If IAssume we have a set of $N$ coils stacked vertically on the surface of the Earth, up to a height of $h$, and Ithat we drop a magnet from the top of the coils, and let the magnet fallwhich then falls due to gravity through the coils, to the ground, how much energy is produced by the induced current?
. The induced EMF induced by dropping the magnet through the coils, which has units of volts, is given by Faraday's equation:
$$V = -N \frac{\phi}{\Delta t}.$$$$V = -N \frac{\delta \phi}{\delta t}.$$
WeRather than calculate the derivative itself, we can approximate thisthe 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 longtall 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, and the change in flux $\phi$ would be the flux induced by the magnet through a single coil (we are assuming. If we assume that the flux of each coil begins at 0).
Note that this assumption actually causes us to understate the induced voltage, since we are dividing bythen the total time of the magnet's fall, as opposed tochange in flux $\Delta \phi$ would be the time it takes fortotal flux induced by the magnet to cross a single coil. Also noteNote 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 / secondseconds), also referred to as power, 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,
$$W = \frac{V^2}{\Omega} = \frac{N^2\phi^2}{h\Delta t^2}.$$$$W = \frac{V^2}{\Omega} = \frac{N^2 \Delta \phi^2}{h\Delta t^2}.$$
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,
$$E = Wt = \frac{N^2\phi^2}{h\Delta t}.$$$$E = Wt = \frac{N^2 \Delta \phi^2}{h\Delta t}.$$
The total amount of energy required to lift the magnet to a heighheight 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 nearNear Earth
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, meaning its acceleration will be less than g. As such, let's assume that $\Delta t = \sqrt{\frac{2h}{g - f}}$, for some value f that accounts for the interactions between the magnet and the coils.
Note that we can always adjust the radius of the coils to achieve any value of f we like. For example, consider the extreme case where you're standing in the center of a football field, and the coils are wrapped around the field goal posts, forming a perimeter to the football field. In this case, the coils have a diameter of roughly 100 meters. Now imagine dropping a tiny, ordinary commercial magnet in the middle of through the football field. The rate of acceleration in that case will be indistinguishable from g, and thus $f$ will be basically 0column. In this case, the magnet is small, and the coils are huge, and as such,Assuming the magnet is just another falling object, and the magnetic field of the magnet has no real impact on its fall.
Now imagine that we gradually shrink the radius of the coils. For any given magnet, at some pointsufficiently strong, the magnetic fieldit will matter, producing a non-zero value forquickly reach some terminal velocity $\phi$, but not enough to completely eliminate the acceleration due to gravity$v$ during its fall. That is, for any given magnetTherefore, we can always choose a coil radius that ensures that the force of gravity is always greater than the force of interaction between the magnet and the coils. This ensures that the magnet accelerates the whole way down, but at some rate of acceleration less than $g$, which we representapproximate its fall time as $g - f$$\Delta t = \frac{h}{v}$.
Putting these equations together, we can solve for the value of $h$ that satisfies the following:
$$E = Fh => \frac{N^2\phi^2}{h\Delta t} = mgh.$$$$E = Fh \Rightarrow \frac{N^2 \Delta \phi^2}{h\Delta t} = mgh.$$
$$h = \bigg(\frac{(Nϕ)^2 \sqrt{g−f}}{mg\sqrt{2}}\bigg)^{\frac{2}{5}}.$$$$h = \sqrt[3]{\frac{N^2 \Delta \phi^2 v}{mg}}.$$
Putting all of this together, it follows that for any magnet, we can find a height from which we can drop thatthe 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.
Note that it is obviouslyAs a practical matter, not all of the caseenergy 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 necessarily reach some terminal velocity, givenbe lost in that case to inefficiency, heat, friction in the radiusmotor, etc. As a result, only some percentage of the coils is unboundedenergy from the current can be converted back into mechanical energy to lift the magnet back up to its original height of (remember$h$. Since $E$ represents the football field example)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 sucha result, any answerif we want the current generated by the falling magnet to this question inpower the negativemotor 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 showdraw some of the energy from the current for other purposes. Assume that it$\alpha E$ is not possiblethe amount of energy we want to have non-zero valuesdraw from the current, for both ϕ and g−fsome $\alpha< 1$. It follows that the amount of energy available to power the motor that lifts the magnet is $E - \alpha E = E(1 - \alpha)$. Since we can adjustthat remaining energy will be converted into mechanical energy, it will still need to be adjusted by $\epsilon$ to reflect the radiusinefficiencies of the coilsconversion from electrical energy to mechanical energy. Therefore, common sense suggestsit 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 actually could make this work.find,
Note$$h = \sqrt[3]{\epsilon (1- \alpha) \frac{N^2 \Delta \phi^2 v}{mg}}.$$
The equations above imply that if this is actuallysuch 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 possibleno 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, thenwhere 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 would be an exampleright? While I don't think it violates conservation of energy, the theoretical arguments above suggest the possibility of free energy, which seems a system that could power its own motionrather unbelievable conclusion.
