Questions about magnetic levitation of a Thompson Ring

[jaumzaum]

The Thompson jumping ring experiment consists in an AC supply connected to a coil with a metal steel core inside. A metal ring is placed around the core. When the AC supply is turned on, the ring suddenly jumps and after a time it starts levitating in a fixed height.

I'm a little confused about this experiment. I know the AC supply will provide varying magnetic field in the core. The varying magnetic field will provide varying magnetic flux in the ring, and so by the Faraday Law it will induce a varying current. Ok.

1) My first question is concerning the magnetic field lines. For the ring to levitate there must be a radial magnetic field component in the ring, so that this could result in a net vertical force. I know the magnetic field lines inside the core will be vertical and almost parallel, and outside the core the magnetic field would be very close to zero. Is this "almost zero" magnetic field that levitates the ring?

2) How is the field lines outside the core, in the height of the ring? Are they ascending, descending, or in a radial direction?

3) As the current is alternating, I would expect the net force to be zero, as it could sometimes be upwards, sometimes downwards. Why is there a net force in some height above?

Thank you!
I would appreciate any help

 

[kuruman]

To answer the first two questions, remember that the magneitc field lines generated by the primary form closed loops. There is a radial component to the field even inside the primary. The farther away you move from the coil, the more tipped with respect to the vertical the magnetic field lines become. If the steel core is isotropic, there is no reason for the magnetic field lines to remain axial.

Of great importance to this are the boundary conditions. Imagine inside the steel the magnetic intensity H vector that is proportional to the current in the primary. Furthermore, let's say that it makes an angle of 45^o (to make things simple) w.r.t. the vertical axis. Then $H_{in,z}=H_{in,r}$. We apply the boundary conditions remembering that $\vec B=\mu~\vec H$.
1. Continuity of the radial (normal) component of $\vec B$ says $B_{out,r}=B_{in,r}=\mu H_{in,r}.$
2. Continuity of the radial (axial) component of $\vec H$ says $H_{out,z}=H_{in,z}~\Rightarrow~B_{out,z}=\mu_0H_{in,z}=\mu_0H_{in,r}.$

Taking the ratio, $$\frac{B_{out,r}}{B_{out,z}}=\frac{\mu}{\mu_0}.$$ The ratio depends on the material. For example, it is 200 for steel and 20,000 for mumetal. However, the fact remains that the radial component of B is greatly enhanced upon exit of the field lines from the steel core right where we want them, in the ring.

Your reasoning in (3) is incorrect. The induced current in the ring will always oppose the proposed change and so will any forces and torques that are generated as a result. That's Lenz's law.

 

[jaumzaum]

Your reasoning in (3) is incorrect. The induced current in the ring will always oppose the proposed change and so will any forces and torques that are generated as a result. That's Lenz's law.

Thanks Kuruman! I really need to study more how a core influences the magnetic field.

But regarding the last question, when I use Lenz law (at least the way I know it), as the current is alternating in the primary, the magnetic field will also be sometimes downwards, sometimes upwards , and the flux will be also alternating (sometimes increasing, sometimes decreasing), this would also led to changing current in the ring (sometimes counter-clockwise, some times clockwise), and a force sometimes being applied upwards and sometimes downwards, am I wrong?

 

[kuruman]

You are not wrong, but you are oversimplifying. You know from the experiment that the ring jumps which means that the average force is not zero. This article is one of many found on the web that does a detailed analysis.

 

[jaumzaum]

You are not wrong, but you are oversimplifying. You know from the experiment that the ring jumps which means that the average force is not zero. This article is one of many found on the web that does a detailed analysis.

Thanks kuruman. This experiment was actually in a high-school test question. I was wondering if it's possible to explain to a high school student why wouldn't the force be zero. I really can't understand the article itself, it's far too complicated for my current knowledge :/

 

[kuruman]

Can you post the actual statement of the test question? If the current in the primary is not AC, e.g. generated by discharging a capacitor, you don't have to worry about the force changing direction.

 

[agrumpyoldphysicstec]

Thanks kuruman. This experiment was actually in a high-school test question. I was wondering if it's possible to explain to a high school student why wouldn't the force be zero. I really can't understand the article itself, it's far too complicated for my current knowledge :/

School textbook writers and examiners seem to be attracted to this type of question where the magnetic field is spread all over the place and there is nothing that you can actually put numbers to. The mathematics involved is way beyond the level expected in school.
Rather, the examiner is looking for an answer that indicates a knowledge of the rules that are applicable. If these are recited quickly, in order, they form a reasonably convincing argument as to how the experiment works.
But, as you have found, more careful thought shows that the ring just rattles up and down. This can be done by considering the apparatus as a transformer where the ring forms the secondary and the resistance of the ring forms the load on the secondary. Applying Fleming's LH Rule separately to each quarter cycle shows that the forces cancel out over a complete cycle.
It requires even more careful thought to realize that the transformer is far from perfect and not all the flux links both primary and secondary and hence the transformer has leakage inductance. This can be modeled as an inductance in series with the secondary resistance.
Now, the secondary current is no longer in phase with the induced voltage. Applying the LH Rule again, shows a net upward force.
The examiner is, of course, looking for his own answer, inadequate as that may be.