 We have two concentric coils one having a radius much smaller than the other and our goal in this video is to calculate the mutual inductance of the system. So how do we do this? Well we've seen before that you can either calculate the mutual inductance of the outer coil with respect to inner or you can calculate the mutual inductance of the inner coil with respect to the outer. Now both of them would should give us the same answer because there's only one mutual inductance for a pair of coils. However only one of them is really easy to calculate the other one is a complete nightmare. So the job really is to figure out which one should we do. So let's first try out both of them and then let's see which one we should do. All right so let me show this to you from the side view. All right so first let's try finding out what's the mutual inductance of the outer coil with respect to the inner coil and do that will pass a current through the inner coil and you can now write the flux equation and if you're like me you will always forget what the flux equation is and so what I like to do is I remember there is a beautiful analogy flux is very similar to momentum just like how objects hate changes in momentum coils hate changes in flux and so I know momentum is mass time velocity and so from there I can say ah flux flux of the outer coil will equal the inertia of the outer coil and that is mutual inductance of the outer coil with respect to the inner coil times instead of velocity we'll get current because velocity is you know here objects are moving here charges are moving how how fast charges are moving kind of tells you how much current is and then and always with mutual induction this is the deal you always calculate the flux due to the current in the other coil if you were to calculate flux due to the current in the same coil then you would get self inductance okay so this is one way to do that and before we try calculating let let's also go ahead and look at what is the other way of doing that and the other way of doing that let me show it over here we can pass current through the outer coil and try to calculate the flux through the inner coil so in doing that flux through the inner coil that would be just like this it would be the mutual inductance of the inner coil with respect to the outer coil multiplied by the current in the outer coil and now to calculate the mutual inductances all we need to do is calculate the flux and once you do that we can plug in and cancel stuff out and that will give us however which one should we go for we should go for the one in which the flux calculation is the easiest right and now would be a great time to pause the video and think about in which case do you think the flux calculation is the easiest and why so think about how the magnetic field is think about this condition that's given to us and from that try to see so pause the video and give this a shot okay so how do we calculate flux in general well if you have a nice and flat area a through which let's say there is a nice and uniform magnetic field which is perpendicular to it b then the flux through that area would be just b times a and so to calculate the flux we need to first figure out how the magnetic field everywhere is over the area so let's look over here here I want to calculate the magnetic flux through the outer coil so I need to check what the magnetic field looks like over here but who's generating that magnetic field that magnetic field is being generated by this coil this is the one that's generating the current having the current right so let's draw the magnetic field generated by this coil and see how it spreads over this area all right so we've seen the magnetic field it looks somewhat like this now look at how that field is spreading over that area is it nice and parallel and uniform not at all let me help you see this if you only concentrate on what the field looks like everywhere on the surface here's what it would look like look at that the field is not at all uniform it's highly concentrated over here but this is like quickly diverging and the field over here is in a different direction altogether and if you had to look at the field everywhere it's like completely different so non-uniform field it will be very very difficult to calculate this flux so this is that living nightmare I was talking about will not do this okay so that leaves us with only one option let's look at what happens over here here since we are calculating the flux through this area now we need to figure out what the magnetic field looks like over here but who's generating the magnetic field it's the outer coil so let's draw the magnetic field over here due to the outer coil it's going to be very similar to this but it's going to be much larger because of the outer coil there you go you might say isn't it the same case over here you know here also we have non-uniform fields true but we are calculating the flux through this one this one so I only have to worry about the field that is acting over here and so again if I only look at the field over there and let me draw some arrow marks oh you can see the field over there is pretty much uniform that's because that area is very tiny since that area is very tiny we can pretty much assume that the field over here is uniform and guess what this field is pretty much the same as the field at the center and we already know we've worked out before what the magnetic field due to a coil at the center is going to be so it's very easy to calculate the flux because I can just view magnetic field times the area and I know what the expression for the magnetic field is and so I can plug in and figure this out so if you are as excited as I am why don't you pause the video and see if you can now figure out what the flux is substitute and calculate it okay let's do this so I'm going to continue somewhere over here so flux equals b times a so over here the magnetic field who's generating the magnetic field the outer coil so let's call that as b1 magnetic field b1 times the area which area are we talking about we are calculating the flux to the tiny coil so area a2 and that should equal m2 1 times i1 okay now all we have to do is figure out what the magnetic field at the center is and that's something we have seen before but again if you're like me you may have you may not remember I usually don't remember formula and what I want to show you is that you know if you remember some basic laws you can do quick derivation so there's not really a need to remember a lot of formula so what's the magnetic field at the center of the coil I don't remember so I do a very quick derivation I know it can be derived from biosawar's law and I know how to derive it so let me do a very quick derivation the way I like to do that is take a tiny current element dL and then the magnetic field due to that tiny current current element dL I know that is mu0 by 4 pi biosawar's law times i mu0 by 4 pi idl sin theta divide by r square now theta is the angle between dL and r which is clearly 90 degrees so this vanishes so let me get rid of the sin theta which is great and r over here is r1 the radius of that big coil so we can substitute that as well so this is r1 r1 square sorry there's an r square the current is i1 over here so this is i1 and I quickly remember that if I do that if I integrate over the whole thing I get 2 pi r so I know I get 2 pi r over here and so the 4 pi 2 pi cancels 1 r cancels over here so what I end up with is sorry there's a 2 over here what I end up with is mu0 i by 2 r1 so let me just write that over here so magnetic field is going to be mu0 i1 by 2 r1 times a2 so I'm writing this over here what is a2 a2 is the area let me bring this back a2 is the area of that tiny coil and that is pi r square which r well r2 r2 squared and that should equal m21 times i1 and so notice i1 cancels out and we get our expression so immediately m21 turns out to be m21 equals what it gives you mu0 times pi r squared pi r2 squared divided by 2 r1 2 r1 and there we go that's our expression so because our secondary coil second coil was so tiny we could assume that the field everywhere is the same as the field at the center and that's why this thing was given to us and again what's important is m21 is exactly the same as m12 so even if I had done all that hard work somehow I don't even know how to do that but if you had done that I know for sure I would have gotten the same answer and I know it looks unbelievable right how can this complicated stuff give us the same answer well we are not going to prove it that turns out the proof is also slightly complicated but we're going to just accept that that we always have only one value for mutual inductance whether you calculate m12 or m21 it should be the same and so whenever we have problems like this the whole idea is to figure out which flux calculation is the easiest and then doing just that