 We have two coils of radius r, same radius, 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 would be the direction of the two magnetic fields at this point. Can you pause and try that? Alright, 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. 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. So that is B1, magnetic field to the first coil. And what will be due to the second one? Well, the current direction is the same. So 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 to this coil will also be in the same direction. So it's going to be like this. So 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 and is the number of turns, i is the current, r is the radius. Now, hold on a second. I used to always wonder how do I remember this monster formula? Well, the turns are 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 formulae would be Biot-Sawar law and Ampere circuit law. Now in the previous video, we used Biot-Sawar and derived 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 Biot-Sawar being used. Mu naught by 4 pi i d l sin theta divided by r squared, where this is r r, sin 90, because the angle between d l and this is 90. And then that magnetic field is along this direction. When you use your d l cross r, you get that. Then before you integrate, you take the actual component of it, because that's the only component that gets added up when you consider all the d l's. 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 d l 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, and if you remember what Biot-Sawar is, you can quickly derive this. So I'm just going to keep that over here. You can pause and, you know, pond it upon this. If you've done that, now we know what the magnetic field is. So can you plug in the values for B1 and B2, 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, you get 2, and it's just one turn, 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 4r squared, and if I have to take 3 over 2, so I take the square root first. 4r squared square root is 2r, and then I take the cube, I get 8r cubed. So this will give me mu naught, i divided 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. 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 squared, l is 0. So it becomes r cubed. 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. I'm 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 aids 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 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. It's 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 what B1 and B2 are. So B1 is going to be mu naught over 2. And is 1i 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? I can write this directly. See B2 has everything. B2 is this one. Sorry. This is the coil. B2 has everything the same. Same radius. Same radius. Same 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 do this? Well, 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, you know what I'll do? I'm just going to call B2 as three times B1. Just to make my calculation simpler. So this is going to be B1 squared plus B2 squared, which is three B1 the whole squared. And so the total magnetic field is going to be, let's see, I can take B1 common and I will get root of 1 plus 9, that's root 10. So that's B1 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. It 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.