Recent content by wellmax

I see you are right and my professor didn't even notice :biggrin:

Then it will point in the +x and -y direction because there is a minus sign in front of the entire vector H = \frac{-I}{2\pi\sqrt{x^2+y^2}}[\frac{y}{\sqrt{x^2+y^2}}ux + \frac{-x}{\sqrt{x^2+y^2}}uy]

Well I plotted it in Mathematica and I rotates clockwise around the z-axis as expected so I still don't see the error

I don't see this error and my professor didn't see it as an error either when i first handed it in so could you please elaborate a bit more?

Yes indeed a typo :redface: forgot the unit vectors and H points in the minus \varphi direction so i think it's alright

Thank you very much, this solves my question

Homework Statement I have a vector field (which happens to be a magnetic field) H = -\frac{I }{2 \pi r}u\varphi u\varphi is the unit vector which is in the cylindrical coordinate system with only the \varphi component nonzero so it circles around the z-axis. r is the radius of the circle...

Thank you very much, this cleared up a lot of my misconceptions :approve:

Alright, I just considered r and r' to be vectors pointing from the two origins to a certain point in the H-field (inside the hole). So r points from the middle of the wire to a certain point inside the hole and r' points from the center of the hole to the same certain point inside the hole. Now...

Yes I see I can just use Ampères law again Now I still don't get what I'm doing wrong with the problem about the hole in the wire

Yes but I only had to find the H-field inside the hole so I thought this was sufficient? Now I have the same question but with two coaxial cylindrical shells with equal current and thus different current density. I thought I would be able to solve this one after solving the one with the hole in...

Ok but how do I know the direction of the total H-field? I can't use cylindrical coordinates because I would have two bases. So I take the cross-product of the current density and the position-vector of the point I want to know the H-field of (with respect to the center of the hole or wire) and...

I think that is exactly what I did. I just subtracted the area of the hole from the area of the wire and did the entire equation using this current density. The thing I was not sure about was the cross-product I used to turn the scalar into a vector.

Alright I think I got it but not sure if it's correct. I replaced the magnitude of the initial current density by |J| = I/(πh2-πb2) and I got two H-fields Hθ1 = 1/2 r1J -> H1 = 1/2 J [cross] r1 Hθ1 = 1/2 r2J -> H2 = 1/2 J [cross] r2 where r1 and r2 are the distances to the centers of the...

I have one (or maybe two) more question on the same subject and setting so I just post them in this thread. The next exercise goes like this " Subsequently, let us consider a new situation in which there is a cylindrical hole of radius h centred about the line r0 + zuz parallel to the...