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(28 Points) A long current carrying pipe consists of two conducting layers. The first layer has an inner radius A and an outer radius B. It carries non-uniform curr...

Question

(28 Points) A long current carrying pipe consists of two conducting layers. The first layer has an inner radius A and an outer radius B. It carries non-uniform current density J=3ysin 4r+ 58 where and B are constants_ The outer layer has an inner radius B an outer radius C, and has uniform current density such that the magnetic field for r > C is equal t0 zero_ What is the total current carried by the inner conducting pipe? What is the current density in the outer conducting pipe? What is the

(28 Points) A long current carrying pipe consists of two conducting layers. The first layer has an inner radius A and an outer radius B. It carries non-uniform current density J=3ysin 4r+ 58 where and B are constants_ The outer layer has an inner radius B an outer radius C, and has uniform current density such that the magnetic field for r > C is equal t0 zero_ What is the total current carried by the inner conducting pipe? What is the current density in the outer conducting pipe? What is the magnitude of the magnetic field strength for r < A? What is the magnitude Of the magnetic field strength for A < r < B? What is the magnitude of the magnetic field strength for B < r< C?



Answers

Figure $29-87$ shows a cross section of a hollow cylindrical conductor of radii $a$ and $b$, carrying a uniformly distributed current $i .$ (a) Show that the magnetic field magnitude $B(r)$ for the radial distance $r$ in the range $b<r<a$ is given by (b) Show that when $r=a$, this equation gives the magnetic field magnitude $B$ at the surface of a long straight wire carrying current $i$; when $r=b$, it gives zero magnetic field; and when $b=0$, it gives the magnetic field inside a solid conductor of radius $a$ carrying current $i .$ (c) Assume that $a=2.0$ $\mathrm{cm}, b=1.8 \mathrm{~cm}$, and $i=100 \mathrm{~A}$, and then plot $B(r)$ for the range $0<r<6 \mathrm{~cm}$

For the given figure in our inner areas as B out there is our out of it. A and at some distance inside this it is our right. So what is the magnetic field for such a case for circular path of radius are concentrated with conductor fun said who led? But Mhm hun sen trick with conductor you know the formula as integral of VDs. In this case it becomes stop by our in to be. So what is me? Which is me or not. I enclosed and what is I enclosed here? It will be new. Not I and what? How much I is enclosed It is there enclosed area which is pi R square minus B squared upon total area which is by is where minus the square. So this much is enclosed. Those be we can see us by the formula mu not higher point. So since this question holds true. So we will beam unit I bye is cancelled from both ever went down and we will have do not I upon to buy our and inflows this our spare minus the square upon. Yes fair minus the square. So this is our mean formula. Right. Mhm. So this is proved in part B. It is saying at our request to what will be this formula converted to so part of me at r equals two. Yeah. Yeah this will be converted as new. Not I upon to buy a yes where minus B squared upon is where minus the stress. So this time will be canceled out only we will have Bs community open to you. Yeah. So this will be our answer and articles to be We will have had articles to be these proportional to artist permanence be spared. All right. So we can write it as no, not by a point to buy our our despair minus of what is the main formula artist permanent B squared by Yes. Where minus the square. Yeah. So this This term will be zero. So for this case we will have to differentiate and we will get the answers Be not equals two. Do not I upon to buy is square two. I spread upon our we just um you know I had a point by a spirit. This is since this is not valid because our experiments the spare valuers. So we will use this formula. Yeah. Yeah this is that articles to be in ATM equals to zero B equals to zero. We will have this formula And articles to be the value of zero. Yeah. So if we make the God This will be like this one. It will start at 10. Wait will start at some point and go on reducing in the field at zero are less than the and is equal to our better than he is equal to result of part Yeah, so this is me, this is it have the value zero and greater than it is also zero and graph is like this one hopefully this will probably be pretty mm

For part a off the problem. The field at the center, Off the pipe it a point c is due to the wire alone with a man shoot off B. C. That is a call to you. Nod. I know for wide divided by two by times three are this gives us mu nod I off a wire derided by six by our For the while we have B B off a wire greater than BC well for a while. Thus, for B P is equal to B C, which is a call to BC off a while. Ah, you off for a while. Must be into the page so we can write. The B p is equal to B p off a wire minus minus Saudi. Here. Ah, BP off a pipe that isn't called immune on io for wire divided by two by our minus mu nod I divided by to buy into two, are setting. B C is equal to minus B p. We hope then I awful wire to be three. I divided by eight, which is our three driver. Eight times eight times tended ball minus tree and Pierre use us three times 10 to the power minus tree. And Pierre for about B off the problem, the direction is into into the page, okay?

Hello, everyone. This question is asking us to find various parts off electrolytic maids, Um, that are present in a conducting material conducting cylinder. Okay, so first of all, let me just say a little bit about the coordinate system that we have going on here. So because we have the cylinder Kohlmeier, it is useful to use cylindrical coordinates in which Zet is pointing, uh, along these directions. So it's across the horizontal are is pointing always perpendicular to this away from this access, as you can see here. So it's going like this. This this etcetera. So it's going in all radio directions around the set access, and five is going in whole circular directions around this axis. Okay, so that's what I've just represented here by this little diagram over here. So he has that head here, our head away and five head going around the Z axis. Now we're told, or we're asked to find first what the electric field is and the edge off this wire or this Conduct conducting material given that it has a radius. Okay, so it has radius are single today. Okay, That means that the cross sectional area of this wire is going to be pie attempts a squared. All right. We're also told that the wire hasn't resisted 80 off row. And what we know from our journalism from before is that this resistive ITI is equal to the magnitude of the electric field divided by the magnitude of the current density that the viral carries. Now, the, uh, current density that the buyer carries is equal to the current divided by the area, the cross sectional area of the wire and lastly for later convenience. I'm also telling you hear that their resistance off such a material in length A. So if this is the length of the wire, then the resistance given that it has course section area, pious squared and London l is equal to row times l divided by pi a square. So this will be important later. And these guys are what are important right now for part A. Okay, so have resisted is able to be over J and J is equal to I over a given this, we can rearrange, uh, this equation for or definition for grow to find that the electric field is equal to row times the current density And then we can substitute into this the value of J, which is the current divided by the area. We know what the area is. We know that is pi squared. So we know that E is equal to Rome times I over pi squared and then knowing that the current is pointing in the direction like you can see over here, we can say that the electric field is equal to row types. I over a pie, a squared pointing in the Zet had direction. Okay, so this is the answer to part A. We have that the electric field is able to grow I or pious work Morning in the head direction. Okay, moving on to the expert, the rest to find what the magnetic field is. Okay, so the next question is, what is B at the edge of the wire. Okay, so for this, if you don't remember how to calculate the magnetic field at the edge of a straight conducting wire, I'm telling you that you can use, um, God's law or are the bs of our law. So for emperors law, rather, this is This is, you know, the simplified version of the BSO our law. So you have emperors Law and so here, giving that the current is pointing in the set head direction. You know that the magnetic if you point your thumb your right thumb along the direction of the off the current, then you know that that's gonna give rice to mean that he feel that's pointing in the direction that you curl your fingers around this around around your thumb. Right? So given that your thumb is pointing this way, the magnetic field is gonna be pointing this way. Okay, so this is gonna be be it's gonna be giving you these circles. It's close circles around the wire. And so that means that the it's parallel to defy hat direction. Okay, so you set up closed comparing loop around this wire, which is this size, are just on the edge. So just just a circumference off the off the wire are conducting material. So you have that the into over disclosed loop off the element off the line off the loop, thought it with the magnetic field and closes, or is equal to mu zero time Staying close current. Okay. D l like I said, is pointing around or is going around the surface off the wire. Still, like, you know, little segment like this. This would be the L. And so this has, uh, of course, a veg. This wedge has the radial part of a and the circular component is defy. You can't really see that. But so the point is that the l is going to be a times defy because you're going around circles. So if I if I drew this head on, So this is the cross section of my wire. This is a And I'm saying that this is the yell, right? This little parts of this little segment on the edge, that's deal. And so this is gonna be pointing in this direction. Okay, so then we see that d l is equal to a times defy times five hat and giving the B has some constant magnitude be on this pointing. And if I had direction to we know that the product off the L and B is going to be be times a times defy, okay? And now it's just a simple matter of integrating um, around the entire circle. This D l, which in terms of fi just corresponds to taking the fi variable from 0 to 2 pi That just corresponds to going once around the circle. Okay, so there is no other argument in this integral or in this instagram. So you just get to pie from this integration And so you find that two pi times a times B is equal to zero. I enclosed, but I enclosed It's just I the same current as we had determining the magnitude of the electric field and so putting everything together dividing by two by a we find it be b is equal to zero. Sometimes I divided by two by a and he's pointing in the fire had direction as we've already established a debating. Okay, so that's part B. And we found the magnetic field. That part C is asking us to find appointing Rector. Alright, What is s now, as is equal to the cross product off E will be divided by mu zero. Okay, so now we know what the magnitude of each of these things are. But we also know in what direction they're pointing. So we have one over, um you zero times the magnitude of the electric field which is the road times are divided by pi squared times the magnitude of the magnetic field mu zero times I or 258 and the cross product of the directions which is said had crossfire hat. And given that in the cylindrical coordinate system are cross five is people to zed. Okay, you can deduce that zed crossfire hat is equal to minus our head. Right? So this just comes from moving everything to the left, once our one to the other side. So it's just it's just a cyclic permutation of directions here. And you find that five head, Of course. Said hat would be our head. And so switching, um, the two around so that you have that coarse five gives you minus ahead. Okay, so now we find that ass is row times. I squared over two pi squared a cube since the mu zeros canceled and is pointing in the negative, our head direction. So this is what s happens, or this is what s is okay. And then lastly, we're also as to find what the energy flow rates into the wire is. Okay, So what's the energy flow rate around the wire? Well, we know that the pointing vector is giving us the energy flow rates over unit area into some area where the electronic waves are penetrating and is being pointing into the S had direction. Okay, so now rearranging this equation for Delta over Delta T, which is the energy Florrie, which is also known as the power. Okay, so P is equal to s study with D A. Oh, that's a And now, integrating over this or summing up all the possible areas gives us this integral were some closed surface, and so we have that the power is equal to integral. Over the close surface off D a dot S or D A s d uh, is the vector off the area element that we're considering? Okay, So to do calculate this part, we draw the cylinder again, and we point out that, you know, as hat is pointing in the negative, our head direction. So as hat is pointing inwards, it's pointing into the cylinder. At the same time. There are two surfaces that you can consider. Um, one of them would be d a one, which is the surface of the edge off the conducting material and the other one would be the actual cylindrical part off the area, which is, um, which is coating the cylinder. Right. And that was pointing in the head. Direction was the first one is pointing in that direction. So consequently, the dot product of the first one with the edge area is gonna be zero because, uh, that had and are perpendicular to each other. So this is zero, but D A to the S hat is equal to s de to their SDA two times minus one. The money's from Comes about because our hat or sorry d A two is pointing in the positive our head direction, whereas S, is pointing a negative or head direction. And so our hat dot minus our hat is equal to minus one. Okay, And now, just just a quick note here is that d A two is gonna be gonna be the a little element on this surface, right? So and that's going to be eight times defy for the part that's going around. And it's going to be desired for the part that's going around here. Okay, So in other words, you could say that this d A two is equal to this part for this. This this component of the area is eight times defy disc Irving part. And this part is he said I have. Okay, so that's how you get, like, a small patch on this in there. And it has an area of eight temps. De five temps sees it. Okay, Now, given that the two is this and and that the DOT product is given by this part right here, we can calculate the power or return to the calculation of the power as being equal to minus the size or magnitude of s times the area of the closed surface over which, or with which the dark correct off s is non vanishing. And so we find that the magnitude is so we're gonna get minus for the productive. The two are vectors, and we're gonna have the magnitude off the specter, which is row times I spread or to buy spread a cute. And then we're going to get the area off this coat of the cylindrical coat around the conducting material, which is to buy a types l now thes things. Cancel again. One of the pious cancels or the ace cancels so What we're left with in the end is that this is equal to minus road times l divided by pi A squared times big I squared. Yeah. And now, if you remember what I told you in the beginning is that the resistance of the wire resisting the T. Rowe length l and cross sectional area pi squared is equal to row l divided by pious word, which is precisely the combination that appears here. Okay, so we find that the power, um, in the wire is exactly minus I squared. Times are okay, which is equivalent to what is known as the heat dissipation off the off the wire. So the radiated energy under radiated power is usually taken to be I squared Times are Yeah. And so we see that these two Onley different by the minus sign. And so the only thing that this minus sign represents is that, um, theme the energy radiated outwards bite, admire in terms of heat is I squared. Art. Where is the power radiated inwards from the trinity? Bates is minus. I squared are and so is this invert outward. Uh, switch is corresponds to the minus sign, but we can see that the two are you going to each other

Hi there. So for this problem we have a cylindrical conductor with a circular grass exception. This has a radius of a and a resistive itty role and also carries a constant current of I. So the first question for this problem is the magnitude and the direction of the electric field that you're at a point just inside the wire at a distance a from the access. So the electric field are a distance A from the access. So um in a wire the electric field is related to the current density by the following equation, the racist activity times the better of the current density. So that in this case the direction of the electric field is parallel to the access of this lender because the current is traveling in that way. So the better of current density will just have the same direction. So we will have the following and the direction is the same as the current. And the magnitude in this case is just the rest festivity times the magnitude of that. Um Better of current density. So that will be this. The density of the current is defined as the current over the area. In this case the cross sectional area and that contraception all area in this case is a circle. So we know that the area of a circle is pi our square but in this case we wanted in the when the radios is eight so we will have a square. So this is the direction and magnitude of the electric field. The direction is the same. Us the current and the magnitude is this in here. Now, for part B of this problem we are asked um the magnitude and direction of the magnetic field better at that same point. A. So to calculate the magnetic field we use in paris law. So in paris law and in this case I'm going to draw the contraception a view of the skin doctor. And it is something like this. Mhm. If we took the current becoming out of the page we will have that the current the magnetic field sorry, should be perpendicular to all the points in this conductor. This should be the math the magnetic field magnetic field demon. Never feel so applying and purse law to a circle of radius. A. So we will have the following. We'll have that damn radius R. Times the bet were being times the deep held in this case is art. So it is the cross product um. Sorry the dot product between the better tour the magnetic feel better and the differential in the radios. And this is equal to the circumference of this cross sectional area. And that will be the magnitude of the electric field and the manager of the magnetic fuel. Sorry. And the circumference, which is of this value. Now the the current that is enclosed by this emperor temporary in surface is just the current of the conductor I. So that we will have the prodded is equal to by um paris law should be equal to absolute zero. Um Music zero times the enclose current. So we will we found that that expression is beat times two pi A. The R. As you can see, goes away with the one at the other side. We can see and hear that this and desk goes away so that our result doesn't depends on the I'm perry in surface. So from here we'll have absolute zero so that the magnetic field is the subzero I two pi A. And this is what we expected to obtain for the magnitude of the magnetic field produced by a cylindrical hole conduct word. And finally the description of B is counterclockwise wild around the circle. Just as I I draw in this picture, I'm going to put that in here. The direction the direction oh bean is counter cloth wise around the circle the circle. Um that describes this conceptual area. Of course. No, for part C of this problem, we are asked um the magnitude and direction of the pointing better S at the same point. So this is a straightforward since we know that the point better tour is defined as one over mu subzero, the cross product between the electric field and the magnetic field. So to see did redemptions of the magnetic field and the electric field, we can draw picturing here. So with this figure from part of this problem, we know that the direction of the electric fuel is parallel to the direction of the current. So if we said that the um current is coming out of the page or coming out of the screen, in this case we will have that that battery could be represented like this with a point. So we're going to do that for some points of the circumference. This should be the vector of the electric field. Now. And partly of this problem, we said that the magnetic field has um counter clock wiles direction in this um perception, all areas. So we will have something like this for the better this. Yeah, um this and this. So all of these are the mhm magnetic field factors. So um we can see that this is just the cross product between the um electric field and the magnetic field. Um china's a constant. So using um the right hand rule, we can get the eruption of S. We put our hands in the in the direction of the electric field so we go from there to the better be so that our tamp should indicate eruption of C. So in although in all of those cases we can see that the pointing that toe is pointing to the center. Yeah of that process optional area. So with that set we obtained that the direction of the point in Bhaktapur is ravioli in war. Right? You really and work. And to obtain the magnitude, we just simply um do the product between the magnitude that we found for the electric and the magnetic field in part A. M par be of this problem. So it should be like this. Um We obtained for the electric field this is row I over the R. E. M. And for the magnetic field is subzero the current over two pipe A. Yeah. And when we did is product we obtain that. The magnitude of the point in bed. Tour is the resistive itty. The current squared over two pi square a cube. And this is the magnitude of the pointing back door. Finally, for this problem, we are asked to use the result in parsi to find the rate of low of energy into the volume that is occupied by a length L. Of the conductor. So now to solve this, since the pointing that the magnitude of the point in metro is a constant over the surface of the conductor. The rate of energy flow that we are going to call peen is given by at times the surface of a length out of the conductor. So it will be like the point in bed door times the area. So in this case we said that that should be um two pi A. L. Accounting for that mm length. Elp. So that we obtain the value for the point in better. Which the mandatory for the point in better, which is receive festivity current over two points where a. Q two pint A. L. And this should be equal to row. Yeah. And and this is the result for this problem. This is the energy flow that or the rate of the flow of energy into that volume occupied by a length L. Of the conductor. So this is it for this problem.


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