- #1
cacofolius
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Homework Statement
We have an infinite cylinder that, from radius 0 to a, has a volume current density ##\vec{J(r)}=J_{0}(r/a) \hat{z}## , then from a to 2a, it has a material with uniform linear magnetic permeability ##\mu=(3/2)\mu_0##
, and at the surface, it has surface current ##\vec{K}=-K_{0}\hat{z}##. I have to find i)the H and B fields everywhere, ii)the magnetization, and iii)the magnetizing volume and surface currents. However I just have a question on the first part of i)
Homework Equations
Modified Ampére's Law ##\oint_C \vec{H} . \vec{dl}=I_{enc}##
##\vec{H}=\frac{\vec{B}}{\mu_0}-\vec{M}##
The Attempt at a Solution
Current from 0 to r, with r<a: ##\int_{0}^{r} \int_{0}^{2\pi} J_{0}(r/a) r dθ dr = \frac{2\pi J_0 r^3}{3a}##
Total current from 0 to a: ##\frac{2\pi J_0 a^2}{3}##
H field, using Ampére's Law (cylindrical symmetry):
r<a: ##H(2\pi r)=\frac{2\pi r^3}{3a} \rightarrow H=\frac{J_0 r^2}{3a}##
a<r<2a: ##H(2\pi r)=\frac{2\pi J_0 a^2}{3} \rightarrow H=\frac{J_0 a^2}{3r}##
My doubt is that my teacher just wrote the answers, and I got the same except the last one, where he wrote ##H=\frac {J_0 r^2}{3}##. Wouldn't that imply that the field is increasing with distance squared from the center, however without the radius enclosing anymore current? Where did I go wrong ?