 Now there is another type of focusing also that is known as age focusing. What does it mean? So far what we have done actually when a charged particle arrives here it suddenly enters into the magnetic field here and it suddenly exits the magnetic field here. This was our model so far and here you can see when we take the sector dipole magnet sector dipole magnet in this case this is the age of the magnet at the entry side and this is the age of the magnet at the exit side. So in this kind of orientation of this magnet which is drawn here particle trajectory is coming at 90 degree of the age and it also exits at the 90 degree of the age here. So particle trajectory enters and exits perpendicular to the edges of the magnet. This orientation is known as sector magnet. Now what happens in this case? In this case we have seen that a trajectory which is deviated or coming in for positive x having some deviation non vanishing deviation from the descent trajectory it will have a longer path in the magnet so it will bend more and the trajectory which is coming for negative x it will have a shorter path length in the magnet it will bend less and so positively deviated trajectory and negatively deviated trajectory both trajectories comes closer to the design path after exiting the magnetic field. So this kind of focusing was the geometrical focusing. Now suppose instead of this orientation we have magnet in this orientation means magnetic edges now is here and entering of the trajectory is not at the right angle. Still you can see we have chosen the magnet such that the design trajectory having exactly the same path length as in the case of the sector magnet. So design trajectory is not changed due to this orientation because we have taken the same length of the descent trajectory inside the bending magnet. So now this magnet have different effect at the edges than the sector magnet while we can see very easily here that this trajectory in the case of sector magnet it starts bending from the here. But in the case of this orientation of the magnetic edge it starts bending from here. So in this case at the exit also bending stops here while in the case of sector magnet bending was taking place up to this point. So in the case of sector magnet bending was more for this trajectory while in this orientation bending is exactly same or path length is exactly same for this trajectory as of the design trajectory means both trajectory bends with the same angle. And similar is the case if particle is deviated in the negative x direction you can see here in the case of present magnet trajectory enters here and starts bending from here. Instead of sector magnet particle was going straight up to this line and then bending started from here. And similarly in the case of exit edge in the case of sector magnet bending stops here. And in the case of present magnet bending takes place up to this point. So bending is more than the sector magnet if particle is deviated in the negative direction negative x direction. So again this trajectory is also having the same bending angle because of the same path length as of the design trajectory. So these trajectories which are coming parallel to the design trajectory exit parallel after the magnet also. So geometric focusing is now absent. So orientation of the magnet how the edges are oriented with the design axis it defines what is the amount of the focusing at the edges effect of the focusing on the edges. So geometrical focusing is affected by the orientation of the magnet. Now this edges we will see what is the effect of these edges on the focusing. So this will produce different effects so we will quantify those effects. So now suppose this is the magnetic edge this is the magnetic edge and design trajectory is exiting at certain angles from the design from the magnetic edge. Now we take a trajectory which is deviated from the design axis and it exits from here from the magnetic edge. The equivalent sector magnet. Equivalent sector magnet means a sector magnet. Sector magnet means the orientation of the edge is such that the design trajectory exit set the right angle to the edge. So this is the equivalent sector magnet. Equivalent sector magnet means the bending angle is same as we have the presently chosen magnet. So path length of the design trajectory is same for the sector magnet. We call that magnet as the equivalent sector magnet. Now in this case you can see that this is the deviation x at the sector edge in this path from the design axis and this angle take this angle between the current age with the sector edge as alpha. Had it been a sector magnet the bending will occur up to this point. So trajectory will come out this point in the case of the sector magnet. So the angle between the present magnet and suppose this is a sector magnet then trajectory comes out this this angle is shallow. So shy defines the extra kick in this case due to sector magnet. This is the extra additional kick in the case of sector magnet and the shy can be calculated very easily with using our previous formula of the kick that is B delta L upon B rho. B rho is just q by p. It is we discussed earlier that B rho is just a convention to write in the P by q in another way. So this will be delta L by rho shy. This is just length arc length divided by radius of curvature and in this case you can calculate the alpha very easily that delta L this delta L assuming this triangle and in the case of sector magnet this will be 90 degree. So delta L is x 10 alpha in this triangle. In this triangle you can see very easily that delta L is x 10 alpha. Now put this delta L here. So you will get shy is equal to 10 alpha by rho x and 10 alpha is just the orientation of the edge with respect to the sector edge. So this alpha defines this shy and in the case of alpha 0 means sector magnet this shy will be 0. So there will be no additional effect compared to the sector magnet. Now again you can see here that alpha is fixed if we have chosen certain orientation of the magnet when we have kept that magnet in that orientation alpha has become a fixed number. Rho is also a fixed number means for a particular magnetic field and a particular charged particle of certain energy this bending radius is fixed. So shy is proportional to x for even magnet in this orientation. So shy is proportional to x means this extra kick this extra kick is proportional to x and if kick is proportional to x means this is equivalent to a focusing action means this orientation of the edge has an effect similar to the quadrupole magnet means edge can be replaced by a quadrupole in a model of the magnet and focal length you can see from here focal length is rho by 10 alpha. Now in the case what we considered here we consider that this is the edge and the trajectory is going like this means up to this point we have curvature in the design trajectory and after this point design trajectory is straight means we are changing the curvature like this step function means here the magnetic field is some finite value and suddenly it drops to 0 and this kind of model we were using so far so this kind of model is known as hard age approximation means this was our model suddenly magnetic field rises and it becomes constant because dipole magnetic field is constant over the space so magnetic field goes like this and at the exit again it drops suddenly 0 means at the entrance and the exit we have a step function in the magnetic field actual magnetic field cannot be of this kind use just Maxwell's equation to prove that step function cannot occur in the magnetic field so in reality magnetic field will rise gradually at the edges it will become B0 after certain distance and then at the exit also it will drops gradually to 0 so how correct is our model if we are using step functions now in step function what length should we take means there is a gradual rise and gradual fall so what length should we take for the magnetic model we will see all these things now now because the angle from any magnet is integration of B dot if we take B0 B0 is the maximum magnetic field in the dipole magnet here at the center so B0 and we take some number L and this combination should gives us the integration of BDL here BDL is the actual distribution of magnetic field means magnetic field with some gradual rise and gradual fall if this integration is equal to this then this length which we have chosen is known as effective length and in hard age approximation we should take this L effective as our magnetic length so by taking this L effective we have corrected the bending angle which we will use if we have distribution of magnetic field with slowly rising magnetic field and slowly fall in it so hard age approximation in hard age approximation the magnetic length should be taken this so far we have studied the effect of ages means what is the orientation relative orientation between the magnetic age and design trajectory in the horizontal plane in which the bending was taking place what about the vertical plane we have seen that fringing can be used for focusing the charge particle in vertical plane in the lecture of cyclotron where we use azimuthally varying field for generating the fringing and that fringing leads to focusing in the vertical plane so how this fringing leads to focusing in the vertical plane we will see now now we have a join in 3d for a given magnet here you can see this is the north pole this is the south pole and this is the median beam means exactly in the middle of these two poles a charge particles trajectory which is a design trajectory is coming from here this is the design trajectory and design trajectory is completely on the median beam now it is the entry age these are the entry age of the magnets these are entry edges so at entry edges we have fringing in the magnetic field means magnetic field will not be have perfectly straight field lines rather than it will be fringing like this like this like this the top view of this picture is given here here you can see that this is the design trajectory and inside this magnet this design trajectory will move like this if this is a positively charged particle and magnetic field line is vertically down and this is the x in our coordinate system so this orientation where the design trajectory is entering perpendicular to the age is for the sector means this age in this orientation will behave like the sector magnet now suppose we displaced the trajectory vertically up here design trajectory was that on the median plane now we have another trajectory which is deviated in the vertical direction from the median plane because we want to see the effect in the vertical direction so it is vertically shifted by some amount this is the vertical shifting so this trajectory when it comes you can see it enters in the magnetic field where the magnetic field is not perfectly perpendicular to the median plane rather than it has certain angle because fringing is there if we take tangent at this location to this ranging this will give direction of magnetic field at this location so at this location you can see that there will be a vertical component as well as horizontal component or bs now this component is exactly in the direction of motion exactly in direction of motion means v cross v for this component will be zero and only this component will have some effect on the particle motion this component will not exert any force on the particle motion because the magnetic field is in the line of velocity so we have seen that in the case of orientation which leads to sector type magnet means perpendicular to the edge vertical displacement gives no extra effect in the vertical peak there will be no vertical peak the vertical component of the magnetic field will exert a horizontal force so there will be no extra vertical force in this case so sector magnet has no any effect in the vertical plane due to fringe now we change the orientation let us see now in median plane the design trajectory is coming right from here means if we see from top the design trajectory is coming here and the edge is in this direction so now edge is not perpendicular to the design trajectory at the entrance so again we can see that because design trajectory is at the median plane is on the median plane and on median plane we have only vertical component of the magnetic field so on design trajectory there will be only horizontal component of the force so design trajectory will not fill any vertical force now we deviate another trajectory another another trajectory of the particle which is deviated in the vertical plane now when it will enter here the magnetic field is in this direction so this magnetic field will have vertical component and a component around the magnetic axis means component around this axis while the entering trajectory has velocity vector in this direction so now there is an angle now there is an angle between these two components and these this angle between these two axis means axis of the magnet and trajectory of the particle we have a vertical component of the force means if orientation of the edge is different than the sector magnetic orientation then in vertical plane fringing field will exerts a force and we now will calculate what is this force now along the magnetic axis if we are taking the step function then our magnetic field will look like this what is uz here uz defines the step function so uz will be 1 if z is greater than 0 and uz is 0 if z is less than 0 so at this point i am considering z is equal to 0 so here b is 0 and here b is b0 so at this stage we have sudden transition this was our hard edge model so hard edge magnetic field can be represented by this expression now remember the Maxwell's curl equation for the magnetic field in current free static region because there is no current on the design trajectory no coil is there so this is a current free region so del cross b will be 0 and one of the component of del cross b means x component you will get del b y by del z minus del b z by del y is 0 this is the this is from the curl equation now we have magnetic field along the viruses here so we will put del b by del z and that del b by del z will give you at the location of step function this derivative derivative will give you a delta pulse a step function when it is differentiated so at the location of this step you will have a delta function so bz will have y b0 that's right using this equation now using this equation you will have this so we now you can see that bz z is in this direction we were considering only b y and taking a step function now here we see that if we apply the Maxwell equation we automatically get a bz means only vertical component is not possible in the magnetic field you will have some bz component also so the magnetic field which you assume is not correct you have to add some component along the z axis also then it will satisfy the Maxwell equation we will see it now you have b y which you have taken and this is the correction term which you have to add for satisfying the Maxwell's curl equation now we see the divergence equation what will happen now if you will take this b y and this bz this divergence equation will not be satisfied so what you will do to satisfy this equation again we have to add a term in the magnetic field when we will add some magnetic field term to satisfy the divergence equation we will see again that curl equation is not satisfied then we will have to introduce one curl one correction term for the curl equation again divergence equation will not be satisfied so we have to again insert some term which will satisfy which will cause the satisfy of Maxwell's divergence equation and in this fashion an infinite series will be there and in this infinite series only this term is linear in y means this term bz is directly proportional to y other term will have y square and higher order now we are actually restricting ourselves to linear accelerator physics linear accelerator physics means we will neglect y square or higher order term or x square or higher order term in this course so if we will restrict ourselves to linear domain we will have linear effect up to this point and this linear effect bz is directly proportional to y is like a quarter polar effect because magnetic field is varying linearly with the coordinate means again we have seen that in vertical plane also this fringing field reduces a quarter polar effect now you can calculate this series and you will get a focal length of minus rho by tan alpha now you see the beauty here in the horizontal plane also you got the similar focal length with plus sign and in this orientation and in this orientation in the vertical plane you get the same focal length with minus sign so you can see that if a quarter polar is focusing in the vertical plane it will have a defocusing strength and similar strength in the opposite plane so this is the case here also that this age effect is similar to quadrupole so it will have certain focal length in the horizontal plane and because that focal length gives you the defocusing effect the same focal length but with focusing effect will be in the vertical plane so this is the age effect so in age effect now you can see that in certain orientation you can get focusing effect in the vertical plane why i am saying that certain orientation if you will make this alpha negative means you orient your magnet in such a manner that design trajectory comes from here then alpha will become the negative and you will get the defocusing effect in the vertical plane due to ages so age orientation with respect to trajectory defines whether your age is going to be focusing in the vertical plane or defocusing in the vertical so it is the interplay between the angle of trajectory and magnetic age so in the case of azimuthal variant magnetic field in the cyclotron we have hills and valleys and hills and valleys at the junction of hills and valleys we have fringing field and that fringing field is used for focusing the charge particle in vertical plane so now you understand how this hills and valley leads to focusing in the cyclotron and we have reached a vertical focusing in this case how really these magnets look so this is dipole magnet i have already told you in previous lecture this is another type of dipole magnet this is a quadrupole magnet this is a quadrupole magnet so you can see that this blue color this is the ferromagnet which is yoke and these are the four coils wrapped on the yoke so these are four poles and direction of the current is flowing through these coils so that this becomes the multiple if this becomes novel this will be the south pole now you can see one interesting thing here you can see a simple magnet in joint is really very complicated when it is fabricated now you can see some certain tubes here what are those tubes in such type of magnets when we want to focus very high energy beam then it requires a huge current in these coils this current may be in the order of thousand times so you have to cool down the coils with flowing current so these coils are actually hollow inside and this is a water pipe which sends water inside those tubes and this water before sending to magnets must have been treated so this have to be low conductivity otherwise shorting will take place inside the magnet so before switching on the magnets one has to have one have to have these cooling water channels inside the magnet and it should be so in real and then you can see certain points here and these points are for fixing the alignment with the coordinate system so a simple drawing of quadrupole magnet when we realize it it is a very complex job it has a mixed engineering job electrical engineering mechanical engineering and alignment engineering all these are interdisciplinary work and then a magnet can be fabricated and right now we can see that in these magnets which we use in rrcat for the biggest accelerator in India that is industry all these magnets have been fabricated indigenously in our nation and this is you can see another type of quadrupole magnet which is from the triumph which is a machine in the canada and again references are seen however the way in which i told you about the vertical focusing due to fringing is given in this paper and it is a very simple paper so you can go through this in textbook this has been treated in some other ways so in next lecture we will see how we can solve the equation of motion we have obtained equation of motion in which you have d2x by ds square plus kx is equal to zero which is similar to the simple harmonic oscillator equation however there is subtle differences what are those differences and how we can solve this equation for obtaining the charge particle trajectories we will see in the next lecture