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For a spherically symmetric charge distribution, $\rho(\vec r)=\rho(r)$, all higher-order multipole moments except the monopole term vanish.

Does a similar thing happen for the multipoles of a spherically symmetric current distribution $\vec{J}(\vec r)=J(r)\hat{r}$ or distributions with some other high degree of symmetry?

I do not think so. Because the monopole moment always vanishes. If all higher-order moments vanish too, then the vector potential (thus, the magnetic field) will be identically zero. But is there any simplification?

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  • $\begingroup$ You'd probably want a cylindrically symmetric current distribution as the lowest order term in a magnetic expansion is usually the dipole term for reasons you've already stated. $\endgroup$
    – Triatticus
    Commented Nov 18, 2021 at 15:37
  • $\begingroup$ @Triatticus In that case, do the moments of order higher than the dipole vanish? $\endgroup$
    – Solidification
    Commented Nov 18, 2021 at 15:53

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The magnetic field for a spherically symmetric current is in fact zero.

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  • $\begingroup$ Welcome to Physics! I think this answer is correct, but you should probably elaborate on why it is true and be more explicit about how it relates to the question. It's a little confusing on its own. $\endgroup$
    – Michael Seifert
    Commented Jul 31, 2022 at 13:38

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