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

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SUMMARY

The discussion focuses on calculating the magnetic field of a long cylinder with a magnetization defined as M=ks² in the azimuthal direction. The bound volume current is determined to be Jb=3ks in the z-direction, while the bound surface currents are Kb=ks² on the top surface, Kb=-ks² on the bottom surface, and Kb=-kR² in the z-direction on the cylinder walls. The solution involves applying Ampère's Law to find the enclosed current by integrating Jb over a circular path of radius S, simplifying the problem by neglecting the effects of the upper and lower surfaces due to the cylinder's length.

PREREQUISITES
  • Understanding of magnetization concepts in electromagnetism
  • Familiarity with bound currents and their calculations
  • Knowledge of Ampère's Law and its applications
  • Proficiency in vector calculus, particularly curl operations
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  • Study the application of Ampère's Law in cylindrical coordinates
  • Learn about the calculation of magnetic fields from bound currents
  • Explore the implications of magnetization in different geometries
  • Investigate the use of vector calculus in electromagnetic theory
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Students of electromagnetism, physicists working on magnetostatics, and anyone studying the magnetic properties of materials with complex geometries.

JFuld
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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= \nablaXM
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
 
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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,
Ienc=∫(Jb2πs ds) to be integrated from 0 to S
 

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