Induced EMF polarity of inductor from bar magnet

Question

I can't understand the polarity of an induced emf of an inductor from a bar magnet. I understand inductor voltage polarities in normal closed circuits in that an inductor produces a voltage drop when current increases since di/dt is positive. Then when the circuit opens, di/dt becomes negative which flips the polarity of the voltage. But I can't understand the polarity in an open loop. Consider the following scenario.

So clearly the current will run up the coil to create a field which opposes the incoming north pole (Obviously no current actually flows). From what I've read online we must use the induced EMF formula since considering the rate of change of current doesn't really work since it's an open loop. Nevertheless, I took the bottom of the coil to be ground. Then by the EMF formula it is clear that the EMF will be negative, let's say it's -9V. So this means the voltage at the top of the coil is 9V below ground thus meaning that the current is flowing from + to -. But from what I've seen online the current should flow from - to + or in other words from low potential to high potential. So what's wrong with my interpretation?

Answer 1

Try this image from here: -

The image is explaining two things; polarity of induced emf and the effect of Lenz's law. The yellow/orange N and S are to do with Lenz's law in the case that a current was flowing due to the induced emf.

The important thing to note is the induced voltage polarity when the north of the magnet is moving towards the coil. In the case above it is positive but, if you wound the coil in the opposite direction it would be negative.