5

Q6: An infinitely long straight wire carries current I. and radius R carries a current Iz a Wire loop of a5 shown in the figure: If L1=4 Iz, and if the magnetic fie...

Question

Q6: An infinitely long straight wire carries current I. and radius R carries a current Iz a Wire loop of a5 shown in the figure: If L1=4 Iz, and if the magnetic field at the center of the loop is zero net how far is the center of the loop from the straight wire in ters of R (the distance y)2. I,JSall LS Ze,lji sO)b;ia 4b; 1o*bubydgL AiLJlI fxjc J#Llfiyj44} ; oublii Jn 47 2s 1.25 R AiLlI__biuiai 4Yy B) 0.86 R 2.55 R D) 1.27 R

Q6: An infinitely long straight wire carries current I. and radius R carries a current Iz a Wire loop of a5 shown in the figure: If L1=4 Iz, and if the magnetic field at the center of the loop is zero net how far is the center of the loop from the straight wire in ters of R (the distance y)2. I,JSall LS Ze,lji sO)b;ia 4b; 1o*bubydgL AiLJlI fxjc J#Llfiyj44} ; oublii Jn 47 2s 1.25 R AiLlI__biuiai 4Yy B) 0.86 R 2.55 R D) 1.27 R



Answers

A circular loop has radius $R$ and carries current $I_2$ in a clockwise direction ($\textbf{Fig. P28.72}$). The center of the loop is a distance $D$ above a long, straight wire. What are the magnitude and direction of the current $I_1$ in the wire if the magnetic field at the center of the loop is zero?

High. In the given problem here there is a circular current carrying conductor which is carrying current in a clockwise direction. And below this is circular current carrying conductor. There is a straight current carrying conductor, but the direction of current in this straight conductor has not been given to us. The current in this loop is I. two. And the current in this sphere conductor is I want and it is given that the net magnetic field at the center of circular loop is zero. No, the distance between the center of the loop and the straight conductor is given us D. So this magnetic field to be zero. The magnetic creates due to the loop and the straight conductor should be in opposite direction, means magnetic fields due to the loop, and straight conductor should be or pose it in direction and equal in magnitude. No. As this face of the loo carrying the current in clockwise direction will start behaving like a south pole. So the magnetic field will be entering into, it means the direction of magnetic field at the center of the loop is into the plane of paper, magnetic field due to the circular loop is entering into the plane of paper so that due to straight conductor should be coming out. And using right hand tom rule the direction of current in this conductor so that its magnetic will may come out of the plane of paper. That current should be towards right means I won should be towards right. I one shoot B. The words right in the straight conductor. Now, in order to find the magnitude of the Taiwan, we will put the two magnetic fields equal in magnitude means magnetic field due to loop should be equal to magnetic field. You do straight conductor. Now the expression for the magnetic field at the center of the loop is new, not I two divided by two R. And its radius is given as capital are is equal to the magnetic field due to the straight conductor Using batons ever slot. That has given us new note by four pi Into two I won by the distance which is capital D. So canceling this mu note from both sides, an expression for Ivan here comes out to be and cancelling this too. Also. So here this island will come out to be four by B. Into I. two divided by for our or we can say this is actually I one equals two pi. D. I two divided by our, which becomes the answer for this given problem. Thank you.

In this problem we know that basic will to um you know what I bye. Who are he to buy 3 15 As he has the value of metaphysical to three points By two, which is equal to 2 70 degree. Therefore I can write the value B is equal to you know what I by two? My application three pi By two x 25, Which is equal to three x 8. Do not I by our So according to the option in this problem option, the age of sunday age direct answer. Obstinately it correct. And for this problem I hope you understand the solution of this problem.

In this problem. I'm doing the diagram first so just look at it carefully. I'm doing it update them your current passing his eye. The distances are this is center. See this is odd. This length is deal. This is quarantine. This is our this is guaranteed. So here I can write the value of D. B is equal to um you not hi everybody or by bl across our bye our cube. So integration of D. B. Is equal to Munich by but by integration of ideals signed a 90 degree by ari square on solving it further I get the value of physical to unit I by four pi R square integration of deal which is equal to immunity. Buy food by artist square fire which is equal to do not die by Fora since the contribution of a straight line personal zero is also the magnetic induction due to the entire world choir so obstinacy each correct answer option ch got against it.

All right, so in this problem, we have a length of wire, um, of length D here. That's carrying a current I and we want to find in part a what the magnetic field will be at point P, which is a distance are away from the middle off the wire here. So we'll use be oh, so far for this. So we set that up like B equals Myu knots, I over for pie on the integral of deal, and I'm gonna write this is cross our vector and then over our vector cube. So sometimes you could write this Justus a unit factor, but this will be a little easier just working with the full factory here. So this is our equation. So let's start filling in, um, some of the variables and coordinates we're gonna use here, so all kind of mark in arbitrary little spots deal here. So let's say this is our little deal. So that steal there. And that means that this factor here would be vector are so the distance from D L two point p. And we'll set up our coordinates here, So I'm gonna call this Middle Parts will call that y equals zero. And so that would put this the top parts as negative d over too. And the bottom is d over too. Yes, you could switch those if you wanted to, um, upwards to be positive here, but shouldn't really make a difference at all. Okay, so next time we can start filling in some of the steel cross are here, so we're gonna head of you, not I over for pie and then are integral Will air in a role we don't know. Can should run the full length of this wire. So that's gonna be from D over too. Thio negative d over, too. And then, um I guess I've been calling this the Why direction now? So instead of d l will say d, Why now? So we have Do I cross our our vector and to do this cross product, it'll be easier if we writes the our vector insurance was components, since that could make doing cross products easy since, um, parallel ones will just be zero. So it looks like then at's, um, this arbitrary points are our vector. We could just writes as whatever Why positions it's at. So it has some vertical heights. So why J hat and then Plus are I had So we're just saying it has this much vertical components and this much horizontal component. And then we have this over, um, are cubes so by the thuggery and theorem here are should just be the square roots of big r squared plus y squared. So then we want this to the Cube. So now when we look at this cross product because we've written in its components, it'll be a little bit easier. So d Y is in the J hat direction so that we can see pretty easily do y cross. Why both under in the J hat direction. So the angle between them a zero. And so the cross practice here between those two. So the only thing that's gonna give us a non zero term will be D y cross our And so, um, something in the J had direction. Cross was something the I had direction should give us something in the negative K hat direction so we can just pull all that out right now. So we'll say we have you, not I over for pie. And then we'd have this are two, which is the constant So we can pull that off the integral as well. So we have times negative are que hats. And now we have are integral do you ever to negativity over to. And so it looks like we'll just be left with d y in the numerator And now we'll have divided by we will right this a little more cleanly here are squared plus y squared to the three halfs. And so now this is a relatively simple in a girl that we can just evaluate so you can look this up in a table. And, Wilk, it's we still have all of this. And now this integral should evaluates to Looks like we'll have Why over r squared times are squared plus y squared to the 1/2 and we evaluate that from dear iTunes and negative d over too. So now we just plug in those limits integration, keep all of these constants out front. And, um, so I'll do the, um, this attraction for you here. But we shouldn't end up then with negative you not I over two pi r and then valuing that air goal. Then we get d over four R squared lost eastward to the 1/2 And then remember, this is all in the K hat direction, and we have the negative sign out front. So it's the negative 1/2 down there. So there we go. So that is the magnetic field at point P. And then Part B asks us, um, to check of this reduces to what? We found an example. 11. So in that example, we're seeing what happens if g goes to infinity. So if we had an infinitely long wire So looking at this, um, this second chairman, the parentheses here if we had d going to infinity Well, this four r squared in the denominator doesn't really matter. So we just have d over D Square to the 1/2. So this is just gonna become d over d or one. So this whole term just converges to one has de approaches infinity, and so then we be left with just negative you not I over two pi r in the cave, that direction, which is exactly what we found. An example 11 or what we'd find using any number of methods for the magnetic field for an infinitely long wire at a distance are away


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