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This is a youtube video with the derivation, see 5:55

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- #1

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This is a youtube video with the derivation, see 5:55

- #2

BvU

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Hesch

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The 90 degrees comes from the cross product in Biot-Savart's law.

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- #5

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r is not pointing straight out of the yz plane, so why is it perpendicular?

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For all other ##\vec {dl}## (in the yz plane) you can see they are ##\perp## 'their' ##\vec R## and also ##\perp## to the x-axis, so ##\perp## the plane in which their ##\vec R##, the x-axis and also ##\vec r##.

Bringing in Pythagoras was unnecessary - a mistake on my part.

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- #8

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So just because yz is perpendicular to xz? There is no way dl is perpendicular to r because r extends from the yz plane

For all other ##\vec {dl}## (in the yz plane) you can see they are ##\perp## 'their' ##\vec R## and also ##\perp## to the x-axis, so ##\perp## the plane in which their ##\vec R##, the x-axis and also ##\vec r##.

Bringing in Pythagoras was unnecessary - a mistake on my part.

- #9

BvU

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No, not just because yz ##\perp## xz. Because ##\hat y \perp## xz for the situation as drwn in the video. In general:

Point P goes around the ring. At all such P $$ \vec {dl}\perp \vec R \quad \& \quad \hat x \perp \vec {dl} \quad \Rightarrow \quad \vec {dl}\perp (\vec x - \vec R) = \vec r $$

Hesch: jeg forstar lidt Dansk

Point P goes around the ring. At all such P $$ \vec {dl}\perp \vec R \quad \& \quad \hat x \perp \vec {dl} \quad \Rightarrow \quad \vec {dl}\perp (\vec x - \vec R) = \vec r $$

Hesch: jeg forstar lidt Dansk

- #10

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I lost you at dl perpendicular to x - R, why did you do that? And why is it equal to r?No, not just because yz ##\perp## xz. Because ##\hat y \perp## xz for the situation as drwn in the video. In general:

Point P goes around the ring. At all such P $$ \vec {dl}\perp \vec R \quad \& \quad \hat x \perp \vec {dl} \quad \Rightarrow \quad \vec {dl}\perp (\vec x - \vec R) = \vec r $$

View attachment 100340

Hesch: jeg forstar lidt Dansk

- #11

BvU

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dl is perpendicular to ##\vec R##

dl is perpendicular to ##\hat x## so dl is perpendicular to ##\vec x##

that means dl is perpendicular to the plane in which both ##\vec R## and ##\vec x## lie.

That plane is described by ##a\vec x + b \vec R##.

The vector ##\vec r## is ##\vec R -\vec x## so it is in that plane.

Why is ##\vec r=\vec x -\vec R## ?

i don't know how to say that. Perhaps a picture helps ?

dl is perpendicular to ##\hat x## so dl is perpendicular to ##\vec x##

that means dl is perpendicular to the plane in which both ##\vec R## and ##\vec x## lie.

That plane is described by ##a\vec x + b \vec R##.

The vector ##\vec r## is ##\vec R -\vec x## so it is in that plane.

Why is ##\vec r=\vec x -\vec R## ?

i don't know how to say that. Perhaps a picture helps ?

- #12

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Okay, I see where x - R comes from.dl is perpendicular to ##\vec R##

dl is perpendicular to ##\hat x## so dl is perpendicular to ##\vec x##

that means dl is perpendicular to the plane in which both ##\vec R## and ##\vec x## lie.

That plane is described by ##a\vec x + b \vec R##.

The vector ##\vec r## is ##\vec R -\vec x## so it is in that plane.

Why is ##\vec r=\vec x -\vec R## ?

i don't know how to say that. Perhaps a picture helps ?

View attachment 100362

So this is how I'm comprehending it:

dl is pointing outwards (imagine a wire pointing out), so any r from its surface is perpendicular to it.

Before, I was confusing dl with the z vector. The angle that r makes with z will not be perpendicular.

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