 All right, so we were revisiting the effect of external magnetic field on the charge or on the external or on the current, okay. So this is what we have discussed till now. This one, then we have discussed these, all right, after that we are we have arrived till here, okay. Now any other formula that comes in your mind, any other thing, effect of magnetic field on the current, any other formula that comes in your mind apart from these that you have learned in this chapter. TORQ on a coil, TORQ on a coil or on a magnetic dipole moment, which is m cross b, okay. This is the TORQ and potential energy on a dipole is minus of m dot b. So magnetic field can do the work, okay. It can do the work on dipoles, but on moving charge it can't do the work, fine. So remember that it can do the work on the dipole. So m is equal to NIA, we already had this, we have discussed this dipole moment. Any other thing that you think we are not writing here? Anyone, feel free to highlight everything we have discussed. So there was one concept, not a formula as such, but I found it very useful, the force on any, the force between, the force on any shape of wire between two points would always be the same. Okay, like that. See what Trippan is saying is suppose between the two points A and B, okay. If you have let's say this kind of wire, this is what you are saying, right? You have this kind of wire I between A to B, I will take C and D because B is magnetic field, this C and D, right? And let's say you have a magnetic field like this, this way, this is a magnetic field. The force on this will be equal to this, if this length is L, the red color, the red length if it is L. If you connect those two points C and D, if that length is L, this L is, this length is perpendicular to magnetic field, the force on the wire will be I L into B, as if it is not a curved wire, it is a straight wire. So there will not be any difference between a situation of C and D having a straight wire and a curved wire. They will both get the same force which is this. How it comes? You need to consider length as a vector, okay? For example, you take a length over here, this is your DL length, take one of its component will be like this, one of its component will be like that, there are two components of DL. So on this component of the length, the force will be zero. Only on this one which is maybe 90 degree with B, the component of force will be there. The force will be there on this, the force on this component of length will not be there. So like that, you can keep on looking at the horizontal component of the length everywhere and if you add up all the horizontal components of the length, it will become a straight line like this. Okay? So that way it comes. So it is immaterial, what is the shape of the wire, all you have to check is what is the length perpendicular to the magnetic field, a straight line length, okay? So that is, yeah, that is useful. Anything else you think we have left, I haven't discussed? I think that's it, right? Let's take up the questions, if anything comes up, we'll discuss it then and there. So I think this is what the entire chapter is about, all right? One by one, you answer, first one, you answer first and then the second. Which direction the charge will go, it will move up like this or down like that. Which one? One or two, then it goes inside. How you get that? How you get up, Q, V cross B is the force, okay? Force will be towards the center because this is going in a circle. So when you align your hand in the direction of velocity and fold it in the direction of field, thumb will point up. So center will be along this line. So that is why it will move up, do the first part. It will try to move in a circle. Now the radius of the circle, if it is less than D, then it will create a semi-circle or not? And come out from here, do you all agree? If the radius of the circle is less than D, then it will move in a semi-circle and come out, everybody understand this? Radius is what? Radius with velocity D naught, magnetic field is this. Radius is M V by QB, right? So definitely it will move in a circle and then come out. This is M V by 2 QB in the second case, okay? So R is equal to M V by QB. Radius is lesser than the distance D, which is 1.5 times M V by QB. So it will spend time which is equal to half the revolution. For full revolution, this 2 pi M by QB, this much time. So half of it will spend pi M by QB, this much time it will spend, okay? So this is the option number one answer, second one. Will it be able to move in a semi-circle in second one? This is D, yeah, Trippan got something, Siddharth got. It can't move in a semi-circle, right? So it will be moving, it will try to move in a circle and it will try to bend, but then before it turns back, this thing ends, magnetic field ends. So it will go away like this, tangentially it will move out. And if this is a center, let's say this is a center, if I know this angle, then I will know exactly how much it has moved in a circle, fine? So can you tell me what is this angle theta? This distance is M V by 2 QB. This distance is M V by QB, theta is how much? So to do a small construction, if you drop a perpendicular like this, you will see that this is M V by 2 QB and this is M V by QB, right? So, yeah, sine theta, sine theta is equal to perpendicular divided by, yeah. Pappandika divided by habitanus, so that is M V by 2 QB divided by M V by QB. So sine theta is half, so theta is 30 degrees, okay? So theta is 30 degrees, we have got, this is equal to pi by 6, okay? So for 2 pi, which is full circle, the time period is 2 pi M by QB, okay? So for pi by 6, the time period will be 2 pi M by QB into pi by 6 by 2 pi. Proportionately, you have to divide it, right? So you will get 2 pi, 2 pi is gone, so pi M by 6 QB is the answer. Everybody understood, type in quickly, also find the sideways deflection. Sideways deflection is this, probably they mean this value, which you can find out. If you know theta, this distance you can find out, total distance is M V by QB. This is done, this one, upwards or downwards, it will be upwards, right? And you are able to see that it will make a quarter circle over here, will it be a quarter circle or not? Sir, isn't it a negative charge? You're asking, telling what? Sir, no sir, in the diagram it's written a negative charge, so it will be opposite. Ah, correct, see there is a negative charge here, right? So whatever the direction of V cross B, it will be opposite of that, so yes, you're correct. It will be moving downwards, like this. All of you change your answer, it will move like this. And this will be circular path, you have to find this distance now. But it should be written on the question also, we can't just rely on the diagram. So they must mention particle of negative charge. Sir, we can assume it enters from the middle? Yes, assume it enters from the middle. That also should be mentioned, some key details are missing. No, no, no, it is written, right? It enters midway, see? It is written, it enters midway. Yes, sir, sorry. It has become mathematics. Are you sure, Swamiak? Everybody type in your answers, what do you got? Okay, it enters from here, all right? Its radius is, its radius is R, but this distance, this distance, if it would have been equal to R, it would have been a quarter circle, right? But this distance is R by 2. It's not R, understood? Okay, so it will take a bigger distance, radius is R, and it will not hit perpendicularly. It will not hit perpendicularly, it will hit like this. What happened, Shashank? This total distance is D, which is equal to radius. Now, we need to find out the distance where it hits. If you put a line like that, over here there will be centre, okay, which will be R by 2 below only. This is R by 2 below, okay? And if you draw a line like this, and put a tangent over there, this is 90 degree, okay? This distance is also R. We need to find out what is X. Okay, now can you do it? Everyone, can you do this? I think now nothing left in the question. This angle is how much? This angle? Theta. The cross of theta is R by 2 divided by R. So this is half, theta is 60 degrees, okay? So X is equal to R sine theta, R root 3 by 2, where R is MV by QB. Got it? Everybody understood? Type it quickly. Let me know if you have any doubts. It is very easy to make silly error here. You can assume it is a quarter circle and say that X equal to R, but that is not the case here. You know, you can say that wait for some time like that also you can say, because I will not know whether you have understood completely or not. Okay? Even if you don't have doubt, you can say hold on for some time. I will wait. Don't just sit for nothing. Depressed means that depression. Depressed means you have slightly moved it down by a distance of X. That is what it is. No one got the answer till now. You can tell the answer in terms of variables. It is fine. Okay? You can take this as I1. This distance you can take it as A. This you can take it as I2. Gravity will be vertically down. Nobody got it? No one answered. Okay, I will do it now. Fine. So now long horizontal wire AB free to move in a vertical plane. So AB can move up and down. Okay? It is in equilibrium. Over a long parallel via CD which is fixed. So CD is fixed. AB can move up and down. Right now it is in equilibrium. So let's say this distance A, everything is in equilibrium. So MG force is balanced by the magnetic force. They will ripple each other unlike currents ripple. Okay? So the force of repulsion is mu0 I1 I2 L by 2A. Okay? This is the force of repulsion. L is the length of AB. Okay? Which is not given. Right? It is not given. This is equal to MG. So you can get MG by L in terms of known quantities. Mu0 I1 I2 by 2A. Everybody probably got this. Right? Easy. Now when you depress it, when it slightly moves down by distance of X. Okay? The force which was this much gets modified to mu0 I1 I2 L divided by 2 times A minus X. Now what do we get? Right? So now the change in force is your restoring force. Okay? Because F1 was balancing MG anyways. So whatever is the extra force F2 minus F1. This will create the acceleration, mass time acceleration. Everybody understood till now? Type in. Only one person understood. Others? Okay. So F2, let's modify F2 a little bit. We'll be using binomial approximation A minus X raised to power minus 1. Then I'll take A outside. It will become mu0 I1 I2 L divided by 2A. A to the power minus 1 comes out. Right? 1 minus X by A to the power minus 1. So this will become mu0 I1 I2 L divided by 2A. Now this minus 1 comes inside binomial approximation 1 plus X by A. Okay? So, you know, the X is very small. Okay? It is slightly depressed. So that is why I'm able to use this approximation. So F2 minus F1. Now subtract, you're going to get mu0 I1 I2 by L times X into 2A square. Okay? So you can see it is proportional to X. It is restoring force. It will be against X. So minus will come if you write in vector form. This is equal to mass times acceleration A. Now A is the distance we have taken. So let's say... Sir, shouldn't it be 2 pi A? What do you mean? Make it the force. Pi. Pi. I keep forgetting that pi somehow. Okay? So acceleration A, there is a pi. Acceleration A will be equal to minus of mu0 I1 I2 divided by 2 pi A square L by M times X. Okay? So L by M you can get from here. All right? M by L is mu0 I1 I2 by 2 pi A into G. So probably a lot of it will get cancelled away. All right? So should I do that? A will be equal to minus of mu0 I1 I2 divided by 2 pi A square divided by mu0 I1 I2 by 2 pi A. This is M by L into G. Okay? That times X. So mu0 I1 I2 is gone. 2 pi is gone. 1 A is gone. So the acceleration is equal to minus of G by A times X. This is what we get. So G by A is omega square. So time period is 2 pi by omega. 2 pi under root A by G. Is the time period. All right? Everyone understood? It doesn't depend on I1 I2. Because I1 I2 do not... I mean it is the distance which matters here. Okay? Restoring force comes because of the distance. Increase happens because of the distance. Not because of the current. Current is not changing. All right? This is... Okay? This is... Okay? Since we are doing everything. Let's do this one also. Find out. Here also you can find in terms of the variables. Don't worry about the values. Assume proton mass is known. Proton mass is M. Small M. And charge is also known. Let's check Q. Same diagram you have to make. You have to draw a line perpendicular to the exit. So that these two lines intersect at the center of the circle. This angle will be how much? Everyone? This is alpha. Can I say this is alpha? Yes, sir. Okay? And the same construction you have to do. Drop a line like this. This is R. This is D. Okay? So sign of alpha. It has become repetitive now. We have done many questions on this. D by R. So from here you get the value of alpha. And D is given. R you have to find out. R is Mv by Qb. Mv is momentum. Which is root over 2 times M into kinetic energy. Divided by Qb. Kinetic energy can be written in terms of potential difference. Which is Q into V. Divided by Qb. So we can further simplify it. You can get it as root over 2Mv by Q times 1 by B. This is your radius. In terms of potential difference V. Okay? This you can substitute here. This will be your sign of alpha. So what is this magnetic force? This one. We have learned that a loop doesn't experience any net force. How is it that here they are talking about net force? Anyone? In a uniform magnetic field. Net force is 0. Okay? Any loop. Any current loop. If it is under uniform magnetic field. Net force is 0. But here field is not uniform. It is changing. Okay? Let's name it A, B, C, D. All of you tried, right? I don't need to wait now. Throughout CD, magnetic field is not uniform. But the good thing is that whatever small length you take here corresponding to it, length over there is there. So can you tell me direction of force on this small length that you are considering? Tell me. On this small length, what is the direction of force? Everyone answer it. Direction of force on this small length, the white patch I have made over there. Towards the left. Right? Towards the right. You have only two options, right? So tell me. This is the length direction of the current. Okay? The magnetic field because of this current over there will be inverse or outwards over here. Near this DL length. Inwards or outwards? Inwards. Outwards. It will be outwards. Align your thumb in the direction of current. Fold your fingers on that side. Your fingers are coming out or not? Sana was that right? Yes, I got confused. I got it. Okay. So the force is the force on that element IDL cross B. DL cross B. It will be left-hand side now. Align your right hand in the direction of DL. Fold it towards B. It should come out. The thumb will tell you that it has to be like this. This is the force on here. If you take a length over there on this side, how the force will be? Now your DL is upwards. Magnetic field is again outwards only. So it will be like this. This way. All of you understood? So for every corresponding DL, there will be a DL here. Force this way and force this way. These forces will be different in magnitude. As you go away, the magnitude of forces will decrease. But every time there is a force on this side, there will be a force on that side also. So the net force horizontally is 0. Not only force, the net torque is also 0 because of these forces. Now talking about AD or AD. Tell me what is the force direction on AD? Direction of force on AD. Magnetic field is upwards. Sorry, not upwards, coming out. So DL cross B is upwards. All of you able to understand this? Force is up like this. On the top, the current is in opposite direction, so it will be down. Now throughout AD, is magnetic field uniform throughout AD? B is uniform? Yes. Yes, it is uniform. B is uniform. So the force on AD minus force on CB will be the net force. So the net force is mu naught i divided by magnetic field near AD is mu naught i divided by 2 pi A, the magnetic field near CB. This is AD, that is CB is mu naught i divided by 2 pi into A plus W. So the force is IL cross B and I know magnitude I have to subtract. Once you know the direction of forces, don't rely on cross product to get the directions. You already know, you have to subtract. So forget about the direction. Use a formula only for the magnitude. So IL into B, now this current is I2, that current is I1. So this is magnetic field, the current is I2. So IL cross B, I1, I2L divided by 2 pi. This will be common in both the cases. 1 by A minus 1 by A plus W. This is the net force. The torque is still 0, net torque is still 0. So torque is 0, it will not rotate. But what it will do? It will try to move away because this force is stronger. So it will try to accelerate. This force, you can write it as mass times acceleration. You will get initial acceleration. But immediately when it moves a little bit, acceleration will change. So acceleration is not uniform. When A and W are the distances, acceleration is this. So you can write in terms of x. If you write x instead of A, you will get acceleration in terms of x. So instead of A, you can write x. So you will get force in terms of x. And then acceleration, you can write as V dV by dx. And then you can do some integration, get velocity as a function of x. So like that you can deal with it. Alright, so one more similar question that came into my mind is this. Suppose you have an infinite wire, which has a current I. An infinite wire. And you have a finite wire over here. This is a finite wire. Somehow you are maintaining a current in it. So let's not debate on how current is there. Current is I. This distance is A. The near end is at a distance of A. The blue wire and the white wire, they are perpendicular to each other. The total length is L. And over here, I need to find out the net torque on blue wire. The infinite wire is fixed. Anyway, it has infinite mass, infinite inertia. It won't move. So I don't need to tell you that it will be at rest. Torque, yeah, torque about certain point you can say. Torque about one of its ends. Torque about one of its ends. Talk about center of mass. That would be better. You can use it to write it as torque will do I alpha. About center of mass, you find out the torque. If you find out torque about center of mass, you can equate that to I alpha. I see him into alpha. The hint is take a DL length like this. The DL length is take DX. You can say this is X. This is DX. Which direction will experience a force? The force will be in which direction? A ports. Magnetic field is downwards. Sorry, into the plane like this. So L cross B upwards. Good. So now force magnitudes, magnitude of the force is decreasing as you move away, because magnetic field is becoming weaker. Isn't it? So this rod will try to turn like this. Because forces on the left hand side is more than forces on the right hand. Left hand side forces are more than right hand side. So it will try to gain an angular acceleration like that. That is I am finding that torque from the center of mass. So the force on it will be equal to Nobody got the answer. So the force on it dF is I DX into B there. So this is I B is mu naught. That is also I divided by 2 pi X 2 pi X. This is B into DX. This will be mu naught I square by 2 pi into DX by X. Now you need to understand here that we are finding torque. So I will be finding small torque due to this force and then integrate that to find complete torque. So what is the torque due to this dF about the center of mass? So from the center of mass what is this distance? What is the distance from the center of mass? L by 2 plus A minus X. L by 2 plus A minus X. How? There you go. X minus A is this distance. This is X minus A. Right? And from here to here total distance is L by 2. Right? So that distance is L by 2 minus this which is this. Okay, that's how it comes. All of you understood that? Everyone? Type it quickly. We understood the distance of dF from the center is this. So that understood okay, so d tau is force that is mu naught I square by 2 pi. Force into this length is a torque isn't it? L by 2 plus A minus X into dX by X. Okay, so this you have to integrate. So I leave the integral here. The X will go from A to A plus L. Total torque. All right? This you can equate it to I alpha to get the value of alpha. So probably something like this could come in advance. Let me see what else I have. Solve this. Okay, Trippan got it. Others. All right, Siddharth got something. The hint is that you consider, you can see here the magnetic field keeps on changing throughout the width, right? Distances increasing. So you have to consider a strip like this. A thin strip. Let's say this distance is X and this length is dX. Why we consider thin strips anyways whenever we solve physics question? Because most of the time in physics we know the formula when things are constant. Things are not changing. Okay? For example, I know the magnetic force between the two wires which are at a fixed distance from here to here. It is changing the distance of this point and that point. It is changing. So I will try to use whatever I know by using a thin strip to get the force between this and that and then integrate it. So this strip of length dX this width is given, right? This width is B. So this one is I1 and I2 is spread. It is spread for the width of B. So I2 by B is current per unit length. So for dX you can say di the current is I2 by B into dx. This is di. Alright? So the force between this and that is dF which you can write it as mu0 I1 I2 I1 into I2 by B into dx. So you have to consider current of this strip. You are finding force between strip and I1. I1 I2 divided by 2 pi into x. All of you are able to understand this. Type it quickly whatever I have done here. Everyone understood dx by x. Now you integrate this. Your x will go from A to A plus B. So this is force per unit length by the way. dF by dL you can say per unit length. So force per unit length you are integrating. This is equal to actually this is dF by L not by dL. dF by total length L. So 1 by L. So this is equal to mu0 I1 I2 divided by 2 pi B log of A plus B by A. All right. So let's proceed further. Let's see what else I have. This is done. This one we haven't done. You have two questions here. Solve both of them. The speed is 2 into 10 by 6. Yeah. 2 into 10 by 6. Xiaomi got something 10 pi Newton that is large amount of force. Even if you get like one Newton force on an electron it's tremendous amount of force on an electron one Newton. Imagine how much acceleration that will create. Best is to draw the diagram first. This is the loop. This is the axis. Okay. Many of you have answered everybody else should I wait or start solving. Okay. I'll give one more minute. Those who are done with the first one start attempting the second one also. All right. So the first question the magnetic field at the center will be along the axis. So magnetic field will be equal to mu naught i divided by 2 times radius. This is the magnetic field and you have electron that crosses the plane 30 degree with the axis like this. This is the velocity V and this is the angle theta. Okay. So clearly velocity and magnetic field the angle is theta only. So the force is Q V cross B. So that is Q V B sign of theta Q V cross B. So that application of the formula this is. Okay. So this is how you do the first question. Do the correct calculation you'll get the correct answer. Okay. Second one anyone got the second one something like this we have done something like this we have done earlier. You need to find out magnetic field in terms of velocity of the charge. Remember function of velocity before taking the ratio modify B O Sawa law. Okay. I'll do the second one the electric force straight forward it is 1 by 4 pi epsilon naught Q square both the charges are same Q square by let's say distance is D D square. Okay. Now magnetic force according to B O Sawa law it is mu naught by 4 pi I D L D L cross R vector divided by R square over here R is let's say D that is what you have considered in electric force. Now I is D L I is DQ by duty D L is V DT right. So you can say it is mu naught Q V divided by 4 pi D square this is the magnetic field sorry this is the magnetic field due to a proton okay because of its motion. So the magnetic force is Q V cross B so you will get mu naught by 4 pi Q square V square by D square do you all understand electric force and magnetic force how it is coming type in quickly have you understood till now so electric force divided by the magnetic force stop hold on electric force divided by the magnetic force is 1 by mu naught epsilon naught times 1 by V square this is what you get right 1 by mu naught epsilon naught is what you remember that EM wave 1 by mu naught epsilon naught is what C square right so C square by V square this is the ratio and this is valid you know every time so electric force divided by magnetic force will be speed of light square divided by the velocity give me a second I will share this solution to your seniors also they are writing advanced test this Sunday finally they are writing something okay where I was this is done this is done this is done this is magnetic field anyone close to the answer will the charge move in a circle why it won't move in a circle magnetic field is not uniform smart if I throw it like this which direction it will try to move up or down magnetic field is into plane it will try to move down right okay so what is the condition for minimum separation have you cracked that condition for minimum separation is okay velocity will become parallel becomes parallel to the wire so it goes like this tries to twist and then it goes becomes like that can velocity magnitude change is it possible for magnitude of the velocity to change all of you can the magnitude of velocity change type it quickly no sir it can't change because magnetic force can't do any work so kind of edge remains constant okay so when the horizontal velocity becomes vertical that is a situation for the minimum distance now do it see you may or may not be able to solve it but you must attempt it okay let me know once you are done trying it okay should we discuss it now everyone alright let us discuss alright so you can see here that the velocity let's say my x and y axis are like this you can take any direction of x and y axis not a problem so if my x and coordinate are in this plane entire motion will happen in the plane only because there is no force perpendicular to the plane and velocity is also in that plane so velocity at any moment you can write it as vx i cap velocity along x axis i cap velocity along y axis j cap okay and i know that the whatever is the velocity vx square plus vy square the magnitude of velocity won't change it will remain whatever is the initial velocity let's say v0 and at any moment if let's say the distance from the straight current infinite current is x okay then magnitude of the magnetic field is i divided by 2 pi x which direction it is it is minus k cap okay if inverse i am taking negative so upward i should take as y axis okay so this is your magnetic field b fine force is what q v cross b okay so q vx i cap plus vy j cap we are doing this kind of question for conceptual clarity only this will take a lot of time to solve so probably exactly this one will not come so q mu0 i divided by 2 pi x into what v this is vx i cross minus k which is j right so vx j cap am i missing anything nothing right vx j cap then minus of vy i cap all of you understood till now everyone understood this any doubt in this everybody understood okay so net force along the x axis is mass times acceleration along x axis which is equal to q mu0 current is this i and that i are different okay this i is current let's keep it as capital i q mu0 capital i divided by 2 pi okay so vy velocity along y axis is coming on minus of vy by x this is m into ax so ax will come out to be minus of q mu0 i divided by 2 pi m vy by x okay and x along y axis similarly will be q v0 i divided by 2 pi m vx by x okay now ax can be written as ax can be written as vx dvx divided by dx right similarly ay can be written as vy dvy divided by dy okay now can anyone tell me what to do next any guesses anything so we have to integrate vx dvx from v0 to 0 no but vy is there right vy how will you replace that vy so when the velocity is parallel to the wire then vy is equal to v0 right vy is equal to v0 correct so which one you find the minimum distance we have to create the vx equation right from v0 to 0 and x from x0 to x there is a variable vy how will you get rid of this okay we can use the first equation there and write vy as v0 squared y as vx squared we can do that we can say vy is v0 squared minus vx squared under root of that but it becomes complicated anyways let me it won't be very complicated if you take the other side both sides it will be easy to integrate say it again so if we write it as if we write vy as root v0 squared minus vx squared then right side we can integrate easily left side also we can integrate easily okay let's try that so minus q mu0 i divided by 2 pi n this will be equal to root over v0 squared minus vx squared divided by x is equal to vx oh yes you have vx squared we can take it as some variable so vx dy dx vx dvx will be something so minus of q mu0 i divided by 2 pi m dx by x is equal to vx dvx divided by root over v0 squared minus vx squared okay nice so initial velocity along the x axis is v not only final velocity along x axis should be 0 if it is a minimum distance x was x0 and then final is d so if you are able to solve this integral i think it can be done easily denominator you can take some variable so you will get the value of d from here okay i will not go beyond this let's take something else this is the field 1 this is also field oh i am done with all this thing how much time we have 10 minutes okay i will not add unnecessarily something else i will try to find out the question on this only over here probably did we solve the b part of this i remember solving a part only no we solved the b part you first have to find out torque and in a vector form you have to find out direction of torque will give you direction of the axis along that axis you have to find moment of inertia and equate that to i alpha there are many center of mass axis is right torque equal to i center of mass into alpha all right now that center of mass axis can be many ones all it has to be like passing through the center of mass did you find the torque proceed b cos 45 i cap plus b sin 45 j cap okay so b by root 2 times i cap plus j cap i need to find the you know i am dealing here in vectors so b also in vector form area a number of turns rigid square frame i naught if number of terms are not mentioned we will take it as 1 so the length is l right the length is l correct it's a square so i into l square and it is in k direction you can see right hand rule you can have curl your fingers in the direction of current the thumb is outwards so torque is equal to this is m m cross b so b i l square by root 2 m cross b so k cross i is j k cross j is minus i so it will try to be along this axis it will try to rotate okay so we need to find moment of inertia about this red line what is given to us is moment of inertia passing through center perpendicular to the plane right so this torque is about the red line so do we have moment of inertia about the red line anyone how to get that perpendicular axis theorem so it will come out to be half of this all of you agree it will come out to be half of this so i x plus i y is equal to i z as in i x and i y need not be along x and y axis only they have to be perpendicular that's all okay so these two red lines are perpendicular to each other if you add them up you will get moment of inertia which is given and you do symmetry these two axes moment of inertia are equal so you will have 2 by 3 m l square as moment of inertia about the red line so i alpha which is this is equal to b i l square take the magnitude of the torque now minus i magnitude is root 2 so root 2 by root 2 cancels away b i l square is this so alpha is equal to okay l square is also gone alpha is equal to b i 3 b i divided by 2 m 3 b i by 2 m is alpha we need to find out in a short duration delta t what is the angle theta if delta t is short i can assume alpha to be constant it's a very small interval so i can use theta is equal to half alpha t square so 3 b i by 2 m into delta t square so that is 3 b i by 4 m everyone understood this type it out quickly yeah little bit some tricks are there here but then if you force yourself to understand this or if you somehow understood this then you will be able to solve many other questions also okay so you need to get comfortable with the difficult question so easy question will become cakewalk to you alright guys so i guess that's it from my side 1 or 2 minutes are remaining we can leave now okay so i hope you have learned a few things today at least it has refreshed you your concepts in this chapter which is one of the biggest chapter lot of weightage is there for this chapter okay so you can spend probably one more day with this chapter and then just park it for the very end do it that way that you don't need to do it again and again so next week we will be taking up electrostatic should we take electrostatics next week everyone type it right so next week we will be taking chapter number one of your textbook will not take entire electrostatic will be just take electrostatics related to electric field related to basics of the charges and yes related to force and all we will not discuss the potential and everything else okay we will restrict ourselves with the field chapter number one basically of your NCRT find them so yeah Gauss law also Gauss law very favorite topic of comedy exams many question from Gauss law comes alright guys so bye for now we will meet next week thank you sir thank you thank you