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A thick wire with circular cross-sectional area of radius Rcarries a uniform current density j. the magnetic field inside thewire, at a radius r < R from its cen...

Question

A thick wire with circular cross-sectional area of radius Rcarries a uniform current density j. the magnetic field inside thewire, at a radius r < R from its center is:derive equation for B

a thick wire with circular cross-sectional area of radius R carries a uniform current density j. the magnetic field inside the wire, at a radius r < R from its center is: derive equation for B



Answers

a thick wire with circular cross-sectional area of radius R carries a uniform current density j. the magnetic field inside the wire, at a radius r < R from its center is: derive equation for B

For this problem on the topic of the magnetic field, we have shown a long cylindrical conductor which has a radius R. And carries a total current of I. The current density is not uniform over the cross section of the conductor, but as a function of the radius jay is equal to be times are webby is some constant? We want to find an expression for the magnetic field be firstly at a distance R one which is less than the radius and then at a distance R. Two, which is greater than the radius measured from the axis. Now we'll use mps law here and that the closed surface integral of B dot Ds is equal to new note times I. So for column density J this becomes new note times the integral of J dot D. A. Let's look at the first scenario in a At a point that's within the Radius R. one is less than big are. So from here we can see that B into two pi are one is equal to um You not times the integral from zero to our. Of br times two pi R. D. R. Which is the area element. And we get mm Magnetic Field B two B. You're not times B Terms are one squared. So the integral from zero to our one. All over three for our one which is less than the radius R. Or inside the cylinder. For part B. We are now outside the cylinder When are two is greater than the Radius R. Again mps Law gives us two pi R. Two times B is equal to um You not times the integral over the radius of the cylinder From 0 to our of br times two pi R. They are. And so solving this integral, we get this to be two pi. You not be r cubed divided by three or inside the cylinder and radius R. Two greater than the radius of the cylinder. Capital. Are We get the magnetic field B two B may not be R cubed of uh three are too.

Mayor Dick feel expression for our equal or less than a and recognize a B function off our that is, you know, I enclosed divided by two by our this will be mu nod. Divide by two by are integral zero to our writing are in terms off a current density are then two by R D R We hoped in an expression there is, um you know j Braden Stijn o r squared divided by three A. For about a off the problem, it's odd, is equal to zero. You have 10. B is equal to zero about me at all. Physically able to we have a B o far is equal to mu nod Jean Odd R squared. Divided by 38 studio values. We get the northern constant times J, which is 310 pairs for a meter. Square times are, which is 3.1 times 10 to the power re 100.1 times 10 to the power minus three meter over too. This square divided by three times 3.1 times gentle about minus tree. This gives us be off our to be one times 10 to the baller minus seven. Tesla for part C off the problem. It all is equal to a we substitute to be off our is a cold, eh? Then our expression becomes you not Jane Odd A divided by three. Then substituting your values we get here B to be four times 10 to the power minus seven Tesla.

This problem we're gonna talk about compares law. So we need to remember is that if we have a distribution of current as I'm showing here on the screen such that the current density pure across session progress sectional area, actually is this vector J Then if we draw a circuit around, uh, any place Yeah, we obtained that the integral over this circuit off B. So be times the l d. L is the element off length is equal to mu zero. The magmatic constant times uh, the integral over the area enclosed by the circuit. So this area here off J Yes, where the s is the the element off area. This is in Paris law. And in our problem, we have ah, cylindrical conductor that has a radius are so like this. The radius is our but J The, uh, current density is equal to b. Times are so we don't have ah, uniformed current density on our goal With this information is to find what is the value off the magnetic field? B a When our is smaller, then capital are and B when R is greater than capital are. Yeah. Okay, So what we need to do first is soon, uh, notice that J's pointing in this direction. It's, uh, flowing along the are cylinder and in question A we're going to use that the integral of b D l and I'm gonna choose, um, when shoes this circuit right here. Okay, that is a ah, circle around the wraps around the cylinder with a radius are but actually, uh, in our case for a question A They should be inside the circuit. Okay for question. We were going to make it outside. So this radio series are so b d l over integrated over. This circuit is equal to the integral of J. D s. Hey, I noticed that I just got rid of the organization because B must be perpendicular to D l. Since D l is in the same direction is the current and be by the right hand rule. Must be, um, pointing particularly to the current at all times here. Okay, actually be will make a circle around our our senator. I mean, okay, um, furthermore, we have at the J and yes, are all also in particular because ds is actually I'm sorry. There there are particular. They are they are parallel and and as a matter of fact, D l is also parallel to be deal exactly the element of line along the circuit. I'm sorry for the confusion. So this is the element of line as well. It's parallel to be okay. So that's why I got rid of the vector notation. Now the line integral off. Be over the l since uh is equal to two pi r meat Notice that I can do this because we chose a circuit are along with along which be is constant So be is the same for any point in the circuit Eso Since it's constant, I just can pull it out I can just pull it out off the integral And this is the integral of JJ's be times are times the element off area in the element of area is two pi are the arm integrated from zero to our right. So I have two pi r b is equal to two pi b Time is the integral of r squared d r from zero to our the two pies and sold out. So have that b is equal to be are cute over three are so this is B R squared over three. So this is our answer that in question we you have to do the same thing. The Onley difference is that when integrating the right hand side of the equation Okay, I'm going to start here. But when integrating the right side of the equation is thought of integrating up to are this little are we have to integrate up to Capitol are because after capital are there is no current density. So two pi b r is equal to the integral zero from zero to capital are of BR times two pi r dear, so too are the two Pipkins a lot quick So you have be easy what you want over our times be, um Times Capital R cubed over three So this is be capital r cubed over three are and this concludes our exercise

Hello, everyone In this problem, there is a long cylindrical conductor. Getty decorated. I asked certain difficult. The current density is not people. It where is really according to the relation being What so ended? Got it. Dead Street. It's not report. It varies. Really According to this relation your response stint and R is the distance. Yeah, these constant that is given to us and art Is the real distress from the excess? No, we have to calculate magnetic field the because after this place Sorry. Just a moment, please. After his face, I went less than Capitola be at United States. I do greater than not distressed. Three million from the axis we start solving when waving for first we will consider Element had said This is the element that said off radius R and thickness here. So here I am writing and the medical cylindrical ship. Oh, for eight years are I'm thickness. Yeah. So current flowing through inch is doing Cooper. You are via engaged, given br it becomes by, uh, current through this elemental inglesby group by our square B b Yet to calculate the magnetically be used em peers law on According to NPR's now, B. Dawg Ideas movie Fatal. You know, tonight what? Very American in density. I can be read in it, Jane. Curious. Now we can solve it. That's great foot. Are you in less than our That is be. But God idiots, you not jeans Given being to, uh you can are directly use this one or you can use this one. In what way you concern here. You cannot to right to be a integration of being. You consent to do the value here. So what may you can solve it? There is no difference between that on area in close by this purpose built by your dia this since Magarri clean lines are symmetrical boogaloo. It will be to buy odd what you have to integrate for the limit. Do you don't do our what new not be constant by constant. And this becomes integration off r squared here for the limit. You don't know what what My belief feel inside and you will get, you know be are really square by three. This part that would be applied for are meant to be equal or less then really is of this trigger in second. But be able to find out. Paid off it. Okay. Are you is greater than our B dawg DS. It's goingto mu not then. And to be a for the limit due to what? So integration off in this one will be too tired. You know, it is being what on this is Cooper Yardena subsequent level and invigorated. You will again. Magarri. Queen, baby. Um, you not be capital. Are you? Upon? We are. So this part of life is applicable. Only part are to be equal or more than the radius of nothing. So this is the answer Portis pulled up. Thanks. What?


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