# Bound Bulk Current Density and Surface Current Density

• Mr_Allod

#### Mr_Allod

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 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##
I think these are correct.

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.
Consider a cross section of the cylinder (perpendicular to the axis of the cylinder). The total bound bulk current is the total bound current through the cross section. You can calculate this using your expression for ##J_b##.

The total bound surface current is the total bound surface current through the circular boundary of the cross section. You can calculate this using ##K_b##.

Do you mean something like this?

##\int \vec J_b \cdot d\vec a = \chi_m \frac 3 {2\pi a^3} \int_0^a s(2\pi s) ds ##
## \oint \vec K_b \cdot d \vec l = -\chi_m \frac 1 {2\pi a} \int_0^{2\pi} a d\phi##

They give me the expected result but I wasn't sure if I approached it correctly.

Do you mean something like this?

##\int \vec J_b \cdot d\vec a = \chi_m \frac 3 {2\pi a^3} \int_0^a s(2\pi s) ds ##
## \oint \vec K_b \cdot d \vec l = -\chi_m \frac 1 {2\pi a} \int_0^{2\pi} a d\phi##

They give me the expected result but I wasn't sure if I approached it correctly.
Yes, that looks good.

------------------------------

If you want to, you can show the result more generally (without using the explicit expressions for ##\vec J_b## and ##\vec K_b##):

##\int \vec J_b \cdot d\vec a = \int \left( \vec \nabla \times \vec M \right) \cdot d\vec a = \oint \vec M \cdot d\vec {l} ##

where the last equality follows from Stokes' theorem.

For the case of the circular cross section, ##\vec M \cdot d\vec {l} = M_\phi dl = -K_{b, z} dl##, where the last equality follows from ##\vec K_b = \vec M \times \hat s##.