Assume we have a set of $N$ coils stacked vertically on the surface of the Earth, up to a height of $h$, and that we drop a magnet from the top of the coils, which then falls due to gravity through the coils, to the ground. The EMF induced by dropping the magnet through the coils, which has units of volts, is given by Faraday's equation:
$$V = -N \frac{\delta \phi}{\delta t}.$$
[Faraday's Equation][1]
Rather than calculate the derivative itself, we can approximate the derivative by calculating the total change in flux $\Delta \phi$, and dividing by the total change in time $\Delta t$. If we assume the magnet has a length of $h$ (that is, the magnet is just as tall as the stack of coils), and is dropped vertically through the coils, then $\Delta t$ would in this case be the amount of time the magnet spends falling from the top of the coils to the ground. If we assume that the flux of each coil begins at 0, then the total change in flux $\Delta \phi$ would be the total flux induced by the magnet. Note that because the magnet is the same length as the stack of coils, each coil experiences a change in flux once during the fall of the magnet, and therefore the current has a single direction.
The wattage (Joules / seconds) generated by the induced current is equal to the square of the induced EMF divided by the resistance. We assume a resistance of 1 Ohm per meter of coil height for simplicity, for a total of $\Omega = h$ Ohms of resistance. This gives,
$$W = \frac{V^2}{\Omega} = \frac{N^2 \Delta \phi^2}{h\Delta t^2}.$$
[Watts from Voltage][2]
The amount of time that the current flows is presumably also equal to $\Delta t$. Therefore, the total energy of the induced current (Watts $\times$ time) is given by,
$$E = Wt = \frac{N^2 \Delta \phi^2}{h\Delta t}.$$
The total amount of energy required to lift the magnet to a height of h is approximately equal to the work function $Fh = mgh$, where $g$ is the gravitational acceleration near the surface of the Earth, and m is the mass of the magnet.
[Force of Gravity Near Earth][3]
If the magnet were in a true free fall, then $\Delta t$ would be $\sqrt{\frac{2h}{g}}$. However, this is not the case, because the magnet will interact with the coils as it falls through the column. Assuming the magnet is sufficiently strong, it will quickly reach some terminal velocity $v$ during its fall. Therefore, we can approximate its fall time as $\Delta t = \frac{h}{v}$.
[Equations for Falling Bodies][4]
[Electromagnetic Braking][5]
Putting these equations together, we can solve for the value of $h$ that satisfies the following:
$$E = Fh \Rightarrow \frac{N^2 \Delta \phi^2}{h\Delta t} = mgh.$$
Solving for $h$, we find,
$$h = \sqrt[3]{\frac{N^2 \Delta \phi^2 v}{mg}}.$$
Putting all of this together, it follows that for any magnet, we can find a height from which we can drop the magnet, and generate a current with an amount of electrical energy $E$ that is equal to the amount of mechanical energy $Fh$ we put into lifting the magnet.
As a practical matter, not all of the energy generated by the current will be convertible back into mechanical energy. For example, let's assume we take the current generated by the falling magnet, temporarily store it in a battery, and then feed that current back into a motor that then lifts the magnet back up to its original height of $h$. Some of the energy from the current will be lost in that case to inefficiency, heat, friction in the motor, etc. As a result, only some percentage of the energy from the current can be converted back into mechanical energy to lift the magnet back up to its original height of $h$. Since $E$ represents the energy of the current generated by the falling magnet, let $\epsilon E$ represent the portion of that energy that is ultimately converted back into mechanical energy to lift the magnet back up to its original height of $h$. Note that $\epsilon < 1$.
As a result, if we want the current generated by the falling magnet to power the motor that ultimately lifts the magnet back up to its original height, it has to be the case that,
$$\epsilon E = Fh \Rightarrow \epsilon \frac{N^2 \Delta \phi^2}{h\Delta t} = mgh.$$
Solving for $h$, we find,
$$h = \sqrt[3]{\epsilon \frac{N^2 \Delta \phi^2 v}{mg}}.$$
The equations above imply that this system would constitute an example of a system that needs no external energy to operate, save for the initial lift of the magnet.
Now assume that we want to not only have the motor power itself, but we also want to draw some of the energy from the current for other purposes. Assume that $\alpha E$ is the amount of energy we want to draw from the current, for some $\alpha< 1$. It follows that the amount of energy available to power the motor that lifts the magnet is $E - \alpha E = E(1 - \alpha)$. Since that remaining energy will be converted into mechanical energy, it will still need to be adjusted by $\epsilon$ to reflect the inefficiencies of the conversion from electrical energy to mechanical energy. Therefore, it has to be the case that,
$$\epsilon (1- \alpha) E = Fh \Rightarrow \epsilon (1- \alpha) \frac{N^2 \Delta \phi^2}{h\Delta t} = mgh.$$
Solving for $h$, we find,
$$h = \sqrt[3]{\epsilon (1- \alpha) \frac{N^2 \Delta \phi^2 v}{mg}}.$$
The equations above imply that such a system could generate electricity indefinitely, with no external source of power, save for the initial lift of the magnet. Obviously, there are practical engineering problems that will need to be solved to build such a system, such as lifting the magnet outside of the column of coils so as to ensure there is no drag due to the magnetic field. This particular problem could be solved by having the magnet follow a circular or elliptical path at the end of a rotating arm, where the fall happens through a column of coils on one side of the path, but the lift happens outside of the column on the other side of the path.
How could this be right? While I don't think it violates conservation of energy, the theoretical arguments above suggest the possibility of free energy, which seems a rather unbelievable conclusion.
[1]: https://en.wikipedia.org/wiki/Faraday's_law_of_induction#Quantitative
[2]: https://en.wikipedia.org/wiki/Watt#Examples
[3]: https://en.wikipedia.org/wiki/Gravity_of_Earth#Estimating_g_from_the_law_of_universal_gravitation
[4]: https://en.wikipedia.org/wiki/Equations_for_a_falling_body#Equations%7D%7BEquations%20for%20Falling%20Bodies
[5]: https://arxiv.org/pdf/physics/0603270.pdf%7D%7BElectromagnetic%20Braking
Q&A (4871)
