Magnetic field due to two current loops: Numerical

we have two coils of radius r same radius having carrying a current i in the same direction kept co actually apart at a distance of root 3 by r our goal is to figure out what the magnetic field strength is going to be at this point p at the center of coil 2. how do we do this well i remember there is a formula that we derived for calculating the magnetic field due to a current carrying coil we can just use that we just have to be careful because there are two coils over here so the way i'm thinking about it is since i want to calculate the magnetic field at this point there will be two magnetic fields one generated by this coil over here and second one generated by this coil over here so before i start plugging in and using the formula magnetic fields are vectors so let's think about the direction of these two magnetic fields because if they're in the same direction we'll add them up if they're in the opposite direction we'll subtract them and how do we calculate the direction we'll use our right hand thumb rule so why don't you pause the video and think about what were the direction of you know the two magnetic fields at this point can you pause and try that all right so let's look at this one the way i like to do this is you take my right hand use my encircling the four fingers to show the direction of the current here the current is coming out of the page out of the screen then going into the screen that's how it is so if i use my right hand thumb rule it's going to look like this it's going to look like this so everywhere on the axis everywhere to the left of it to the right of it the magnetic field is going to be to the right so the thumb represents the magnetic field right so let's do that so the magnetic field over here due to this coil which i'm going to use yellow to represent it's going to be this way let me call that as b1 maybe if we do the first coil and what do we do to the second one well the current direction is the same so the my right hand is going to look exactly the same and so again the magnetic field will be in the same direction so that's great we just add them up so the magnetic field with this coil will also be in the same direction so it's going to be like this this will be b2 and now we just have to figure out what these two are and we just add them up so let's do that let me get rid of that hand all right so what's the formula for the magnetic field this is the monster formula and i remember it and i'll show you in a second how i remember it but here it is so the magnetic field on the axis of a circular coil let's say b1 is given by the constant mu naught over 4 pi 2 pi n i r squared divided by r squared plus l squared whole to the power 3 over 2 where l is the distance from the center n is the number of turns i is the current r is the radius now hold on a second this is a i used to always remember i wonder how do i remember this monster formula well it turns out in physics if you remember some fundamental formulae then you can derive the rest of them that's the fun thing about it so for magnetism the fundamental formula would be b.o sawar law and ampere circular law now in the previous video we used b o sub r and derived the expression for the derive this expression so if you have not seen this before highly recommend you to pause and go back and watch that video but if you want to do a quick dirty derivation here it is i'm just going to show you a quick snapshot of how i do this yeah you may want to pause over here and just look at this but this is the basic bo sub r being used mu naught by 4 pi i d l sine theta divided by r squared where this is our r sine 90 because the angle between dl and this is 90. and then that magnetic field is along this direction because when you use your rdl cross r you get that then before you integrate you take the axial component of it because that's the only component that gets added up when you consider all the dls and so when you consider the actual component you add another cos theta and cos theta from this triangle becomes r divided by this adjacent side divided by the hypotenuse and then you add the additional cos theta now this formula starts to become familiar so now when you integrate you get an integral of dl that gives you 2 pi r and so if you now put 2 pi r over here you'll get exactly this formula so even if you forget this formula you can and if you remember what bu sub r is you can you can quickly derive this so i'm just going to keep that over here you can pause and you know ponder upon this if you've done that now we know the magnetic field is so can you plug in the values for b1 and b2 and at and then see what the total magnetic field is going to be can you pause and try doing that okay let's do this let's calculate so b1 is going to be if i simplify i get mu naught by the you get 2 and it's just one term i is i r is r divided by r squared plus what is l here well l is this distance for the first coil this is where we are calculating magnetic field so this distance is root 3r if i square that i get 3r squared whole power 3 over 2. so if i simplify i get mu naught i r squared by 2 this is 4 r squared and if i i have to take 3 over 2. so i take the square root first 4 r squared square root of 2r and then i take the q i get 8 r cubed all right so this will give me mu naught i divide by 16 and r squared and r cancels and i'll get r so that's b1 for me let me just quickly separate this okay now let's calculate b2 same formula but for b2 what's l well since i'm calculating at the center for b2 l is 0. so it's going to be mu naught by 2 i r squared divided by r squared plus l square l is zero so it becomes r cube you can just check that and that gives me i just simplified over here so i just get b2 as mu naught i by 2r and before i continue i just do a sense check so i'm seeing that b1 is 1 over 16 b2 is just 1 over 2 so b2 is higher than b1 and that makes sense because b2 is right at the center and so b1's contribution is very small and b2's contribution should be very close so that makes sense all right so now the total magnetic field the total magnetic field is going to be just add them up we're going to take the common things out mu naught i by r is common i get 1 over 16 and 1 over 2 common denominator is 16 so 1 plus 2 8s are 16 and so if i just because there's not much space over here all right so there we go 9 by 16 times mu naught i by r there we go that's the total magnetic field let's try one more problem this time we have the two coils kept perpendicular to each other with their centers aligned and our goal is to figure out the magnetic field at this common center the total field and they don't have the same currents this time so a very similar formula a little a similar problem a little bit different so why don't you pause and give this one a shot okay just like before we first look at the direction of the magnetic field let's start with the yellow one use my right hand rule it's going to be this way so let me draw that the magnetic field over here let's call it b1 is going to be in this direction this is my b1 and the magnetic field due to this one is flowing this way so the magnetic field if i use my right hand it's going to be like this so b2 due to the pink one let me get rid of the hand pink one will look somewhat like this this is b2 so this is the second coil and this is my first coil okay now since i'm calculating magnetic fields at the center the l value would be zero for both of them so i can just quickly go ahead and substitute and figure out where b1 and b2 are so b1 is going to be mu naught over 2 n is 1 i r squared divided by just r cubed and yeah the current is i so this is going to be mu naught i by 2r okay what's b2 going to be well i can write this directly see b2 has everything b2 is this one sorry uh this is the coil b2 has everything the same same radius same same radius same you know l value is also zero but the current is three times more which means the magnetic field b2 would be three times more than this right because i have 3i over here so it's just going to be mu naught into 3i divided by 2r okay and so now the total magnetic field is going to be well the total field this time they're not in the same direction so i have to use vectors so how do i do that so i calculate how do i how do i do this well i take i take the parallelogram law this will be my total magnetic field and i can use pythagoras theorem for that so my total magnetic field is going to be this squared plus this squared so b1 squared plus b2 squared because there are a lot of numbers over here what you know what i'll do i'm just going to call b2 as 3 times b1 just to make my calculation simpler so this is going to be b1 squared plus b2 square which is 3 b1 the whole squared and so the total magnetic field is going to be let's see i can take b 1 common and i will get root of 1 plus 9 that's root 10. so that's b 1 times root 10 and there we go so that's going to be root 10 times mu naught i divided by 2r and what's the angle what's the direction well let's call this angle as theta then i can use tan theta so let's do that we go a little bit over here to make room okay so tan theta is going to be the opposite side which is b2 divided by the adjacent side b1 and we know b2 is 3b1 so that's 3b1 by b1 even cancels out so theta is going to be tan inverse of 3. and i'm not going to simplify this any further i'm just going to keep it as it is so i know that the let me just fit everything in the screen so i know now that the total magnetic field is this much and it makes an angle of tan inverse of 3 with respect to the horizontal with respect to the second coil and there we go