Surface current of superconducting sphere in magnetic field

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SUMMARY

The discussion focuses on the surface current of a superconducting sphere in a magnetic field, utilizing the continuity of parallel components of the magnetic field and spherical coordinates. The magnetic field inside a superconducting material is zero, leading to the equation ##\mathbf{K} \times \hat{\mathbf{r}} = B / \mu_0 \, \hat{\mathbf{z}}##, which simplifies to ##\mathbf{K} = B / \mu_0 \sin \theta \, \hat{\boldsymbol{\varphi}}##. Additionally, the relationship between magnetization and magnetic field is established with ##H_m=-M_i/(3 \mu_o)## and ##H_o=M_i/(3 \mu_o)##, indicating a factor of 1/3 or 3 due to spherical geometry.

PREREQUISITES
  • Understanding of superconductivity and its properties
  • Familiarity with magnetic field concepts, particularly in superconductors
  • Knowledge of spherical coordinate systems
  • Basic grasp of vector calculus and magnetic field equations
NEXT STEPS
  • Study the implications of the London equations in superconductivity
  • Explore the derivation of magnetic fields in spherical geometries
  • Learn about the behavior of surface currents in superconductors
  • Investigate the relationship between magnetization and magnetic fields in materials
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Physicists, electrical engineers, and researchers in superconductivity and magnetism will benefit from this discussion, particularly those focused on theoretical and applied aspects of magnetic fields in superconducting materials.

Jan05
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Homework Statement
Suppose a sphere of superconductiong material is placed in a uniform magnetic field ##\mathbf{B} = B \, \hat{\mathbf{z}}##. What is the induced surface current distribution?
Relevant Equations
##\mathbf{B}_{above} - \mathbf{B}_{below} = \mu_0 (\mathbf{K} \times \hat{\mathbf{n}} )##
My idea was to use the continuity of parallel components of the magnetic field and the spherical coordinate system. Because the magnetic field in a superconducting material is 0 and the current is completely confined to the surface, there only is a ##\mathbf{B}_{above}## component. The equation then reduces to ##\mathbf{K} \times \hat{\mathbf{r}} = B / \mu_0 \, \hat{\mathbf{z}} ##. Then evaluating components and using ##\hat{\mathbf{z}} = \cos \theta \, \hat{\mathbf{r}} - \sin \theta \,\hat{\boldsymbol{\theta}}## we obtain ##K_\varphi \, \hat{\boldsymbol{\theta}} = B / \mu_0 \sin \theta \, \hat{\boldsymbol{\theta}}##. So ##\mathbf{K} = B / \mu_0 \sin \theta \, \hat{\boldsymbol{\varphi}}##.
 
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