The coaxial cable shown in (Figure 1} consists of a solid inner conductor of radius Ri surrounded by hollow; very thin outer conductor of radius R2 The two carry eq...

The coaxial cable shown in (Figure 1} consists of a solid inner conductor of radius Ri surrounded by hollow; very thin outer conductor of radius R2 The two carry equal currents but E opposite directions. The current density uniformly distributed over each conductor Pan A Find expressions for the magnetic field within the inner conductor Express your answer in tenns of the variables Ri, Rz, I and appropriate constants. AZd B(r < Ry) = Submit Request Answer Pan B Find expressions for the magnetic field the space between the conductors Express YOur answer in tens of the variables Ri, Rz, I, and appropriate constants: Azd B(Ri < r < Rz) = Figura 1 of 1 Submit Request Answe[ Par € Find expressions for the magnetic field outside the outer conductor Express your answer in tems of the variables R1, Rz, I, and appropriate constants. Azd B(r > Re) =

All right. So we have current on the inside pipe here, Um, I in the pipe with the radius of a and then we have a hollow space, and then we have current going in the opposite direction. Um, in the wall of C thickness there. We would like to know what the magnetic field is when we are inside the in er pipe. So in our is less than a B is equal to, um, mu zero I It's usually over two pi r. Now, this are, um is to calculate the whole, uh, current. So that would be a and then. But, um, we're only a certain spot in there, so we need that ratio. So we have our over a making that a squared if our is less than a because it is a uniformed current. Yes. Okay, now, if we are between the walls, So we are between the diameter A and B. Um, we have, um basically, we're outside off the inner current, so therefore, I would have mu zero I over two pi r alright. Had to rethink that, but still agree with myself. The outside, uh, conductor does not affect either of these because you're inside. Ah, conductor. And there's no magnetic field inside the conductor because I is zero. So the portion affecting these two magnetic fields that I created by my I hold on, I got it. My this I is zero. It has zero effect on it because you're inside of it and containing none of that current, so that will not affect either of these answers. So those answers I still agree with, all right, trying again. We're going to go all the way outside this time. Where are is going to be greater than B plus c putting us outside both currents. Now, if we're outside, we should have the formula off mu zero I over two pi r. In this case, though, we would be taking into effect both currents which happened to be the same size in the same direction, and therefore they would add up to zero. So you would end up with a magnetic field of zero outside of the whole conducting thing

This problem was the concept of the Mps law. And we are going to use this equation to solve that problem. So first the region inside the conductor, we consider ampere lou. Okay, so the consider and pierre Luc inside the conductor. So first minute to calculator. I True. So the eye through equals the area covered by the empire lou. That is by our square upon the area of the conductor. That is fine. Ri square into the current through the conductor. Okay, but you can write I threw is our square I. Upon R. I square. Now using NPR slow we can right You mean to buy us equals may not times I. True. That is R squared I upon ri squared of the magnetic field inside the conductors. We're not uh minute upon ri square may not open to ri square sudden may not upon to buy our I squared into the current I times the distance from the center axis. Okay. And from the right hand girl room we can say the direction of the magnetic field is the longer, counter clockwise direction. So we can say counter clockwise direction. This is for the region are less than ri Or greater than zero. No, if you consider I'm paying blue between the conduct turned out this office. So let's consider this imperial rule. Okay, This is the Radius R. For this one. So for our greater than ri and less than are not. We can right from the and be a slow being to buy our equals may know times I other magnetic fulfilled between daughter selling the conductor is may not I upon to buy us and the direction is again longer. Counterclockwise direction. Well, Consider 3rd Imperial Blue covering the outer shell. So the slow. Okay. And in this case the areas are is greater than are not so for the region are greater than or not from the NPR. Look, you can right, we don't do by our equals, may not into it through. And that is I minus the current and out of self since the net current is zero. Therefore the magnetic field outside the conductor will be C.

So first using amperes law will calculate the magnetic field between the conductor's and MPs. Don't tells us that the interview both be don't d l is equal to new not and the enclose current I and we can see. But for this case, this is the magnetic field around its circumference off the cable. So be times two pi r is equal to unite I and hence we can find an expression for the magnetic field. Be easy put too. You know what I I was the current in the cable over two pi r We are is the radius off the cable So ends we have an expression forming it fuel Second, he and infinite Tess Imo trucks Deify is equal to be d a a is the area off the cable and we know that this that he's given by you know what? I only two I are and eight immediately is us area is the length attends the wood intimate are so the length of the cable is out ending with elements tr so infant asthma flux is getting by. You know, I over two pi r times out er next we wish to calculate the total flux across the cable. We can do this by integrating the expression. About what if we integrate for me to eat off all his tracks. Elements defy we get you know it times I am in l all over two pi. I'm sitting to go a to be one of our on and the ends we get at the total flux over the cable, isn't you? Not times I and momentum to cable over two pi and the natural law O B over a, which is what we get really integral case. And that's an expression for the daughter of flux across the surface, off cable Next induct INTs we know define. It's in has a magnetic flux. Well, I where is the member of tons? And this is simply oh times and you know it over two pi using the expressions above tens of natural Look off me. Oh, okay. As you're quiet finally and distort by the minute field you Is he going toe 1/2 comes out times I swear, Mrs Ah half and we substitute our expressions about times l times. You know what? Over to high and the natural. Look off being over a Okay. Friends of kind I squid. And if we simply find us, we get and distort the magnetic field Be You know what? Oh, I squid over four pi the natural look. Oh being so, the magnetic field between the conductors is due only to the current.

Well, uh, we consider a collection Campbell, which consists off a solid conductor, betrayed. He is a It's supported by insulating disks on the axis off conducting to within a radius be an outer radius. See well the central conductor. And to Katie equal currents. I, in opposite directions, will not beat. The current is distributed uniformly over the conductor's Let us to party. Well, first, we need to find the magnetic field. It is less than our in already is less than B. Well, applying in beer's law to a settler bad of radius R We have lying in people off be don't d l is equal to r B l right which is equal to being too to my are length is equal to circum fans and here in close current physical toe I So we also know the line integral off be don't d l is equal to you know, it's times I enclosed right And therefore we have be into to my r is equal to Maybe not I no solving for being so magnetic field be magnetic field we use equal to me or not Times I divided by do by our well no, we do part B. Well, we need to find the magnetic field. A Teoh C is less than art or artists created and see. We know that if we are going around the path in a clockwise direction of the current out of the page, ah will be positive and the direction into the page will be negative. So then close current in these cases, so I enclosed is equal to I minus I that is equal to zero and therefore magnetic field B is also