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I have two powerful rare earth magnets, that are separated by a distance of 1 mm. I applied energy to bring them closer to each other, hence increasing the potential energy. Now, when one of the magnets is released(other is fixed) there is acceleration, with time force = 0 since the gap(D) is increased, yet there is massive kinetic energy. Assuming from position X( where the air gap is 1 mm) we could calculate the repulsive force, knowing m of course.

From the initial acceleration we could calculate the time + velocity, since the distance is known as well. Can the KE of that moving magnet being repelled be calculated?

This magnet is easily moved around (attached with wheels & low friction), and the force is known over certain positions, beyond those points F = 0 since D is very large, and there is no magnetic force being applied yet we know acceleration = 0. But there is kinetic energy? I'm a bit confused as to what might happen. Can anyone explain with great detail of the outcome of such an experiment?

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  • $\begingroup$ A comment on safety: Please be careful when playing around with powerful magnets. If they do not repel straight, one may flip and the attraction may result in collision, with danger of bruises and sharp splinters. $\endgroup$ – mikuszefski Jan 23 '14 at 15:16
  • $\begingroup$ Thanks, but I already learned that... the hard way unfortunately. $\endgroup$ – Pupil Jan 24 '14 at 6:41
  • $\begingroup$ Related: physics.stackexchange.com/q/17309/2451 , physics.stackexchange.com/q/47612/2451 and links therein. $\endgroup$ – Qmechanic Dec 19 '14 at 11:15
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If you bring two magnets together in such a way that they repel each other, you store potential energy $E_\mathrm{pot}$ . This is correct. The force will be the according derivative $F=-\mathrm{d} E_\mathrm{pot}/\mathrm{d}x$, which does not depend on the mass, of course. In detail the stored energy does depend on saturation flux, distance, orientation, and shape of the magnets. The gap, hence, is not enough information to calculate the energy (although there are some extreme cases where this can be done approximately). If one magnet is fixed and the other one can move (without friction), it will accelerate (according to $F=m a$) after release and the potential energy will be transformed into kinetic energy. This is the standard transformation of one energy form into the other. If the distance is very large and there is not interaction energy anymore, then the moving magnet just stays with its kinetic energy. With constant potential (including constant zero) this stays true and there is no acceleration as $\mathrm{d} E_\mathrm{pot}/\mathrm{d}x=0$

By the way, as magnetic forces are long range, it will actually never be zero. If the gap is much larger than the magnets extensions you may approximate the interaction by point dipoles.

Note: If you have rectangular magnets with parallel oriented edges, there are exact formulas to calculate the energy, sophisticated ones, though.

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  • $\begingroup$ mikuszefski: my apologies, I flagged your answer before you edited it (the downvote isn't mine) $\endgroup$ – John Rennie Jan 22 '14 at 9:35
  • $\begingroup$ @JohnRennie no problem. I accidentally hit enter, then edited. Fortunately, one can undo things (impatient downvotes, etc.) $\endgroup$ – mikuszefski Jan 22 '14 at 9:39
  • $\begingroup$ @JohnRennie. Like that you got basically the same answer in integral form. $\endgroup$ – mikuszefski Jan 22 '14 at 9:44
  • $\begingroup$ mikuszefski: yes, I posted my answer before seeing your edit. I think the two answers are complementary so I think we should leave both. $\endgroup$ – John Rennie Jan 22 '14 at 9:47
  • $\begingroup$ @mikuszefski What do you mean that the magnetic forces are long range? There are massive forces between two magnets at distances = 1 mm, then increase it to 30 mm the force drops significantly. $\endgroup$ – Pupil Jan 22 '14 at 16:14
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If you measure the force $F$ as a function of distance $r$ as you bring the moving magnet towards the fixed one, then the work you've put in is simply:

$$ W = \int_{r_1}^{r_2} F(r)dr $$

Where $r_1$ is the initial separation and $r_2$ is the final separation. If you let go of the moving magnet it will end up with a kinetic energy equal to the work put in i.e.

$$ \frac{1}{2} m v^2 = \int_{r_1}^{r_2} F(r)dr $$

and you can use this to calculate the velocity $v$.

At long range you can estimate the function $F(r)$ by taking the force between two magnetic dipoles. However as the separation becomes comparable to the size of your magnets the force will depend on the shape of your magnets and gets hard to calculate from first principles. To get an accurate answer you'll have to measure the force experimentally.

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  • $\begingroup$ It seems to be that the work done is always a small value, due to the immense forces applied over extremely small distances, I mean the force would drop from 1000N to 10N by increasing the air gap by 3 cm. That kinetic energy can't even be properly converted into another form... I deem it as a useless system. $\endgroup$ – Pupil Jan 22 '14 at 16:09
  • $\begingroup$ @Key: I don't understand what you mean by That kinetic energy can't even be properly converted into another form. $\endgroup$ – John Rennie Jan 22 '14 at 16:12
  • $\begingroup$ Due to the small distances, the KE energy gained from the conversion is not useful. The energy due to magnetic dipoles always seems useless due to the large forces applied to extremely small distances. That's what I meant. $\endgroup$ – Pupil Jan 22 '14 at 16:42
  • $\begingroup$ I also don't get the useless part. When calculating the energy for two $2\times2\times 2 \;\mathrm{cm}^3$ NdFeB magnets, the stored energy is enough to accelerate one to above 30 km/h. (starting in contact and not talking about friction and the weight of any vehicle one may need for wheels or similar). In dipolar approximation I still get 10 km/h. So what is meant by useless? $\endgroup$ – mikuszefski Jan 23 '14 at 14:19
  • $\begingroup$ Well sure, I saw powerful Neo. magnets that can generate repulsive forces(facing an equal magnet) of 4000N+(when the gap is 1mm), and it's mass is about 4kg, the acceleration is significantly high, that is strong enough to cause a lot of damage when in motion... I was in confused state when I stated it's useless, it can be used if various interesting things. $\endgroup$ – Pupil Jan 24 '14 at 6:47

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