Current density of rotating spherical shell

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SUMMARY

The current density of a rotating spherical shell with radius a and surface charge density sigma is expressed as J_phi = (sigma)*a*Sin(theta)*(omega)*d(r'-a). This formula incorporates the delta function to confine the current density to the surface of the sphere. The velocity of a point on the shell is determined by v = a*Sin(theta)*(omega). The discussion clarifies that the restriction of theta to the domain 0 PREREQUISITES

  • Understanding of spherical coordinates
  • Familiarity with angular velocity concepts
  • Knowledge of surface charge density
  • Basic grasp of delta functions in physics
NEXT STEPS
  • Study the application of delta functions in electromagnetic theory
  • Explore the derivation of current density in rotating systems
  • Learn about the implications of surface charge density on electric fields
  • Investigate the dynamics of rotating rigid bodies in classical mechanics
USEFUL FOR

This discussion is beneficial for physicists, electrical engineers, and students studying electromagnetism or classical mechanics, particularly those interested in the behavior of charged rotating bodies.

icelevistus
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Express the current density of a spherical shell of radius a, rotating with angular velocity omega, with surface charge density sigma



Delta function will be denoted d(x). Spherical coordinates will be used



I have concluded that for a given chunk (if we restrict to the 0<theta<pi/2 domain), the velocity will be given by v=a*Sin(theta)*(omega). It is clear that the current density will only have a phi component. I have concluded:

J_phi = (sigma)*a*Sin(theta)*(omega)*d(r'-a)

Where the delta function is used to restrict the current density to the sphere's surface.

Can anyone confirm that this reasoning is correct?






 
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It looks good, except I see no reason for you to restrict theta to [itex]0<\theta<\frac{\pi}{2}[/itex]...what is wrong with your expression for the lower half of the sphere?
 
gabbagabbahey said:
It looks good, except I see no reason for you to restrict theta to [itex]0<\theta<\frac{\pi}{2}[/itex]...what is wrong with your expression for the lower half of the sphere?

you're right, as I was writing it I thought I was using cosine and it would introduce a sign error. Whole shell it is!
 

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