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Problem 40 Medium Difficulty

(II) Consider a straight section of wire of length $d,$ as in
Fig. $48,$ which carries a current $I$ (a) Show that the magnetic field at a point $\mathrm{P}$ a distance $R$ from the wire along its perpendicular bisector is
$$B=\frac{\mu_{0} I}{2 \pi R} \frac{d}{\left(d^{2}+4 R^{2}\right) \frac{1}{2}}$$
(b) Show that this is consistent
with Example 11 of Sources of
Magnetic Field for an infinite
wire.

Answer

A) $-\frac{\mu_{0} I}{2 \pi R} \frac{d}{\left(4 R^{2}+d^{2}\right)^{1 / 2}} \hat{\mathbf{k}}$
B) $\lim _{d \rightarrow \infty}\left(\frac{\mu_{0} I}{2 \pi R} \frac{d}{\left(4 R^{2}+d^{2}\right)^{1 / 2}}\right) \approx \frac{\mu_{0} I}{2 \pi R}$

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Video Transcript

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

University of Wisconsin - Madison
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