What is the Magnetic Dipole Moment of a Spinning Spherical Shell?

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SUMMARY

The magnetic dipole moment of a spinning spherical shell with radius R and uniform surface charge density σ is calculated using the equation m = (1/2) ∫ (r' × K(r')) da'. The surface current density K is defined as K = σv, where v = ωR, with ω being the angular velocity. The angle between the velocity vector and the radius vector varies, but the azimuthal radius accounts for this variation in the integral. The polar angle ranges from 0 to π, which is crucial for accurate calculations.

PREREQUISITES
  • Understanding of magnetic dipole moment concepts
  • Familiarity with surface charge density and its implications
  • Knowledge of vector calculus, particularly cross products
  • Basic principles of rotational motion and angular velocity
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  • Study the derivation of magnetic dipole moment in rotating systems
  • Learn about the implications of surface charge density in electromagnetic theory
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Physics students, electrical engineers, and researchers in electromagnetism seeking to understand the behavior of magnetic dipole moments in rotating charged systems.

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Homework Statement


Find the magnetic dipole moment of a spherical shell of radiu R carrying a uniform surface charge sigma, set spinning at angular velocity omega.


Homework Equations


\vec{m} = \frac{1}{2} \int_{S} \vec{r'} \times \vec{K} (\vec{r'}) da'

The Attempt at a Solution


So we got to figure out the surface charge density (since it is a spherical shell)

K = \sigma v

and v = \omega times R
this is where i am doubtful...
the angle between v and R varies from 0 to 2 pi

so this cross product is not unique...
or am i thinking about this the wrong way??

please help!
 
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You are right that the angle changes but v will always point in the correct direction. The changing angle (sin(\theta)) accounts for the "azimuthal radius", that is, distance from the spin axis z to the shell measured parallel to the xy plane, that changes with polar angle. That distance is also the r you need to use in your integral.

BTW, polar angle only varies from 0 to \pi
 

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