Angular momentum in ElectroMagnetic fields(Feynman's Disk Paradox)

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In the discussion on angular momentum in electromagnetic fields, a calculation from Griffiths' "Introduction to Electrodynamics" regarding the angular momentum density in a scenario resembling Feynman's disk paradox is examined. The initial conclusion that the angular momentum points solely in the z direction is challenged by the inclusion of an s component when considering the r vector in both s and z directions. It is argued that while the angular momentum density should indeed have an s component for z not equal to zero, the total angular momentum remains solely in the z direction to prevent tipping of the cylinders. The conversation highlights a potential oversight by Griffiths in neglecting the s component. Ultimately, it is noted that the xy component of the total angular momentum cancels out through integration.
Henriamaa
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In Griffiths book, "Introduction to Electrodynamics" example 8.4 he calculates the angular momentum density for a set up that is a version of Feynman disk paradox. His answer for the angular momentum points in the z direction. But if we you assume that the r vector has component in the s direction and z direction(I am almost sure this is correct) \vec{r} = s\hat{s}+ z\hat{z}, then the angular momentum density has both a z component and s component. The s component is not constant. The total angular moment on the other hand has to end up with only z component or the cylinders would tip over. Where is the error in my reasoning?
 
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It seems to me you've made no error. The angular momentum density should in fact have an \hat{s} component for z \neq 0. It seems Griffths neglected this. However, there is no xy component of the total angular momentum of EM field; it cancels out in via integration.
 
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