriented. That means the material is process such that the permeability is maximum along this direction of flux. In this case it is like this or in this case it will be in the horizontal direction. Whereas in case of rotating machines you know you have you can see here the flux direction is not fixed along the you know core and you know as the and since the three phrases are 120 degree displaced in time this flux direction continuously changing with time and there is no fixed direction of the flux. For example here you can see the flux is not flux is at an angle here then it becomes straight then it turns and this then it goes like this and turns and like that. So there is no point here in using grain oriented material because then most of the time the flux will not be along the grain orientation and then will not be able to exploit better properties of material along the grain orientation. So that is why in case of rotating machines you use non-oriented material. And finally you know you have permanent magnets which are called as hard magnetic materials. What we saw in this slide grain oriented and non-oriented and this hysteresis loop typically is shown for soft magnetic materials. The soft magnetic materials are characterized by a low value of hc. That means the value of hc required to demagnetize core is small whereas in case of permanent magnets which are called as hard magnetic materials you have hc very high and the loop is very broad and of course you are aware that permanent magnets had has got lot of they have lot of applications these days in rotating machines and we will see one such example later when we get into finite element analysis. Next topic is boundary conditions. So magnetostatic first we will see magnetostatic fields and probably next lecture we will see time varying fields also. So without going you know giving you derivation and directly stating b1n normal component is continuous like we had d1n is equal to d2n in case of electric fields we have here normal component of b is continuous and if you substitute b equal to mu h as done here the normal component of h is discontinuous and h1 normal upon h2 normal is equal to mu 2 upon mu 1. And then by using another Maxwell's equation that is you know integral h dot dl is equal to i and then if we assume absence of surface currents which is valid for you know static and low frequency field I will explain you know when we talk of time varying fields we will bring in this surface current and see how this gets affected but little later. So in absence of surface currents the tangential component of h is continuous and correspondingly then you can express this equation as given here. So bt is discontinuous across the interface. Now for example here if you see this is the case of a transformer these are two windings and this is the leakage field and this is the core. So leakage field is you know I have shown only one flux line here right. So the flux is impinging the core and it is coming back and it is forming a closed part. Now here since core the permeability is in thousands as we saw in the previous slide 1000 to 5000 so you know mu 2 is very high so b 1 to b 2 2 will be very small because b 2 t will be also very high because mu 2 is high so b 2 tangential will be high. So effectively you know this once the flux enters the core the tangential component will immediately dominate. So that is why you know we will find if you see the you know leakage field plot in case of a transformer you will find that the flux is seen to be almost entering normally to the core and then it is just immediately turning in. So the normal component of incident field is high and then it turns inside and the tangential component becomes high because of this bonding condition right. So more about these bonding conditions for time varying fields later. Next is the inductance. So flux linkage we know it is n psi, psi being the flux. So l is equal to n psi by i flux linkage upon current and that you know flux can be represented as b into s right and then i is h l upon n that we have not had seen but this is mmf, mmf is equal to h into l that is equal to n into i. So mmf is equal to h l is equal to n i and then this can be you know simplified to mu n square s by l. So this is another formula famous formula for computing the inductance and the third formula is if you can replace l by mu s because s we are denoting here area. So then this becomes n square upon reluctance. So remember this r is drawn special. So this symbol you should know for reluctance. So reluctance is l by mu s right. So these are the three formulae that are very commonly you know used for calculating inductance. Now you should remember that reluctance in magnetic circuit and reactance in its you know corresponding you know representation electric circuit they basically do not go hand in hand although they both look similar but they they are not going hand in hand. So if the reluctance of magnetic circuit increases for example in this you know magnetic circuit with a gap if this gap is increased because now we know from our previous discussion that air gap has mu r of 1 relative probability of 1 and mu r of this is say 1000. So effectively what we are saying is 1 mm of air gap is equal to 1000 mm of ion core is it not 1 mm of air gap is equal to 1000 mm of core. So effectively you know that is what is written almost entire energy of this magnetic circuit is stored in this air gap only right and you know it can be proved that suppose you know you had same gap in electric circuit maybe you know it if the gap means it is a capacitance there. So if we actually compare the stored energy by a capacitor and stored energy by some gap in the magnetic circuit suppose you know you take this flux density here as 1.5 Tesla for example because that will not saturate definitely this magnetic material right. So at 1.5 Tesla if you calculate you know this energy for some number of turns and all that and the source and the corresponding you know energy which is stored in the you know capacitor for example if this was you know an electrostatic case you know this the energy stored in the magnetic gap is something like you know 1000 times more than 1000s of times more than the corresponding energy stored in the capacitor which is half CV square and where there the limit is the breakdown of this air gap is it not. In case of electric field and the capacitor how much energy you can store is a function of the limiting case would be breakdown. So if you there you consider 3 kV per mm as you know breakdown for air. So if you take that 3 kV per mm and then calculate the corresponding stored energy we will find that that energy is much less as compared to the magnetic energy stored in this gap and that is the reason you find whenever you require larger amount of forces in practical applications like plungers and what not you always use the magnetic energy. You use electrostatic forces and all that only in very you know in those application where the forces required are very small like in case of MEMS, microelectromechanical devices and all that where forces required are very small their new use make use of this you know electrostatic forces but this is a thing that I thought I should just mention to you. So I think what we will do is we will you know stop at this and continue in the next lecture where we will discuss more about magnetics and then we will go on to time learning fields.