fine so today we are going to do problem practice on this chapter electromagnetic induction short form of that is EMI and you might have seen how this chapter is the theory for this chapter is very less but when you look at the numericals the varieties are many the reason why such a small chapter has so many numericals because this chapter assumes that you have all the knowledge of magnetism you have all the knowledge of electrostatics and on top of it something related to how the how the EMF is induced and then you have these you know mechanics topics are also mixed with the concept of electromagnetism that creates a good mix of many concepts together because of that multiple numericals are getting generated all right so we'll quickly brush up the topics related to only this chapter of course there are few things that it takes from the previous chapters for example you should know what is a magnetic field due to the various shapes and structures so that is from moving charges and magnetism chapter anyways so this chapter is about EM induction okay as in how you can induce magnetic field from electric field and vice versa okay so basically if I can write it's about how you induce E and B from each other okay now you have you have a relation for the magnetic flux okay magnetic flux can be written as a integral of B dot da okay now when I'm finding the flux do I need loop or some area my question is can I find out flux through this structure which is called structure from A and B will it have any flux so it needs to be closed needs to be closed it needs to be in a loop okay so when it comes to circuits flux you can find only when there is a loop okay so that is the basic need of this once you find the flux which is equal to let's say area vector is let's say area vector is a constant and B vector is also direction of that is constant so you can no sorry let's say B and A let's say you're talking about a very small area you can write on flux as B into A into cos of theta okay here I'm assuming angle is constant area is constant and B is also constant otherwise you have to integrate this whole thing to get flux so this is the magnetic flux okay now this is I mean I'm saying B A and theta are constant in space okay at a particular time everything is constant but things can change with time okay so when you differentiate it there can be change in that with time with space it is not changing but with time it can change and with time if it changes it creates an emf which is equal to minus of n times d phi B by dt okay so n is number of turns and this is rate of change of flux all right and once you get emf this emf is induced in a loop and if loop resistance is r current will be equal to this emf divided by r okay now the current in the loop will be generated how will you find out the direction of current by using languages correct so direction of current is such that it opposes what the call the thing is created what what creates it there are change in flux correct it opposes change in flux change in flux remember that okay and Lange's law goes beyond just talking about the flux Lange's law can be applied directly I'll tell you when I'll go there in that concept so yeah so what I was saying was when I was changing flux with respect to time I can have magnetic field changes changing with time and d B by dt into a into cos theta plus n B dA by dt into cos theta minus of n B a sin theta d theta by dt okay of course there is a minus sign but that minus sign is having just physical significance it doesn't create any you know mathematical significance because it just an indicative of the Lange's law of fact that it will oppose so that's why just a minus sign would put it over there so this is the emf remember that any three things can change magnetic field area and theta all three can change some may contest that even number of turns can change with time that why can't that create the emf but the formula for emf is what minus of number of turns into rate of change of the flux in one of the terms okay so the derivative is not for number of turns then we have the current which is dq by dt to be equal to emf which is d phi by dt divided by the resistance okay so from here you can see that dt dt you can cancel out and you can get dq is equal to d phi by r all right so if you integrate it over a period of time this integral is anyway independent of the area sorry independent of the time so doesn't matter whatever time frame you take the amount of charge that has flown through the loop which has undergone a change of flux of delta phi you can relate them like this okay these are few things which sometimes get asked so you should be clear so this is about the flux rate of change of flux and yeah small equation on the amount of charge that has flown for a given change in the flux now to induce the emf you don't need a loop okay there can be a situation of emotional emf but the formula you have is valid only for the loop this formula is equal to minus of n d phi by dt is valid for the loop but it is very much possible that if a rod of length l moves with velocity v in a magnetic field b it will create an emf of blv okay length is l but all three should be perpendicular b should be perpendicular to the length should be perpendicular to the velocity the length velocity and b should be perpendicular if they are not perpendicular take the perpendicular components okay over here can you tell me which one is at a higher potential point number one or point number two right you can use right hand curl rule and look at the direction in which the positive charge will feel the force so electron will feel the force in opposite direction so electron will tend to accumulate there and the positive charges will tend to accumulate there so they will keep on accumulating till the electric field that is developed inside the rod balances the attraction between them okay so you can see over here the electric field is getting induced inside the rod which creates an emf of this much so this is one of the situation in which charge is static but then also there is an electric field so yeah this is the emf emotional emf sometimes you have a situation in which you have like a wire like that moving with velocity v in a magnetic field b in that scenario you have to take you know there is a derivation to it i think we have done it but not required and right now so you can take a projection in the direction perpendicular to the velocity take the furthest point from here furthest point there this length is let's say l the emf will be equal to blv you don't need to worry about the twists and turns in the wire just take the projected length which is perpendicular to the velocity like this you can get the emotional emf okay now once you know about the emotional emf this sort of thing which i'm drawing right now is very popular and i'm sure you have seen it already so this kind of structure in which there are two rails okay here they can connect resistance inductor what not they can connect anything and you have a rod like this this rod is sliding with velocity v okay in a magnetic field b which is in two this kind of thing is very popular in problem solving so here you can find out the emf between point one and point two by using emotional emf and you can also find out emf same emf you'll get by considering the flux inside this loop you can get d emf okay so using both the scenarios you'll get blv only as emf blv you'll get as emf and because of that you'll get current which is blv by r now because there is a current in the loop and there's a rod okay so this rod because of this external magnetic field b will experience a force which direction that force will be backward why of course why why say it again so because of lenses law because of lenses law so you can apply lenses law directly you don't need to find out the direction of current and from direction of current use right hand thumb rule to get the direction of force no the basic cause of the emf induction is movement of the rod moment of rod should be opposed how it is getting opposed by doing anything and one of the thing is force so force should oppose the velocity so that's the reason why the force is in this direction okay this is the magnetic force and according to that the direction of current will be you can find out that this is the current the direct the magnitude of forces il cross b okay and you can see that current sorry length and magnetic field they are perpendicular to each other so the force will be b square l square v by r il cross b now sometimes you need to be paying a little bit more attention to a direction of force the direction of force should be perpendicular to both you should remember that magnetic field and length of the rod both okay I have seen some cases becomes little tricky so this is the force using this force you can also get the power that is developed to power that is delivered to the rod power that is delivered to the rod is forced into velocity which is also equal to i square r so whatever power you deliver whatever work you do here it is getting lost as heat from the resistance because there is no energy storage device in this circuit so entire work whatever you do instantly gets converted into heat and it gets released so this is the emotional emf then whatever we have revised till now regarding emf and the rod and everything many students more than 70 to 80 percent student think that chapter is only about this okay but the reality is there is at least you can say 30 percent more chapter is remaining which talks about inductor okay so inductor is a big chunk of the chapter which is ignored and it is known to everyone okay it's an open secret so if I have to give you a question on this chapter I will create some inductive circuit and ask you to analyze that fine and yes lately many questions are getting asked from this topic so do not ignore it so inductance you know is a concept just like you can say q is equal to some constant times v this concept is called capacitance similarly flux can be written proportional to the current and the proportionality constant is the inductance okay and as the definition of induction suggests the geometrical features just like geometrical feature of capacitor is what two conductors separated by insulator similarly it requires a flux so inductance is only possible for loops a loop is required okay otherwise you can't define flux okay and emf is d phi by dt okay so this is equal to l di by dt and according to Lenge's law minus sign will come now minus sign means that whatever emf induced will oppose the cause what is the cause of emf induced in the inductor everyone changing current change in the current so inductor does not inductor doesn't like current to change okay if current is decreasing let's say you have an inductor like this like that if current is going like this and it is decreasing it will try to increase the current now if it tries to increase the current which one is a higher potential point number a or point number b point a no b it is b so this is minus this is plus so this will push the current this way okay it becomes like a battery if current is increasing then it will be reversed all right in both the cases you can blindly write down that the change in the sorry the potential drop is minus l di by dt the signs will automatically take care if current is decreasing di by dt will be a negative quantity so when you multiply by minus one it become positive minus two plus okay so do not think too much on it when you have inductor you'll apply creature of loop rule just write down minus l di by dt whenever you're crossing along the direction of the current don't need to worry much and this inductance is such that sometimes the flux is because of its own current then this concept of personality is called the self inductance and sometimes the flux is because of the current on the other this thing other thing it other thing can be loop or one can be wire or anything so when that happens the concept of personality is called mutual inductance all right and you might have seen that inductor when it is connected across a resistance like this when I close the switch immediately after closing the switch what is the current everyone 0 0 it doesn't like the current to suddenly increase it will slowly increase so when it is when it is switched off like this then you can apply Kirchhoff loop rule e minus l di by dt minus ir is equal to 0 we have solved this differential equation and we have got the expression for the current also which is e by r 1 minus e to the power minus rt by l this is what you get and l by r is a time constant for inductive circuit these are just some basic things so these kind of questions are becoming more popular in which you have a loop in which resistance and inductors are there and the DC source is there okay I'm not talking about the alternating current so these kind of questions are getting popular I mean every year I see more and more questions okay now for solenoid in our syllabus we have derived the expression for the self-inductance and mutual inductance the self-inductance is mu 0 mu r n square area of procession into l okay and mutual inductance between the two coils between the two solenoid mu 0 mu r n 1 n 2 pi radius of the smaller square into l okay so I don't know maybe this may be helpful but usually direct formulas are not helpful in the J exams and in neat neat sometimes it is useful neat is slightly more theoretical anyways so the energy stored in this solenoid sorry not one or inductor you might be knowing it is half li square right and because of that you can derive the expression for the density of the energy in a magnetic field b b square by 2 mu 0 okay density of the energy in the electric field e is half epsilon not e square so wherever the magnetic field is b b square by 2 mu 0 is joule per meter cube energy stored in that zone okay so after this there is a small topic of ac generator which is nothing but uh it's an application of lenge's law so nothing special about it just that it is an instrument so flux there is a coil that rotates between the magnetic field so flux is b a cos of omega t right so emf will be equal to minus of d5 by dt so this will be equal to b a omega sin of omega t and from here you can get the current also the amplitude of emf is b a into omega okay in fact there is number of turns also n n also we should write in dc generator and dc ac motor we there is in portion yes ktg everything is there okay as long as you know it is an application of the lenge's law in fact they don't need to tell you theory for the ac generator because what is it rate of change of flux only lenge's law is there in our curriculum so indirectly ac generator they can ask questions but they can't ask you like in dc generator what is the function of the uh you know split rings and brush and all that so they can't ask you in detail if there are both len m potential drop is lm and a yes the total flux is equal to li plus m12 i this is li1 i2 plus m13 i3 you can keep on writing the flux like this flux because of its own current flux because of the adjacent current i2 flux because of adjacent current i3 so totally emf will be derivative of this so yes theoretically that is a case but then when it comes to numericals mutual inductance are ignored actually so only self inductance are there unless it is specified exclusively so when doing solenoid we were treating like length is much greater than radius but in modules there was some case where it's not greater and they they took like two angles cos theta one cos theta two would you explain that what they're talking about self inductance no sir like solenoid only we took l is greater than r right much greater while driving what sir the what do you say magnetic field magnetic field yes sir okay so that is that we will do in fact i'll after the class i'll send you that it is not part of emi okay okay sir in fact when they mentioned solenoid you take it as an ideal one only in fact you're asking to solve a numerical all right it is a part of a numerical don't treat it like a derivation can a rod have emf if magnetic field is varying if a magnetic field is varying can a rod no rod won't have an emf if magnetic field is varying it requires movement of the rod systematic there should be a systematic movement of the free electrons inside the rod you what happens is that electrons are moving randomly okay so because of the magnetic field they will experience force in random direction so net net the force get cancelled out and nowhere accumulation happens but when rod is moving systematically in one direction then you can have velocity of electron as random velocity plus systematic velocity so force due to random velocity is ignored only systematic velocity force is there and because of that electrons get accumulated over there and there is an emf that get generated so systematic movement of rod is required understood which is yes but then yes time varying magnetic field do generate electric field for that you don't need a rod it can generate in open space so we have learned that in em wave also right and em wave will get generated so let me quickly tell you that the induced electric field induced electric field emf is equal to integral of e dot dl this is equal to rate of change of magnetic flux okay there's a minus sign both sides so you get cancelled away and if induced sorry if circular symmetry is there circular symmetry is present then integral e dot dl can be written as e into 2 pi r this is equal to rate of change of magnetic flux so electric field can be written as this 1 by 2 pi r d e 5 b by dt and the induced electric field will create a closed loop if there is a circle symmetry the electric field will be circular like this which is which is never the case when electric field is created by the static charges so only induced electric field create the closed loop and they are non-conservative in nature non-conservative so if you move a charge from here and come back to there total work done is not zero so you can't have potential energy of induced electric field so okay so this is the summary of this will not work there what then we will catch off loop rule not work in this kind of system no it is about conservation it is only true for the conservative forces in fact did didn't you like I thought you should be wondering the case when we talk about a loop like this in a magnetic field get shrunk and its area is reduced emf is generated and current is there so emf between which two points have you have you ever wondered between which two points emf or between which two points potential difference no right here also the electric field that get generated it is having this nature okay but then you can ignore that nature because this is analogous to having a battery over there but in this case you if you go around in this loop I mean there is no physical loop as there's no current as such so you can't ignore the non-conservative nature here so more you go in depth I mean more there is there are things to explore so it is this topic induced electric field is something which is not touched in your ncrt textbook they have ignored it all right so let us let us start solving numericals I have around 60 to 70 numericals let us see how many we can solve okay we'll take from simpler ones to the difficult ones you are looking from the side of the magnet you're standing on the magnet this is you anti-clockwise current will flow like this yes okay over here current will flow in opposite direction yes okay because the induced current is not opposing the magnetic field induced current is opposing the change in flux okay so wherever the flux is increasing it will try to reduce that wherever it is decreasing it will try to increase it is also simple only the current in loop it decreases with time so this flux this flux is decreasing it will try to increase it by having a current like this okay will it attract or ripple it will attract the color of this law it will try to bring this closer so that it can increase the flux okay so directly when apply the languages law and b is exact opposite of the a b is ripple party first answer and party it's like two parts right party there are three wires connected like this oh yes party what is the answer how much current I mean which direction the current will be this way or this way on this connector so both so it'll add up is here it'll add up they won't be current yeah there won't be any current over there okay so if you consider it as if it's part of this thing one here one there like that if you consider so current is diminishing which is directed away so going in diminishing it will try to increase it so one current will be like this other current will be like that so when you add it up this will become zero so that is how the current will be like this over here so so for that question can you say that since the field is along the circle if I take that path then the potential difference is zero across those two points between a and b yes sir it has to be actually see the resistance of this wire is also zero right so even if it wasn't the potential is zero correct potential is zero this one no current here you're telling no current why the loop is not complete right loop is complete boss it is like I have like this a rectangle which is twisted to make a doughnut kind of thing so but the thing is not touching this we're still a closed loop closed and I this is closed so magnetic field inside this zone is diminishing it'll try to increase it so it'll go like this and come out like that getting it yes this one no current anywhere no current on this no current on this okay okay but what about current on these two so it'll be both like this yes opposite here they don't have option here they have option so both will be clockwise like that okay over there so be like clockwise in the bigger circle clockwise bigger yes yes because bigger can have larger flux so that's the reason why it can oppose more net net it should oppose the smaller one will compromise smaller one will be like supporting it but the bigger one is opposing so net net opposition is there there's no option because of the twisted this thing mathematically what you can see is that if entire this thing is there you have twisted it so that I mean this area has been rotated by 180 degrees so its area vector is downwards its area vector is upwards they are making 180 degree with each other so like that you can imagine mathematically physically if you analyze it'll be easier so there'll also be a force on the smaller one trying to flip it okay there will be see this is a uniform magnetic field so the force will not be there okay it's a loop yes so that's why like it's a dipole so if you rotate it a little bit then it'll try to go to stable equilibrium huh there'll be this is a situation in which right now this we can consider as if two dipoles are there this one and this one two dipoles are there this one is in stable equilibrium is it this is a down direction magnetic dipole moment and magnetic field is also down yeah this is stable this is unstable so this will there'll be torque which will try to flip it over okay magnetic torque hmm so I think should be simple figure conducting loop straight wire straight spelling current shown if current is increasing continuously because of this current the magnetic field is into the plane like this it increases so it will try to decrease it by generating current like this isn't it so this is the answer and it will try to it will get rippled or attracted net net rippled rippled right you guys are not getting bored because of simple questions soon I will turn the table I have to take all levels of question including your CT level to advanced level you can't make mistake in these questions okay all of you should get it right one all of you one minute giving you 20 seconds okay the answer is one actually minus one those who have not get got the got the wrong answer you know it's a serious serious thing all right so emf is minus of n d phi by dt 3 t square minus 2 into 10 raise to power minus 3 then put t equal to 2 over here you get emf is equal to minus 1 in fact minus has just symbolic meaning over there even one is the correct answer if you say one that is also correct minus will just tell the direction nobody knows what is plus direction so how can you say minus this one so you have another opportunity to check whether you can trust yourself with calculations do the correct calculation get the correct answer and be proud of yourself okay Ruchir Singh has answered something which is not correct Rajdeep has answered which is again not correct any other answer param calculate okay okay sir Chaitanya got it correct Ruchir Singh finally got it he got it only these guys have got till now see I am not testing whether you know the cal know the concept here I am testing your calculations okay I can see more than 70 percent of you did not get the correct answer what does it say the correct answer is emf equal to 6.75 volts okay so how to do such calculations first of all do not put the values in detail okay at the end you put the values phi 1 is what pi into r square into b second time the flux is a square into b is a side length of the square so we are talking about average emf so phi 1 minus phi 2 or phi 2 minus 5 and forget about the signs okay sign doesn't matter phi 2 minus phi 1 divided by delta t so this is equal to pi r square minus a square into b divided by 0.1 is the time okay in fact 0.1 also you can put it there but it's okay now the circle is becoming square right so the perimeter should match 2 pi r should be equal to 4 a so a is equal to pi r by 2 now you put the values you will get a pi r square minus pi square r square by 4 divided by 0.1 times b okay so simplifying that further you can take r square common pi r square in fact 1 minus pi by 4 into 10 now you put the value of r do not unnecessarily find out nobody is asking you to find out what is a okay so now you put the value of a your calculation will be at least two times simpler than if you would have substituted in the middle anyways the answer is 6.75 move ahead emf induced straightforward blv the current blv by r x0 force required for constant speed if it is a constant speed net force should be 0 magnetic force is this direction so your force should be in opposite direction il into b b square l square v by r power is force into velocity b square l square v square by r and power dissipated is same as power delivered right you can see i square r this is i e i square r also if you do you get the same thing as power that is in line with the conservation of energy because energy can't be stored anywhere so it has to get released get the answer in terms of a and b don't put the values get it in terms of variables hello if you try it you have to subtract these two fluxes or not their orientations are different 180 degrees so the flux will be subtracted flux is a square minus b square total area into b0 sine of omega t right so from here you will get d5 by dt a square minus b square into b0 omega cos omega t okay this is the emf current is emf divided by the resistance direct direct question this is formula substitution that divided by rho times four four a plus four b into cos of omega t so if you like more crisp answer a square minus b square divided by a plus b is a minus b so a minus b b0 omega divided by four times rho into cos of omega t so this is the current okay amplitude they asked right find the amplitude of the current amplitude is this this is the amplitude the bracket down all right okay you don't need to type in the answer once you're done message that you're done no no no picture on the whatsapp it's okay all of you do this message once you're done okay everyone attempted all right karthik is it done okay so all of you agree that there will be a magnetic force which will be il into b current is blv by r so it will be b square l square v by r now this is the formula for the instantaneous force okay it means that at whatever time whatever is the velocity magnetic force will be this much so even the Newton's second law is also the formula for the instantaneous values net force is equal to mass time acceleration so f minus magnetic force is backwards right it'll try to oppose it so f minus b square l square v by r this is the applied force minus the magnetic force this is equal to mass times acceleration which is equal to m dv by dt all of you understood till here yes so m now you have to I mean now this is mathematics actually this is not physics from here f minus b square l square by r into v this is equal to dt integrate this from zero to t velocity goes from zero to v so now minus of m r by b square l square log of f minus b square l square by r times v divided by f this is equal to t so from here you just simplify the expression you will get velocity to be equal to f r by b square l square 1 minus e to the power this this is how you solve this question it proceed further in case you have any doubts you can type in at any point in time do this okay here and param got something rucha sing param modified the answer your graduation day will happen online so they haven't told anything but it seems like that so you never get to go to school in class 12 the school got over in 11th itself all right sure I don't type correctly why don't you type correctly okay do it now so again here the velocity is like this velocity is perpendicular to the rod and magnetic field both so I can safely write down emf is blv right velocity is perpendicular to length as well as magnetic field all of you can see that yes current is equal to blv by r this is a current okay so magnetic force is b square l square v by r which direction the force magnetic force will be if I draw an inclined plane looking from side and let's say this is the wire that goes inside this red dot so which direction the magnetic force will be up the slope up the slope yes this is b this is a direction of the velocity okay it'll be up the slope correct magnetic force will be up the slope this okay this is your magnetic force what if the magnetic field would have been like this then which direction the magnetic force would have been if magnetic field would have been like this magnetic force would be in which direction so it would be in the same direction up the slope no it will be left it will be perpendicular to the magnetic field as well as length perpendicular to magnetic field is this direction that direction will be the magnetic field force magnitude wise it will be same but direction will be like this into the slope is the wire okay wire is going into the screen okay the magnetic force is like this all of you understood this point yes now see it is a straightforward situation magnetic force is upward and if the what do you have to find steady state velocity if velocity keeps on increasing magnetic force keeps on increasing when velocity becomes so big that the magnetic force matches the mg sin theta the sin alpha then the velocity will become constant there is no other force along the incline right so mg sin alpha minus b square l square v by r should be equal to zero velocity you can find out mg r sin alpha by b square l square fine these are the standard questions you must get it so what is the answer if you take projected length perpendicular to the velocity which is this length 2 r okay so emf is okay param got something kirtika got ktg rucher k rucher correct kirtika correct only two people got it correct abhiram why you came so early wait sir just give me one minute one minute i'll take to solve also tell me which component of well the magnetic field will be used here horizontal or vertical why because the magnetic field should be perpendicular to the length right so magnetic field that is vertical should be utilized here what is given to us is horizontal component of magnetic field airplane is flying horizontally so i have to take the vertical component of the magnetic field so magnetic field vertical divided by magnetic field horizontal this is equal to tan of 60 which is root 3 so vertical component of magnetic field is root 3 times the horizontal component 2 into 10 is so minus 5 okay and then you can get the value of emf which is blv so root 3 times 2 into 10 raise to power minus 5 bl which is 20 into velocity 250 you will get around 0.173 volts as the answer understood may all clear yes i got the answer good i'm taking a lot of school level numericals also sir is there a figure show in figure copper rod angle lost to omega it's simple shown in figure hotow okay in a uniform magnetic field that is going into the board okay it is rotating in the board derive it okay you may know the answer derive it very good so i can see i hope you're not remembering and answering you have a rod that rotates with angle lost to omega so all the length is perpendicular magnetic field but the problem is velocity magnitude is different at different points so i cannot use this formula blv directly so i have to take the help of calculus i'll take a small strip here at a distance of r dr thickness this entire dr will be moving with a velocity of omega r so a small emf induced in dr is b into dr into velocity which is omega r so total emf will be integral of that so it'll be b omega l square by 2 i think it's direct okay now going forward we are not going to take direct formula substitution questions okay some of you answered the chair sing got it i think paarek also got it chetanya what is by root 8 what does it mean okay should i solve fine omega is the constant acceleration with which the rod is moving upward starting from point o okay so this distance y i can write it like this y is equal to half omega into t square initial velocity is zero okay so the length of the rod which you should consider for emf is this one right this is the length of the rod which you should consider okay and this distance is your x coordinate for the rod this is also a x only so the length is two times of x so this is point y so whatever is a x coordinate is this distance i'm saying and due to symmetry the total length is 2x okay so emf is d into length of the rod which is 2x into the velocity that is dy by dt velocity we could have written omega into t directly because v is equal to u plus a t this into omega into t okay now i want to write as a function of y only i don't want x i don't want t so x i can write as a function of y like x is equal to root of y by a and t from here you can write t as root over 2 y by omega okay so you simplify this get the answer understood all of you type in how i answer okay i think this is again in direct formula based see what is given at the question what is shown in the diagram are different so a is this this is a if you read the question okay so the emf is blv in which all three are constant but over here with the length magnetic field is changing velocity and length they are same but magnetic field is changing so i'll again consider a strip in the direction of the change so that i can say the strip is thin and the change is negligible inside that strip so at a distance of r a strip of dr is taken the magnetic field is mu naught i divided by 2 pi r so emf is blv so mu naught i by 2 pi l is dr this is 1 by r dr this is velocity v so emf is integral of that and when you integrate the r will go from a to a plus l so this will be mu naught i v by 2 pi natural log of a plus l divided by a 1 plus l by a this is our emf okay i think this should not be difficult so i don't think they're given the direction of the field direction of field take like that what else you want so that's all everything else here if something is not given in exam you consider it to be the most obvious one because if you attempt then only you probably get the marks for wrong question okay ruchir Singh got something Raul mishta others hmm can i write it like this a battery of some emf with r equivalent r equivalent is r plus r1 r2 divided by r1 plus r2 i can do that right emf is blv the current is blv divided by this r equivalent okay clear to everyone these two resistance are in parallel and this resistance is in series when they are the compare because battery is here only clear to others who did not answer type in clear but how can we just do r equivalent is equal to this so because in each of the things like both of the loops are trying to drive current in opposite directions along that line when you do the left side it will be same direction one person speak ha param what do you say one minute sir let me just i'll bring up the next question here meanwhile no sir i've made a mistake do this ruchir Singh got something discuss now yes sir Rajdeep you got the answer so we have a square frame of a side a long wire located frame translates to the right with constant velocity and the emf induces a function of x okay so what is the emf what formula comes into mind when you see this thing there's a loop there's a magnetic field they're asking to use emf so this formula okay so for somehow i should get d5 by dt over here let us see how we can do that here i need to find the flux because the magnetic field is changing with the distance along the distance i'll take a strip at a distance of r i'll take a strip of width dr okay so for that a strip the flux in it magnetic field is magnetic field is going into the area is also parallel to it as an entire the angle made by area and the magnetic field is zero so b into a only for that a small this thing will be mu naught i divided by 2 r to pi r into a dr like this okay fine so over here i will integrate and get the flux first d5 is equal to mu naught i by 2 pi mu naught i a by 2 pi dr by r integrated r will go from x to x plus a and this is from zero to five total flux so i'll get flux to be equal to mu naught i a by 2 pi natural log x plus a minus log x and then you differentiate with time you'll get emf okay the dx by dt is velocity v so the differentiator will get the answer any doubts emf as blv yes why not we can take that yeah that would be simpler you know doing all of this it if you take blv there are these two rods only this rod and this rod so these two rods can be considered as two batteries like this like that so this em this emf minus that emf with the answer even minus e2 there is no emf induced in this rod and that rod because velocity is parallel to the length okay so yeah you can get it as mu naught i by 2 pi x bl is a into v blv minus mu naught i by 2 pi x plus a into a into v so this minus that even that way you'll get the same answer only clear yes okay Rajdeep clear so the induction inductor starts here which flux do we take which flux do you take tell me i don't know sir you take the cross section area where the maximum amount of flux will be passing through when you calculate the self inductance that is why there is a hint given in the diagram itself taken a rectangular plate like this right so through that you take the flux find out the flux through the rectangular plate okay this is equal to l times current what is the magnetic field due to the outer current over here somewhere in between zero zero right so magnetic field will only be because of that how the magnetic field will be circular okay what you consider the ampere circuit okay you consider ampere circuit and you can find out the magnetic field okay did you find the flux magnetic field is mu naught i by 2 pi r as if it is a current carrying wire outside it okay you can apply ampere circuit law b into 2 pi r is equal to mu naught i same thing this is the magnetic field so this same magnetic field is throughout this strip d orange one so the flux through that a small amount of flux through that is mu naught i by 2 pi r b into area that is length into dr total flux will be integral of that when you integrate you get mu naught l by 2 pi log of b by a to integrate from b to a that times current so this bracket term is your self inductance l then anyone okay which I saying got something others get the expression okay get the expression in terms of l naught capital l mu naught don't need to tell the final answer all right I'll do it now how many meters of the thin wire required to manufacture a length l naught and inductance l if solenoid cross section is considerably so inductance l should be equal to mu naught n square what is the formula into area that is pi r square to length l okay so the length of the wire amount of wire should be how much how we calculate that length number of turns into 2 pi r 2 pi r into small n into l this is what it amount of wire it will be have one loop will have 2 pi r length that into number of turns is the total length okay now r you can get it from there so r is equal to square root of l divided by mu naught n square pi l just put the value here get the answer which is sin parek others k tg yeah others the current is decreasing which one is at higher potential this side of the inductor or that side right or left right hand side is at higher potential it will try to increase it correct yeah so v a minus current 10 ampere into 2 plus 12 plus 5 millihenry l di by dt this is equal to vb see if you want to use it is decreased right so actually di by dt is minus 10 is part 2 so you could have written minus l di by dt in the formula then substitute di by dt as minus 10 is part 2 because it is decreasing or if you don't want to be depending on the mathematics you know that this is lower this is higher you should add the potential different so like this also you can do it so you will get vb minus v a as equal to minus 8 plus 0.5 right 7 point minus 7.5 volts undershirt rasdeep chattanyah here k tg yes sir who is that rasdeep chattanyah okay looks like you sang a song do you sing a song no time we miss the classroom do this anyone close the answer get the expression expression you want in school what is going on so we are done oh great fine so let's do this after this we'll take a break the inductor a resistance and you have a battery le so when you close the switch we we know it like when you derive it you will get the growth of the current like this current is equal to v by r 1 minus e to the power minus of rt by l time when current is 1 ampere approximately so if you can substitute the value 1 is equal to v is 12 by 6 substituting values because values are straightforward they are integers so probably it gets simplified so we have e to the power minus rt by l to be equal to half take log both sides you will get rt by l to be equal to natural log 2 so from here t is equal to l by r ln 2 what is ln 2 you remember 0.693 0.693 0.693 so l by r into that is the answer okay so let's take a break we will meet after that fine so come back and time do this current 0 at equal to 0 you have to find time dependence current after time t what it is no one one minute yeah just write down the Kirchhoff loop rules you look at the answer done differential equation fine so let's see this is current i take this as i1 because di by di1 by dt terms will be there is i minus i1 so we have e minus i minus i1 times r minus i times r is equal to 0 and in the second loop over here you have minus of l di1 by dt plus i minus i1 times r is equal to 0 now what to do anyone you should eliminate i minus i1 no i okay so from the first equation you will get i to be equal to e plus i1 r divided by 2 r just check that yes then substitute that i over here you will modify entire equation into i1 okay so if you do it properly you will get a differential equation like this d l di1 by dt plus i1 by 2 times r is equal to e by 2 anybody got this differential equation yes yes okay so when you solve this you know take this term over there and then separate the variables you'll get i1 is equal to e by r 1 minus e to the power rt by 2 l okay now there is a simpler way of doing it without differential equation can you solve this can you think of any other way of doing this question so could we look at the final and initial states and try predicting so we can find the effective time constant and then substitute in that equation how will you get that see i think we have two basics see wait i want this this circuit to look like this if i make this circuit look like this then i know the direct formula isn't it can i make it look like this l and some r equivalent and some e equivalent can i do that let me modify the circuit a bit can i say the circuit is this this then another emf can i say it is like this r r e over here emf is 0 can i say that yes yes so e by r plus 0 by r is equal to e equivalent by r equivalent where 1 by r equivalent is equal to 1 by r plus 1 by r so r equivalent is r by 2 so you will get e equivalent to be equal to e by 2 and r equivalent to be r by 2 so now the simple that is r by 2 e equivalent is e by 2 so wherever in the equation r comes put it as r by 2 you know that equation of single inductor and resistance is e by r 1 minus e to the power minus rt by l so whenever e comes put e by 2 wherever r comes put r by 2 you'll get this directly without differential equation solving you consider this as internal resistance of this i have taken more questions on inductor because i know what you will ignore how you do this question anyone how you proceed because like do we find out the energy that will be stored in the inductor and then after that that will get dissipated as heat why that will get dissipated so because the switch is disconnected and then it's just discharging of the circuit no it will get discharged here as well as there only right both places it will get discharged there ask you how much heat generated in the coil over here how much how will you know the coil will lose all its energy because current at infinity will be zero current is current at deeper to infinity will be zero and before it was i so half l i square will be heat so we can say that whatever the heat is dissipated in both of them it'll be proportional to the ldi so the heat dissipated by the coil will be one third of the total thing because it's resistance not resistance so one third of the total heat will be the coils i mean that is tricky to say because heat is i square okay so but i will have to be the same it's a closed circuit for both of them what what i will have to be the same for both of them because it's a closed circuit correct correct i has to be same correct so uh it is proportional to r yeah that way you can solve that is one way other way suppose you are not very confident about that scenario which is with you know i am also not so confident but then yeah we can always verify whether that is fine and that should be fine actually because current i is flowing through r and r not so it should be proportional to r only total energy is half li square that will only get dissipated because resistance can't store the energy so one third of that you said right one third should be the answer okay this is one way suppose i want to solve it by using brute force first principle how will you go about it what do you find the current and then do i square r and integrate so what is the current current is how much we'll get current as a function of time with this how it's e by r e to the power you remember all of this if you get a differential equation you solve it you'll get this yeah so this is the current over here it slowly decreases to zero so the energy that is dissipated is i square r dt integral and zero to infinity okay clear to everyone sir i didn't get why only half li squared won't work what say it again sir i didn't get why only if we do half li squared we won't get the thing see half li square will dissipated it will get dissipated in both the resistance okay so what param said is that it will be proportional to the resistance whatever heat get generated so that's a one third of the heat will be here over here and two third will be liberated over there so i am convinced that yes it should be like that but you know i still want to check whether i will get that only so param can you check it after the class okay so fine okay so param here like that and i'm otherwise param will not do it param will check okay both ways you get the energy and let us know whether you get the same answer okay sir it'll be interesting and i think it should be same otherwise it'll be against conservation of energy let's look at the integer types i have many questions actually i've collected more than required okay ruchir parek got something not correct chattan i got it correct okay others i think it's a simple straightforward question ktg got it wait sir can you can give me one minute at least i've already given two minutes take one more minute answer will be an integer no ruchir singh that is not correct i will misshra got it misshra ji got the answer okay param also got it okay so the effective length how much you should take this length is how much so 0.3 0.3 that is 2 into 0.3 into sine of 30 effective length is 0.3 meters only okay so the emf is blv b is five blv is four blv is six volt there's six volt the power is what e square by r equivalent that should be the power that is 36 by r equivalent is six and three what is the r equivalent six and three sir two so but even the resistor has resistance like that like the the thing that's moving okay this has one ohm so six and three two two n plus one three so 63 36 by three 12 watts now once you get when you're very close the answer take one second pause one second pause and see what they're asking three x what this is three x so x is equal to four if you jump your gun like yeah i got 12 so let me write the 12 it'll be it'll be some some mistake which you'll remember the rest of your life that you did like this in the exam okay one three four all kinds of answers others shaytan is saying three praptit is saying four or praptit is here okay okay okay here why it is four i think it is the same case as when a rod is moving about its axis with some angular velocity why it is having rotation right it's moving forward also the disc is rotating right pure rolling is happening so rod is moving forward or not yes sir you did not consider that yes sir praptit why it is four so even i consider a transition param why it is four right so that i'm working on it now abhay i did it and then i realized that i'm yeah but the correct answer is four only now tell me why it is four all of you okay so you need to consider the rod email develop as a function of time will be this this at this moment when rod is like this the center of mass velocity is along the rod okay and the entire velocity that is perpendicular to rod is omega r only okay so if you consider the velocity that is perpendicular to rod so the parallel to rod velocity doesn't affect so that is only omega into r in this position but i am wondering that suppose it becomes vertical then um then the situation will be different velocity of this point will be vcm minus omega r like that it will be so it will switch polarities right after five i do and the rod is like this and when rod is like that there will be a different polarity because right now the perpendicular velocity is like that then the perpendicular velocity will be like this so polarity is will should polarity should get reversed so four should not be the answer why four is given the answer much we need to properly solve it actually so you can consider the rod like that this angle is let's say omega t so over here the vcm is like this so its component perpendicular to the rod is how much this is v omega t so v sin omega t is perpendicular to rod so perpendicular to rod the total velocity will be omega r plus v sin omega t isn't it all of you are able to understand what i am trying to do here yes yes so take a distance of r thickness dr so d e is b l v so b l this is this is v actually l is dr so integrate it when you integrate that it will become b omega r square by 2 plus d v r into sin of omega t d omega is omega omega into r is v so b v r by 2 plus b v r into sin of v by r into t okay what happened but it is rotating the other way now i am considering that way only the rotation but yeah i have considered this rotation starts when it is no no yeah i have considered rotation the other way actually that is why after time t it has gone like this it should be i should have considered down actually i have considered it up at omega t so what changes it will have it will yeah it will be omega r minus of v c m into omega t so that will come after integrating you change it to cos huh it will become plus b v r into cos no not we are integrating with dr right yeah yeah yeah so the answer should not be four it should be three yeah book is giving four so wait i don't understand why you took it from the bottom because after time t the rod will become like this right the the wheel is rotating like this okay okay okay so okay then you should be thinking minus b no that is a minus because v c m like this it's component perpendicular rod is opposite of omega r yeah next integer types again for this you need to know magnetic field due to the loop along the axis i'll write the expression because i am sure some of you do not remember that solve it completely sir it's a into r square yes check it yourself yes sir okay chat and i got something this smaller loop is very small so you can consider entire magnetic field to be uniform throughout this smaller loop hmm so rucher parek got sing got chat and i got mahit got that's d is this so the flux is mu naught i r square divided by two times z is how much here z is root three okay so root three meter capital r is one meter so root so this becomes uh three plus one four four two it becomes eight right eight this into pi r square this is the flux okay current is equal to what you can know the value of a what a i is equal to ta is that every ampere is ampere oh okay fine i thought that was some constant okay yeah all right so current is one by r d five by dt okay so when you do d five by dt uh only current is changing because of which flux is changing so when you do d five by dt d i by dt will come that is equal to one so mu naught r square by 16 pi r square this is your current what we have to find out current only right so when you correlate you will get this is our capital r is one only so gone okay when you correlate you will get the value of x to be one x is equal to one is given to us okay all of it here so the capital r resistance yes that's resistance yes i am sorry this is resistance this is the radius pi ohm's pi is yes it's three point one four pi like this r is one huh now yeah i was wondering correct now we can correlate better x equal to one all right next up let's look at some weird questions they have drawn circle instead of rod the rod is going into your screen so it'll look like a circle right final answer i do not know this question it's entire plane the z coordinate they have not mentioned so the rod is sliding on this plane the rod is this this is your rod which is sliding down will rod rotate also while coming down no it requires torque to rotate and anyway radius is not given right length l starts sliding from y equal to a uniform magnetic field emf when it reaches the point of x equal to a by root 2 x equal to a by root 2 so y is equal to how much this is y a by 2 a by 2 so from a to a by 2 it has gained a velocity okay is velocity perpendicular to rod no sir the velocity is not perpendicular to rod anything that happens in the plane is perpendicular to rod because rod is perpendicular to entire plane velocity is perpendicular to rod but is it perpendicular to magnetic field no sir that's what i meant that's what you meant huh so like i didn't listen that is the most important thing you have to do when you write example read the question properly attention to detail this is the magnetic field in the j directions is velocity perpendicular to magnetic field no you have to take perpendicular component of magnetic field so if this velocity is this velocity tangential to the curve yes it is tangential so if this is theta you have to take cos theta component of the velocity which is perpendicular to magnetic field so tan of theta there is a very nice way of finding out tan of theta slope of the curve dy by dx that is 2x by a right so x is how much a by root 2 so it becomes root 2 tan of theta so cos of theta will become how much 1 by root 3 1 by root 3 so once you get velocity from u2 plus k2 equals to u1 plus k1 perpendicular component also u2 is mg into a by 2 y coordinate is a by 2 plus k2 half m v square u1 is mga k1 is 0 okay so m is getting cancelled out that is why it was not given to us v is equal to root of ga v is equal to this so perpendicular component is root of ga into cos of theta so root of ga by 3 so emf is blv so b0l into this n is equal to 3 clear to everyone type it may look different but it's not difficult done anyone close to the answer yes sir one so which resistances are with sir is it not clear what is the doubt so for the horizontal plane are both of them one one yes okay i'll do it now so could you lead out what m a is equal to 2 pi mu naught i naught by pi square anyways listen here all of you these one ohm suppose there is a current let's say i over here okay these one ohm will have same current or not yes this one will have different current okay so current in two ohm will be half of the current on the one ohms isn't it let us find out first total current in order to find total current let's find out equivalent resistance so one by equivalent distance is one by one plus one by one plus one by one plus one by two so r equivalent is how much two by seven two by seven ohm okay so current is how much v by r that is seven ampere okay now i know that these three currents are equal one ohm currents are equal and this is half of it so i can say three times of equal currents plus the i one by two should be equal to seven okay so i one will become how much i one is two clearly okay now go back to the figure here the magnetic field because of this horizontal ring will be how much at the center so zero zero and magnetic field due to these these two currents they will also try to cancel out each other so you treat it like half of the circle magnetic field will be mu naught i one divided by four radius is a minus mu naught into i one by two which is one divided by four of eight this is the magnetic field everybody understood this net magnetic field we have got clear right we're not by four a this is the magnetic field and then you have a magnetic dipole moment m so torque is what m b sin theta m b sin of theta is torque which is equal to i alpha now here you have to see that the diametric axis they are taking for which moment of inertia is m r square by two uh okay small m and small r but dipole moment is also small m mass is also small m so in the end if you simplified you just need ma and then let us take the dipole moment as mu mu dipole moment is actually current into area only is current on the loop is given yeah yeah yeah so why do we use all of this we can directly write it as i naught into pi r square this is equal to m r square by two into alpha right so alpha is equal to r square r square is gone alpha is equal to two into current into pi into b divided by m into theta okay because angle is very small i'm considering the oscillations here so this thing is omega square all right okay so once you get the omega square runder root of that is omega time period is two pi by omega and everything else after this is mathematics only that you are anyway good at so isn't m r squared only for the ring oh one person speak what you're saying so is it moment of inertia only m r squared for the ring no no it is rotating about diametric axis okay yes so about the diameter moment of x is m r square by two sing what you're saying so what's the answer given four answer i think it's four it's m r square by two then it's four all right if you are able to solve this you're done with the chapter param got some answer no that is not the answer if you're getting 36 that's not correct are you counseling entire magnetic energy getting dissipated as heat both of you yes that's not correct do you think i'll give you that kind okay i'll give you a hint could you give one moment sir are both of the carolines in the same direction as shown so because for the second like the one with inductance l they're pointing in both direction ignore that so then k-chops it doesn't work k-chop won't work okay all of it right richard parek done no that's not the answer okay one moment one more minute i'll give a hint no the hint is become zero you have assumed current will become zero why it will become zero if i start reasoning that out do you have answer to that it is an ideal inductor okay done one moment one minute okay i'll not keep everybody waiting for that all of you listen here so now we have initial energy i know initial energy is half l i not a square plus half two l into i not a square okay this is initial energy final energy i have to find out how to get that let us assume current won't become zero but if current remains can there be current through the resistance till eternity is it possible can there be current over here after long time current is not possible there because if current is passing through this energy will be lost energy loss happens if current goes through resistance so if there is remaining current if current remains it will be on this part of the loop only okay nowhere else because it can remain there till eternity there is no energy loss happening there this l this is two l consider that the loop has been formed and current is moving like this this is the way let's say current is going just an assumption current is i which is now once current i starts flowing like this can it sustain or can that be can that go away after some time what do you think all of you if somehow current remains like this will it sustain forever yes right current is not changing so inductor is nothing it's like a resistance less path so till current becomes uniform till that becomes uniform i have l di by dt okay let's say current over here is i1 and over there it is i2 di1 by d2 ldi1 by d2 should be equal to 2l di2 by dt did you get this equation they are connected parallel right so dt dt if you cancel you have l into di1 is equal to 2l into di2 i can cancel out l also so change when you integrate it will become change in current in the first one should be equal to two times change in current in the second one this as long as this is satisfied Kirchhoff law is valid and everything is as per the laws so change if the final current is i what is the change in current in this delta i1 will be how much everyone i minus i1 i0 plus i i0 plus i okay they are reversing the direction okay this should be equal to two times now which direction you're taking positive downward you have taken positive right no upward is positive final minus initial upward is positive now delta i2 is how much over here changing current is what final current is downwards which is negative minus i minus of minus i2 are you getting this not i2 i0 are you guys able to understand what i've done here two times i0 minus i okay so i will be equal to i0 by 3 so this much current will be sustained all of you type in is it something which you have understood so shouldn't you take like like let's say i shouldn't you base your sign on the loop direction like shouldn't i've taken that current to be positive you can take that any sign you you're taking sign on the left hand side as well as right hand side if if your convention is proper both side can become negative both side will become positive so sign convention doesn't matter no so shouldn't it be like shouldn't upwards be positive on one side and downwards be positive on the other side because i want to keep positive around the loop no not like that you just look at the resistance only and look at the change in current there so if final current is i the final energy is half li square plus half 2l into i square everybody understood yes all right param fine so like i didn't really get it i don't know how like why it's negative on that side because whenever we loops don't take the positive direction which is along the loop over there okay you mean to say that why i'm taking this one like this and i get it you you're trying to say clockwise take one direction anti clockwise take another direction like that you're telling me yes but then i'm not looking at the loop i am just i i just care about what is happening to this inductor okay what's happening to this inductor and right now current is i not over here like this later on the current is how much current is i so change in current is i minus i not there okay let's say that is a change in current now this is the change in current and for the other one if i write like this the way you're telling me for the other one if let's say the final current is this way i and initial current is this i not then this will be equal to what final current is upwards right so final i have taken over here i have taken downward to be positive so i'm saying is shouldn't you take downward positive on this side and take upward positive on that side because whenever we do things in a loop that's what we do right no no no no here it's delta i1 equal to delta i1 you don't do that here because here their emfs are equal when you consider loop rule over here you subtract it emfs this emf minus that emf you do it okay even though current will be like this only but to emf you subtract or add so the deltas will get subtracted if you really want to do that way you need to subtract their deltas and equate that to zero okay so okay just do it again your own way properly and then we can discuss it further you know more specific manner whatever you're saying yeah what so would we get it if we find the power across the resistor and integrate that across time dissipated in the register yeah of course you can do that if you can okay yes have you done like that okay so can you share it after the class yes see that okay here we just i think you have become the loop rule only right along this loop and then that loop and then yeah this this current plus that current is equal to this current you probably have done that way you should get the same answer actually but when you do i square r current will current will go to in current will go to 0 and t tends to infinity so it should take care automatically but initially current what you have taken initial current do i not yeah you should get the answer should get it you want me to check the answer for this yes we'll try it and send after class again okay meanwhile you do the next question i'll check the answer for this and let you know because ruchir probably have solved that way this answer is 32 ruchir 32 yes aja wait wait let me go there let me answer the param param here see when you're writing it like this you're considering both current down right yes so when you consider both the current down in that case only you have written like this for that case yes or no yes now you got it yeah okay yeah so downward is positive for both otherwise you write one is equal to minus of the other good question param's question get the answer here wait sir can you go back again abhi ruk jaa pehle tui solkar get the answer we'll go back don't worry okay what is the answer annush got something sir just give me one more minute one more minute i'm giving others annush checked the silly errors no one got correct answer till now so so i send the answer on what sir the way ruchir said it no you just tell the integer right integer just tell me the answer for k so 30 no not this one message previous all right so first tell me the answer for this one what is k message it see we can't we cannot hold 25 students for just one question which we are discussing individually right you have to move ahead we'll discuss that appropriate time okay so ruchir also answered mahit got it correct only mahit got it why others are not getting it is it difficult this one has 10 ohms this one has 15 ohms this emf blv that is four mayhul got something is this correct everyone whatever i have drawn whatever i have drawn is correct circuit or not let's say polarity is 4 millivolt polarity is ulta for this getting it so i think ruchir you did that mistake right e equivalent by r equivalent is equal to 4 by 10 minus 2 by 15 and we have to take minus because polarities are reversed 1 by r equivalent is 1 by 15 plus 1 by 10 how much you get e equivalent here e equivalent comes out to be how much mahit this is what you get e equivalent okay the current will be equal to e equivalent which is 24 by 15 divided by the divided by the resistance total resistance is how much current in the 5 ohm resistance so 10 and 15 from here r equivalent will be how much r equivalent is how much 10 and 15 sorry 10 how can it be 10 six ohms sir six ohms click six ohm yes sir yes five sir not sleeping sir you sound you sound as if you're sleeping what is that be more attentive in the class what is k then k is eight okay okay so i'll just go there briefly previous question huh what are you saying param so don't we get this equation by using the loop only because no no no no by using loop you get minus of l di 1 by dt minus of 2l di 2 by dt is equal to 0 so using loop equation get 1 is equal to minus of the other are you getting it so then your whatever you're saying now that will work but you get the same equation but it i have taken plus how did this equation come in so like if we take the loop the loop right then we get l like l di 1 by dt minus why minus 2l in so because we're going opposite to the current no no no current to find ultimately current will start flowing in the same direction you're considering the same direction right for the current like this clockwise anti-clockwise so l di 1 by dt plus 2l di 2 by dt is equal to 0 that is what you will get i have considered to be in the same direction down so that is why for me it is 1 minus other is equal to 0 okay so yeah fine that makes sense yes so how can we consider them in the same direction sir i have considered i have like i have considered because i wanted to equate their potentials only this potential equal to that potential that's all okay yes but then yes i mean i can understand the difficulty but i'm had so you can consider in the clockwise only and do the way you just go by the rules you'll get the same answer okay don't worry about it multi-correct okay more than one option so is this multiple correct no one is getting others okay you guys need more time so one minute one cat two minutes but get the answer okay only ruchir parek chattanya and ruchir sing nobody else here what is your answer param what is your answer i'm getting some weird things for all of them other way tiga so let us say the velocity of the rod is v okay so if velocity of the rod is v doesn't matter what force and whatnot the emf should be blv and if emf is blv charge stored on the capacitor is c times emf which is c blv okay understood this is the charge now if this is the charge current is dq by dt so c bl into acceleration a so option one is correct okay and if this is the current there will be a magnetic force of il into b okay so c b square l square into a is a magnetic force this minus magnetic force is mass times acceleration so acceleration is f divided by m plus c b square l square so even three is correct everyone understood yes okay let me see let us consider let us see last question for today if it takes more than three four minutes will it i think it will take this last question do it all of you hmm anyone chattanya got something no chattanya that's not correct yeah yes i made a mistake i'll take your time we have a lot of time okay no need to rush no one is close to the answer anyone i got a differential equation that i don't know how to solve you got a differential equation do you need a differential equation yes sure i think i can answer you don't need a differential equation chattanya that's correct okay so this is equivalent to two circuits this battery supplying double power to power the capacitor as well as the inductor both the potential difference are v not only right yes sir yes cover as the super excited you're excited because class is getting over or what this is normal yes we are having one more r x that's fine okay so v this is equal to v not by eight into one minus e to the power minus of r t by l okay and i2 is equal to v not by eight again one minus e to the power minus t by rc all of you understand this everyone so but how why can we explain it that easily why like why can't one component influence the other it is rich of loop rule only v not minus four d i1 by dt minus eight times i1 is equal to zero this will lead to you this and v not minus q by c minus i2 into 8 will lead you to this isn't it the loop equation over here is similar to loop equation there loop equation here is same as that getting it yes sir but i thought that we would have to like eliminate v not and then solve the combined thing because one can influence the other one what do you mean by one influences the other like if the capacitance is like because the capacitor will start to remove the now i understand what you're trying to say see the battery is electromotive force if this is acting like a force it will not let the potential difference across these two points to go deviate from v not it will push it to v not every time it tries to influence so that's but like i2 will decay right and because of that i1 will increase and the inductor doesn't no no no i2 i1 won't increase because of i2 getting decayed the the pressure on battery will be lesser than battery will have to do lesser work okay battery will what is the function of battery to maintain constant potential difference that's all whatever happens but we know that there is no influence and like that there can't be any influence shouldn't be because battery is there now battery will push battery will maintain a constant potential difference across these two points it is like this what you're asking is there's a pump okay there's a water pump which is pumping the water at a constant rate okay you're telling me that some water will push the pump itself so this rate should get influenced my answer is whatever it is the pump will adjust its power in such a manner that the flow rate coming out of the pump is constant okay similar kind of analogy here the the battery will adjust itself to maintain the potential difference construct and it is all driven by potential difference nothing else okay so so i1 by i2 should be what value of i1 by 2 at t equal to ln2 so just put t equal to ln2 r by l whatever you is values that are what are values are given you put it you'll get it as three three is the answer okay all right so that's it with respect to my what so could you just see the answer i put on the group which answer so that 32 are you send me after the class so you send it on the group is it yes this one can i reply afterwards it's a big two-pager and this right i'll reply on the group itself all right so it looks like some of you might have taken emi for granted so my purpose was to shake you up a little bit i hope i have successful in that and you know what all you don't know now so make sure you strengthen your topics which are already strong make it so strong that whatever question comes you should be able to answer it fine so that's it we'll meet next week and probably start class 11th now bye thank you sir thank you sir okay sir thank you sir here as the one hour extra hold bye sir good night


When you're shooting an open star cluster, I think classic diffraction spikes like you get with a Newtonian reflector telescope look really, really good. The diffraction spikes effect is strongest on bright stars. And so it helps bring those stars in the cluster out from the rest of the scene. And the reason you get the classic sort of X style or plus style diffraction spikes is due to the design of a Newtonian reflector where light enters the front of the telescope, hits the primary mirror in the bottom before bouncing back off the secondary mirror and then off to your camera on the side. And the diffraction spikes come from this thing holding the secondary mirror in place called a spider. And the actual metal pieces here are often called the spider veins. A camera lens or a refractor telescope focuses light differently and it's a straight through design. Sometimes you'll see diffraction spikes with a camera lens if you stop down the aperture internally because the internal iris, which controls the aperture isn't perfectly round. But these diffraction spikes are usually a little bit crazier and less predictable. And to me, they're not quite as aesthetically pleasing. I know some people love them though. But anyways, I decided recently that I wanted to start testing cheap ways to add diffraction spikes to lenses. And I wanted to do this physically, not with software. I know there are paid software plugins that can achieve a really nice look these days. But I'm just interested in how to achieve it in camera, as we'd say. I do all my initial astrophotography testing from my light polluted home near Boston, Massachusetts, right here where I'm filming. And then only once I figured out something, to my satisfaction, do I really use a new technique or a new piece of gear at a dark site? Because typically if I'm driving, I don't want to waste too much time with testing when I'm at a dark site. So anyways, last night it was clear and I had a few tests in mind to do right here in the backyard. The first test is I've been working to get this old Soviet telephoto lens that I bought off eBay for a hundred bucks to focus properly. It's called the TR-3S Photo Sniper. And I saw some photos taken with it online. I thought it was worth a gamble. When it arrived, I quickly found it wouldn't focus to infinity with any camera I adapted to it. And the Iris, which is again, the blades to control the aperture were permanently stuck at f22. Luckily, this lens is very easy to take apart and repair. And so I've been working on it and I thought I had it working. And so then last night I confirmed it was working. So it's really a pretty nice value at 300 millimeters f4.5, but if you do buy one, expect to put a bit of work into it. I might do another video about all that in the future. So last night I finished testing the lens. I realized my reconstruction of it had worked. That went by pretty quickly. So I thought I'd try another test. And so I taped some yellow thread to the front of the lens in the same X pattern that you'd get with a spider on a Newtonian. And my goal here was to see if I could mimic the diffraction spikes that spider veins usually give you, except with a wide open lens. I think I probably should have tested it on a really bright star, but instead I was really hankering to photograph this thing called the coat hanger cluster or also known as Brocky's cluster, which is just right off the beak of Cygnus. And here's the result. And I think that this worked reasonably well. It's a subtle effect, but I think it looks pretty natural. And I think the spikes would have come out a little bit better with all of the light pollution. I Googled around to see what others have used to make spikes like this on refractive optics. And I see many people recommending fishing line or guitar strings. So I have some other things to test, but I thought for a first try, thread worked pretty well and you can't beat the price. I think I estimated if this much thread in a little tape, maybe four cents for this experiment. Well, that's it for this five minute Friday this week. Next week, I'm gonna be leaving for my annual trip to visit my family. I'll be hanging out in Minnesota and Wisconsin. So all of a video about how I pack for airline travel as an astrophotographer. Till then, this has been Niko Carver at nebulaphotos.com. Clear skies.


And technology is still evolving too fast. I think that this is one of the fastest evolving areas of web development, performance and speed. So it's very technical. It limits your website. So once you start thinking about speed, it's no, no, no, no, no. Like everything is a no if you wanna be fast.  So it's not clear. And everyone else is ignoring it. So let's just ignore it, you know? Like everyone else. So my talk is to make your website fast. And it's gonna, I'm gonna try it because I have tried to do a lot of optimizations and I have followed all the possible path you can imagine. Even trying to redo the WordPress core engine was one of the paths that I tried to do. And it was not a very good one, by the way. I'm not an expert on performance. I think it's extremely hard to be an expert in this. You can talk for one year about the first 0.2 milliseconds that the website takes to load. So it's very deep and technical stuff. I know I don't think I'm an expert, but at least I have tried to do it several times. It's almost impossible to hit 100% in Google so don't become obsessed with it. I'm not gonna try to be very technical because I know that a lot of people here are not developers. And especially the story that I'm gonna try to share is the one trying to optimize the website of Forkies Academy. So why do I care about speed? I think this is one of the biggest crisis in WordPress right now. We should be like, O and U should be talking about this, about speed in the internet in general. And this is in numbers how big the crisis is. Per industry, you have all the biggest industries and how fast are they right now loading the website? The yellow line you see there, it's the industry standard, like the best practice that you should follow. And the white line, it's what Google asks for you to become like a real, to have an okay, like a good okay. Yeah, so that you hit 100%. I thought it was easy and I, like after a while I just gave up. I managed to hit 100% but I gave up so much limitations and so much functionalities that I went back to 95 or 90% because 100% it's crazy. So 70% of visits will come from 3G connections in 2020. So you do care about speed. Google Chrome has in the inspector, if you open the inspector in Google Chrome, it has a way of throwing your connection so you can simulate a 3G connection if you want. It takes, for every two seconds of loading time, you have 20% more bounces. And it's now a ranking factor since like five years ago, but it's bigger every day. The influence on your ranking factor for city changing optimization. 40% of the visitors expect a web to load in two seconds, in less than two seconds and 40% lips the page after. After it takes more than three seconds and the industry is nine, eight, seven. So basically it's a mess, right? And it affects your business everywhere. This is one of these key metrics that is gonna affect everything. And eventually your revenue, of course. So what's the goal is to have the fastest possible website without modifying the worker score. Like, don't go into the worker score because it's gonna be a battle that you're gonna end up losing. You don't wanna sacrifice workers. If you wanna sacrifice workers and redo the core, then you should use Laravel or another framework. And what is the core? Why do I keep talking about this? Well, the plugins, Gothenburg themes and the REST API lives on the worker score.  What are the enemies of the speed? Well, requests in the internet, a request is the enemy of speed. A script is the enemy of speed. Fonts, styles, databases, images, plugins, plugins and plugins. Those are the biggest enemies. So basically, everyone is an enemy of the speed. Basically, the best website, it's just a PDF, maybe not even a PDF, like an HTML hello world. That will be the fastest website in the world. So it's hard. It's extremely hard. It's easy to make a slow website. All you have to do is not to try to do a fast one. And this is, I put this here because that's exactly what happens. Like, you don't think about it. We should start thinking about it. This article is amazing. If you're technical, if you're a developer, you should read it. This guy tried to do something similar with React. He tried to redo the React core and he shows the entire path. It's like three blog posts on website optimization. I think this post inspired me a lot and this guy, it's very hilarious. Like, his way of writing, it's also super cool. So top 10 things to make your site faster. Pick your theme, thinking on speed. When you go to Google, search top 10. Instead of searching top 10 themes of 2018, search top 10 themes of 2018, fast, whatever, like with good performance. The hosting is also extremely important. Don't just use cheap hosting. Try to use cheap, good hosting that has speed. Use a CDN, this is a must, like all the others.  This is extremely hard, so you don't have to do the seventh right now, but I think it's important. Use PHP 7 instead of PHP 5 or 6 or 5 because there's no six. Make your site compatible with AMP. This is like Google standard. So I'm gonna speed up now because I don't have much time. For the page rank, for the page speed rank, you should go to searching Google speed test. And right now, you can search if your website is fast or not in Google speed test. Google page speed test, not Google speed because that's for the internet connection. But Google page speed.   But don't worry, it's easy to be a lot better in speed if you think about what I'm saying right now, if you pick a good theme, et cetera. What I was saying, search for the fastest WordPress themes 2018. Think about that. Oh, I'm going to the other side. Account delivery never, it's also important because it's a layer that goes in between your website and your users, in between your server and your users. And it's only for speed purposes. So they are gonna load your website in a different server that it's faster than yours. So you should use a CDN. And it has a lot of other perks. So like it protects you from security stuff. So it's also very good. The catching plugins are extremely important. Here in the bottom are the fastest ones. The best ones, I think. To compress your images, this plugin is amazing. Compressing the code is hard. But if you use this plugin, this plugin is one of the ones that don't have many issues because compressing the code is so hard that most of the plugins break your website if you try to compress the code. But with this plugin, I haven't had any issues and I have used them a lot. AMP, it's a standard that Google is trying to propose and a lot of websites are using it. And if you use AMP, like right at the moment you start using it, you're in a different page, in a different part of the page, in the top results. They have a special page for AMP if you're searching in Google in a mobile device. So I think it's extremely important to install the AMP WordPress plugin that was developed by Automatic. And beware of plugins, please, don't use plugins. Like I think I have a key metric that I tried not to have more than 15 plugins in a website. At the beginning it seems easy, but as your website starts growing, you're starting installing and installing and installing, trying not to have more than 15 plugins, maybe you should hire a developer and try to incorporate that in the theme instead of using another plugin. Because plugins, you don't know how, what the quality of the code sometimes and they actually sacrifice a lot of the speed. This is a to-do list of things you should use and recommendations on my site. The good news, believe it or not, is that you only need to be faster than anyone else and everybody is doing pretty bad. So if you really start thinking about this, it's gonna be good for you. It's gonna give you a lot of good things. Thank you very much. Any questions? Okay, there. Yeah, the slides are in the same place of all these speaker slides. I don't know the URL. Maybe you know the URL for the slides. I don't know it often, but they will post that after the event. They'll send out a note where all the slides are. They do it every year, so. Yeah. Well, when you install AMP, it's a plugin. So they basically take over your website on mobile devices. Like now you're gonna, you have to be very integrated with the WordPress core. So WordPress knows how to grab your posts, pages, or custom posts and publish them in a different way, in a WordPress way. Because when you're browsing mobile, you don't need much of the stuff. And you should think about that when you're building the posts, because you know that the content of the post is the only thing that is gonna be published.  So if you, for example, when I have a call to action for marketing purposes, you should put it in the text. That way, if you have a call to action or something that, if you're trying your users to click somewhere, you should put it in the text. Because that's the only thing that is gonna show on the AMP version of the website, on the AMP version. Any other questions? A page builder? Well, I don't know about what I have used with cell composer and some others. And I think if you use AMP, they are gonna remove all the circuits. All the circuits, all the extra stuff is gonna be removed because it's not gonna be used. So imagine when you are, you can actually, in any page builder, you can switch to the HTML version and you're gonna see the circuits.  But the good thing is that you can test live. Like you can actually build your website and test your AMP version. Yeah. How to keep the web? For the pictures, actually the best plugin that I've used is this big bear. Yeah. It's a tiny PNG. I think it's amazing. It's like 60% compression on your images and you don't lose any quality at all. So I will install that one. Okay, so Alejandro will be at the happiness bar if you have additional questions. We do have to wrap it up now. So if I can have a final round of applause for a great presentation.