Inductance problem, puzzles professors

[fhqwgads2005]

I posted this question a few months ago, but got no responses, especially after it got moved to the homework help forum. Since then, I have asked several professors, many of whom I expected to be experts in this area, and I've been given several different shaky sounding responses. So, I'd like to have a discussion about it. This is not a homework problem. It is simply a thought experiment.

Here's the picture:

http://pics.bbzzdd.com/users/theman/inductorbulbs.gif

In the picture, the inner circle represents a solenoid producing a decreasing (over time) magnetic field pointing uniformly out of the screen. A wire loop with two lightbulbs in series is placed above the solenoid (the outer circle). The changing B-field produces an EMF in the wire as shown. Next, a segment of wire is connected to this wire loop as shown by the thick black line.

The question is, what happens when you connect this wire segment?

I've gotten answers from, both lightbulbs going off, to nothing happening, to one lightbulb going off, to both bulbs dimming, and now I'm just confused and I'm starting to think I don't understand inductance at all... if anyone has some words of wisdom on this one, it would be a load off my mind.

thanks.

 

[AJ Bentley]

It depends on precisely how you connect the wire.

As you've drawn it, the wire is a short-circuit across the left hand bulb, which would go out. (A very small current would continue to flow through it because it's still part of a loop round the field, but most of the current would flow through the short).

I have to say, I can't see what confusion could arise - it looks very simple.

 

[DaTario]

It is quite an interesting problem in deed. IMO the answer will depend on the orientation of the thick part of the drawing. Note that it can rotate around the vertical axis belonging to the screen. Suposing this thick part is rotated in such a way to belong to a a plane which is perpendicular to the screen. In this case I would say that nothing happens.

In the very way it is drawn, I would say that the left bulb would be sort of shorted and so it would not shine with the same intensity as the right one. I think the left resistor would shine a little for there is still a loop (the left loop) with which induction could take place, but the right resistor have more induction effects on it (there is no competition with simple wires), resulting in a higher intensity (more current).

I am not prety well convinced, but I would sign the text above.

Best wishes

DaTario