Magnetostatics problem: find B and H

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SUMMARY

The discussion centers on calculating the magnetic fields \(\mathbf{B}\) and \(\mathbf{H}\) around a long wire carrying current \(I\) within a hollow cylinder of inner radius \(a\) and outer radius \(b\), made of a linear material with permeability \(\mu\). Using Ampere's law, the magnetic field \(\mathbf{H}\) is expressed as \(\mathbf{H} = \frac{I}{2 \pi p} \hat{\phi}\) in cylindrical coordinates. The solutions indicate that for regions \(0 < p < a\) and \(b < p\), \(\mathbf{B} = \mathbf{H}\), while for \(a < p < b\), \(\mathbf{B} = \mu \mathbf{H}\). The discussion clarifies that if the cylinder is made of steel, the high permeability would significantly alter the magnetic field calculations.

PREREQUISITES
  • Understanding of Ampere's Law
  • Familiarity with cylindrical coordinates
  • Knowledge of magnetic permeability (\(\mu\))
  • Basic concepts of magnetic fields (\(\mathbf{B}\) and \(\mathbf{H}\))
NEXT STEPS
  • Study the implications of magnetic permeability in different materials
  • Learn about the differences between SI and cgs units in electromagnetism
  • Explore the effects of conducting materials on magnetic fields
  • Investigate advanced applications of Ampere's Law in complex geometries
USEFUL FOR

Physics students, electrical engineers, and anyone involved in magnetostatics or electromagnetic field theory will benefit from this discussion.

summerwind
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Homework Statement



A long wire carries a current I and is centered in a long hollow cylinder of inner radius a and outer radius b. The cylinder is made of a linear material with permeability [tex]\mu[/tex]. Find [tex]\mathbf{B}[/tex] and [tex]\mathbf{H}[/tex] everywhere.

Homework Equations





The Attempt at a Solution



The only free current in the problem is the current in the wire. Therefore, by Ampere's law,

[tex]\mathbf{H} = \frac{I}{2 \pi\ p}\ \boldsymbol{\hat{\phi}}[/tex]

(I'm using cylindrical coordinates with the wire at p = 0 and the current moving in the +z direction.)

For 0 < p < a and b < p, we have [tex]\mathbf{B}=\mathbf{H}[/tex]. For a < p < b, we have [tex]\mathbf{B} = \mu\ \mathbf{H}[/tex].


If this solution is correct, would anything be changed if I said that the hollow cylinder with permeability [tex]\mu[/tex] was also a conductor (made of steel, say)?
 
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Your solutions look right to me, but you use the cgs system & I use SI so to that extent I don't know, but as I say they all look right.

The fact that it would be a conductor, e.g. Cu, is of no consequence. Cu is non-magnetic.

But - you said steel - steel has a very high permeability so to that extent your numbers would be quite different inside the steel.
 

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