This question is about a system involving a horizontal row of length L of equally spaced pivotable magnets, each with a pole at either end. These magnets will often be referred to as units.

So each unit when rotated causes it's neighbours to rotate in the opposite direction to itself.

When the first unit is quickly rotated 1/4 of a turn and fixed in place rotationally, each will likely turn less than the previous at first, taking longer to complete the full 1/4 of a turn. They will all eventually rotate the full 1/4 as the far end of the line is still not fixed.


Turning the first magnet will require energy. For a line of 1000, while this will be more energy than that for a line of for example 2, it will be less than 1000 times the energy required for a line of 1. This is because:

  1. Momentum means that every unit doesn't need to be rotated the full distance to turn the first unit, it is not as if they are connected together by rods, the first can be turned all the way before the others have had time to rotate much at all.

  2. More distant magnets contribute less force, so most units are contributing very little force to the first unit directly. Another way to think about this is that the line of separated magnets forms a compound magnet that will result in more force than each component magnet has individually, but not as much as 1000 times more, for example if metal becomes attached to an end it will take less than 1000 times the force of an individual magnet to remove.


Coils around each unit slow rotation but produce N electricity after 1/4 of a turn, however long the 1/4 turn takes.

The amount of electricity produced appears to be L * N, for an input power that is increment by a diminishing amount for the growth of L, an example of this sort of growth could be sqrt(L).


Question: This has the potential not to add up energy wise for long lines, what's going on?


$\def\vm{{\bf m}}$If the magnets are free to rotate they will find the lowest energy state. This occurs when all the magnets are aligned head to tail along the line. $$\begin{equation*} \rightarrow\rightarrow\rightarrow\cdots\rightarrow\tag{1} \end{equation*}$$ If now the magnet at one end is rotated a quarter turn and then fixed, it does not follow that all the other magnets will align themselves in an alternating pattern. $$\begin{equation*} \uparrow\downarrow\uparrow\cdots\uparrow\tag{2} \end{equation*}$$ In fact, the magnets will align themselves something like this $$\begin{equation*} \uparrow\searrow\rightarrow\cdots\rightarrow\tag{3} \end{equation*}$$ since the potential energy of this configuration is lower than that of (2).

Some details

The potential energy of the magnetic dipole-dipole interaction between nearest neighbors $i$ and $i+1$ is of the form $$H_{i,i+1} = -J(3(\vm_i\cdot \hat x)(\vm_{i+1}\cdot \hat x)-\vm_i\cdot \vm_{i+1})$$ where we take $\hat x$ to be the unit vector pointing from dipole $i$ to $i+1$. Letting $\vm_i=(\cos\theta_i,\sin\theta_i)$ we find $$H_{i,i+1}=-\frac{J}{2}(3\cos(\theta_i+\theta_{i+1}) + \cos(\theta_i-\theta_{i+1})).$$

The potential between nearest neighbors is minimized when $\theta_i = \theta_{i+1}=0$ or $\pi$, in which case we find $H_{i,i+1}=-2J$. Thus, the energy of configuration (1) is roughly $$H = -2Jn$$ where $n+1$ is the number of magnets. (Here we ignore longer range effects.)

On the other hand, if $\theta_i = -\theta_{i+1} = \pm\pi/2$ we have $H_{i,i+1}=-J$. Thus, the energy of configuration (2) is roughly $$H = -Jn.$$

Notice that the configuration $$\begin{equation*} \uparrow\rightarrow\rightarrow\cdots\rightarrow\tag{4} \end{equation*}$$ has potential $$H = -2J(n-1)$$ since $H_{1,2}=0$. For large $n$ this is vastly less than the energy of configuration (2).

Some results for particular values of $n$

For two magnets the configuration will be as claimed.

$\hskip2.8in$enter image description here

For three magnets the energy can be minimized algebraically. Already we see that the magnets like to be aligned. We find $$\begin{eqnarray*} \theta_1 &=& \pi/2 = 90^{\circ} \\ \theta_2 &=& -\cos^{-1}\sqrt{2/3} \approx -35^\circ \\ \theta_3 &=& \cos^{-1}2\sqrt{2}/3 \approx 19^\circ. \end{eqnarray*}$$

$\hskip2.6in$enter image description here

For $50$ magnets the energy can be minimized numerically. We find, to a tenth of a degree, $$\begin{eqnarray*} \theta_1 &=& 90^{\circ} \\ \theta_2 &=& -30.0^\circ \\ \theta_3 &=& 8.2^\circ \\ \theta_4 &=& -2.2^\circ \\ \theta_5 &=& 0.6^\circ \\ \theta_6 &=& -0.2^\circ \\ \theta_7 &=& 0.0^\circ \\ &\vdots& \\ \theta_{50} &=& 0.0^\circ. \end{eqnarray*}$$

$\hskip1.5in$enter image description here

  • $\begingroup$ I'm not sure that ↑↘→⋯→ is stable, although that is as I imagined for after a short amount of time. For example ↑↘ will become ↑↓ when left. I'm not certain, but wouldn't ↑↘→ not be any different, it will take longer but end up ↑↓↑ as the force drops of more rapidly with greater distance, nearly all of the influence of each magnet being only it's neighbours. The same occurring with a longer line but taking more and more time. $\endgroup$
    – alan2here
    Jun 11 '13 at 13:00
  • $\begingroup$ @alan2here: You're welcome. The system will find its lowest energy state. The energy of state (3) is less than state (4) which is vastly less than state (2). If you have some compasses you can easily convince yourself that the system will not look like state (2) after the twist. When thinking of neighbors make sure to account for the neighbor to the left and right. $\endgroup$
    – user26872
    Jun 11 '13 at 17:44
  • $\begingroup$ Cool, I'm not sure, remember that only one end is fixed in place, but I can imagine this could be the case. I may award you the correct answer despite my doubt as I can't test this and your maths is much better than my own. I hope to wait for a reply to my comment on the other answer first. $\endgroup$
    – alan2here
    Jun 12 '13 at 14:10
  • $\begingroup$ @alan2here: Yes, the left end is fixed at $90^\circ$ as I've indicated above. Since physics arguments don't make an impression, I strongly advise doing this simple experiment. This will not only convince you, but it will also improve your physical intuition. $\endgroup$
    – user26872
    Jun 12 '13 at 18:37
  • $\begingroup$ TY for all your input :-) Since your last edit I now understand, with slightly longer range influence still having enough importance to cause my intuition to be wrong. I've marked your answer as correct and you should get the bounty. $\endgroup$
    – alan2here
    Jun 13 '13 at 11:38

You don't get the same amount of work, N, from each coil. Units further away give you less work. Therefore, total work recovered by the coils is not L*N. (by the way L and N have very specific meanings in electromagnetism so you should avoid using those variable names)

In fact, total energy of the system is preserved at all time because the potential energy is temporarily transformed into kinetic while the far away magnets are taking their sweet time to move.

To convince you that N is not constant, think of an alternator/generator and remember that work is the time integral of power.

For a given torque, trq, @ 0 rpm the alternator gives 0 power, therefore it preforms 0 work after time t. For the same given torque, trq, @ rpm > 0 the alternator has >0 power, therefore it preforms nonzero work after time t.

Therefore, the alternator's power depends on the speed of rotation. As the does the work you'll recover from your magnetic units. Those slowly moving units far from the initial magnet give you less work than that closest to it. (this is not the most elegant way to explain it, but perhaps the most intuitive)

  • $\begingroup$ Thank you for the reply. I wasn't able to follow all of it. Does this mean that a magnet that makes 1/4 of a turn inside a coil slowly produces less electrical power than one that turns in the same way but more quickly? $\endgroup$
    – alan2here
    Jun 11 '13 at 12:45
  • $\begingroup$ Therefore the coils further along the line have produced less power after all the movement has finished? $\endgroup$
    – alan2here
    Jun 11 '13 at 13:26
  • $\begingroup$ This now seems insignificant compared to Oen's reason. $\endgroup$
    – alan2here
    Jun 13 '13 at 11:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.