Magnetic field of a cylinder with magnetization M=ks^2

[JFuld]

1. Homework Statement [/b]

this is griffiths 6.8 fyi.

a long cylinder of radius R carries magnetization M=ks^2 $\hat{∅}$, k is a constant and s is the distance from the axis. $\hat{∅}$ is the azimuthal unit vector. find the magnetic field inside and outside the cylinder.

Homework Equations

bound volume current Jb= $\nabla$XM
bound surface current Kb=MX$\hat{n}$

A(r)=μo/4π∫Jb(r')/(script r) dV' + μo/4π∫Kb(r')/(script r) da'

script r = r' - r

The Attempt at a Solution

first thing i did was to find the bound currents.

Jb = 3ks $\hat{z}$,

Kb=ks^2 $\hat{s}$ (on the top surface)

Kb= -ks^2 $\hat{s}$ (on the bottom surface)

Kb = -kR^2 $\hat{z}$ (on the walls of the cylinder)



Now I am stuck. I was thinking I should plug the bound currents into the equation for A(r), and find B by taking the curl of A(r). However, I am confused on what r' or script r would be. If anybody could help me out there I would apreciate it.

I am not sure if amperes law would work here, no single amperian loop would incorporate every bound current so maybe I should break the problem into multiple pieces?

How would you go about this problem? thanks

 

[pabloenigma]

They key point in this problem is "its a long cylinder".So you are not really obliged to worry about the the upper and the lower surfaces.They are far away.

Amperes Law does the trick for the rest of the problem.Take an amperean loop of radius S and find the current enclosed by it by integrating Jb over a circle of radius S centred about the axis.
Mathematically,
I_enc=∫(J_b2πs ds) to be integrated from 0 to S