Motion of Thompson's jumping ring Thompson's jumping ring experiment is set up as follows:
 
There is a force acting on the ring $F(x)$ where $x$ is the vertical displacement. The force is due to the $90^\circ$ out of phase flux lines caused by eddy currents induced by an alternating magnetic flux that is passing through the core due to a coil of wire connected to an AC power supply. The force acting on the ring is a function of its displacement which is in turn a function of time.
Therefore it forms the differential equation:
$$F(x(t))-mg=m\frac{\mathrm d^2x}{\mathrm dt^2}$$
Therefore solving this electrodynamic equation for $x(t)$ should describe exactly the motion of the ring. I am going to guess from doing the experiment and by looking at the differential equation that the solution for $x(t)$ will be some sort of dampened harmonic oscillator.
How can you find out what $F(x)$ is in order to solve the equation for $x(t)$?
 A: The mean relative permeability, $\mu_r$, of the core over the a.c. cycle will no doubt be in the order of $10^3$. Therefore the magnetic field affecting the ring will be almost entirely that due to the core, rather than that due directly to the coil. We shall assume a sinusoidal variation of flux in the core, ignoring the variation of $\mu_r$ over the cycle.
Suppose that at height $z$ up the core, the flux in the $z$ direction through the core is $$\Phi=\phi(z)\ \cos(\omega t).$$ 
The emf induced in the ring (if it is at height $z$) is $$\mathscr E =-\frac{\partial \Phi}{\partial t}=\phi(z)\ \omega\ \sin(\omega t).$$
So the current in the ring (in an anticlockwise sense seen from above) will be $$I=\frac{\phi(z)\ \omega R\ \sin (\omega t)}{R^2 + \omega^2 L^2} -\frac{\phi(z)\ \omega^2 L\ \cos (\omega t)}{R^2 + \omega^2 L^2}$$ 
in which $R$ is the ring's resistance, as a turn of a coil, and $L$ is the ring’s inductance.
[We assume the ring has thickness $h$ in the z direction, an inner radius $r$ only just bigger than that of the core, and a very small radial thickness, $b$.] 
Flux will be escaping out of the sides of the core, all along its length, so
$\frac{\partial \Phi}{\partial z}$ will be negative. But (using $\text{div}\vec B =0$)
$$-\frac{\partial \Phi}{\partial z}=2 \pi r B_r$$ 
in which $B_r$ is the radial flux density just outside the core at height z, due to flux escaping from the core.
The upward motor effect (Laplace) force on the ring will be
$$F_z=-B_r I 2 \pi r=\frac{\partial \Phi}{\partial z} \times I=\frac{d \phi}{d z}\ \cos(\omega t) \times\left[\frac{\phi(z)\ \omega R\ \sin (\omega t)}{R^2 + \omega^2 L^2} -\frac{\phi(z)\ \omega^2 L\ \cos (\omega t)}{R^2 + \omega^2 L^2}\right]$$
Over a number of cycles the mean of $\sin(\omega t)\ \cos(\omega t)$ is zero and that of $\cos^{2}(\omega t)$ is $\frac12$, so the mean value of $F_z$ is
$$\langle F_z \rangle =-\frac12 \frac{d \phi(z)}{d z} \times\frac{\phi(z)\ \omega^2 L}{R^2 + \omega^2 L^2}$$
That is
$$\langle F_z \rangle =-\frac14 \frac{d}{d z} [\phi(z)]^2 \times\frac{\omega^2 L}{R^2 + \omega^2 L^2}$$
Since the magnitude of $\phi$ diminishes with $z$, $\frac{d}{d z} [\phi(z)]^2$ is negative so $F_z$ is positive; the ring jumps rather than dives! 
$L$ and $R$ can also be evaluated or at least estimated. The force problem has therefore been shown to be equivalent to the simpler-sounding one of finding $\phi(z)$ for a steel rod with a coil wound round one end. This is essentially a magnetostatics problem, which may well already have a standard solution. 
It's easy enough, though, to determine $\phi(z)$ experimentally. We run a.c. through the coil at the bottom of the rod and measure the emf induced in a 'flat' search coil placed around the rod at various values of $z$. If the search coil has $n$ turns then $$\mathscr E_{search\ peak}=n \omega \phi(z)$$
A: First of all, lets ignore all waving nature of this experiment (meaning the quantity $lf/c$ is far smaller than $1$. Where $l$ is the dimensions of the whole circuitry, $f$ is the frequence of the AC source, and $c$ is, well, the speed of light. If that is not true, this solution is invalid.
Here is my very nice simplistic model: 


*

*Waving nature is ignored (regine under $lf/c \ll  1$)

*Lets assume this ring as an adimensional current-loop circuit. 

*Lets assume this primary coil is made of only one current loop. The effective current of it can be computed: $I_{eff} = NI$ where $N$ is the number of spirals.


There are two circuits. 1) In series, connects an AC source $V = V_0e^{i\omega t}$, a resistance $R_1$ and an inductance $L_1$. 2) In sieres, connects a resistor $R_2$ and an inductor $L_2$. Circuit 1 is the primary coil. Circuit 2 is the ring. Their inductors has mutual inductance $M$. We can solve using basic Kirchhoff law for AC circuits:
$$
\begin{cases}
\mbox{Circuit 01:} && V_0 = I_1 R_1 + i\omega L_1 I_1 + i\omega M I_2 \\
\mbox{Circuit 02:} && 0 = I_2 R_2 + i\omega L_2 I_2 + i\omega M I_1 \\
\end{cases}
$$
Now all you have to do, is solve this circuit for $I_1$ and $I_2$. They are complex, so you need to take the real part of those currents to have the real currents in the circuit. Theoretical values for $M$ is the hard part. However, $M$ is extremily easy to measure using experimentation. You can also guess polite values for $M$ making use of the inductive coupling constant $0\le k\le 1$ such that $M = k\sqrt{L_1 L_2}$, which indicates how strong is the coupling between ring and primary coil thru the magnetic media of the core. Now that you have both currents. We now need to calculate the forces.
Every closed circuit $C$ has an associated magnetic dipole moment that can be calculated in the following integral:
$$
\mathbf m = \frac{I}{2}\oint_C\mathbf r\times d\mathbf r
$$
The magnetic field generated by a magnetic dipole with moment $\mathbf m$ can be computed the following way:
$$
\mathbf A(\mathbf r) = \frac{\mu_0}{4\pi}\frac{\mathbf m\times\mathbf r}{|\mathbf r|^2}
$$
Where $\mathbf A$ is the magnetic vector potential. To have the magnetic field, just compute its curl: $\mathbf B = \nabla\times\mathbf A$. The magnetic field of each magnetic dipole will be necessary when calculating its force and torque.
If you have the currents, you have the magnetic moment. Both primary coil and ring will have their own magnetic moments, time dependent. The force and torque of a magnetic dipole $\mathbf m$ inside a magnetic field $\mathbf B$ can be computed:
$$
\mathbf F = \nabla(\mathbf m\cdot\mathbf B), \quad\quad
\mathbf\tau = \mathbf m\times\mathbf B
$$
So, to calculate the force between two dipoles, just put $\mathbf B$ the field generated by the other dipole. And vice-versa.
Now you have all information necessary. I hope you are very patient... This integrals and the procedure will consume many pages of paper. Well, I tried to make your calculations to be as small as I possibly could without sacrificing basic predictive power of the model. Hopely, I had succeeded. =). Any questions let me know.
