Mutual inductance of two coaxial solenoids

say we have a long solenoid of length l1 with n1 number of turns and say the coil area is a1 now there's another tiny solenoid inside so let me make this transfer and then show it to you let's say there's a tiny solenoid inside which is coaxial they have the same axis and it has a2 area and two number of turns and it has a length l2 now if that tiny solenoid is carrying a current i2 and let's say it's changing then there will be an emf induced in the larger solenoid we call this mutual induction and we've seen before that that emf induced in the larger solenoid depends upon the mutual inductance of the larger solenoid and our goal in this video is to calculate just that so our goal in this video is to calculate the mutual inductance of the larger solenoid with respect to the smaller solenoid and a quick spoiler it's going to be a nightmare to try and directly calculate this and we'll see why but we will find a quick a simpler way of figuring this out so i hope i've gotten you pumped up okay but before we begin a quick recap of mutual induction so the emf induced in the larger coil e1 that can be written as negative mutual inductance of the larger coil with respect to the smaller coil times the rate of change of current in the smaller coil that's the whole idea right and if you are not familiar with this equation feel free to go back and check out our previous videos on mutual induction now my question is how do i calculate this because that's what i want to figure out so what i like to do is since this is a constant we can put inside the differential equation we can put inside the differential right so let me go and do that so it's the same equation but why did i do that because i remember from faraday's law emf induced in any coil should equal negative rate of change of flux and so this means this quantity should represent flux flux through which solenoid well since i'm calculating emf through the outer solenoid this should represent the total flux through the outer solenoid and so now i can say that the total flux through the outer solenoid should equal the mutual inductance of the outer solenoid with respect to two times the current in the inner solenoid and so if i can calculate what the flux through the outer solenoid is the magnetic flux is i can plug it in and i can figure this out and i also want to remind you a shortcut way of remembering this is connecting this with momentum just like how momentum which is mass times velocity is something that objects hate changes in they like to continue their state of motion they like to continue their momentum we can say coils hate changes in flux so flux is like momentum and so mass is the inertia over there here inductance represents the inertia of the coil there we talk about speed of the objects here current represents loosely represents the speed of charges we can say and so if we were calculating the flux due to its own currents if this was i1 then this would have been self-inductance when we calculate the flux due to the current in the other solenoid that's what we're doing over here here that's when you get mutual inductance so let's go ahead and calculate the flux so how do you calculate the flux well you've seen before if you have a nice flat area through which you have a uniform magnetic field which is perpendicular to that area then the flux through that area the magnetic flux just becomes the product of the magnetic field and the area and if there are n such turns then you just multiply it by n but the big question is is our magnetic field uniform let's find out let's find out over here so who's generating the magnetic field in this case is the tiny coil and we have to calculate the flux through the big solenoid right so let's go ahead and see what the magnetic field looks like what do you think the magnetic field looks like everywhere well we know inside the solenoid the field will be pretty uniform but as it comes outside it will be nice and diverging and so here's what it would look like something like a bar magnet and you can immediately see the problem the magnetic field is not uniform everywhere so let me get rid of this coil because now we'll just focus on the larger solenoid because that's where we need to concentrate our flux on so if i just for a moment dim this coil so we can completely focus notice if you now try to calculate flux through any of these coils the field is not nice and uniform let me give you some example if i take a coil somewhere over here maybe we'll get some nice uniform field because it is at the center of the inner solenoid and the field is uniform near the center but if i take another coil somewhere up here the field here is not uniform so the calculating flux over here will be very complicated i have to integrate and stuff if i take another example somewhere down here look again the field is incredibly complicated i'll have to integrate so calculating the total flux can you imagine how much work that would be i i don't even have any idea of how where i would begin in other words this looks mathematically impossible for us and that's why we will not do that this is a living nightmare it's a mathematical nightmare we can't calculate this so what do we do do we give up we've hit the road block no we don't give up there is one amazing thing about mutual inductance m12 is always equal to m21 meaning the mutual inductance of the larger coil with respect to smaller coil always equals the mutual inductance of the smaller coil with respect to larger coil so what we can do now is instead of passing a current through the smaller coil and calculating the flux to a larger coil let's do it the reverse let's pass current through a larger coil and find the flux through the smaller coil and now you might ask and you should ask well how does that help wouldn't that also be complicated we'll look at that in a second before we do that can you try writing a new equation now an equation for calculating flux in the secondary due to the current in the primary in terms of the mutual inductance can you write that equation yourself first all right so the new equation will be very similar now it would be we're calculating flux in total flux in the secondary and that would be mutual inductance due to the secondary basically the the tiny coil to the tiny solenoid due to the large solenoid multiplied by the current in the larger solenoid this time we are putting the current in the larger solenoid let's so let me redraw that circuit uh the setup over here here we go now we're going to put the current through the larger solenoid we'll call that i1 and we'll calculate what the flux to the smaller solenoid is going to be so how does that help well now the magnetic field will be generated by the larger solenoid right so let's look at what that field is going to look like here goes tada now we see something very similar to what we saw earlier the field is nice and uniform inside and it tends to diverge as we go towards the edge but the thing is the smaller solenoid is well within the center of the larger solenoid so if i concentrate on the flux that we are calculating over here now it will be much easier again let me help you focus on that since we are calculating flux through the second solenoid let me dim the larger solenoid for a while and you immediately see how nice and uniform the field is through each coil again let me show you some examples if you take some bottom coil over there field is nice and uniform and perpendicular to the area so i can just do b times a over here i take any other coil again field is nice and perpendicular even the top most coil is well within the center and again the field is nice and perpendicular so here the flux will just be b times a through each coil and so the total flux would just be the number of turns multiplied by b times a nba so here flux calculation is much much easier and so we'll calculate what m21 is so now would be a nice time to again pause the video and see if you can calculate the rest of the stuff we know that flux will be just n times v times a we know the number of turns we know the area of the coil all we need to know is figure out what the magnetic field is and we have seen that before this is the magnetic field due to the larger solenoid we know the properties of the larger solenoid as well and we have derived that expression before so can you put it all together and see if you can figure out what the mutual inductance m21 turns out to be pause and give it a shot okay so the total flux through this small solenoid is going to be just nba so which n are we talking about is it n1 or n2 well we're calculating the flux through the smaller solenoid so it's the number of turns of the smaller solenoid so it's going to be n2 times b this magnetic field who's generating that magnetic field hey that material feels generated by the larger solenoid which i've like dimmed over here but it's generated by the larger solenoid right so that's the let's call that b1 times the area whose area are we taking is it a1 or a2 well i'm calculating the flux through each coil of the smaller solenoid so i am calculating a2 hopefully that makes sense right so if you if you ever get confused whether it's two or one it's easy to get confused just keep thinking about which solenoid which flux you're calculating it'll be fine so that's going to be m21 something that we need to calculate times the current i1 all right so the only thing we need to figure out now is what's the magnetic field and we've derived that before so let's quickly try and do that so it's going to be b what's the expression for magnetic field well we've derived earlier from ampere's law magnetic field at the center of a solenoid or inside a long solenoid is going to be mu naught times n times i divided by l so which n is this well this is the number of turns of the solenoid that's generating the field who's generating the field it's the bigger solenoid so this should be n1 so let me call this n one and which current is this this is the current who's generating the field and that's the largest solar center in the field so this will be i1 so this will be i1 times a2 equals m 2 1 times i1 and notice the current cancels out and we have our final expression let me just make some space so we put this up a little bit okay so what's our final expression so we get m21 the mutual inductance of the secondary or the smaller coil smaller solenoid with respect to the larger solenoid that's going to be let's see we get a mu naught we'll put that over here times n1 times n2 times a2 divided by oh whose length is this is this l1 or l2 what is the length of the solenoid that's generating the field in our case the bigger one so it's l1 so this would be divided by l1 and there we go that's the expression for mutual inductance it's basically now you can see what it depends on depends on the number of turns of both of them depends on the area of the smaller one because that's where flux matters and it depends on the length of the larger one because of the field that's generating none of this is pretty obvious and i don't expect you to remember any of this you can always derive it in your head but what's important is m21 is the same as m12 this is exactly the same as m12 will not prove it but it happens to be true for any two coils regardless of how complicated the system is and which means we have answered our original question so i guess the big takeaway over here is whenever you are asked to calculate the mutual inductance between a couple of pair of coils or a system of coils doesn't matter what they've asked in the question you always pick the equation where the flux calculation is the easiest where it's the flux whether the field is going to be nice and uniform over the entire area because m one two will always equal m21 for a given pair of coils you only have one value of mutual inductance