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Coaxial cable is a type of transmission cable which conducts electrical signals using an inner conductor surrounded by an insulating layer (dieleetrie) and enclosed...

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Coaxial cable is a type of transmission cable which conducts electrical signals using an inner conductor surrounded by an insulating layer (dieleetrie) and enclosed by thin outer conducting shield: The whole transmission line then protected by an Outer insulating jacket (sce below).Coaxial cables used t0 carry high frequency electrical signals while reducing power losses They are used in many applications such as telephone lines. broadband intemnet cables. compuler ethernet connections. radio tr

coaxial cable is a type of transmission cable which conducts electrical signals using an inner conductor surrounded by an insulating layer (dieleetrie) and enclosed by thin outer conducting shield: The whole transmission line then protected by an Outer insulating jacket (sce below). Coaxial cables used t0 carry high frequency electrical signals while reducing power losses They are used in many applications such as telephone lines. broadband intemnet cables. compuler ethernet connections. radio transmission lines and many more Coaxial cables were used for the first transatlantic telegraph cable lines in 1858. In an ideal coaxial cable: the electromagnetic field carrying the signal exists only in the space between the inner and outer conductors- This allows coaxial cable runs installed next t0 metal objects without the power losses that occur other types of transmission lines. Coaxial cable also provides protection from external electromagnetie interference This problem involves analyzing the magnetic field inside coaxial cable. Suppose ha coaxial cable which consists of solid inner conduclor 0l radius R, surounded by Very thin hallow outer conductor of radius Rz- The two conductors car equal current but in opposite directions The current density uniforly distributed over each conductor. For the purposes of this problem we will assume there no dielectric in belween the conductors_ (5 points) Using - Ampere` Law find expression for the magnetic field for points within the inner conductor (i.e. for points Rz) Make sure t0 draw picture of the physical situation and indicale where your Amperian loop is localed. b. (5 points) Using Ampere Law find an expression for the magnetic field for points in the space between the two conductors (i.e. for points Rz <r Rz) Make sure draw picture of the physical situation and indicate where your Amperian loop located Spoints) Using Ampere Law, find an expression for the magnetie field for points outside the ouler conductor (Le. for points Rz) Make sure draw picture of the physical situation and indicale where your Amperian loop is located_ (5 points) Sketch graph of B versus from 2Rz assuming that Rz 3R1 -



Answers

A Coaxial Cable. A small solid conductor with radius $a$ is supported by insulating, nonmagnetic disks on the axis of a thin-walled tube with inner radius $b$ .The inner and outer conductors carry equal currents $i$ in opposite directions. (a) Use Ampere's law to find the magnetic field at any point in the volume between the conductors. (b) Write the expression for the flux $d \Phi_{B}$ through a narrow strip of length $l$ parallel to the axis, of width $d r$ at a distance $r$ from the axis of the cable and lying in a plane containing the axis. (c) Integrate your expression from part (b) over the volume between the two conductors to find the total flux produced by a current $i$ in the central conductor. (d) Show that the inductance of a length $l$ of the cable is $$L=l \frac{\mu_{0}}{2 \pi} \ln \left(\frac{b}{a}\right)$$ (e) Use Eq. $(30.9)$ to calculate the energy stored in the magnetic field for a length $l$ of the cable.

So the magnetic field between two conductors is on Lee due to the current flowing through the inner conductor. So from amperes law for the circular part, we know that the loop integration off be dot d r is equal. Do, um you not I And enclosed So from there well and also we know that I enclosed is equal to I hear and ah, the integration is equal to be times d r which is Ah, the circumference So two pi r So from there we can right be times to buy Our is equal to you not i r b z will do you not I buy to buy our for part B The energy density of the magnetic field is given by you is equal to b squared by tomb you Not so for a small cylindrical shell with a thickness d r and in a radius are on da length l So we have the thickness are, um sorry thickness d r in a radius are length l. The volume element will be devi, which is equal to two pi r l d r. And the reason for that is because what we're doing here is we are trying to calculate the volume alien element off this part. So this drink like, uh, thickness is DEA and the inner radius is Are So we have to buy are as wth e in a circumference and we multiply that with the length l And then we want to play that with d r o the thickness which gives us the volume element Devi So once we know Devi, we know that the energy stored in this thing sit under consent Ah, cell will be d'you It is equal toe Ah little you times dv which is it will do be spread by two new not to buy our l d. R. Now if we substitute the expression for being which we got from the previous part we have d'you equals to one over to, um you knock then you not I divided by two pi r squared there times two pi r and the are it is equal to knew not I squared l d r divide that by four pi r So that's the energy stored in this thin set shell. So now in part, see, we have to calculate the total energy stored between the two conductors. So all we have to do is we have toe take the integration off the small energy element. The U So from a to B, we have a mule, not ice. Cre ed l buy for fi r d r So if we now do the integration, we can take the constant outside. So we have new note. I squared l divided by four pi. Then we're left with They are over. Our and integration limits are from A to B. So now, using the limits and solving for the integration we have No, no. I squared, divided by four. Sorry, that's going by. Bye. Ah, Log base e be over eight. So where is the inner radius and Beastie? Our radius. So this is the energy stored between the two conductors? No. For the final part, we have you, which is equal to And I squared by two now also from the previous calculation, we know that this is equal to you know I squared l d r divide that by, um for pilot. Sorry. So this gives us l is equal to new, not and divided by two pi. Ah, log base e be over a And as we can see here that this is the same result as obtaining the previous problems. All right, Thanks for watching. I'll see you next time.

So, firstly, we apply MPs law to a circular part of radius R. An MP s law says that nine integral off be daki l when he's the magnetic field, is he put immune owed on been close charge? And from here we can see that lightning to go off b started with a pot de l. It's simply be tens. The conference off the circle the radius R That's two pi r and from MPs door This is equal to you know what times kind of I the kind eyes the kind that's enclosed second part. Therefore, if you re into this to solve be get the magnetic field within the tar oId sequent in you know it times I over 25 So, yes, we have an expression for the bingo to field. Next we know the engine density which we asked to use and the density u b squared over to you know it Indians in an untenable energy element. Do you is equal to the energy density times and institutions? More. What do you mean a TV? It says you. That's too. I are times l you are So we just expend Devi the volume in it and it's Do you is written as we put our expression for the and you get city from both to be one you know, it's expression for the magnetic field You know what I over two i r all squared times are what you make of it just two pi r and the lend off Some annoyed out times. And if we simply by this we get an expression you know what times I swear there's hell you bet it by for i r You are so expression for the infinitive More change in the total energy. So next we can find total energy But integrating tens more energy changes Do you? And we can solve this thing to go I applied by How can you know that you some about this is You know what ice were times out over four pi which all constant with this victor What's played? An integral from pain. We are my arm. This is something You know it times I squared times l over four. High rebellion going to go. Get that the natural law off B minus natural. Okay, that's an integral off one of our our You see what to a normal bar and we can go one further and write this as you note and I squared that l over four pi and the next moment you fight it by since we know local B minus a number eight music with the novel over A So ends, we have an expression for a total energy. Lastly, I would like to captain it. Our inductions. Okay. And we know the energy started. My doctor, you is equal to 1/2 l i squid. They're full. Huh? How I squared is you go to our expression for the energy about off You know it I squared Oh oh, for pie times A natural local over a We can rearrange this. And so with the inductive out in the inductive l for solid neue meant little l can be shown to be You know it friends Oh, bullet to I times the next in a long off me over a. So the value for the inductive is that we obtained from these energy considerations Agrees being value for the inductions as calculated in problem that t 0.48

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.

Hi obr. Even here it is. Given a coaxial cable usedto transmit the high frequency Sing the inner radius is our one outer radius is radios up Inner conductor Harto radios up outer conductors The space between the the material filled with their space blasting Cody In the first part we have to find expression for inductions, perimeter off coaxial cable and apart. And in second part, we have toe find conductors per unit length for are one Toby 10.5 m millimeters Sorry. Or are to Toby three millimeter. Let us see it. We have to calculate the flux associated to the inner. Quite due to current flowing in the outer Quiet. Do you fight with the skull toe? We talked to you. That is B l dear. So total flux will be integration off bl do your unlimited will change from our oneto magnate feet. Do not I upon toe fire. This is n They are for the limit, Armento. So flux. You will get you Not I l upon to buy a loan off. Hard to upon urban Andi inducted is defined its flux per unit Current from you not I l upon to buy. Learn off our two upon hard one divided by I. So conduct inspiring that length We will get you not upon toe by loan off. Are you upon urban? No beeper. Self inducted. Sorry. Induct inspiring great length mu not is four pi and to the par minus seven upon to five. Learn off three millimeter upon 30.5 millimeters. So it is to be 3.6 and toe turn toe the par minus seven. Henry permitted our 0.36 micro Henry permit. Er that son, Thanks for watching it.


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