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L2Two steel conductors are bent into rectangular prisms with square bases of lengths and where /-Za: If the thin prism has length of Lz-10a and the thick prism has ...

Question

L2Two steel conductors are bent into rectangular prisms with square bases of lengths and where /-Za: If the thin prism has length of Lz-10a and the thick prism has length of L2-400; compare the current densities of the two conductors (given that the same DC current passes through the conductors in the direction shown above):The thinner conductor has lower current densityThe thicker conductor has lower current densityThey have equa current densitiesd We cannot answer the question with the informa

L2 Two steel conductors are bent into rectangular prisms with square bases of lengths and where /-Za: If the thin prism has length of Lz-10a and the thick prism has length of L2-400; compare the current densities of the two conductors (given that the same DC current passes through the conductors in the direction shown above): The thinner conductor has lower current density The thicker conductor has lower current density They have equa current densities d We cannot answer the question with the information provided



Answers

Two conductors of the same length and radius are connected across the same potential difference. One conductor has twice as much resistance as the other. Which conductor dissipates more power?

All right, if we're talking about being a lot lot less than w we would be using, we're going to calculate the magnetic field b dot E d s is equal to mute zero. I go, um, d s would be around the bar. Eso therefore, if it was very thin bar w must be much larger than t. Um, you would have w plus w if you're going around it like the circumference to would end up with two w divide by two w and you get mu zero. I divided by two w for the magnetic field. Now for way out a way, way from it. Way, way, way Then the conducting bar would be just like a wire. Um, and then the DS would be the, um, magnetic Where, um I'm sorry, the circumference. So therefore, you would have mu zero I, uh, two pi r two pi r is the circumference of a circle. So, um, if it was, like, treated like a wire and then it would the magnetic field would make a circle around it. Um, n w the width would have no negligible effect on it because of how large are would be

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.

This question. We have our cool exile cable. Okay? The inner cable is, uh, has reduced. Uh, a The current is coming out of page in this drawing. Okay. And then the of the ventricle show okay, has in a radius B or the radio C and then the current is going into the pitch. Okay, so we want to find a magnetic view in the region between a b and r. Agreed and see Okay, outside the tube. Okay, so this is a question on NPR slaw. Okay, so NPR's losses that the the line integral of dot dll is a close look, integral of u dot dll is equal to You're not. I am close. Okay, so in a he, uh, NPR Lou, we are going to use It's going to look like this. Okay. It's going to be a circle of radius R, uh, surrounding the, uh uh huh. In our cable. Okay. So, uh, using and personal. Okay. He got the hour equals two. You know, I and close Uh huh. Okay. Site. Okay, so the B along the circumference of the NPR loop is uniform, so we can actually, uh, right e take me out And then the d O is in the same direction as be okay. And so we actually obtained be times two pi r. Yeah, And then I am close. It's just I So the expression becomes the times two pi r equals two. You know, I and so b equals two. You know, I over two pi r. Okay, so this is the 22 of the magnetic view in the region between a m b a and then in prime be, uh, yeah, looking at region are greater than C. Okay, so, uh, NPR Lou, maybe I've just recall the situation. Okay, simplify one. The current coming out and then you have Ah, yeah, it's going to go conductor shell outside. But this time the current is going in, okay? And then the emperor and there will be drawn this way. Okay, so if we calculate the eye and close Hey, this is equal to zero because that's one current coming out and one current going in into the pitch King. So, um, the NPS law, the right hand side. Yes. Zero. Okay, so the conclusion is equals to zero. Okay, So is this the answer for part B? And that's all for this question.

Hi in the human problem four. The hollow conductor. Although cylindrical conductor outer radios is given as let it be capital are and that is 3.0 by two centimeters only. This is millimeter because the diameter the outer diameter is given ass 2.0 100 millimeters. So it's really it's outrageous. Will be half of the diameter means this is 1.50 millimeters. Or we can say this is 1.50 into 10 for minus three m. Similarly in the videos, smaller that will be half of inner diameter, which was 2.0 millimeters divided by two means this is 1.0 millimeter or we can say this is 1.0 into 10 for minus three m. Land will be same for both the conductors. One hello and another. Solid conductor radius of the solid conductor is given us our p We have to find it provided resistance of both. The conductors is saying means resistance of solid conductor is given to equal to resistance up. Follow cylindrical conductor provided the length is same and both of the conductors are made up of the same material. The resistive, it is also saying. So here it becomes roll into L. By area of cross section. For the solid conductor is equal to roll into L. By area of cross section. For the hollow conductor canceling these things, we get area of cross section of the solid conductor should be equal to area of cross section of the Hollywood. And for the expression for the 80th cross section, this is why R B square is equal to area of cross section of the outer conductor, which is by r squared minus body small are square. Or we can say this is by our square minus one are square canceling this pie. Here square of this army which is missing comes out to be artist where means this is one 0.50 into 10 dish par minus three square. Or if we want to find out final answer. In terms of millimeter only the radius of the solid. A solid conductor in millimeter only. So no need to and work these radio of the polar conductors in two m. This is 1.50 square minus 1100 square. It comes out to be 2.25 minus 1.0 means this is 1.25 So finally radius of the solid conductor will be given by square root of 1.25 milli meter, which finally comes out to be 1.12 millimeter, which is the answer for a given problem. Thank you.


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