 Good morning and welcome to 9th lecture of this course. In the previous 2 lectures we saw magnetostatic field and in that we saw magnetic vector potential, magnetic scalar potential, magnetic forces and boundary conditions in case of static magnetic field. We will continue that discussion for one or two more slides now and then we will go into time varying field. So now it is appropriate time to just summarize the analogy between electric and magnetic fields or circuits. So here this table shows the corresponding parameters of the two types which basically are analogous to each other. So conductivity and permeability, magnetic field intensity and magnetic field intensity, current which is open surface integral J dot ds, flux which is open surface integral B dot ds, right. J current density is current upon area which is sigma E which is point form of ohm's law, right. Whereas B is flux upon area which is also equal to mu x. V in case of electric field, MMF in case of magnetic fields, resistance is L by sigma s or rho L by s because rho is one upon, resistivity is one upon conductivity. Here reluctance, note the you know symbol for reluctance, reluctance is equal to L by mu times area. V is equal to E times L is equal to I into R whereas here it is MMF is equal to H into L which is flux into reluctance, right. The KVL for electric fields and the corresponding circuit C, summation V around the loop minus summation Ri is equal to 0. Here summation all the MMF sources minus summation over of reluctance s into flux s is equal to 0, right. So this analogy is important for you know understanding various electromagnetic devices. Now we will go over to magnetic energy. So in this course also you know we will be dealing with magnetic energy for at least 2 lectures if not more and we will also be talking of co-energy later when we are calculating the forces. So magnetic energy E, this symbol I have purposely drawn it like this to differentiate with respect to this E. So energy magnetic is we know half Li square and that is also equal to half integral mu H square dV, right or half integral B square upon mu dV where V is the volume. In this you know expression since this is in this energy is in joules it is obvious and we are taking integral over volume is obvious that energy density is this square upon twice mu is also equal to half mu H square, right and this is the unit is joules per meter cube. Now here for a linear material where in B and H characteristic is linear, right we have what is known as energy here and co-energy given by this triangle. Now let us understand you know this energy co-energy term little bit. Later on we will talk more when we see the calculation of forces in one of the lectures when we start looking at finite element application to calculation of forces. So energy is given by integral H dV dV and there are two volume one is volume integral other is integration over B because there is a variable dV here, right. So it is basically you know when we so the area of this triangle, right. So this is the triangle, triangular area that we are integrating with respect to B as a variable. So this the area under this curve will be given by this area is it not. So that will be in fact H dV, right and this you know integrated over the volume that means this B and H remember those are at a point in space. So if you may have core material and then there are so many you know points of interest and at each point you will have B and H. So you have to integrate this integral H dV and then that value you have to integrate over the whole volume. So there are two integrals in one, right. So now you know why H dV is called as energy and B dH is called as co-energy, let us understand that. So co-energy as I said is integral B dH over you know H and then again integrated over volume. So energy if you see we all know is energy is integral Vi dt, right because V into I is power, power into time is energy. So now if you actually if I you know sort of express V as N d phi by dt which is Faraday's law which means that we are going to say in the next immediate slide probably most probably next slide. So N d phi by dt is voltage and I is H into L by N, HL is MMF divided by number of turns will give you I, right. N I is equal to H into L and then dt. Now if you actually substitute phi, okay. So just to make consistent writing I think this should be N phi by dt because we are calling flux as psi, is it not? So let us make this as N d phi by dt so that we use consistent N d phi by dt into HL by N into dt and psi is V into area and volume is equal to area into length, right, S stands for area. So now if you substitute these two expressions in this equation then you will get you know energy density as integral H dV, because you know voltage in case of all these inductions and all that voltage is always d phi by dt so it is related to dB by dt. So that dB in the energy density term comes from that you know d phi or dB term, is it clear? So that has the physical significance, the energy has the physical significance. Whereas co-energy is this given by this area which is the area under the curve, this same curve straight line in this case which you know H has the reference axis because it is V dH so that is this area, area under this curve and that is called as co-energy. For linear material energy and co-energy are equal, right, but for non-linear materials they are not equal, right. So we will see more about this when we see the one-to-one work approach for force calculation. Now here you know hysteresis, now we are you know ready to understand little bit more on hysteresis of aeromagnetic materials which you know we covered in the previous lecture. So typical hysteresis loop is as given here. So if we now take you know the first quadrant we all understand that the area of this hysteresis loop gives the what is known as hysteresis loss. So let us understand you know why this is loss. When we say loss it is some input and some minus some output that leads to loss. So let us understand here input to the magnetic circuit, what is you know given back to the source and what is the difference which is you know dissipated as loss. Now here if you see when actually in the first quadrant, it should be first quadrant, in the first quadrant you have this you know when the excitation is increased and the curve follows like this. So now our expression for energy is integral H dV dV. So when from here to here the energy that is given to the magnetic circuit will be the area under this whole curve. That means this is that curve and area under this whole curve which is O A P C D O is it not because it is integral H dV dV as the reference axis. So that is the input to the magnetic circuit when the excitation goes from A to P. Now when you reverse the excitation you know the curve does not follow the same path and that is why we have this you know memory effect or hysteresis effect and then it follows this. Now the area that is represented by this P C D P this small area right that is the energy returned back to the source. Why it is energy returned? Because now what is the direction of dV or the sign of dV is negative is it not because we are going from P to D or you know we are going from C to D while integrating. So dV is negative. So that the sign of the energy sign of the energy or sign of the power is reversed. So that means when the power sign is reversed that means power is flowing back to the source right. So that means in this when the curve is traversed from P to D the area corresponding to this P C D P that represent the energy that is returned back to the source. That means in the first quadrant what is the energy lost? O A P D O that is the energy lost right. And if this is for the first quadrant you can then do the similar way for the all 4 quadrants and areas of all these 4 areas will give you the total hysteresis loss. Basically you know this lost per cycle for the first quadrant is as I just explained to you it is given by is proportional to O A P D O and that is in joules. Now this when you add all these you know 4 areas corresponding to 4 quadrants right you will get loss area that is energy lost per cycle right. That if you multiply by F right it is cycles per second. So so much energy is lost in so many cycles suppose it is 50 hertz. So 50 cycles per second so into 50 by seconds. So that means it will be when you multiply this by F you will get the hysteresis loss because this energy into F cycles per second energy upon second is power. So that is why it is the hysteresis loss. So that is why hysteresis loss is proportional to F because later we will see eddy current loss is proportional to F square. So now we will see time varying field. The most important law in electrical and electronics engineering which makes this whole world move is Faraday's law and it basically establish coupling between electric and magnetic field. The law says that for any closed circuit if the flux linking that is changing with time then voltage will be induced and that voltage is called as EMF electromotive force and that is given by this famous equation V EMF is equal to minus double d lambda by dt where lambda is the flux linkage that is nothing but N d psi right. So lambda is N psi so here you get minus d psi by dt as induced voltage. Now this is as written here it is field view point. Why it is field view point? Because of this you know Lenz's law because Lenz's law says induced current field opposes the cause. So in that closed circuit the corresponding current that will be flowing that will flow in such a way that the flux produced by it will oppose the cause which is this changing flux. So that induced current field opposes the cause and that opposition is represented by this minus sign. Now the circuit view point is you know in circuit view point we do not consider this minus sign because in circuit view point we consider induced EMF as just N plus N d psi by dt and then N psi is nothing but L i because L is N psi by i L is flux linkage upon current right. So then you know you have N d psi by dt is equal to L dr by dt and that is we take it as induced voltage. Now here this induced voltage is opposing the you know source voltage. Now let us consider a transformer under no load condition. Induced voltage even opposes V1 in the primary circuit and the primary current depends on the difference between the two voltages. So here the induced voltage is opposing the source voltage right. So that is why that opposition is inherent in this representation that is why we do not put minus sign here okay. Moving further this induced voltage we also know that it can be expressed as close circuit integral E dot dl right and then by using Stokes theorem you can get it as surface integral open surface integral del cos E dot ds and then you know here this d psi by dt psi can be represented as integral V dot ds and we are taking N as number of terms as equal to 1 and then this further we can simplify as minus integral dv by dv dot ds with minus sign here. Now here the derivative with respect to time d by dt is made to act only on V because we are talking of here stationary circuit. So ds is not changing with time. So this in a way this is a transformer emf because in case of transformer the flux that goes to the core that you know everything is stationary there. So that is a transformer emf and then now this when it is equated to this you get famous Faraday's law del cos E is equal to minus dv by dv by dt. So we go further now after seeing transformer emf we will now see motion emf. Motion emf gets induced when conductors move in a static magnetic field right. So for the corresponding force, mechanical force due to this induced emf and corresponding induced electric field intensity is given by this fm is equal to q u cross b, u is the velocity, b is the flux density. So the induced electric field intensity due to the emotional emf or due to the motion is given by fm by q that is nothing but u cross b just rearranging the equation. Now induced voltage will be again given by integral em dot dl and if it is closed circuit that is why it is closed line integral for closed circuit. Then you know you substitute in place of em you substitute u cross b and then if u cross b is in direction of dl then this reduces to simple product blu where integral dl will be l and this l will be the length of the total closed circuit. Now let us go further and understand what happens to a conductor placed in static magnetic field in such case where b is vertically down and conductor is placed there right. Now moment it is kept there you will have you know this b will interact with this current through this equation. So we know j cross b is equal to force density right. So j is in minus a y direction right because y is into the paper whereas j here j and current is coming out. So j in minus a y a y hat direction into cross b b is in a z direction minus a z direction because it is vertically down z is up that will give you force density in x direction. Now this force density will move the conductor in x direction with velocity u. Now you can see here earlier we saw how we can utilize this moment in this direction because moment you have this current it will have its own field around it and that on this side the field will get enhanced on this side field will get reduced. So it will have some kind of stretching action of the conductor in this x direction that is why the current would this conductor would move. And remember this force is really acting on the free charges in the conductor right because this force is finally on the Q right. So the force is acting on basically free charges which constitute this current i. So going further we already had you know derived this expression in I think one of the earlier lectures for this j cross b can be further you know when you integrate it you will get total force as d i l the charges are moving in y direction minus a y direction and force is in x direction. So this force in x direction means not going to change the kinetic energy of those free charges right and that is why it is only going to change the direction but it is not going to impart any kinetic energy. So if the energy associated with that conductor is not going to change that means there is not going to be any work done by the magnetic force on the charges. Then the question arises then who is doing the work of you know this moving this conductor. Now let us go further this conductor when it starts moving with velocity u. Now u cross b as per this formula will induce emotional electric field intensity into this conductor. So now that em electric field intensity induced one will act in direction to oppose the current by Lenzer's law. So now current is coming out so em direction will be going in so that is why em is shown by cross here. Now this moment is em occurs and it opposes the current if we assume that i is constant because when we are actually explaining some phenomena by some you know intuitive explanation we have to assume something as constant. So here if we assume i as constant that means the source which is supplying you know power to this conductor if that is going to maintain current constant then what will happen moment is induced electric field intensity will appear and the corresponding voltage will appear that voltage will oppose this current flow and it will try to reduce the current. So this source then has to come into action and it will try to maintain this current constant in in other words it has to supply the corresponding power V emf into i. Earlier when this V emf was not there it was only the you know i square are lost kind of power that was you know supplied by the source but now this will be the additional power due to the induced emf that has to be supplied by the source. So V emf into i is the extra power that will be supplied by the source which is being connected which is connected to this conductor and that will maintain current constant and then now we know this V emf we have we had derived here the induced voltage is Blu. So we are substituting this Blu here and then this becomes Blu into i and then rearranging this becomes BiL into u. BiL is nothing but mechanical force which we have already saw we have seen it before and u is d by t distance upon time d is the distance in this direction divided by time. So now you can see here this is the f mechanical BiL into distance upon time so is work upon time. So this work is really done by the corresponding power source which is supplying the conductor and that power is nothing but V emf into i additional power drawn from the source and that is numerically equal to the work done, mechanical work done by time in moving this conductor forward in the x direction. So this explains the work is done in this case by the source which is maintaining the current constant. So and here of course it is explained the direction of this induced electric field intensity can be again given by this u cross B and I have just explained why it should be in a y direction by using this cross product u cross B. Now if suppose we want to say that this current is not constant current is you know let us not assume that the current is constant then also it can be proved that basically it is a source that is doing work. How? When suppose the because of this induced voltage the and the corresponding opposition to the current what will happen the current will reduce and the kinetic energy associated with these particles the free charges in the conductor that will go down and that reduction in kinetic energy can be associated or can be equated to the mechanical work done and the corresponding kinetic energy gained by this conductor while moving it forward in the x direction. So that way also it can be explained if we do not make the assumption that current is constant. But remember this is all little bit intuitive explanation and some assumptions are involved and you know for more detailed discussions and more comprehensive discussions on this topic you can refer Griffiths book on introduction to electrodynamics fourth edition and you will find comprehensive discussion on this whole topic there with many you know case studies being explained there. A final point in this which I want to say is this that here what we sort of understood intuitively is that magnetic forces do not act on or do not do work on free currents. Free current means the currents which are because of moment of free charges as in this conductor. But if magnetic monopoles were to exist, if magnetic monopoles were to exist if somebody finds in future magnetic monopoles then the magnetic force can act on those magnetic monopoles individual magnetic charges as in case of you know electric field acting on a charges electric charges. So that is one. So magnetic forces can act on monopoles if they exist and do work on them. In this we actually saw a case of closed conductor that means this conductor was part of a closed circuit. Now we will see what happens when we have a conductor which is like open circuit that means this conductor is a bar conductor is not connected to any circuit closed circuit and it does not form a closed circuit. So now here let us assume that the conductor is placed like this and B field is into the paper. It is different than here. Here the B field was like this vertically down. Here the B field is into the paper and let us assume that you have this you know conductor moved in x direction and with velocity u. Moment that happens you have again u cross B and that u cross B will result into the corresponding induced electric field intensity and the corresponding force in vertically upward direction here. So that will what it will do because there are many free charges here. So this E field will act on positive charges here and that positive charges will move vertically up and negative charges will move vertically down. So you will have charge accumulation at the two ends and this process will continue till this you know the potential difference due to this charge accumulation which is vertically down and the corresponding force will balance the corresponding force which is vertically up due to this charge separation on account of this moving conductor. So this will continue till there is a balance between the two forces. So here I wanted to bring forth this point that emotional EMF is generally in direction opposite to the corresponding the emotional EMF and the corresponding electric field intensity is in direction opposite to the electric field intensity due to the accumulation of charges. Now you know we little bit see more on magnetic vector potential because this vector potential is quite you know most of our course when we are solving magnetic field problems by using FEM those formulations will be based on a magnetic vector potential. So it is you know worth spending little bit more time on understanding magnetic vector potential. So now del cross E is given by minus double B by double T, paradise law in point form. Now B you can replace as del cross A and then you can rearrange and then it gives del cross E del cross E plus double A by double T is equal to 0. Now we know curl of gradient is always 0 curl of gradient is always 0. So now this E plus double A by double T you can express as gradient of something we have expressed it as gradient of B. Now this choice is not arbitrary it could have been any other scalar potential but we have purposely taken V here because when this reduces to static case D A by D T will be equal to 0 and then E will be minus double E will be equal to minus del V and that is consistent with electrostatic is it not. So E has to be del V gradient of V minus gradient of V in case of electrostatic that is static field. So that is why we have choosing here E plus double A by double T as minus del V okay and then you can rearrange now this as E is equal to minus del V minus double A by double T. So now this is the total sort of definition of E. So the question is why we are considering this E plus double A by double T as minus del V because this has to be consistent with the electrostatics that we have already covered. In case of electrostatics you have E is equal to minus del V. So when here it is a static case when then then D by D T is 0 D by D T of A is 0 because there is no current flowing if there is no current there is no A is it not. So there is A by double T will be 0. So then this will reduce to E is equal to minus del V which is consistent with electrostatic that is why we have chosen it to be like this. So now here this E is minus del V minus double A by double T. So this del V term is the charge is representing the charge accumulation right and this double A by double T term is representing the induction right. So in most of the cases both what we analyze we generally analyze either of the two cases either we are for calculating purpose when we solve some problem either we are dealing with electrostatics then we are only considering E is equal to minus gradient of V or when we are talking of magnetic field analysis we do not consider charge accumulation is it not. We are purely considering the current flowing induced voltages, forces, losses and all that. So there we are actually not considering the charge accumulation right. So either we are that is why considering either del V term or minus double A by double T term whenever we do deal with any practical problem in FE analysis. So a little bit more on magnetic vector rotation it is physical interpretation E is now if you take no charge accumulation case then del V is equal to 0 right. So then E reduces to this. Now actually if you you know also know you also know E is equal to F by Q F is equal to Q E. Now this F we have already seen that F can be on this F can be on static charge or moving charge. So that is why I can definitely even in time varying case when Q is moving I can use this expression E is equal to F by Q right. Now here we know also know force is basically rate of change of momentum this by in mechanics we have seen this force is rate of change of momentum. So that is you know force you can express as D by DT of MU where U is the velocity right. So D by DT of MU by Q. Now you you basically compare this expression E is equal to minus double A by double T and this expression E is equal to this then you can easily now infer that A is nothing but momentum per charge. That is the reason you know Maxwell originally called A magnetic vector potential as electro kinetic momentum vector. This is the correct definition of A unfortunately what has happened is magnetic vector potential this term has got established over the last few decades and you know but it is not really potential but it is a vector and it is electro kinetic momentum vector. And whenever we are considering particle dynamics whenever you know electrons as they are moving with velocity V and U V velocity U and the electron is having charge E right. Then the total momentum of that electron will be given by MU plus E into A right. MU is of course normal momentum because of its motion and E into A because this A is momentum per charge into charge E will give you the momentum. So this is the electro kinetic momentum and this is the momentum due to the motion. So the total momentum is given by this of course this momentum we are not going to use in this course but you know those who are interested in particle dynamics probably this equation would be required at some point of time. Now we go next to theory of eddy current which is a very important topic while dealing with electrical machine because we are dealing with induced voltages and induced currents and then corresponding losses that are occurring they are important to be estimated so that the temperature rises temperature rise within the electromagnetic device under investigation is not exceeding certain limits right. Now here we start the discussion with you know again Maxwell's equation one of the Maxwell's equation that is again Faraday's law del cos E is equal to minus dou V by dou T and then we take curl of both sides right. And then we you know this del cos del cos E there is a vector identity curl of curl of any vector is given by gradient of divergence E minus del square of that vector right. So it seems again here we are talking of eddy current induction and all that we are talking of conductors so we will not consider here charge accumulator. So that is why you know this divergence E which is nothing but rho V upon epsilon I hope you remember divergence V is equal to rho V so divergence E will be rho V upon epsilon so that is will be 0 because there is no charge we are not considering charge accumulation. So then you know this expression this equation reduces to this right. Now here this J what we have considered this J is we are considering only J induced which is sigma E I will tell you the you know difference what are the other current density is possible and how the then the equation get modified little later in this slide itself. So this is you know induced current density which is given by sigma E so E here represents the induced you know voltage right it is volts per meter but it is representing induced voltage right. So now this you know this expression this equation is a very characteristic equation actually we started with this you know Faraday's law and we arrived at this if you start with some other equation and we you know sort of eliminate other all other variables except some there may be B variable we will also get the same equation in B or H or D. So this equation is satisfied by all these vectors B, A you know H, B and so on what we encounter in electro magnetic. So this is a characteristic equation this is also satisfied by A and this is this is the equation in A that we will mostly use when we do two dimensional magnetic field calculation in this course right. So before we go to that let us you know understand this equation and its context in all its importance and when to use which term when we are dealing with either low frequency electromagnetic or high frequency electromagnetic. So if you take the ratio of conduction to displacement current density right so J by dA by dA by T then J is sigma E and dA by dA by T J omega epsilon E and that the modulus of that is just sigma upon omega epsilon sigma is the conductivity omega is the frequency 2 pi and this is the permeability right. So now there are two cases if sigma is much greater than omega epsilon which is typically you know valid for metals being analyzed at low frequency when I am saying low frequency I am not talking you know like one hertz through the hertz I am talking low means power frequency that means 50 hertz, 60 hertz or even 1 kilo hertz. So these are all low frequencies as compared to you know gigahertz, Terahertz those frequencies in high frequency spectrum right. So here you know for metals whenever we are interested in finding eddy current induced and all that typically you know the conductivity of metals is high and frequency is not that high and epsilon is quite low is it not because 8.8 0.2 10 raise to minus 12 is epsilon not that makes this epsilon quite small number. So sigma is much greater than this omega epsilon. So then you know this the last term here third term in this equation almost reduces to 0 because that is representing the displacement current density is it not. Because this epsilon here is coming from this daba d by daba t term is it not. So this term here daba d by daba t term is responsible for this term right. So basically you know we neglect when we are talking of we are analyzing eddy current we neglect this third term and that reduces this equation to what is known as diffusion equation del square E minus mu sigma daba E by daba t is equal to 0. And as I said this equation if we had started with some other you know Maxwell's equation or did some you know substitutions then we would have got del square A minus mu sigma daba A by daba t is equal to 0 in terms of magnetic vector potential right. Now here if this was in terms of A then there will not be any there is no source current here because this sigma d by if suppose this was in A sigma dA by dt is representing the induced current is it not because we in the previous slide itself we saw E is equal to minus dA daba A by daba t is representing the induced effects is it not. The induced effects are represented by daba A by daba t right. So here if this was this expression this equation was in terms of A then this mu sigma dA by dt daba A by daba t would represent the induced current right and that means there is no source current here. Now you know you may wonder you know if there is no source how can eddy current be there. For example you know if you are given if there is some metal piece and then if you know that there is some time varying excitation on that metal that may be produced by some source and that relation you know and you know that certain distribution of time varying field is there on the conductor surface in that and if you are analyzing the eddy currents in that then you do not require their source current. You what you are requiring only is the boundary condition right which we have seen this you know how boundary conditions can be used to analyze fields you know by properly setting of the boundary. Same thing is here in fact and classical eddy current theory that we are going to see next is you know we will make use of this equation only without source current and only the surface boundary condition. But there could be some cases where you know that you may have source current as well as induced current that means suppose you know there is a current flowing in a conductor in the vicinity of that current time varying current you have some you know metal plate and eddy currents are induced. So then you need to analyze both induced the induced current as well as the source current. So you may want to model the whole thing then you know you have both induced current and source current to be you have to represent in the equation. Then you know that equation will get modified as del square A minus mu sigma dou by dou by dou by dou t is equal to minus mu J s. Now we know already I do not know whether I recovered the Poisson's equation. Yeah I think we covered Poisson's equation in magneto statics is del square A is equal to minus mu J is it not? We have seen this earlier. So if there is no time varying field then del square A this second term will be 0 and it will be del square A equal to minus mu J which is vector Poisson's equation which we have seen earlier. But this equation is the Mohr's general form wherein both source current is there and eddy current is there and this term is representing eddy currents. Is this clear? So this is in presence of source current J s. Another thing to note here is if you transfer this induced current term on the right hand side the sign of that will become plus is it not? So this will become plus. Now the sign of source current and induced current they are opposite and which should be the case is it not? Because induced current generally opposes the source current. So that also is consistent here. The next possibility the second possibility is sigma is much less than omega epsilon which is you know when we you know analyze dielectric for example. Then you know we have this case where you know second term gets neglected has to be neglected and then what we get is wave equation. And this wave equation is the starting point for analysis of you know waveguides, antennas, transition lines at high frequency. So the high frequency electromagnetic basically starts from this wave equation. Well you stop here because we will continue this in the next lecture. We will stop at this point and we will next lecture we will continue from this point. Thank you.