# Is the standard explanation for the ring launcher incomplete?

Related: Faraday's law in a ring

The ring launcher is a standard introductory physics demonstration that I assume almost everyone has seen (if not, YouTube it). The explanation of why the ring is launched is explained on many websites. For example, here is the explanation from the Pasco’s website, which was exactly the same explanation that several of my professors have used:

The changing magnetic field from the AC powered coil causes a changing magnetic flux through the aluminum ring. The induced EMF in the ring sets up a current which produces a magnetic field. The induced magnetic field opposes the field of the coil, pushing the ring up.

Here is where my problem lies – I feel that this is an incomplete explanation and has a serious flaw. Since this is AC powered, I imagine that for one instant in time the current through the solenoid that is creating this changing magnetic flux induces a current in the ring, say, in the CCW direction and this results in pushing the ring up. However, a split second later, the current through the solenoid changes in the opposite direction and induces a current in the opposite direction (CW) in the ring. Doesn’t this imply that the ring would now be attracted to the solenoid instead of being pushed up? However, this doesn’t happen so why is the ring launched from the solenoid?

The key for me came after taking circuit analysis and learned how the voltage leads the current in an AC RL circuit (which the ring launcher is). If the induced voltage on the ring occurs before the “second induced current” (this would be the CW current above) comes in, then the ring has enough time to be pushed up before the attraction comes in from the second induced current.

I’ve never seen this explanation before so I am assuming that I am incorrect. Please correct me! In my opinion, there is no way that such a fundamental concept could have been missed.

Thank you in advance for any help on this question

• Couldn't you calculate how the force depends on the $RL$ time and the driving frequency? The basic explanation seems to assume that the $RL$ time of the loop is much shorter than the driving frequency (or is that backwards?) - How reasonable is that for a loop of copper wire? – BebopButUnsteady Jul 30 '13 at 16:50