What is the magnetic dipole moment of a rotating charged record?

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SUMMARY

The magnetic dipole moment of a rotating charged phonograph record can be calculated using the equation \(\vec{m} = \frac{1}{2} \int_{S} \vec{r}' \times \vec{K}(\vec{r'}) da'\). The surface charge density is defined as \(K = \sigma \omega R\), where \(\sigma\) is the uniform surface charge, \(\omega\) is the angular velocity, and \(R\) is the radius of the record. The current \(I\) can be derived from the surface charge density by integrating over the circumference of the record. The solution confirms that the approach taken is correct, leading to the determination of the magnetic dipole moment.

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Homework Statement


A phonograph record of radius R carrying a unifrom surface charge sigma is rotating at a cosntant angular velocity omega. Find its magnetic dipole moment

Homework Equations


[tex]m = int I \bullet da[/tex]

The Attempt at a Solution


Need to find the current first

well the sruface charge density is
[tex]K = \sigma \omega R[/tex]

[tex]K = \frac{dI}{dl_{perp}}[/tex]

where dl perp is the infinitesimal widthruning parallel to the flow of current
so here dl perp refers to circumference
[tex]I = \int K \bullet 2 \pi r dr[/tex]??

is this even close to being right??
I fear i may have made some geomterical error..

thanks for your help!
 
Last edited:
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ahh i got the answer using the equation

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

[tex]da = r dr d\theta[/tex]

thanks anyway!
 

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