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So for the story I'm writing, I'm designing is a gas dwarf with a very strong magnetic field and a planetary ring with a slightly inclined orbit. However, the rocks within the ring are very sensitive to magnetic fields. How would this affect the shape of the ring, if it is able to form at all?

They would probably be ferromagnetic, since the rocks need to have a strong, permanent magnetic field. As for more details about the planet, I did make another post about it here; the rings are made up of the same moon that created the planet's "Aether" surface.

Mattias
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    Why are they sensitive to the magnetic field? – dot_Sp0T Feb 13 '17 at 05:29
  • @dot_Sp0T I'm assuming they're metal but that probably won't be a problem – Zxyrra Feb 13 '17 at 06:12
  • The best I can offer is that the particles in the ring would turn to face however the magnetic field is oriented (if the particles are ferric) because that's what happens as lava cools ... but I'm not sure if they'd float toward the planet if they had that much freedom to move. – Zxyrra Feb 13 '17 at 06:14
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    There are many ways that something can be sensitive to magnetic field. It can be a magnet. It can be ferromagnetic material. Or ferrimagnetic. Or something else. What kind is it? – Mołot Feb 13 '17 at 08:00
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    Or, what effect you want it to have? Then we could find a way for it to happen, or at least try. – Mołot Feb 13 '17 at 08:04
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    A "gas dwarf with a very strong magnetic field and a planetary ring". A gas dwarf! Not exactly the most well known sort of planet in my experience. An ingenuous conundrum about a curious planet. – a4android Feb 13 '17 at 11:11
  • Mattias I would probably suggest you go the route that @Mołot mentioned and provide the end state you are looking for, as well as a little more information about the bodies in question. – James Feb 13 '17 at 15:19
  • @Mołot They would probably be ferromagnetic, since the rocks need to have a strong, permanent magnetic field. As for more details about the planet, I did make another post about it here; the rings are made up of the same moon that created the planet's "Aether" surface. – Mattias Feb 13 '17 at 15:38
  • Please, edit clarifications directly into your question. – Mołot Feb 13 '17 at 15:40
  • I suspect that the the axis of the ring will align with the magnetic field. Consider the extreme: A ring at a serious angle to the magnetic field will experience fluctuations in the field and it gets closer and further from the pole. These will set up eddy currents in the particles which in turn will react against the magnetic field. I think that the net result will be to nudge each particle into an orbit of constant magnetic field. The only orbit that is compatible with conventional gravitation orbit is one where the plane of the orbit is at right angles to the magnetic dipole. – Sherwood Botsford Feb 19 '17 at 17:55

2 Answers2

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This has indeed been investigated. I found a thesis by Daniel Jontof-Hutter (2012) that did a stability analysis of particles in rings orbiting, among other bodies, Saturn. One important quantity he studied was the ratio of electric force to gravity, $L_*$. When $L_*\gg1$, the electric force dominated (the "Lorentz regime"), while when $L_*\ll1$, the gravitational force dominated. The exact value of $L_*$ determines whether the rings will be stable or not; the presence of the electromagnetic field can lead to instabilities. There's not an analytical stability condition, unfortunately, so we have to look at numerical simulations to answer your question in full.

Technical details

Jontof-Hutter considered the Lorentz force in the case of a rotating magnetic field: $$\vec{F_B}=\frac{q}{c}\left(\vec{v}-\vec{\Omega_p}\times\vec{r}\right)\times\vec{B}$$ where $q$ is the charge of the particle, $c$ is the speed of light, $\vec{r}$ and $\vec{v}$ are the particle's position and velocity, $\vec{\Omega_p}$ is the planet's spin vector and $\vec{B}$ is the magnetic field. The electric component of $\vec{F_B}$ is $q\vec{E}$, where $\vec{E}$ is the electric field: $$\vec{E}=-\frac{1}{c}\left(\vec{\Omega}\times\vec{r}\right)\times\vec{B}$$ $\vec{B}$ is given, for a dipolar magnetic field, as $$\vec{B}=\frac{-g_{10}R_p^3}{r^3}\vec{z}$$ where $R_p$ is the radius of the planet, $r=|\vec{r}|$, and $g_{10}=|\vec{B}(r=R_p)|$. Therefore1, $$L_*=\frac{qg_{10}R_p^3\Omega_p}{GM_pmc}$$ where the planet's mass is $M_p$ and the particle's mass is $m$.

Let's go to a picture. 16,000 grains were simulated, at many different values of $\vec{r}$ and $L_*$. Here are the results:

Chart of numerical results for particle stability

The darker the dots, the more unstable the configuration. Here's an annotated version of the diagram:

Annotated version of stability chart

$R_{\text{syn}}$ is the radius at which an orbit is synchronous with the planet's rotation. Essentially, at small $L_*$, the rings should be stable at almost all radii. In regimes where $0.5<L_*<10$, there are fewer stable regions, and you can't have large stable rings. This does become less of a problem with large $L_*$, but there are still instabilities, more than for $L_*\ll1$.


1If you want, you can view this as $$L_*\propto\frac{\Omega_p}{Q\bar{\rho}_p}$$ where $\bar{\rho}_p$ is the mean density of the planet, and $Q$ - notation I've absolutely butchered - is the mass-to-charge ratio of the particle, $m/q$.

HDE 226868
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The magnetic field is likely not going to be uniform in the space orbited by the ring particles. This means that any variation of the magnetic field will result in an induced current into the particle.

A flowing current means, by Ohm effect, an increase of the temperature of the particle that, being in very low pressure, will result in an increased evaporation rate.

If the evaporation is "slow" the particle will slowly dissolve in space, if it is "fast" the evaporation will propell the particle somewhere.

In both cases the rings system is not going to last for long.

L.Dutch
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