# Magnetic field in asymmetric hollow cylinder

1. Mar 7, 2012

### bobred

1. The problem statement, all variables and given/known data
Cylinder of radius $a$ and a cylindrical hole $b < a$ is displaced a distance $d$ in x-direction. Current density $\textbf{J}=J_z\textbf{e}_z$. Show that a uniform magnetic field inside the hole is

$\textbf{B}=\frac{\mu_0}{2}J_zd\textbf{e}_y$

2. Relevant equations
Using previous result of whole cylinder

$\textbf{B}=\frac{\mu_0}{2}J_z(-y\textbf{e}_x + x\textbf{e}_y)$

and that the cylindrical hole can be modelled as a charge density of $-J_z\textbf{e}_z$.

3. The attempt at a solution

I tried superposition of fields so

$\textbf{B}_1=\frac{\mu_0}{2}J_z(-y\textbf{e}_x + x\textbf{e}_y)$

$\textbf{B}_2=-\frac{\mu_0}{2}J_z(-y\textbf{e}_x + (x + d)\textbf{e}_y)$

$\textbf{B}= \textbf{B}_1 + \textbf{B}_2$ to which I get

$\textbf{B}=-\frac{\mu_0}{2}J_zd\textbf{e}_y$

Any ideas, is this the right approach?

2. Mar 8, 2012

### bobred

Sorted, silly mistake

$\textbf{B}_2=-\frac{\mu_0}{2}J_z(-y\textbf{e}_x + (x + d)\textbf{e}_y)$

should be

$\textbf{B}_2=-\frac{\mu_0}{2}J_z(-y\textbf{e}_x + (x - d)\textbf{e}_y)$