 Okay, so I have got eight people who are here. Is my voice audible? Okay, so let's wait for two, three minutes so that each one of you can join and then we'll start with the session. Let's wait for two minutes, guys. So people are joining slowly. Okay, so what we have done till now is we have completed biosawa law. We have also completed, so under biosawa law we have completed few things which are magnetic field due to a current carrying conductor, magnetic field due to a current carrying conductor and then magnetic field due to a current in a straight wire where we use the formula. So I'll keep on writing the formulas till the time people join. The first formula biosawa law was B is equal to mu0 by 4 pi i dl vector into r vector divided by r cube. So I explained that this dl vector and r vector has been taken so that this can be changed to a vector. So if you see magnitude wise, the magnitude of the magnetic field is proportional to the length here, proportional to the current here and antiproportional to square of the distance between the point and the current carrying conductor. So why I am taking r in the numerator and the magnitude of r in the denominator because I want to show db or magnetic field as a vector. So both magnitude and vector is represented by this particular formula. That is why usage of r in a cross product with dl. So what is this dl? dl is always taken in the direction of the current. Second, so I told that when you take a current carrying conductor of certain length and if that is a straight wire, so suppose I have a straight wire and this is a point. So I have explained that this you take as theta1 and this you take as theta2. And my formula changes to mu0 by 4 pi i by d. And what is the d? So d is perpendicular distance of this particular point p. Suppose this is point p. Here is the direction of the current and d is, remember d is perpendicular distance. d is not this distance or anything. d is perpendicular distance. I use the length in the form of theta. So it becomes cos theta1 minus cos theta2. Now I told that when the straight wire has in very long or infinitely long, theta1 becomes equal to 0 and theta2 becomes equal to pi. When you put these values, d comes out to be mu0i by 2 pi d. So these things we have already done. We have already done these things. After this, what I have explained you in the class is that now let me know. After this, what I have explained you in the class is that that was a previous story when I started the magnetism. I told that I am just revising it for 5 minutes so that you all can be on the same page and then we will perhaps move ahead. So a lot of new things will be there in today's class. What I have also told you guys is that suppose there is a current carrying conductor and there is a charge q. If this charge is stationary, there is no magnetic field applied on this charge. This charge starts moving with certain velocity. So this particular charge will feel some kind of force on it and the magnitude of the force and the direction of the force is given by formula q v into b. So this was one formula that we have discussed in details and the logic behind it has also been discussed properly. Next what I have told you is that whenever an external magnetic field is there so suppose I have a current carrying conductor and in that current carrying conductor I have or like this I am giving it a heading. The heading is magnetic force on a current carrying wire. When I take this particular concept I have told you that the magnetic field over there what I am trying to find out is that if a current carrying wire is kept in a magnetic field and what kind of magnetic field. I am not talking about magnetic field due to this particular wire. I am talking about an external magnetic field. So if a current carrying conductor is kept inside an external magnetic field how much force would be applied on that and I told you I took this particular formula that f is equal to na dl into minus e vd into v and I have done this derivation in the class and from here I found out that the force vector would be i vector l vector into b vector. Now this is a cross product of l vector and b vector. I have told you in the class that l is always taken in the direction of the current and with what rule you will find out the direction of force. I have told you that any cross product the direction of the resultant would be found out by right hand rule. So what is the right hand rule? I have told that the first vector or the left hand side vector your finger should be in the direction of the first vector and your palm should be in the direction of the second vector. The position of the thumb will give you direction of the resultant vector. So all these things are clear right? I see a lot of people leaving the chat and just repeating it. I am saying a lot of new things would be there. So don't leave the chat and roam around here and there. You will miss a lot of things. Then you will come back to me saying that I didn't do something. So on this particular concept I am trying to give you a question. All of you just try to solve the question. I will just write the formula. Here in the corner f is equal to i l vector into b vector. So I am making a triangle. This triangle is p q r and a current i is the triangle is made of current carrying conductors. This is current i which is flowing through different. I am sorry one second. So direction of the current has been given to you. This is the direction of the magnetic field. Now what kind of current, what kind of triangle it is? So triangle p q r is equilateral triangle. Is it fine? Now find out force on three wires p q q r r p. The hint is magnitude of the current which has been given is. No don't take magnitude of the current just find it in terms of i and b. Start solving guys. Find out force on p q, find out force on r q, find out force on r p. I will give you two more minutes to solve it. Should I solve it? There would be no negative sign. On p q, yes, it is zero. Okay, let me solve it for you. In online classes, I cannot give you much time because yes, shawmeek on p q, it is zero monsee. It is not negative. No where it will be negative. I'll tell you how it happens. First of all, I have told you to find out force. I am not bothered about direction. I'll tell you later. Let's in this class. Let's focus about magnitude because I have discussed enough. I have discussed enough right hand rule with you. Let's see p q. So p q i is like this in horizontal direction and B is also in horizontal direction. Why I am so much. Why I am using horizontal direction so much in my discussion? Because when you find out only magnitude of the force, magnitude of the force would be equal to i, L vector, B vector, magnitude and sine theta, where theta is angle between L vector and B vector. So as both are in same direction, so f p q would be equal to i, L, B, sine 0. Why I am using sine 0? Because L vector and B vector are in same direction. So f would be equal to 0. Now let's go to... So did you all understand this? Okay, so I hope you all understood this. Now let me go to q r. Now q r see the direction of current is like this and the direction of magnetic field is like this and this is the B vector and this is the L vector. So this angle is how much? This angle is 60 degree. So this angle will be how much? 120 degree. So if you extend this B vector here, this will make 60 degree from here and if I extend i vector here, this will make 120 degree here, fine? So if you take i from here, i2 would be 120 degree from this side, from this side you see here. So f would be equal to i L, B, sine 120 that is equal to root 3 by 2 i L into B. Is it okay? And on PQ, PR vector again, the B is like this and i is in this direction. So again force would be equal to i. This angle is 120. So i L B sine 120. So this comes out to be root 3 by 2 i L B. Is it okay? So I have given you magnitude of the forces on three of the things. Now you can find out the magnitude, sorry direction, okay? Now let me give you a second question. Okay, so what I am saying is that again a wire is bent in equilateral triangle form. So this is P, this is Q, this is R. Triangle PQR is equilateral. Now this is the direction of the current and magnetic field is going inside. So what I have been making is to show you that magnetic field is going inside. Current i is same in all the conductors. i is equal to 5 ampere. As it is an equilateral triangle, length of the sides are, magnetic field is of 10 centimeter and magnetic field is of 2 tesla. Find out resultant force on all three wires. No Rohe, you should take 120 degree. I'll explain later why you should take 120 degree. Solve the question here. Good, Shomic, the result is 1 Newton on PR. What about PQ and QR? Okay guys, let me solve this question for you. So if you keep your fingers, suppose I find it out for PQ, so Shomic is exactly right that for all of them, angles are equal that would be 90 degree. Why it would be 90 degree? Because the magnetic field is in direction which is going inside the plane. What do you mean by going inside the plane? Take a paper on the paper, make an equilateral triangle and put your pen of the plane vertically getting inside the paper. So for both PQ, QR and RP, the angle value would be 90 degree. Why? Because the magnetic field lines are going inside the plane. Hence they would be making 90 degree angle with the plane. So anything drawn on the plane will make equal angle 90 degree with the magnetic field. I hope you understood this. So if angle is equal, so F value is equal to ILB sin theta. Now try to understand as it is an equilateral triangle, L is equal as current is equal and B is equal for all of them. Suppose force F1 is working. So if you put your fingers in this direction and palm inside it, the F1 would be like this. Suppose this is F1, this is F2 and this is F3. This would be direction of F1, F2 and F3. So F1 would be like this, F3 would be like this and F2 would be like this. Magnitude of F1, F2, F3 will be equal. Why? Because I is equal, L is equal and B is equal. So what is I? I is 5 ampere, L is 0.1, B is 2 and sin theta is 90 degree. So how much it would be 10 into 0.11 into sin 90 is 1. So that comes out to be 1 Newton. So all F1, F2, F3 would be 1 Newton and what is the resultant force? See here, it is an equilateral triangle. So equilateral triangle, the resultant force would be working on the center. So this is F1, this is F3, this is F2 and if this kind of force works over there we have done it in the class what would be the resultant would be? How much? See here, let's find out resultant of F1 and F3. Resultant of F1 and F3 would be 1 square. So it is like this. If vectors A and vector B make theta degree with each other the resultant is given by A square plus B square plus 2AB cos theta. So it becomes 1 square plus 1 square plus 2 into 1 into 1 cos 120. Cos 120 is minus 1 by 2. So what happens is this becomes 1 square plus 1 square plus 2 into 1 second into minus 1 by 2. So this comes out to be 1 square plus 1 square is 2 so I am writing it over here which is 2. 2 minus this is how much 1 so it comes out to be 1. So this will be 1 as the magnitudes are equal the angle from it it would be directly in between F1 and F3. So this comes out to be 60 degree and this is 120 degree. So F2 which is of 1 Newton is exactly opposite to resultant of F1 and F3 which is also 1 Newton. So the complete force working on this would be 0. How many of you got 0? Should I explain once again? If anyone wants me to explain it once again you can let me know because after you find out the magnitude it all becomes vector. Should I explain it once again? Aditya I am explaining it once again. I am going to the different page. Mansi I am doing it again. I told that this is P this is Q this is R so this is I, I, I and it's going inside so suppose this is F1 this is F2 and this is F3 so once you I found out the magnitude of F1, F2, F3 is equal to 1 Newton so now where would the force be acting the force will be acting at the center so this is F1, this is F3 and this is F2 so what you can do is that this angle would be as the magnitude are same and they are working at the center of the equal lateral triangle angle between all of them would be 120 degree so I am writing it here 120 degree. Now find out the resultant of F2 and F3 so this is how much 1 Newton this is how much 1 Newton and this is 120 degree so try to understand if you have okay I am explaining it once again don't worry about it suppose let me teach you vectors so first let me teach you vectors and then I will come to this. Suppose there are two vectors this is vector A and this is vector B and angle is theta degree so how do you try to solve this so to try to solve this or to try to find out the resultant of this or what I do is that finally I have to find out A plus B so what I do is that I try to make a parallelogram so to make a parallelogram I draw two perpendicular lines two parallel lines this line is parallel to B vector and this line is parallel to A vector and the diagonal of it would give me vector A plus vector B what would be the magnitude of vector A and vector B so try to find out magnitude of vector A and vector V would be I will prove this in the class don't worry guys magnitude would be A square which is D square plus 2AB cos theta fine so this is the magnitude of the vector or let me do one thing let me prove this formula for you on the next page so what you can do is that what you can do is that look at here I am doing it on the next page I am proving this formula so what I did this was my vector B this was my vector A so what I did was this was theta so I made a parallelogram the parallelogram was such that I draw a parallel line to B and I draw a parallel line to A the diagonal gives me the resultant of A and B vector so this diagonal would be equal to vector A plus vector B now what I do is that to find out the resultant there is some construction that needs to be done so I draw a line which is perpendicular to B vector so if this angle is theta and this is a parallel line so this vector B becomes transversal and this would be corresponding angle theta which would be equal so what I can do is that I know that this particular vector is A vector what is this which vector is this this vector is B sorry this vector is what is this angle ok one second guys if one second one second this is my A vector so this vector would be A cos theta and this line would be A sin theta are you guys understanding it let me take this is P this is Q this is R so PQ by PR is equal to sin theta now PR is so PQ is equal to PR sin theta PR is what PR is A vector so this is A sin theta so that's why I have written PQ is equal to A sin theta now what is QR so I know that cos theta is equal to QR divided by PR so QR is equal to PR cos theta PR is A vector so QR is equal to A cos theta as simple as that so that is why I have written QR as A sin theta PQ as A sin theta and QR as A cos theta now try to understand this is my V vector if I take PQ and S I am assuming this particular point to be S so in right angle triangle PQ S perpendicular is PQ and that is equal to A sin theta and base is equal to SQ that is equal to this SQ would be equal to SR plus RQ so SR is how much B plus QR is how much A cos theta so hypotenuse PS can be written as equal to under root PQ square plus SQ square so let me make some space for you I hope you all are understanding this thing this is S so PQ is how much under root let me write out PQ is A sin theta so A square sin square theta and this square of this would be I am doing it here only square of this would be B square plus A square cos square theta plus 2AB cos theta so this would be equal to B square plus A square cos square theta plus 2AB cos theta so this is equal to A square sin square theta plus cos square theta plus B square plus 2AB cos theta now this sin square theta cos square theta is equal so I can write it down as A square is equal to 1 so I can write it down as A square plus B square plus 2AB cos theta so for any two vectors making angle theta with them A plus B would always be given as vector A plus vector B magnitude will always be given by A square plus B square plus 2AB cos theta now I am stopping here let me know if any one of you did not understand this let me know if any one of you did not understand this I will move ahead in 30 seconds so in 30 seconds let me know if any one of you did not understand it I will explain it once more as nobody is replying I am moving ahead now question of Ruhi that how do you find out what would be the angle between now try to understand this complete angle was theta now what would be this angle so suppose this angle is theta 1 so this angle theta 1 I am writing here so that you can understand it angle theta 1 is between resultant vector A plus B vector at base which is B here is it okay and this can be given by tan theta 1 is equal to this was A sin theta and this was A cos theta so it can be given by perpendicular A sin theta divided by base B cos theta sorry A cos theta plus B so this is the angle by which you find out okay Ruhi I am going back to the question so how do you find out now the question the question is this try to understand so the question is I have to find out F2 and F3 resultant how do you find out F2 F3 resultant so resultant of F2 3 would be F2 square plus F3 square plus 2 F2 F3 cos 120 F2 is 1 so 1 square F3 is 1 so 1 square 2 into 1 into 1 would be 2 cos 120 is minus 1 by 2 so this comes out to be 2 this comes out to be minus 1 so F2 3 comes out to be 1 so which is equal to 1 now what would be the angle so for angle I told you suppose I am assuming F3 to be base angle so if I assume F3 to be base angle suppose I am calculating this angle theta so F3 if F3 is base angle this tan theta would be equal to opposite size which is F2 sin theta divided by F2 cos theta and what is theta this complete 180 degrees so not theta I am writing it theta 1 here where theta 1 is equal to 120 degrees plus F3 so how much it is F2 1 into sin 120 sin 120 is how much sin 180 minus 60 which is sin 60 so 1 divided by root 3 by 2 F2 cos 60 which is 1 by 2 minus 1 by 2 sorry plus 1 because cos 120 would be negative so I am writing it minus 1 by 2 so how much it is tan theta is coming out to be root 3 by 2 divided by 1 by 2 1 minus 1 by 2 is 1 by 2 and this root 2 root 2 gone so tan theta is equal to root 3 which means that this particular angle is equal to 60 degree this is by formula you should know that if 2 vectors are of equal magnitude the resultant vector would be directly in between it so if the complete angle is 120 degree the resultant vector would be the angular bisector hence from both sides it will make 60 degree angle so what is the magnitude of the resultant 1 Newton and it is making 60 degrees from both the sides now try to understand F1 is just opposite to it so F1 is just opposite to it how F1 magnitude is 1 Newton from F2 it is making 120 degree and resultant F2 3 is making 60 degree from here this is 60 degree so F2 3 and F1 is making how much angle they are making 180 degree angle 120 from this side 120 this one so they are making 180 degree angle so they are exactly opposite to each other they are exactly opposite to each other fine so F1 and F2 3 are exactly opposite to each other so hence magnitude are equal F1 minus F2 3 because they are exactly opposite would be 0 1 minus 1 would be equal to 0 so cancel out okay guys so this is what the question was now let me move to a few concepts and the concepts are like this any doubt you can post I am waiting for 30 seconds I am not getting any doubt anyway attendance is very less I don't know why people are still celebrating Diwali how many of you are there can you again give me your attendance so that I can find out how many of you are not present let me write down their names let me write down their names how many of them are not present who is still present can you give me attendance once again I can see a few names giving me answer one is Shamik then Ruhi is there Aditya C is there Momika is there Aditya M is there Mansi is here great Shri Hari okay anyone else I am not getting any other name so either people have just switched on their youtube and enjoying our pitai is here so how many of you 1 2 3 4 5 6 7 8 Kirtana is here great 9 only see 9 names 13 people are okay fine I will have to speak to all others so most of Ananya is here Ananya are great don't skip guys I know that few of you are skipping the class even when you are present on youtube you are not focusing it should not be the case Ruhi I have written your name okay now let me start with the concept and the concept is force between carrying conductors okay I want you all to try out I am giving you conditions the condition is I have only got the name of 10 people still 12 are showing it means that 2 of them are Neha is there okay good so 11 I got Neha I have written your name sorry I couldn't find it out okay so 11 people are there good let me talk to other 12 13 people why they are bunking on the classes force between 2 parallel current carrying conductors so this is a long wire try to understand the question guys and the current is I1 long wire when I say long wire you know what should be the formula of magnetic field in the vicinity of the long wire there is another wire kept where current is I2 take this small length DL on this I2 and find out the force acting on small DL second wire due to current I1 first wire solve it out I have given you enough concept that you do this question take 2 minutes and solve it out no Shri Hari this is not an integration based problem this is very simple problem should I solve it okay let me solve it for you guys as none of you are giving me the answer I think there is some problem in understanding so let me give you the answer so guys understand now stop writing and listen to me very carefully stop writing listen to me very carefully where I want to find out the force I want to find out the force on this length DL why this length DL on this second wire suppose this is wire 1 and this is wire 2 why would wire 2 or any part of wire 2 will feel the force first so what is the formula of force on any current carrying conductor it is I into L vector into B vector now is there a current in second wire yes so current is there length I have already taken what about this B when I was giving you this particular formula I told you that yes Ruhi I already told that it's a long wire so I have written it on the screen that it's a long wire so what about this B I told that this B is external magnetic field what do I mean by external magnetic field it means that field created by some other magnetic object so see here when this wire W2 would be kept in the vicinity of second wire W1 which is a current carrying conductor this wire W1 its current I1 will create some kind of magnetic field so what would be magnetic field as this is a long wire magnetic field due to wire W1 I am writing it W1 would be equal to mu knot I by 2 pi D so suppose I forgot to give you suppose this distance is D between them is it okay so mu knot I by 2 pi D now what about force on second wire force on second wire would be I length I am assuming at DL and B is how much so I is IT in second wire DL and B would be due to first wire which is mu knot I by 2 pi so if I write the force on second wire FW2 that would be mu knot I1 I2 divided by 2 pi D DL is it okay so if I find out force per unit length of W2 wire force per unit length would be force divided by length of the wire DL and that comes out to be mu knot I1 I2 divided by 2 pi D so write down this formula and try to do understand or try to write down this formula by understanding it so don't look at the screen just try solving it out by yourself what is your question what are you writing how okay great what about the angle between the length and the width good so angle between if I write it down it will be ILB or IBLB due to first wire and sine theta apply right hand rule so your finger should be in the direction of the first wire so and your palm should be in this direction so if finger is in this direction and palm is in this direction the magnitude the direction of magnetic field would be going inside the plane on this particular wire and if it is going inside the plane the sine theta value would be 90 that is why I am not writing 90 degree fine so sine theta would be 90 degree so it is IBLBW1 and BW1 I have already written it here so FW2 is mu1 I1 I2 DL divided by 2 pi D and per unit length the force would be mu1 I1 I2 divided by 2 pi D now is there any doubt in this question now so try to understand guys there are two formulas I have given first formula I have written as force on any particular length DL would be mu0 I1 I2 divided by 2 pi D and DL if I make this DL 5 centimeter don't keep it there where until and unless I am asking force per unit length don't use the second formula until and unless I am repeating again until and unless I am asking force per unit length don't use the second formula only first formula would be used now write down a question on this concept the question is two long straight wires I am again saying this is W1 W2 both long straight wires how much current is there current is 5 ampere here and 5 ampere here distance between them is this distance is 2.5 centimeters now I am asking if I take a DL length of 10 centimeter find out FDL on W2 wire where DL is equal to 10 centimeter okay Ruhi I will repeat the last two what I was doing in this question you will understand Ruhi don't worry about it in this question I will do it from derivation so don't worry just apply the formula I can see 16 people now attending in the last 10 minutes what happened at 4-5 more people joined it anyway I have names of only 10 people so rest all who are writing it sorry rest all who are attending it I don't know their names one more minute guys start giving me answers if you want formula the formula is F is equal to mu naught I1 I2 DL divided by 2 pi D some mistakes we hurry Momika your answer is right Aditya see your answer is right Ruhi I take your attendance Shamik you check your answer it's not 10 to the power minus 6 I think it's 10 to the power minus 5 Shamik you have to take total length I told that whenever I am not taking per unit length I have to take per unit length why will I give you length that is why I have I explained it very clearly that if understand guys if per unit length I have to find out it means that the value of length given is 1 so length itself would not be given in case of per unit length now let me solve it for you guys so first of all I know that 4 pi is equal to 10 to the power minus 7 so if I have mu naught by 2 pi I take 1 2 from here to the other side it becomes 2 into 10 to the power minus 7 so I write it down it comes out to be 2 into 10 to the power minus 7 so mu naught by 2 pi is done I1 is 5 I2 is 5 dl is 10 cm it means that it is 10 to the power minus 1 in terms of meter and this is 2.5 into 10 to the power minus 2 so how much it is this is 2 so 2 into 2 is 4 4 into 5 is 20 20 into 10 to the power minus 8 minus 7 minus 1 is minus 8 and this 10 to the power minus 2 will go above and become 10 to the power plus 2 so this becomes 20 into 10 to the power minus 6 or 2 into 10 to the power minus 5 Newton that's the answer did you guys find it out okay so what I am doing is that it's almost time I am wrapping up the session because I have to start with class class 9 session if any doubt is there guys please ask me from next lesson I will start with one thing is left out which is circular and after circular I would be mostly done with circular and solenoid 2-3 things left out and then I will start with permanent magnets and the syllabus would be over so it only is a matter of 2 to 3 class now till the time we find out till the time we do our syllabus 2 to 3 class that's it so any doubt guys ask otherwise we will wrap up the session we are not getting mu naught by 2 pi into 100 how what do you mean by 100 so somebody was asking that how did I do this I will explain once more so I have to find out force at this dL force to find out force at dL b due to first wire would be mu naught by mu naught i1 divided by 2 pi d because it's a long wire force would be equal to force at wire w2 would be equal to i2 dL vector into v vector now I have already explained that the angle between them is 90 degrees so it would be i2 dL vector and this would be due to d1 and bw1 is mu naught by 2 pi i1 divided by d so force due to second vector force on second wire would be mu naught by 2 pi i1 i2 divided by d into dL now if I want to find out force per unit length I will divide it by dL and that comes out to be mu naught by 2 pi i1 i2 divided by d so this is how you do it okay yes Ruhi your answer is right okay fine so guys thank you so much for attending the class I hope you enjoyed your Diwali and Diwali break is over so break is not over but Diwali is over so I hope you will start focusing on your studies now because next two months is going to be very very important we are as it has been communicated to you in PTM also that we would be taking your level to something much beyond then class 10th level so let's gear up for that and from next relation onwards you will see me going in that direction so thank you so much for attending the class and if you have any doubt now I am wrapping up the class if you have any doubt you can just ping me individually so thank you so much