# Magnet moving between $2$ separate coils, connected to a circuit

I was thinking about a question regarding a magnet entering a coil when the following thought came: what if we unravelled the middle of the coil such that it splits into $$2$$ equal parts, and connected the middle unravelled part to the top of the circuit?

Below shows the magnet as it first enters the circuit. Assume all both galvanometers, coils and all resistors are identical.

As the magnet goes into the coil, an EMF will be induced and current will flow ACW (viewed from the left side of coil), and the parts of the coil closest to the viewer will have an upwards current. So both galvanometers will point right.

When the magnet is in the middle of the first coil, no current will be induced by the first coil, because a tiny motion of the magnet will cause an increase in magnetic flux to the right of the coil, but this is offset by an equal decrease in flux to the left of the coil. So magnetic flux linkage through the coil is constant and EMF induced = 0.

However, EMF will be induced in the 2nd coil, so both galvanometers will point right. However, I'm confused about the following:

If I removed the middle resistor and replaced it with a galvanometer, which direction will it deflect in? Would it be zero?

Lastly, when the magnet is in the middle of the both coils, the south pole is moving away from the first coil, so an EMF will be induced such that the current will flow CW (viewed from the left), and the parts of the coil closest to the viewer will have an downwards current . This means that the both galvanometers will deflect away from each other.

If at this point, I removed the middle part of the circuit, it would be exactly the same scenario as when the magnet was in the middle of the coil in picture 2, so no current will be induced and the galvanometers will not deflect.

Am I correct in saying all of the above?

As far as I can tell, you are correct in your description of what happens in each case.

If I removed the middle resistor and replaced it with a galvanometer, which direction will it deflect in? Would it be zero?

The current through the galvanometer would be in the upward direction. You can see this by writing Kirchhoff's voltage law for the loop on the left. Since there is no voltage drop on coil 1, KVL can only hold if the current through the galvanometer is upward, i.e. the voltage of the upper node is lower than the voltage of the lower node.