How much current is induced by a free falling magnet?

Question

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

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

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

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

Electromagnetic Braking

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.

Answer 1

For a stack of coils, as you say, this is actually a famous result, but not quite in the way that you've phrased it.

If you drop a magnet through the air from some height $h$, it'll generally hit the ground with a speed $v$ that obeys $\frac12 mv^2 = mgh$ --- that its, its gravitational potential energy gets converted into kinetic energy. However if you drop a magnet through a conducting pipe, or through a series of conducting rings as you suggest, it'll strike the ground with a much smaller kinetic energy: instead of being converted entirely into kinetic energy of the magnet, some of the magnet's inital gravitational potential energy is converted to electrical energy in the conducting pipe.

Having done this experiment as a classroom demo (and Youtube doesn't really do this justice --- magnets and pipe are cheap at the hardware store, and you should try to see it on your own) it's pretty clear that what happens is that the falling magnet achieves some terminal velocity. We can try to estimate what that is, but we'll have to correct some errors in your treatment (v32).

You seem to be imagining that your stack of coils is connected like a solenoid, so that all of the current can be extracted from the coils as the magnet falls. That only works if the falling magnet is long compared with the solenoid (as you specify). I'll show the result for both long and short falling magnets.

First, there are a couple of issues with your expression for the induced voltage. You give the induced voltage as a function of the total fall time $\Delta t$. However until we have an equation of motion for the falling magnet, we can't make any assumptions about the total fall time. The voltage produced by a magnet moving into or out of a solenoid depends on the instantaneous velocity $v$ (as my students verify by hooking a solenoid up to an oscilloscope and shoving magnets through them). If the flux produced by putting the magnet through a single coil is $\Phi$, and the stack of coils has $N$ coils and height $h$, the flux through the solenoid increases by $\Phi$ every time your long magnet penetrates another coil. Let's define the coil density $n = N/h$, in which case the height of each coil is $\delta h = h/N = 1/n$. This gives

$$ |V| = \frac{d\Phi_\text{solenoid}}{dt} = \frac{\Phi_\text{coil}}{\delta h / v} = n\Phi_\text{coil} v \tag1 $$

Note that $\Phi_\text{coil}$ depends on both your magnet and your coil. The constant property of a permanent magnet is its magnetic moment, which has units of $\text{current}\cdot\text{area}$. If the magnet mostly fills the coil, you can measure the surface field $B$ near the pole of the magnet, assume that the field within the magnet is pretty uniform, and assign a coil flux $\Phi_\text{coil} = B\cdot A$ proportional to the area $A$ of the magnet. However if there is much empty space between the falling magnet and the sides of the coil, the return field will reduce the flux seen by each coil. I'm going to assume that the magnet mostly fills the coil, so that we don't have to worry about the falling magnet tumbling.

If you connect the two ends of the solenoid across some load resistance $R_\text{ext}$, and the solenoid has internal resistance $R_\text{int}$, the induced voltage (1) will drive a current $I = V/(R_\text{ext} + R_\text{int})$. You get the most current out of the solenoid by shorting the ends together, $R_\text{ext}=0$. We'll call that total circuit resistance $R_\text{ext} + R_\text{int} = R$.

Your statement for the power absorbed by the stack of coils (as of v13) is not correct, because its units are not correct. (Your current version (v32) isn't much better: you have $h=1\,\Omega$ in some places and $h$ a variable height in other places.) The power dissipated by your solenoid will be

$$ P = IV = \frac{V^2}R = \frac{(n\Phi_\text{coil}v)^2}R \tag2 $$

Now, the work done on an object is the scalar product of the force and the displacement, $W = \vec F \cdot \Delta\vec x$, and the power exerted by a constant force is $P = \vec F \cdot \vec v$. The heat deposited in the circuit by (2) must be stolen from the kinetic energy of the relative motion of the solenoid and the falling magnet: the solenoid will, for finite $R$, exert a braking force on the moving magnet. That braking force obeys

\begin{align} P = \vec F_\text{mag} \cdot \vec v &= -\frac{(n\Phi_\text{coil} v)^2}R \\ \vec F_\text{mag} &= - \frac{(n \Phi_\text{coil})^2}R \vec v \tag3 \end{align}

This magnetic force will be present when one end of the magnet is is moving through the solenoid and the other isn't. In the magnet-falling-through-pipe examples above, a short magnet gets a force $\vec F_\text{mag}$ from the current induced where each section of the pipe tries to oppose the increase in flux as the leading edge of the magnet approaches, and a parallel force $\vec F_\text{mag}$ from the opposing current induced where each section the pipe tries to maintain the flux as the trailing edge of the magnet departs. In a solenoid, those opposing currents are connected in series, and so the braking force vanishes to first order if the magnet is moving within the solenoid. (I think there's a second-order braking force that appears if you consider the long solenoid as a set of short solenoids connected in series, but your question is constructed so that we don't have to worry about that.)

In that case, a free-body diagram for the falling magnet gives a net force of

\begin{align} \vec F_\text{net} &= \vec F_\text{grav} + \vec F_\text{mag} \\ m \vec a &= m \vec g - \frac{(n \Phi_\text{coil})^2}R \vec v \\ \frac{d\vec v}{dt} &= \vec g - \frac1\tau \vec v & \text{with } \tau &= \frac{mR}{(n\Phi_\text{coil})^2} \tag4 \end{align}

where the constant $\tau$ has units of time. The equation of motion which obeys the Newtonian constraint (4) is

$$ \vec v(t) = \tau \vec g + \left( \vec v_0 - \tau \vec g \right) e^{-t/\tau} \tag5 $$

where $\vec v_0$ is the velocity of the falling magnet at $t=0$, and the falling magnet exponentially approaches a terminal velocity $\tau \vec g = \vec gmR/(n\Phi_\text{coil})^2$ with time constant $\tau$. This does not give a constant acceleration $\frac{d\vec v}{dt} = \vec g + \vec f$, as you assume in your question.

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Let's consider how energy conservation works in some limiting cases. The only parameter in our equation of motion is the time constant $\tau$, so the interesting limits are where $\tau$ is very brief or very long compared to the time $\Delta t$ that's required for the magnet to complete its fall through the tube.

First the case where the time constant is very brief, $\tau \ll \Delta t$. That is the limit when the flux from the falling magnet $\Phi_\text{coil}$ is very strong, when the mass $m$ of the falling magnet is very small, when the coil density $n$ of the solenoid is very large, or when the load resistance $R$ is very small. In that case, no matter what the falling magnet's initial velocity, it rapidly approaches a constant velocity $\tau\vec g$ and maintains that speed for the duration $\Delta t \approx h/\tau g$ of its descent through the tube. At the top of the tube the magnet has only potential energy $U_\text{top} = mgh$; at the bottom it has kinetic energy

$$K_\text{bottom} = \frac12 m (\tau g)^2 = \frac12 m (\tau g) \left( \frac h{\Delta t} \right) = U_\text{top} \frac \tau{2\Delta t} \ll U_\text{top} \tag6 $$

If the fall speed $v = \tau g$ is constant, the current generated will be constant, and the energy deposited in the resistor $R$ will be

$$ E = P\Delta t = \frac{(n\Phi_\text{coil}v)^2}R \cdot \frac hv = \frac{(n\Phi_\text{coil})^2h}R \frac{mRg}{(n\Phi_\text{coil})^2} = mgh = U_\text{top} \tag7 $$

That is to say, if the speed all the way down is constant, the kinetic energy doesn't change, and all of the gravitational potential energy gets spent as electrical energy in the resistor. There's no perpetual motion here; it is totally ordinary Newtonian physics.

Now let's consider the other limit, where the travel time $\Delta t$ is much briefer than the time constant $\tau$. I'll point out in particular that this covers the case from your question where the coils are expanded to the size of a football field while the magnet remains small, because that change in the relative sizes means the coil flux $\Phi_\text{coil}$ includes both the strong field inside of the magnet and the oppositely-directed returning dipole field that surrounds it. In this $\Delta t \ll \tau$ limit, the exponential in the equation of motion (5) can be approximated at all times by

$$ e^{-t/\tau} \approx 1 - \frac t\tau + \cdots $$

and so the equation of motion (5) itself becomes

\begin{align} v &= \tau g + (v_0 - \tau g)\left(1 - \frac t\tau\right) \\ &= \tau g + \left(v_0 - \frac{v_0 t}\tau - \tau g + g t\right) \\ &= v_0 + gt + v_0 \frac t\tau \tag8 \\&\approx v_0 + gt \end{align}

which is, as you said, pretty close to free-fall.

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If you wanted to build an oscillatory system ("a system that powers its own motion"), an interesting thing to explore would be an ideal solenoid (self-inductance $L$, internal resistance $R_\text{int}=0$) connected so that the induced current drives a capacitance $C$, chosen so that the oscillatory time constant $1/\sqrt{LC}$ is comparable to the crossing time $\Delta t$. (Many years ago I had a desk toy that worked something like this: a pendulum with a magnetic bottom that swung near a solenoid; that solenoid was powered by a battery that lasted for weeks if the pendulum was oscillating, but died quite rapidly if the pendulum got stopped.) That'd be the subject of another question, though.

Answer 2

Here's the thing about questions that ask "isn't this a perpetual motion machine?": the answer is always "no," unless you've built a lossless oscillator or there's a hidden external energy source. Making the problem more complicated doesn't make it more interesting.

In the early versions of your question, you made the mistaken assumption that the falling magnet would experience constant acceleration, and observed (with some mistaken algebraic details, that's not relevant) that if such a coil were very long, the very high final speed might make the electrical energy extracted larger than the gravitational energy put in. That analysis was correct; the mistake was to assume a constant acceleration rather than finding the equation of motion.

In the most recent version of your question (v33? wow), you give

$$ E = \frac{N^2\Phi_\text{coil}^2}{h\Delta t} \tag A $$

which is again equal to the gravitational potential energy $mgh$ only at some magic $h$, which you find as

$$ h = \sqrt[3]{\frac{N^2\phi^2v}{mg}} \tag B $$

Encoded in these statements are two assumptions that I didn't make. One is that you can vary the height $h$ of the solenoid without changing the number of turns $N$, which I think is a strange way to go about it, but isn't impossible. The other is the idea that "making solenoid longer" involves changing the resistance $R\propto h$. (You use a confusing notation for this that makes it impossible to do dimensional analysis on your results.)

That's something that I didn't include in my other answer. But doing so isn't hard, because my analysis explicitly used the total resistance $R = R_\text{ext} + R_\text{int}$. Your choice $R \propto h$ would correspond to a vanishing external resistance $R_\text{ext} = 0$ and an internal resistance $R_\text{int} = N R_\text{coil} = nh R_\text{coil}$. Following the analysis in my other answer, we find that the definition of the power (2) becomes

$$ P = \frac{(n\Phi_\text{coil}v)^2}{R_\text{ext} + nh R_\text{coil}} = \frac{(n\Phi_\text{coil}v)^2}{R'} \tag{$2'$} $$

That is, at the same speed $v$, the extra resistance in the longer coil means that a little less power is extracted from it. The equation of motion (4) is changed to

$$ m\vec a = m\vec g - \frac{(n\Phi_\text{coil})^2}{R_\text{ext} + nhR_\text{coil}} \vec v \tag{$4'$} $$

The shape of the solution (5) is unchanged, but the time constant now depends, thanks to the internal resistance, on the height:

$$ \tau' = m\frac{R_\text{ext} + nhR_\text{coil}}{(n\Phi_\text{coil})^2} = \frac{mR'}{(n\Phi_\text{coil})^2} $$

Therefore the terminal velocity $\tau g$ also changes with $h$, and therefore so does the travel time. These changes conspire with the reduced power output from the coil at a given velocity to give $E = mgh$; none of the algebra in my (7) on the other answer is affected by substituting $R\to R'$.

That's the issue with your solution (B) for the magic height at which energy is conserved: you've assumed that the terminal velocity $v$ is independent of $h$ and put them on opposite sides of the equation. In fact the terminal velocity, in your language, is

$$ v = \tau g = g\frac{m\cdot (R)}{(n\Phi_\text{coil})^2} = g \frac{h^2m \cdot \left(h \cdot (\rm1\,\frac\Omega m)\right)}{N^2\Phi_\text{coil}^2} \tag C $$

I'm not sure why you have chosen to change the length of your coil by keeping the number of coils $N$ constant, rather than the number density $n = N/h$, and I'm not sure how you square that with a resistance proportional to length, but it doesn't matter. With this velocity $v \propto h^3$, your solution (B) for the magic height becomes

$$ h = \sqrt[3]{\frac{N^2\Phi_\text{coil}^2}{mg}}\times \sqrt[3]{\frac{mgh^3}{N^2\Phi_\text{coil}^2}} \tag {B$'$} $$

which is much less interesting.

You assumed that all the gravitational energy would be converted to electrical energy when you assumed that magnetic drag would maintain the falling magnet at a constant velocity. That's the second time in this evolving question that solving the equation of motion has removed a degree of freedom that you thought was unconstrained.