5

Current flows parallel to the 2 axis in a cylindrical region (shaded in the diagram) between r = a andr = 2Za, where r is the distance from the 2 axis and a is a co...

Question

Current flows parallel to the 2 axis in a cylindrical region (shaded in the diagram) between r = a andr = 2Za, where r is the distance from the 2 axis and a is a constant: (a) Find the total current / if current density in this region is ] = Jok, where Jo is a constant: (b) Find the total current if current density in this region is j = Jo SsinZa k, where Jo is a constant:

Current flows parallel to the 2 axis in a cylindrical region (shaded in the diagram) between r = a andr = 2Za, where r is the distance from the 2 axis and a is a constant: (a) Find the total current / if current density in this region is ] = Jok, where Jo is a constant: (b) Find the total current if current density in this region is j = Jo SsinZa k, where Jo is a constant:



Answers

A long, cylindrical conductor of radius $R$ carries a current $I$, as shown in Figure P30.27. The current density $J$, however, is not uniform over the cross-section of the conductor but is a function of the radius according to $J=b r$, where $b$ is a constant. Find an expression for the magnetic field $B$ (a) at a distance $r_{1}<R$ and (b) at a distance $r_{2}>R$, measured from the axis.

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?

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

Question says that certain cylindrical buyer carries current. We draw circle of radius are around the central access to your 26 test 24 day to turn the current I within the circle Figure 26-24 p shows current as a function of our square. The vertical scale is set by ice of s equals 4.0 1,000,000 peers or 4.0 times 10 to the minus three amperes and the horizontal scale is set by arson. Best squared equals 4.0 millimeters squared OK or four points. Your attempts to the minus six meters squared since one millimeter of equals 10 minus three meters a millimeter squared is 10 minus six meters. So for part a, is the current density uniform of property. If so, what is the magnitude? Well, a circular area depends, of course, on our square. So the horizontal access of the graph and figured 26-24 b is effectively the same as the area except for a factor of pie and the fact that the current density increases linearly in the graph, Um means that j so this is a jay is equal to curry divided by the area. This is a constant therefore Thea answer is yes, since that's a constant since its linear the current density is uniforms so we can say box set in because their solution of party party says since its uniform, what is its magnitude? So this is I over the area which is pi r squared So looking from the graph and given the scale factor, I is 0005 amperes oh divided by pi Times R squared which would be four times 10 to the minus six. Because remember, one millimeters tendonitis three meters. So millimeter squared his tendon minus six meters, I swear. Oh, it's a plug was not using your calculator. This comes out Thio 398 computers per meter square we can box it in is their solution for r P.

Here we have a long someone who conductor of radius capital are carrying eggs given current. We also know that the current density J isn't uniform throughout it, but it said is given by the R, where R is a constant reviews, some kind of constant, and our is our distance from the center and we want to find an expression for the magnetic field at two different radio. One lesson are and one greater than our. So let's start with amperes law, which is which says that the lions agro of BDs is equal to the enclosed current times, you not to the permeability of free space. We also know that I current is the integral of current density body A our area and were given our area changes radius be on. So now we have all the pieces that we need to solve for our magnetic field at different points. The line integral B is b times two pi R once and set the circumference off the region that we're looking at and that's gonna be equal time to immune on time. The integral of Jake D. A. Our current and we know that is gonna be pyre squared and there that d a d are So are circumference is going to meet two pi r Therefore, we can rewrite d a as being two pi r d r No, we can plug that in as well, replacing our current density J with BR we're gonna integrate from zero to our one because that's the region that we're looking at for port A specifically I mean, when we integrate, we get our cube divided by three and we can get eliminated constants to pie to get an expression for B, which you can then further simplify to say is equal to a little B times new, not times R one squared, divided by three. The next part of the problem is asking us to solve for be at a radius greater than are Once again we're gonna start with the same equation be times to be times two pi r two equals mu, not times J. D. A. So from NPR's law, once again, we're gonna rewrite RJ because that expression isn't changing. And neither is our equation for D. A s o old. The only that's different is that now we're integrating from zero to our we're integrating from here to our because remember Now our second remember, our is the total radius of our cylindrical conductor are too is just an appearing in loop so imaginary circle that we're drawing around it. There's not actually any current enclosed in all of our two. So we're only integrating. Aren't the region where we know that a current is flowing once again being integrate and solved and then rearranged to get a final expression for B, which is little B times mu not times are cubed all over three r squared.


Similar Solved Questions

5 answers
1. Solve_(x5 + 1)y' = 1Oxky, ylo) =3
1. Solve_ (x5 + 1)y' = 1Oxky, ylo) =3...
5 answers
(2 points) Consider the surface in R' defined by the equation 2x2 4y2 + 222 = 6. Check that the point (1 , 1,2) lies on the given surface: Then; viewing the surface as a level surface for a function f(x,y,2), find a normal vector to the surface at (1, 1,2) and an equatior for the tangent plane to the surface at (1,1,2)_Normal vectorEquation of tangent plane:
(2 points) Consider the surface in R' defined by the equation 2x2 4y2 + 222 = 6. Check that the point (1 , 1,2) lies on the given surface: Then; viewing the surface as a level surface for a function f(x,y,2), find a normal vector to the surface at (1, 1,2) and an equatior for the tangent plane ...
5 answers
1 32 (probubil RA) (81 Jev0 4 Dacaueonis 2 ALCeiCO 1
1 32 (probubil RA) (81 Jev0 4 Dacaueonis 2 ALCeiCO 1...
5 answers
2. (2 pts/each) Iterated Integrals: Find the integral J5 dmdn R: 0 Sm <2;0 Sn <1 (m +n + 1)3
2. (2 pts/each) Iterated Integrals: Find the integral J5 dmdn R: 0 Sm <2;0 Sn <1 (m +n + 1)3...
5 answers
Prove that(i) $an left(frac{pi}{4}+Aight) an left(frac{pi}{4}-Aight)=1$;(ii) $(sin A+sin B)(sin A-sin B)=sin (A+B) sin (A-B)$;(iii) $frac{sin (x-y)}{sin (x+y)}=frac{an x-an y}{an x+an y}$;(iv) $cos left(frac{pi}{4}-xight) cos left(frac{pi}{4}-yight)-sin left(frac{pi}{4}-xight) sin left(frac{pi}{4}-yight)=sin (x+y)$;(v) $frac{an left(frac{pi}{4}+xight)}{an left(frac{pi}{4}-xight)}=frac{(1+an x)^{2}}{(1-an x)^{2}}$
Prove that (i) $ an left(frac{pi}{4}+A ight) an left(frac{pi}{4}-A ight)=1$; (ii) $(sin A+sin B)(sin A-sin B)=sin (A+B) sin (A-B)$; (iii) $frac{sin (x-y)}{sin (x+y)}=frac{ an x- an y}{ an x+ an y}$; (iv) $cos left(frac{pi}{4}-x ight) cos left(frac{pi}{4}-y ight)-sin left(frac{pi}{4}-x ight) sin lef...
5 answers
Partal fract ions 6) dy Jxax (x+)(zx-1)Sx_ [email protected]
partal fract ions 6) dy Jxax (x+)(zx-1) Sx_ [email protected]
5 answers
2. Find the Average Value of the function g(x) =2r' _ 24x2 +64*+12 o 2<*<6 1Opts
2. Find the Average Value of the function g(x) =2r' _ 24x2 +64*+12 o 2<*<6 1Opts...
5 answers
Part /The units most likely to be used to measure Ihe amount of alcohol to be added t0 a small test tube arecm:MLSubmitRequost Answor
Part / The units most likely to be used to measure Ihe amount of alcohol to be added t0 a small test tube are cm: ML Submit Requost Answor...
5 answers
Write the standard equation of parabola. Find the focus and make a graph.passes through (25,20)$;$ axis horizontal
Write the standard equation of parabola. Find the focus and make a graph. passes through (25,20)$;$ axis horizontal...
1 answers
Evaluate the integrals. $$\int \frac{1}{\sqrt{e^{2 \theta}-1}} d \theta$$
Evaluate the integrals. $$\int \frac{1}{\sqrt{e^{2 \theta}-1}} d \theta$$...
5 answers
8 5 8 2 ? ? 1 1 1 1 ~ 054 =/8 1 9 1 D 5 1 5 1 1 H 1
8 5 8 2 ? ? 1 1 1 1 ~ 054 =/8 1 9 1 D 5 1 5 1 1 H 1...
5 answers
Problem No. 14 70 pts_ 4X1 +X2 + 2X3 = - 1 Xi-2X2+2X3 = 2 4X2 -2X3 = 3 Solve the given system using elementary rOw operations. Do not use matrices Show all your work do not skip steps_ Displaying only final answer is not enough to get credit.
Problem No. 14 70 pts_ 4X1 +X2 + 2X3 = - 1 Xi-2X2+2X3 = 2 4X2 -2X3 = 3 Solve the given system using elementary rOw operations. Do not use matrices Show all your work do not skip steps_ Displaying only final answer is not enough to get credit....
5 answers
V(t)r"(t) v(t)r'(t) Use the equality N(t) to find the unit normal vector N(t) of r(t) = (7t2 , 7t3) at t = 1 Ilv(t)r"(t) v(t)r'(t)|lN(t)eBook
v(t)r"(t) v(t)r'(t) Use the equality N(t) to find the unit normal vector N(t) of r(t) = (7t2 , 7t3) at t = 1 Ilv(t)r"(t) v(t)r'(t)|l N(t) eBook...
5 answers
Use the information provided t0 determine AHFrxn for the following reaction CHA(g) Clz(g) CCIa(g) HCI(g) AH". CHa(g) = -75 kJlmol CCl(g) = -96 kJlmol HCI(g) 92 kJlmola) +113 kJb) -113 kJ=389 kJd) -71k
Use the information provided t0 determine AHFrxn for the following reaction CHA(g) Clz(g) CCIa(g) HCI(g) AH". CHa(g) = -75 kJlmol CCl(g) = -96 kJlmol HCI(g) 92 kJlmol a) +113 kJ b) -113 kJ =389 kJ d) -71k...
5 answers
Determine which of the sets of vectors is linearly independent:A: The set {P1 Pz P3} Where pa(t) = 1, pz(t) = t2, p3(t) =1+ StB: The set {P1 12 Po} where pi(t) = t, pz(t) = t2, pa(t) = 2t + 5t2C: The set {P1P2 Pa} Where p(t) = 1, pz(t) = t2, pa(t) = 1 + 5t + t2Select one: A. B onlyB. C onlyC. all of themD. A and CE: A only
Determine which of the sets of vectors is linearly independent: A: The set {P1 Pz P3} Where pa(t) = 1, pz(t) = t2, p3(t) =1+ St B: The set {P1 12 Po} where pi(t) = t, pz(t) = t2, pa(t) = 2t + 5t2 C: The set {P1P2 Pa} Where p(t) = 1, pz(t) = t2, pa(t) = 1 + 5t + t2 Select one: A. B only B. C only C. ...
5 answers
458458If BC has length 11, find the length of AB and AC _
458 458 If BC has length 11, find the length of AB and AC _...
5 answers
Claim: the proportion of graduates in Illinois, this year; majoring in the social sciences is at least 20% The correct null hypothesis statement for testing this claim is: (choose one)(a) Ho: p 0.2 Ho: p=0.2 Ho: H4 + 0.2 Ho: H4 = 0.2 (e) Ho: p 0.2
Claim: the proportion of graduates in Illinois, this year; majoring in the social sciences is at least 20% The correct null hypothesis statement for testing this claim is: (choose one) (a) Ho: p 0.2 Ho: p=0.2 Ho: H4 + 0.2 Ho: H4 = 0.2 (e) Ho: p 0.2...
5 answers
Question 21 ptsConsider a 10-sided polygon (decagon or 10-gon):If we form a pyramid with a decagon base, then the number of faces isthe number of edges isand thenumber of vertices IsIf we form prism with decagon base, then the number of faces Isthe number of edges isand thenumber of vertices is
Question 2 1 pts Consider a 10-sided polygon (decagon or 10-gon): If we form a pyramid with a decagon base, then the number of faces is the number of edges is and the number of vertices Is If we form prism with decagon base, then the number of faces Is the number of edges is and the number of vertic...

-- 0.020832--