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An infinitely long; solid, cylindrical conductor of radius 2.50 cm carries uniform-density 0.480 current out ofthe page. (The 'diagram shows a cross- sectional...

Question

An infinitely long; solid, cylindrical conductor of radius 2.50 cm carries uniform-density 0.480 current out ofthe page. (The 'diagram shows a cross- sectional view.) A coaxial, infinitely long; thin, cylindrical shell of radius 6.00 cm carries uniformly-distributed 0.180 A current also out ofthe page.What is the magnitude of the current density J in the inner conductor?Calculate the magnetic field, B, at each of the following distances I, measured from the center axis. r= 1.30 cm r = 3.00

An infinitely long; solid, cylindrical conductor of radius 2.50 cm carries uniform-density 0.480 current out ofthe page. (The 'diagram shows a cross- sectional view.) A coaxial, infinitely long; thin, cylindrical shell of radius 6.00 cm carries uniformly-distributed 0.180 A current also out ofthe page. What is the magnitude of the current density J in the inner conductor? Calculate the magnetic field, B, at each of the following distances I, measured from the center axis. r= 1.30 cm r = 3.00 cm r= 4.50 cm r= 8.00 cm



Answers

The figure shows a cross section across the diameter of a long, solid, cylindrical conductor. The radius of the cylinder is $R=10.0 \mathrm{~cm} . \mathrm{A}$ current of $1.35 \mathrm{~A}$ is uniformly distributed throughout the conductor and is flowing out of the page. Calculate the direction and the magnitude of the magnetic field at positions $r_{\mathrm{a}}=0.0 \mathrm{~cm}$$r_{\mathrm{b}}=4.00 \mathrm{~cm}, r_{\mathrm{c}}=10.0 \mathrm{~cm},$ and$r_{\mathrm{d}}=16.0 \mathrm{~cm}$.

Hi the given problem here, this is cross sectional view. So far screen, solid cylindrical current caring conductor was radius. It's given us capital R is equal to 10.0 centimetre. Then in order to find magnetic field at any point within this conductor, supposed that observation point at which we have to find magnetic will suppose that point lies actually stands smaller from the center of this conductor. So first of all, we will find current density. That current density J is given as current passing through the conductor or particular to its area of cross section or unit area of cross section. No, if this is the absolution point, we will consider an empty eerie in lieu passing through this observation point and having a radius R. So three years of imperial loop will be equal to dispense of observation point from the center conductor is here in this general case that dispenses are so area of imperial loop. That will be buy into our square. So finally, but I didn't passing through imperial luke will be given by let it be I dash and that I dish will be equal to current density multiplied by area of the emperor liu. There this is I buy bye arty square into by small artist square canceling this pie. Finally the current linked through the emperor liu will be given by I. R. Squared by capital artist square. Well no using um peer circuit A law to find the magnetic field at that observation point line integral of magnetic field at that point will be given by mu note times the current wearing the imperial blue. So as A B means magnetic field at that point will be constant at all. The points of imperial loop. So it can be taken as a constant out. Then remaining means the integration of deal will be equal to circumference of imperial loop needs to buy our so it will be new not times the I dash which is I into our square by ari square canceling this small. Are finally expression for magnetic field at any point within this solid conductor will be given by um You not by two pi into I. R. By Are square. Or if we multiply it by two x 2. It may also be given as new. Not by four by Into two i. R. Invited by are the square Now we can find a magnetic fields at all. The given points in this problem. Most of all we have to find a magnetic field At our .8. Here the distance of point is zero means we have to find magnetic bill at the center. So this be a means magnetic field at this point will come out with zero. Then in the second case this is Observation Point B whose distance is 4.00 centimetres. So this be its magnetic field at this point B will be given by Yeah, the values are Our vic was to 4.00 cm Radius of the conductor. We have already seen it to be 10 centimetre and current passing through the conductor wars 1.35 M. P. R. So this BB will be given by For me, not upon four by the suspended for -7 into Two times of high into armies between 2 1.35 multiplied by Three years of the imperial which is four centim or foreign to. And it's part -2 meter divided by the radius of the solid conductor square of the radius means this is turned into and it's about minus the square. So finally this be at the observation point here comes out to be 1.08 Into 10. To the power -6 Tesla. Then in the third case, this time the observation point R. C. This is Equal to and zero Centim is equal to the radius of conductor. Is the observation point is at the surface of the conductor. So this magnetic will be C will come out to be Again using the same formula. Um you're not upon four pi into this time, this is capital, are in place of smaller, divided by our square. So canceling this, Finally putting the known values. This is standard for -7 into two into 1.35, invited by our which is 10 cm or turn into 10 for -2 m. So finally this magnetic will now at the surface of the conductor comes out to be 2.7 Into Standish Power -6 Islam. one more answer for this given problem. Until now we have obtained three answers. No. In the 4th case. Now the observation point is outside the street conductor. That distance is given as 4.2 d. This is 16.0 centimetre and we do at any point outside the conductor of the magnetic field is simply given by You, not upon 45 into two, I by our only. So this is tending bar -7 into into 1.35, invited by Our which is 16 cm or 16 to 10 days per -2 m. So finally, this magnetic field here comes out to be 1.72 And it's AR -6 islam. The 4th answer for the same problem, and the direction of all these magnetic fields is counter clockwise using right and and rule using right hand rule, which is a rule to find the direction of magnetic field around the street current carrying conductor directions of all the fields. His counter clockwise around this even conduct. Thank you.

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.

For this problem on the topic of sources of the magnetic field, we are told that a long cylindrical conductor has a radius A. And two cylindrical cavities of diameter. Aid through its entire length. As we can see in the figure. We then have a current I which is directed out of the page and is uniform through a cross section of the conductor. We want to find the magnitude and direction of the magnetic field. In terms of you know what I. R. N. A. Firstly at point P. One and then at point P. Two. Note that. Note that the current I exist in the conductor with a current density J. Which is the current I over the cross sectional area A. Where A. Is equal to hi into a squid minus a squared over four minus a squared over four, which is pie a squid over to. Which means that the current density J. Is two I over pie A squared Now to find the field at either point P one papito P two will find Bs which would exist if the conductor were completely solid. And we do this using Mps Law. Next we'll find B. One and B. Two. That would be due to the conductors of radius A. Over to. That could occupy the void where the holes exist with the news, the superposition principle and subtract the field that would be due to the part of the conductor where the holes exist from the field of the totally solid conductor. So for part A. At P. One we have this field Bs. For the solid conductor to be mu. Not times J. Times pi a. Sacred over two pi R. The one is you're not time's jay times pi times A over to old squared, divided by two pi into our minus A. Over to. And the field B two is equal to um You know, T. J. Times pi times A over to old squared, divided by two pi into our plus A. Over to now the total magnetic field B is equal to B. S minus be one minus B. Two from superposition. And so this is equal to you're not, time's jay times pi hey squid over two pi into one over our minus one over. Fall into our minus A. Over to minus one over for into are plus A over to. And so we can simplify this further as um you note into to I over two pi into four R squared minus a squared minus two R squared all over the common denominator which is four are into R squared minus a squared oh four. And so this finally becomes the magnetic field at point P one is new note I oh uh pi R into two R squared minus a squid over four hours squared minus a squared directed to the left. So that's the magnetic field at point P one. Now we want to do the same for part B but now we want to find the magnetic field at point P. Two. And so the magnetic field due to the solid sphere, bs would be you not tens J times pi a squid over two pi R and be one prime and B to prime due to the cavities is mu not J times pi into a over to old squared divided by two pi times the square root of r squared plus a over to all squared. Now the horizontal components of B one prime and be to prime cancel while the vertical components add. So the total magnetic field B is equal to that of the solid sphere minus B. One prime. Coz I intend to minus B to prime course I data and this is you're not jay times pi a squid divided by two pi r minus two into. You're not jay pi a squared over four of uh two pi times the square root of R squared plus a squared over four, multiplied by our into the square root of R squared plus a squid over four. And so this simplifies two. Um You not G times pi a squared over two pi R into one minus R squared over to into R squared plus a squared over four. Which is new note. And making a place the current density the current to I over two pi R. Into one minus two. R squared of uh four R squared plus a squared. And so therefore we get the magnetic field B at point P to to be um You not I over pi R into two R squared plus a squared, divided by four R squared plus a squid directed to the top of the page.


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