- #1

Mr_Allod

- 42

- 16

- Homework Statement
- A steady current I flows down a long cylindrical wire of radius a. Assume that the wire has magnetic susceptibility, ##\chi_m##, and that the current is distributed in such a way that ##J## is proportional to ##s##, the distance from the wire axis. Determine the magnetic field B both inside and outside the wire. Calculate the bound bulk current density ##J_b## in the wire and bound surface current density ##K_b## on the surface of the wire, and show that the total bound bulk current is equal and opposite to the total bound surface current.

- Relevant Equations
- ##\vec H = \frac 1 \mu_0 \vec B - \vec M ##

##\vec M = \chi_m \vec H##

##\vec J_b = \vec \nabla \times \vec M##

##\vec K_b = \vec M \times \hat s##

Hi there, I've worked through most of this question but I'm stuck on the final part, showing that total bulk current ##I_B## is equal and opposite to total surface current ##I_S##. I calculated ##\vec H## the normal way I would if I was looking for ##\vec B## in an infinitely long cylindrical wire where ##\vec J## is proportional to ##s## and found:

##

\vec H =

\begin{cases}

I \frac {s^2} {2\pi a^3} & \text{inside} \\

I\frac 1 {2\pi s} & \text{outside }

\end{cases}

##

Then rearranging the relationships ##\vec H = \frac 1 \mu_0 \vec B - \vec M ## and ##\vec M = \chi_m \vec H## I got:

##\vec B = \mu_0 \vec H (1+ \chi_m)##

##\mu = (1 + \chi_m)##

Using this I found ##\vec B## inside and outside the wire to be:

##

\vec B =

\begin{cases}

\mu I \frac {s^2} {2\pi a^3} & \text{inside} \\

\mu_0 I\frac 1 {2\pi s} & \text{outside }

\end{cases}

##

Then using the two cross-products I found ##\vec J_b## and ##\vec K_b##:

##\vec J_b = \chi_m I \frac {3s} {2\pi a^3} \hat z##

##\vec K_b = -\chi_m I \frac 1 {2\pi a} \hat z##

Now I am unsure of the right way to find the total bulk current ##I_B## and surface current ##I_S##. I'd appreciate it if someone could point me in the right direction.

##

\vec H =

\begin{cases}

I \frac {s^2} {2\pi a^3} & \text{inside} \\

I\frac 1 {2\pi s} & \text{outside }

\end{cases}

##

Then rearranging the relationships ##\vec H = \frac 1 \mu_0 \vec B - \vec M ## and ##\vec M = \chi_m \vec H## I got:

##\vec B = \mu_0 \vec H (1+ \chi_m)##

##\mu = (1 + \chi_m)##

Using this I found ##\vec B## inside and outside the wire to be:

##

\vec B =

\begin{cases}

\mu I \frac {s^2} {2\pi a^3} & \text{inside} \\

\mu_0 I\frac 1 {2\pi s} & \text{outside }

\end{cases}

##

Then using the two cross-products I found ##\vec J_b## and ##\vec K_b##:

##\vec J_b = \chi_m I \frac {3s} {2\pi a^3} \hat z##

##\vec K_b = -\chi_m I \frac 1 {2\pi a} \hat z##

Now I am unsure of the right way to find the total bulk current ##I_B## and surface current ##I_S##. I'd appreciate it if someone could point me in the right direction.