This is a simple inquiry, based on what my textbook has told me.
"...the amount of induced current also depends on the angle of the conductor in relation to the external magnetic field. The induced current is at a maximum when the plane of the loop is parallel to the external magnetic field. As the loop rotates towards 90° of rotation, the amount of current decreases. Once the loop is completely perpendicular to the magnets (90° of rotation) the current reads zero [in the diagram, the AC generator is connected to a galvanometer]"
The textbook neglects to explain (probably irrelevant, but I'm curious) why the current decreases as the conductor becomes perpendicular to the flux lines of the external magnetic field.
The Attempt at a Solution
I understand that according to Faraday's Law: A conductor moving relative to a magnet will induce a current in the conductor (because of a change in the magnetic field). And that according to Lenz's Law: If the changing magnetic field induces a current, the current's magnetic field opposes the change that produced it (a magnet moving out of a coil will attract, going into a coil will repel).
I understand that the flux lines of the conductor are going in the same "direction" as the flux lines of the external magnet (If N→S, then conductor has a flux line going clockwise).
So if we need a changing magnetic field, at perpendicular (or as we approach it), does the magnetic field not change? So would the conductor be unaffected by the north and south poles of the external magnet because the push and pull forces have reached an equilibrium? That's the best reasoning I could come up with for this.
Any help is greatly appreciated.