electromagnetismlevitationmagnetic fieldsmagnetic-momentnewtonian-mechanics
Imagine I use one static magnet to levitate another magnet indefinitely.
This levitating magnet is in a narrow tube that wont allow it to rotate and get attracted. So it will keep levitating forever, right?
Since gravity uses energy to push us down to earth, and this magnet is counteracting gravity, where is this energy to counteract gravity come from?
You do not need to invoke friction. The magnetic forces are in equilibrium by themselves so if you place the magnets in that configuration, they will not spontaneously begin to move. The reason is that there is a corresponding force on the magnets when they are vertical that matches the ones you've already drawn.
Let me make a simple model. First of all, start by upping the game and including two big magnets, which can only make it better:
If red is a north pole, then each rotating magnet, when horizontal, has a north pole repelling its backside north pole and a north pole attracting its front south pole. Focusing on the big magnets for the moment, the fact that their north poles face each other suggests that we can trade them for a pair of anti-Helmholtz coils. This means the important character of their field is its quadrupolar nature, and we can approximate the magnetic field as $$\mathbf B=\frac{B_0}a\begin{pmatrix}x\\y\\-2z\end{pmatrix},$$ where the $z$ axis goes from one big magnet to the other, $a$ is some characteristic length, and $B_0$ is some characteristic field strength.
Now, for the little magnets, I think it is uncontroversial to model them as point dipoles. If $\theta$ is the angle the wheel spoke makes with the $x$ axis (with the wheel in the $x,z$ plane) then each magnet is a dipole with moment $$\mathbf m=m \begin{pmatrix}-\sin(\theta)\\0\\\cos(\theta)\end{pmatrix} \text{ at } \mathbf r=R \begin{pmatrix}\cos(\theta)\\0\\\sin(\theta)\end{pmatrix}.$$ With this, the potential energy of each spoke magnet is $$U=-\mathbf m\cdot\mathbf B=3\frac R a mB_0\sin(\theta)\cos(\theta)=\frac32\frac Ra mB_0\sin(2\theta).$$
To see how this behaves, here is a colour plot of the energy, with negative energy in red and positive energy in blue.
You can see there is a gradient pointing up on the right and down on the left. However, these are matched by clockwise gradients when the spokes are vertical. A single magnet will settle on the lower left or the upper right; a pair of magnets will settle on that diagonal. For a symmetrical wheel with three or more magnets, the total potential energy is flat at zero, $$U=\sum_{k=1}^n\frac32\frac Ra mB_0\sin\left(2\left(\theta_0+\frac{2\pi}{n}k\right)\right) =\frac3{4}\frac Ra mB_0\text{Im}\left[e^{2i\theta}\sum_{k=1}^n(e^{2\pi i/n})^k\right] =\frac3{4}\frac Ra mB_0\text{Im}\left[e^{2i\theta+2\pi i/n}\frac{1-(e^{2\pi i/n})^n}{1-e^{2\pi i/n}}\right] =0 $$ and there is no resultant magnetic force.
With a generator of some sort extracting energy from the system, the system will stop in all three cases. By the principle of conservation of energy, you know that the total energy neither increases or decreases through any process. If the generator extracts energy from the system, the man will slow down and eventually stop. Intuitively, this is because of resistance in the generator. Even in an ideal world without friction, turning a magnet within a coil of wire (in order to generate an EMF in the coil) requires a constant force due to the retarding electromagnetic forces on the generator. (Consider science museum demonstrations in which you turn a wheel to light a light bulb. The wheel can be really tough to turn, and that's primarily because of electromagnetic force, not just friction.)
Without a turbine, the question is marginally more interesting. In any environment with friction, the man will stop - again, since friction extracts energy from the system, it must slow down and eventually come to a halt. Only in the last case - the frictionless environment without a turbine - will the man perpetually remain in motion. Of course, this is a physically implausible situation, since any mechanical system will have at least some amount of friction.
Best Answer
This is incorrect. Gravity does not use energy to pull in us.
If we started falling, then yes, gravity used energy do make us move. But that is only in the special case where gravity makes us move. In general, gravity spends no energy pulling in us.
In general, a force spends no energy. An apple lying on a table is both pulled down by gravity and held up by the table's normal force. Gravity spends no energy here. The normal force doesn't spend any energy either. This situation of apple-lying-still-on-table will stay like that forever. It will never change, since no energy can "run out" when no energy is spent.
The levitating magnet is the same case. No energy spent. Thus, this will theoretically remain forever. (Unless there are other forces acting as well, that do spend energy.)