# Cylinder carrying volume and surface current, H field

1. Nov 15, 2016

### cacofolius

1. The problem statement, all variables and given/known data
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)

2. Relevant equations
Modified Ampére's Law $\oint_C \vec{H} . \vec{dl}=I_{enc}$
$\vec{H}=\frac{\vec{B}}{\mu_0}-\vec{M}$
3. 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 ?

2. Nov 15, 2016

### kuruman

It appears that your teacher made a mistake and wrote down an expression that is dimensionally incorrect. The dimensions of H, as seen in Ampere's law, are current per length. Your teacher's expression has units of just current.

3. Nov 15, 2016