Troubleshooting the Biot-Savart Law for a Circular Loop and Helmholtz Coil Setup

[Paintjunkie]

Homework Statement

there is no problem really this is not homework its just something I wanted to include in my lab report. the lab was working with the Helmholtz coil. I was trying to come up with the magnetic field at a distance, Z, from the center of a circular loop of radius, R, that has a steady current, I. I thought that I did it right... but I got a negative sign for some reason. where did I go wrong...

Homework Equations

see attached I think I got it all there

The Attempt at a Solution

see attached

 

[Paintjunkie]

I updated my attached file. nothing big, just interchanged theta and phi to what I believe is a more conventional (at least in physics) use for theta and phi.

 

[Paintjunkie]

any guidance on this would be great...

 

[Paintjunkie]

ok this is my last update to attachment file. I just changed a few minor things with the ordering and included the Z hat in my final answer

 

[BvU]

Good thing you added the $\hat R$ and $\hat z$ (even though now I have to edit my reply to remove the comments on those).

I wondered how you got the minus sign in your cross product:
In your figure 5.21 the z component of $d{\bf B}$ is clearly positive.

Can't pinpoint where it goes wrong, but
I suspect $\ \vec \Re \equiv \vec r -\vec R \$ has something to do with it ( $\vec r = (0,0,Z)$ ) .

 

[Paintjunkie]

I don't think I did the cross product wrong... but the values in the matrix might be off...

 

[Paintjunkie]

ok ok I think I see where your going. so if (scriptR) ⃗ ≡r ⃗ −R ⃗ with ( r ⃗ =(0,0,Z)) then...

R ⃗ simply equals (R,0,0)?

making (scriptR) ⃗ = (-R,0,Z) and that's what I should put into my cross product?

srry I don't know latex at all.

 

[Paintjunkie]

well that definately fixed the negative sign... I am going to go with it. thank you for fixing it for me. ill attach a the corrected document if anyones curious...

and again thank you for your time on this

 

[BvU]

But the problem still is that the r component of a vector in r,phi, z coordinates can't really be negative ...

The guys here don't have the problem: their $\alpha$ is $\pi/2 - \theta$

 

[Paintjunkie]

well.. I am not understanding the issue.

 

[Paintjunkie]

I want to accept it just cause I guess. I am looking at pg 9 in my e&m book its Griffiths 3rd ed. it describes the r-r' formula, does not seem to have the limitation where the vector subtraction has to turn out positive. the problem 1.7 on the following page seems to illustrate this for me. is asks for (4,6,8) - (2,8,7) = (2,-2,1).

so I don't see why the resulting vector can't have a negative component.

 

[BvU]

$\vec \Re$ can have a negative component in cartesian coordinates, but not in the r component when working in cylindrical coordinates (then r is the distance to the z-axis and $\phi$ runs from 0 to $\pi$ -- the 2d polar coordinates).

 

[Paintjunkie]

hmm phi is from 0 to 2pi. not that that voids your point. so is my derivation completely bogus? should I just throw it out and go with Cartesian coordinates?

 

[BvU]

yes sorry.

 

[Paintjunkie]

are we certain that r=(0,0,Z) and r' = (R,0,0)...

could it be that r(-R,0,0) and r' = (0,0,Z) so I would start at the outside of the bottom circle and move to the center going a distance of -R then go up Z?

 

[Paintjunkie]

is it ok for me to make my source point on the circle that dl is on?

 

[BvU]

I still have a hard time finding out how to get a positive z component.

I think we don't have the right $\vec \Re$ anymore: the r component is definitely not equal to -R.
More like $\vec \Re = (R, \phi+\pi, z)$. $\phi+\pi$ is multiplied with 0 and disappears (?).

The determinant form looks correct, but gives us a minus sign, whereas $\vec {dl} \times \vec\Re$ definitely points as in Figure 5.21 with a positive z-component, so there's something wrong, but what ?

And (I'm making things worse now): "The first integral equals 0 because of orthogonality" ?
What exactly is orthogonal in the first integral (and not in the seceond) ?

Bedtime for me..

Let's ask @haruspex

 

[Paintjunkie]

that almost makes more sense to me. that the source point is on that circle. attached is an updated derivation.

 

[Paintjunkie]

damn yea huh idk I guess that's not right either! soab I need that first integral to go away. it has to cancel itself out somehow...

 

[BvU]

I like the Griffiths approach in the tekst under fig 5.21: "$d{\bf B}$ sweeps out a cone. The horizontal components cancel..." (well, I don't like the term horizontal, but I understand what he means). But I still have a hard time expressing that in a decent integral of the vector product. Has to do with $\phi$ running for the current loop, but we don't want the $\hat \phi$ and the $\hat r$ of the point (0,?,z) where we want to know the field to run at all.

Something screwy with using cylindrical coordinates for a running $\phi$ on the current loop wrt a point on the z-axis (for which $\phi$ is undefined) ?

 

[Paintjunkie]

I just wanted to do it in cylindrical coordinates cause to me it feels like it should be done that way... I do see that derivation and I could put that down in my lab report instead I guess dl would end up being (rsin(theta), 0, rcos(theta)) or something like that

 

[BvU]

really have to get some sleep now. And work in a few hours. Good luck !

 

[Paintjunkie]

ok thank you for your time

 

[Paintjunkie]

well.. i think i thought of a valid reason that first integral can go away.
(R2π+ Rπ) R ̂=R ⃗-R ⃗

still slightly troubled over the (-R,0,Z) thing...

attached is my updated work..