Hey you guys welcome back to the channel physics surgery and here we are in pathfinder solution series and the topic of magnetism and this is secure understanding problem number six it was a long pending problem because I have already taken up a a ats select series question in which I did discuss multiple methods of solving a particle moving in electric and magnetic field. So if at all it results in a cycloidal path I've given you three methods in that particular video and I gave this as a practice problem for one of the methods. So this is the video I was trying to talk about. So just in case you want to have a look at this particular video which I would insist because all the methods that are there to solve this particular problem have been discussed in that particular video the link is in the description below or in the I buttonable. So it's a very nice little problem with four multiple choice options and the particle was moving in a magnetic and gravitational field and I have discussed the usual Cartesian coordinate system method and then the rolling disc analogy to ensure that the problem gets all quicker and even much quicker by using the field transformation technique which is slightly out of j-cellabous but it's a nice technique which can shorten your solutions okay and also gives you a very good introduction to the topic which you would face in higher level physics courses okay. So let's move ahead and see the formal wording of the question from the pathfinder. A charge particle starts sliding down a frictionless slope of inclination theta in presence of a horizontal uniform and static magnetic field of induction B directed perpendicularly into the plane of the figure and uniform gravitational field of the earth. The intensity of gravitational field is G. After sliding an unknown distance L as you could see in the figure the particle leaves the slope and follows a cycloidal trajectory right as expected because there is magnetic and gravitational field. If on the trajectory maximum vertical displacement of the particle from end to end is H. Find the value of this unknown distance L it had slid on the slope angle theta is also given. So the expected answer obviously dimensional easy in terms of H and theta alone and you cannot use the value of B or G because using B or G you would be not getting the dimensions of length okay. So that's the initial idea if you want to have a try please do give it a try and don't expect me to solve again using all the three methods so you want all the three methods of solving this one please as I said go to this particular video the link is in the description below or in the I button so please help yourself with that and then only come back I will be taking in this problem only the field transformation technique which is least discussed so it will try to it will give you answer much more quicker okay I hope you have done that what's this video I'm going ahead with the solution using field transformation technique. So as I have discussed in the previous video itself right there are two stages first stage is common for any of the three methods you need to know at what point the contact is lost I think losing contact is nothing but normal reaction being zero for this particle after sliding through distance L right so in the normal direction the value of the mg cos theta should be balanced by the magnetic force which is qb u right where u is the speed at which it loses contact let's suppose okay. So I have written down the value of the tue in terms of these parameters I boxed it because I'll be using this in the stage two of my calculation also we know that using work energy theorem right I get that confidence because as it slides through this distance L only the work doing force is the gravitational field right because magnetic force and normal reaction will not do any work because they are perpendicular to the displacements. So mgl sin theta work then by gravity should be increase in kinetic energy up till here starting from rest so half m u square you'll get this so these are the two relations of u versus mg and theta and here u versus g and again a theta will be useful in my next set of calculations okay right let's move to the stage two when it has already lost contact now with respect to ground frame the person has seen a cyclodil path after it loses contact okay now what I'll do in order to ensure this cyclodil path can be converted to a circular path we need to ensure that the gravitational field is cancelled how do you do that you can move with respect to a frame in rightward direction let's say a velocity of v0 you choose that velocity in such a way that the particles mg vector in the downward direction gets cancelled by a QE field vector that would be generated due to the motion of the frame okay remember in actual ground frame there is no electric field in the question but the transformed electric field can be found out as v0 cross b bar as discussed in the previous video where you choose v0 such that you can move with any speed right towards right but you move with that particular speed so that the value of the transform field force QE in the upward direction cancels the mg 10 what happens then the charge particle in this frame feels like it is only under the action of a gravity magnetic field and we all know whenever magnetic field and velocity are perpendicular to each other it executes a circular motion so to please understand in the moving frame this vertical width doesn't change right if you move towards right the width in the vertical direction shouldn't change so whatever H the ground frame observer sees will be seen by the moving frame observer as the diameter of that circular motion okay right so in that moving frame the particle execute circular motion so that QE is mg therefore I will substitute value of e s v0 b okay because v0 is towards right and b is perpendicular so cross product is direct product and I would be judging the choice of the velocity as v0 is mg by QB this is the third box that I will borrow so already two relations I got in the previous page for losing contact and work energy theorem and this is the v0 for the choice of the frame of reference velocity okay and we all should be knowing that the radius of the circular motion in that relative frame should be m into speed of the relative thing that means as seen by this rightward observer a combination of the actual speed u and this v0 vector results in a u relative which will govern the radius of the circle so this is something we know so using these three important relations let me go back the box here the box here and the box here I'll borrow all the three boxes into the third page and I'll try to manipulate them okay our agenda is to eliminate all the unknowns you know in order to get the value of L okay so you start with this you see this is mg by QB into cos theta this was from remember normally action equal to 0 condition this was from work energy theorem and the third one was from the field transformation and if carefully observe with because of luck the mg by QB factor is the same here so I'll substitute the value of v0 here therefore u which is the velocity just by leaving the inclined plane in ground frame is coming out to be the v0 cos theta in manipulation so what how do you calculate the u relative the actual velocity with respect to ground is towards this direction the person or the frame is moving towards right so in his frame he will see relative velocity by subtracting that v0 from u and fortunately the v0 cos theta immediately exactly cancels you because of this relation so the u relative which will be in this direction is this component of v0 which is v0 sine theta okay right and therefore the value of r that I got in the previous case which is m u relative by QB I substitute the value of u relative by squaring it reason for squaring is because of the square here okay now same u v0 cos theta I substitute here and I end up getting so these are the two relations in which the v0 which I chose remember it's not given in the question had to be eliminated so the best way to do that is to square this further reason I'll tell you there is a QB by m here which I will again be looking at this v0 right so I'll eliminate that value QB by m I have eliminated can you see that v0 square has become v0 power 4 because of the equation 3 so head square by 4 which we know r is h by 2 right so h square by 4 is equal to this expression and v0 cos theta whole square equal to this expression we are almost there so from here to here you could see there is a v0 power 4 that I need to do right so even though this is square you square this one further to get the v0 power 4 and divide it with this equation and eliminate the value of v0 so after all those multiple squares and everything only surviving factors would be the h and l and a theta and you take the square root of everything you end up getting this expression that's given in the answer book okay so I think you should understand that most important part even though it looks three pages that is for me to explain it's actually only single step solution of being able to transform this particular cycloidal motion into a circular motion using this technique which I actually also explained in this problem in the past okay right so that's the best way to solve this problem according to me instead of the usual Cartesian system okay right I hope you have liked this video of pathfinder solution series there are many more pathfinder solution series please the link link of the playlist is in the description below or in the ibertenabo and also the rest of the workout series olympiad a a a t s and result series are also running parallel in this channel so please do check out all these links of the playlist are in the description look please do read through the description 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