If you were place a strong permanent magnet into the solar wind, like a 1 inch cube neodymium magnet, would the charged particles of the solar wind colliding with the magnet's magnetic field push the magnet out of our solar system, that being if it does not encounter any planets or objects in its path?
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1$\begingroup$ Magnets tend to align themselves with magnetic field lines. More likely the magnet would turn in the direction of whatever magnetic field it crossed. The solar wind is created by magnetic storms but in and of itself, I don't believe the wind is strongly magnetic. It has both positive and negative charged particles, mostly protons, electrons and alpha particles. Think of a compass needle, the needle turns, the needle isn't pushed away from magnetic north. $\endgroup$– userLTKJun 22, 2018 at 6:38
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$\begingroup$ I think if the magnet is placed in line with the Sun's equator then the magnet should align itself in a north-south orientation in relationship to the Sun. This would mean the the magnet's magnetic field lines would be perpendicular to the flow of charged particles flowing out from the Sun. If so, then the charged particles should be deflected around the magnet's magnetic field and this should produce a drag on the magnet and push it away from the Sun. $\endgroup$– user23327Jun 22, 2018 at 13:31
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$\begingroup$ How about we make it easy and magnetize Russell's Teapot? :-) $\endgroup$– Carl WitthoftJun 22, 2018 at 17:15
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$\begingroup$ @Carl Witthoft, ouch that hurts, but its funny at the same time :) $\endgroup$– user23327Jun 22, 2018 at 17:41
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$\begingroup$ @FanofComets - One would be skeptical about the existence of Russell's teapot but the lid of the coroner's jar is a real thing - also an appropriate shape. $\endgroup$– RobJun 22, 2018 at 18:29
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3 Answers
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No. The magnetic field has two poles, the force on the two pole of the magnet is equal and opposite. This is why a compass needle will align in the North South direction, but is not pulled towards the North or the South.
Two magnets will attract or each other because the North pole of one is nearer the South pole of the other. You can get a net force on a magnet when the field-lines are getting bunched together. But the magnetic field in the plasma is not like this. To a first approximation they are parallel to each other and perpendicular to the equator of the sun. In fact the magnetic field is complex as it is carried with the flow of solar wind.
There would be no net force on the magnet, but it may tend to align with the magnetic field. This might cause a turning force, but no acceleration.
There would be some force due to solar wind particles impinging on the object, but since there is only something like 5 atoms per cm3, the force isn't very big, and it has nothing to do with the object being a magnet.
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$\begingroup$ Yesterday I was thinking about the solar wind density of 5 atoms per cm3 at 1 AU and something occurred to me. If the magnet was put in motion, such as going around in a circle over a certain distance, then it would come in contact with many solar wind atoms. For example, if the magnet is placed at the end of a 10 ft pole and the pole is then spun around, it will travel 754 inches/1915 centimeters in one second. 1915 x 5 atoms = 9575 atoms. So, this means that the 9575 atoms in the path of this magnet will be deflected away and thus produce a force on the magnet, pushing it away from the Sun. $\endgroup$– user23327Jun 24, 2018 at 12:21
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1$\begingroup$ Actually the way magnetic fields sweep out material is generally not the Lorentz force anyway, because the Lorentz force cannot impart kinetic energy. It is electromagnetic forces owing to the fact that the fields are moving, and those forces work in the same direction on both signs of charge, which is how the solar wind can push out ions of either charge. For a magnet, what would be most relevant is Faraday's law, the way an EMF is induced when you try to change the magnetic flux. I suspect moving magnetic variations would push out a permanent magnet, as an application of Lens' law. $\endgroup$– Ken GJun 24, 2018 at 14:37
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$\begingroup$ I realized that I miscalculated in my last post. Since the magnet is a 1 inch cube magnet, that means that the surface area of each side of the magnet = 2.54 cm squared = 6.45 sq cm. So, the number of solar atoms that are in the path of the magnet would be 9515 x 6.45 = 61,372. Also, this magnet would be a N52 strength neodymium magnet for maximum effect. $\endgroup$– user23327Jun 24, 2018 at 15:20
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$\begingroup$ I just redid the math and the answer is 61,759. $\endgroup$– user23327Jun 24, 2018 at 15:41
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$\begingroup$ on the other hand, the magnet is composed of about 10000000000000000000000000 atoms. (I may have miscounted the zeros) . You are trying to move a planet with a peashooter. It doesn't really matter if the planet has a magnetic field, or if the peas are electrically charged. $\endgroup$– James KJun 25, 2018 at 6:46
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If you were place a strong permanent magnet into the solar wind, like a 1 inch cube neodymium magnet, would the charged particles of the solar wind colliding with the magnet's magnetic field push the magnet out of our solar system ...
Technically speaking, yes. Practically speaking, it would take an extremely long time. The solar wind consists of 5 atoms per cubic centimeter per second. Solar sails are usually huge and work by reflection, simply pushing on a tiny object is inefficient, being magnetic helps very little.
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1$\begingroup$ I'm not sure that's true, though I could be wrong myself. The solar wind pressure is very slight and past a certain distance, virtually nil. (exiting the solar system would take a very long time). There's also the Poynting-Roberton effect for an object that size that . 1 inch is a little large for that, so that effect would perhaps be minimal too. planetfacts.org/poynting-robertson-effect $\endgroup$– userLTKJun 22, 2018 at 6:34
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$\begingroup$ At .5 AU, the density of the solar wind is 20 atoms per cubic centimeter per second. So, it the magnet was placed at this distance from the Sun, it would be pushed along with a higher dynamic pressure and its acceleration would be faster. The farther out it goes, the atom density will decrease, but I believe it will maintain the speed it gained between .5 to 1 AU. I'm not sure about the Poynting-Robertson Effect, but I suppose it would have some effect on the magnet's overall trajectory as it travels out of the solar system. $\endgroup$– user23327Jun 22, 2018 at 13:17
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$\begingroup$ As Ken G points out, you need to know whether the magnet started out in an orbit, or stationary and about to fall directly into the sun, etc. $\endgroup$ Jun 22, 2018 at 14:10
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$\begingroup$ My idea is that it would begin its journey as a stationary object placed at a distance of .5 AU. I agree with Ken G that the Sun's gravity would pull the magnet in despite moving against the solar wind. $\endgroup$– user23327Jun 22, 2018 at 15:47
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$\begingroup$ @FanofComets - A metal cup or parabola would likely work better but a 1" cube neodymium magnet is only $10. I thought your question presumed that the journey would start sufficiently far away from the sun because you wrote "... that being if it does not encounter any planets or objects in its path?" - the sun behind it, and sufficiently close, puts the sun's gravitational field in the cube's path. I didn't assume things you didn't write and answered your question as-is - a solar sail is thought to be practical. $\endgroup$– RobJun 22, 2018 at 16:38
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The Sun's gravity would pull in any large object not already in orbit, that would be dominant over magnetic forces from the very weak wind. If the object was already in orbit, the situation would be akin to dust particles leaving a comet-- their path is more affected by light pressure than magnetic pressure, but for a large object, neither of those could rival gravity.
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$\begingroup$ Exactly - the total force diagram is what matters, and the solar wind - be it mag fields or particle momenta impinging on the magnet - are small factors. $\endgroup$ Jun 22, 2018 at 14:09
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$\begingroup$ This is a good point that Ken G brought up and one that I did not consider. I was thinking that the magnet could start off as a stationary object and then the solar wind would begin pushing it away from the Sun, yet its logical that the Sun's gravity would win out and draw the magnet in especially if the stationary magnet is placed at a distance of .5 AU. $\endgroup$– user23327Jun 22, 2018 at 15:43