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A square loop of wire, of side $d$ , carries a current $I .$ Show

that the magnetic field at the center of the square is

$B=\frac{2 \sqrt{2} \mu_{0} I}{\pi d}$

$\frac{2 \sqrt{2} \mu_{0} I}{\pi d} \hat{k}$

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University of Washington

Hope College

University of Sheffield

McMaster University

our question wants us to show that at the center of a loop wire of side D carrying current I that the magnetic field is too, uh, route two times. Muna I over Pi di. Um, Okay, so in order to do that, I went ahead and drew out the diagram of what we have here. Where we have this loop that goes from point A to B to C to D carrying current I going in the of clockwise direction. So if you using the right hand rule, if you curl your fingers in the direction of I, the magnetic field then would be into the page. So since each side is length D to the center, there in the diagram that I drew is a distance d over to. So we can say here is that the length l to the center is the distance d divided by two. And then you have these two angles that it would make here angle alpha and beta for just considering one sign So we can find the magnetic field from each side of this loop and then add them all together to get the total magnetic field at the center So the angle Alfa here that we're considering is going to be equivalent to the angle beta when these are equal to 45 degrees. Okay. So we can use the formula of the magnetic field to it due to a straight wire of finite length for each side of these squares. And then, like I said, the field of all the sides is going to add, so you'll consider all four sides to calculate the total strength. So, in general, the magnetic field due to a finite, A finite length. Um, Well, actually, why don't we write it this way? So the total magnetic field B is going to be the magnetic field of each side, right? The one plus B two plus B three plus before. So now we just need to find the magnetic field due to one side, so we'll start with the one the one. The equation here is, um you know, the magnetic permeability of free space times the current, which is I divided by four pi times the distance out or the length. But we said that l was d over to here, so this is gonna be times D over too. And then this gets multiplied by the sign of Alfa, plus the sign of beta. But we also know that alpha and beta are equivalent and they're both equal to 45 degrees. So we can further simplify this a little bit. Um, and we see that this is equal to let's actually writing on a mother Paige. So we're not pressed for room here. Be one is equal to be having, You know, I still on top, divided by we had four pi times d over two. So this is to Kai Times D. And then we had the sign of 45 degrees. Let's rewrite that d make it look nicer. We have the sign of 45 degrees plus the sign of 45 degrees. But the sign of 45 degrees is route to over two. So we have route to over two plus two route to over two, which is just fruit, too. So this all gets multiplied by the square root of two. We can simplify this even further because route to over to reconsider the square root, too. On top in the two in the gut denominator, uh, route to over two is equal to one over route to So this is you, Not times I divided by group too. Times pi times d So we would need need to do the same thing for B two b three and before. But if you look at the diagram the distances on each side, the length of each side is D Since it's a perfect square here and ah, Alfa Beta are gonna be the same. I is the same. So be one is gonna be the same as B two b three and before just from symmetry of the problem. So be one is equal to be too, which is equal to be three, which is equal to before. So therefore be here is equal to four B one, right? Since if you Adam altogether if they're all equal, it's just gonna be four times be one. So this is four climbs and you, not times I divided by route too. Times pi times D. So we're really close to what we're apt to prove, but four over route to is the exact same thing as too rude too. So this is two times route too. Promise me you're not times I divided by pi d and this is what we were asked to show. So this is the solution to our question, and we can go ahead and box that in