How Does Connecting a Wire Affect Light Bulbs in an Inductive Circuit?

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Connecting a wire to an inductive circuit with a solenoid and light bulbs affects the current flow through the bulbs. The induced EMF creates a counter-clockwise current, and adding a wire loop may short-circuit one of the bulbs, potentially causing it to turn off while the other brightens. However, the orientation of the added wire does not seem to affect the overall circuit behavior, as the magnetic flux through the new loop remains zero. The consensus leans towards the idea that the bulbs will continue to glow with the same intensity, as the rate of change of magnetic flux remains unchanged. Experimentation with the setup could provide further clarity on this unusual situation.
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I was recently given a problem I can't quite wrap my head around. Here's the situation:

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

(you'll have to copy the link, - the s, into your address bar)

Oh-- and i just realized a mistake with the drawing, the dots should be x's to indicate the field is going into the page.

We have a solenoid creating an increasing magnetic field into the screen. A wire loop goes around the solenoid with two lightbulbs connected in series. The induced EMF is counter clockwise. Then, another wire is connected as shown to the top and bottom of the wire loop.

Question: what happens to the lightbulbs when the wire is connected?

My initial thought was that the light bulb on the left would turn off (short circuited) and the one on the right would get brighter.

But then I was thinking, what if the wire were just flipped over and was on the other side. Would the light bulbs then switch, and the right one would go off? I don't see why this would be true. Why should the physical orientation of the added wire affect the circuit?
 
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You are right in arguing that this is a symmetric situation.

I'm not entirely sure but I think nothing will happen i.e. the bulbs will continue to glow with the same intensity. Like I said, I'm not sure so I'm waiting for a concrete answer too.
 
At a first guess, I'd say that nothing will happen.

Consider the closed loop consisting of the wire between the two bulbs. The rate of change of magnetic flux within that loop stays the same. Now consider the loop of wire which you have attached, and the closed loop formed by that on the left hand side of the figure. The flux passing through that is zero, and the rate of change of flux passing through it is also zero, so the attachment of the wire should do nothing.

(Note: This is simply what I remember off the top of my head. It has been more than four years since I studied this in high school, and I (regrettably) haven't touched physics since.)
 
The problem is though, that the attached wire offers a path of lesser resistance through which the current can flow, instead of through one of the bulbs. I think I might actually try and build this soon and see what it actually does. This is a very strange situation indeed.
 
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