0

Inside the Sun there has to be a net current giving rise to the Sun's magnetic field. Because the Sun orbits the core of the Milky Way with considerable speed (about 0.1% the speed of light) the charges in the Sun's electric current will experience a Lorenz force. On top of that has the Sun a small residual charge (lost electrons). Currents and charges experience a force in the magnetic field of the Milky Way. The current in the direction of motion will be greater than the current in the opposite direction giving a net force.

How big will the net Lorenz force due to the currents and net charge be?

1 Answers1

3

The effects of the galactic magnetic field on the net charge of the Sun aren't too hard to calculate:

  • The magnetic field for a spiral galaxy is about 1 nT.
  • The net charge of the Sun is on the order of 80 C.
  • The speed of the sun is about 0.1% of $c$, which is about $3 \times 10^5$ m/s.

Plugging these in to the standard $F = qvB$ equation, we get a total force of about 0.024 N on the Sun. For comparison, an Earth-like planet in the Andromeda Galaxy would exert a force of about 1.4 N on the Sun, or about 60 times larger than this effect.

The effects of the currents within the sun are probably well-approximated by treating it as a dipole. The magnetic field of a dipole at a distance $R$ is on the order of $B_\text{dip} = \mu_0 M/R^3$, while the magnetic force on a dipole due to an external field is on the order of $F = (\vec{M} \cdot \vec{\nabla}) \vec{B}_\text{ext} \approx M B_\text{ext}/\ell$, where $\ell$ is the typical length scale of the variation of the external field. Combining all of these, we have $$ F \approx \frac{B_\text{ext}}{\ell} \frac{B_\text{dip} R^3}{\mu_0} = \frac{R}{\ell} \left( \frac{B_\text{ext} B_\text{dip} R^2}{\mu_0} \right) $$ We can get the order of magnitude by noting that the Sun's magnetic field at the edge of its photosphere is about 400 µT (from Wikipedia); and plugging in the numbers, with $R$ as the solar radius, the above equation becomes $$ F \approx \frac{R}{\ell} \cdot (1.5 \times 10^{11} \text{ N}). $$ The question is then how large $\ell$ is, i.e., on what length scales the galactic magnetic field varies. If $\ell$ is itself about the radius of the galaxy, then $R/\ell \approx 1.5 \times 10^{-12}$, and you end up with a result that's comparable to the Lorentz force on the Sun's charge. It seems likely, though, that local effects might decrease the scale of variation by a couple of orders of magntiude, leading to a force that would be on the order of a few tens or hundreds of newtons rather than a few tenths of newtons. All in all, though, we should not expect either force to be very large at all.

Michael Seifert
  • 48,772
  • 1.4N seems very smal. That is about 1/100 the force the Earth exerts on me. Im not sure the dipole approximation is right. Do you know the speed of the currents in the sun? –  Jul 16 '21 at 17:58
  • @Vielloosoof: The "1.4 N" number is for gravitational forces between the Sun and a Earth-like planet in Andromeda; the number for the effect of the charge is even smaller. As far as the effects on the currents, remember that there is no net accumulation of charge anywhere in the Sun, which means (for example) the force on any left-moving current somewhere in the Sun is canceled out by the force on a corresponding amount of right-moving current somewhere else. (Unless the magnetic fields at the locations of these currents are different, which is where the scale of the variation comes in.) – Michael Seifert Jul 16 '21 at 18:47
  • I know what you mean by cancellation, and this is in fact exactly my point. Arent the currents in the direction of motion of the sun bigger than in the opposite direction? If the current has the same speed as the suns velocity then the opposite current is zero (the velocity of the currennts is smaller but to illustrate my point). –  Jul 16 '21 at 19:02
  • I cant imagine the forces to be that small. –  Jul 16 '21 at 19:14
  • @Vielloosoof: You have to take into account both kinds of charges. Suppose that it's the electrons that are moving around in big circles and the protons are basically stationary with respect to the center of the sun. (This is false, but it's just an example.) If we look at this in a frame in which the Sun is moving like you're saying, there are 2 units of current from the electrons on one side and 0 units on the other. But there are also -1 units of current from the protons on both sides (since they're moving too). The net effect is still +1 unit on one side and -1 unit on the other side. – Michael Seifert Jul 16 '21 at 19:44
  • I'm fairly confident the net charge of the sun is much smaller than 80 C as well, which would make the Lorentz force here even smaller. – honeste_vivere Aug 27 '21 at 13:39
  • @honeste_vivere: The 80 C estimate comes from Neslušan, "On the global electrostatic charge of stars", A&A 372, 913–915 (2001); it was linked in the other message board thread I linked to. But I will admit that I have only read the abstract of that paper, and for all I know its arguments are spurious. – Michael Seifert Aug 27 '21 at 17:22
  • Yes, I am aware and have actually written an answer (e.g., https://physics.stackexchange.com/a/253491/59023) explaining why that old model is likely to be wrong. There is a small charge imbalance caused by the work done by gravity, but it is typically estimated to be much smaller than these hydrostatic models predict (which I should note also predict no solar wind). – honeste_vivere Aug 27 '21 at 17:40