<p>Due to the fact that electrons can escape the gravitational field of the sun more easily than ions, the sun is positively charged by an amount <span class="math-container">$q=77$</span> Coulombs (see <a href="https://www.aanda.org/articles/aa/abs/2001/24/aah2649/aah2649.html" rel="nofollow noreferrer">https://www.aanda.org/articles/aa/abs/2001/24/aah2649/aah2649.html</a> ). You can calculate the Lorentz force from this. The magnetic field of the Milky Way is however even much smaller than 1/1000 of the earth's (not sure where you go this figure from). According to this reference <a href="http://www.scholarpedia.org/article/Galactic_magnetic_fields#Magnetic_Fields_in_the_Milky_Way" rel="nofollow noreferrer">http://www.scholarpedia.org/article/Galactic_magnetic_fields#Magnetic_Fields_in_the_Milky_Way</a> , the galactic magnetic field near the sun is about <span class="math-container">$6\mu G =6\times 10^{-6}G$</span>. With this the Lorentz force on the sun comes out as about <span class="math-container">$110 N$</span>. For comparison, the gravitational force on the sun by the galaxy amounts to about <span class="math-container">$4\times10^{20} N$</span> (see <a href="https://astronomy.stackexchange.com/questions/18866/how-strong-is-the-force-between-the-sun-and-the-centre-of-the-milky-way">How strong is the force between the Sun and the centre of the Milky Way?</a> ).</p>
<p>The fact that there are electric currents in the sun doesn't make any difference here. It is only a non-zero net charge that can create a Lorentz force on the sun as a whole.</p>

CC BY-SA 4.0
2021-07-05T00:35:23.813Z
Oh *this is great!* You are most welcome to post an answer to [What are the experimental limits to the residual charge of the Sun?](https://astronomy.stackexchange.com/q/23319/7982) along the lines "It is not likely to be measurable since predictions put it at only..." Quote a few lines from the paper, such as the paragraph beginning "We can demonstrate that the existence of the global charge is necessary to avoid some serious physical problems." etc.
uhoh
CC BY-SA 4.0
2021-07-10T22:13:10.330Z
@uhoh I added an answer to that other question, using my own arguments though (the one in the referenced paper, although giving the correct result, contains one or two slightly hand-waving arguments in the derivation)
Thomas
CC BY-SA 4.0
2021-07-10T23:16:14.510Z
I see that now and I love it; thank you very much!
uhoh
CC BY-SA 4.0
2021-07-03T11:45:33.713Z
2021-07-04T09:04:48.083Z
2021-07-04T09:04:48.083Z
Thomas
<p>The <a href="https://academic.oup.com/mnras/article-abstract/484/3/3646/5301409?redirectedFrom=fulltext" rel="nofollow noreferrer">LOFAR</a> observations have made an estimate of the magnetic field of the Milky Way. The value in our region is very small. About 1/1000 of the Earth magnetic field.</p>
<p>The Earth and the Sun move at a cosiderable speed around the center of the Milky Way though. Both move with a speed of around 828 000 km/hr around the center of the Milky Way.</p>
<p>Now currents moving in a magnetic field experience a Lorentz force:</p>
<p><span class="math-container">$$F_{l}=-qv\times B$$</span></p>
<p>Where <span class="math-container">$q$</span> is the charge, <span class="math-container">$v$</span> the speed of the charge, and <span class="math-container">$B$</span> the magnetic field the charges move in. The force is always perpendicular to the velocity.</p>
<p>Now on the Sun, as well as inside the Earth, there is a net current giving rise to their magnetic fields. This means there is a net current in the direction of motion around our galaxy.</p>
<p>How big would be the force on the Sun moving through the magnetic field of the Milky Way?</p>

CC BY-SA 4.0
2021-07-02T23:07:37.673Z
"This means there is a net current in the direction of motion around our galaxy." I don't follow this logic.
ProfRob
CC BY-SA 4.0
2021-07-16T20:11:06.983Z
Crossposted to https://physics.stackexchange.com/q/651722/2451
Qmechanic
<the-sun><earth><milky-way><magnetic-field>
How big is the Lorentz force on the Sun?
Deschele Schilder
# How big is the Lorentz force on the Sun?

(Asked by: username_0 on 2021-07-02)

The [LOFAR](https://academic.oup.com/mnras/article-abstract/484/3/3646/5301409?redirectedFrom=fulltext) observations have made an estimate of the magnetic field of the Milky Way. The value in our region is very small. About 1/1000 of the Earth magnetic field.

The Earth and the Sun move at a cosiderable speed around the center of the Milky Way though. Both move with a speed of around 828 000 km/hr around the center of the Milky Way.

Now currents moving in a magnetic field experience a Lorentz force:

$$F\_{l}=-qv\times B$$

Where $q$ is the charge, $v$ the speed of the charge, and $B$ the magnetic field the charges move in. The force is always perpendicular to the velocity.

Now on the Sun, as well as inside the Earth, there is a net current giving rise to their magnetic fields. This means there is a net current in the direction of motion around our galaxy.

How big would be the force on the Sun moving through the magnetic field of the Milky Way?

**Tags:** the-sun, earth, milky-way, magnetic-field

> **Comments:**
> **username_1**: "This means there is a net current in the direction of motion around our galaxy." I don't follow this logic.  
> **username_2**: Crossposted to https://physics.stackexchange.com/q/651722/2451  

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## 1 answers

### Answer by username_3 on 2021-07-03. Score: 4

Due to the fact that electrons can escape the gravitational field of the sun more easily than ions, the sun is positively charged by an amount $q=77$ Coulombs (see <https://www.aanda.org/articles/aa/abs/2001/24/aah2649/aah2649.html> ). You can calculate the Lorentz force from this. The magnetic field of the Milky Way is however even much smaller than 1/1000 of the earth's (not sure where you go this figure from). According to this reference <http://www.scholarpedia.org/article/Galactic_magnetic_fields#Magnetic_Fields_in_the_Milky_Way> , the galactic magnetic field near the sun is about $6\mu G =6\times 10^{-6}G$. With this the Lorentz force on the sun comes out as about $110 N$. For comparison, the gravitational force on the sun by the galaxy amounts to about $4\times10^{20} N$ (see [How strong is the force between the Sun and the centre of the Milky Way?](https://astronomy.stackexchange.com/questions/18866/how-strong-is-the-force-between-the-sun-and-the-centre-of-the-milky-way) ).

The fact that there are electric currents in the sun doesn't make any difference here. It is only a non-zero net charge that can create a Lorentz force on the sun as a whole.

> **Comments:**
> **username_4**: Oh *this is great!* You are most welcome to post an answer to [What are the experimental limits to the residual charge of the Sun?](https://astronomy.stackexchange.com/q/23319/7982) along the lines "It is not likely to be measurable since predictions put it at only..." Quote a few lines from the paper, such as the paragraph beginning "We can demonstrate that the existence of the global charge is necessary to avoid some serious physical problems." etc.  
> **username_3**: @username_4 I added an answer to that other question, using my own arguments though (the one in the referenced paper, although giving the correct result, contains one or two slightly hand-waving arguments in the derivation)  
> **username_4**: I see that now and I love it; thank you very much!  

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