# Flux and emf induced graphs when magnet is dropped through a coil

I've been told that if you drop a magnet through a coil the induced emf and flux graphs would look like this:

I understand that when the bar magnet is in the middle of the coil the emf induced is zero as flux change in top and bottom is in opposite directions but why is flux maximum when emf induced is zero, shouldn't the effective flux be zero as well? And, in the second half of the magnets jounery shouldn't the effective flux be negative as more of the flux linkage is contributed by the top half of the magnet when it is leaving the coil?

• As you can see from the dimension Vs , the flux graph is the integral of the voltage graph, also if the dropping magnet is in the middle one B field in the coil is at its maximum – trula Mar 6 at 11:52

The closer the magnet gets to the coil, the bigger is the magnetic field going thrue the coil and thus the bigger the flux thrue the coil. This is due to the magnetic field being stronger closer to the magnet ($$V$$ is the Volume of the magnet which we integrate over) $$\vec B(\vec r)=\frac{\mu_0}{4\pi}\int_V \frac{\vec j(\vec r')\times (\vec r-\vec r')}{|\vec r-\vec r'|^3}dV',$$ and because the flux $$\Phi$$ depends on the magnetic field inside the coil (thrue the surface $$S$$) $$\Phi=\int_S \vec B \cdot d\vec S.$$ Thus, when the magnet is inside the coil, the magnetic field going thrue $$S$$ is at its maximum because $$S$$ is near to $$V$$ (the magnet). You can understand this intuitively imagining a finite amount of magnetic field lines. When the magnet is inside the coil all of the field lines go thrue the coil. Before or after reaching the coil, only some field lines go thrue the coil. This explains the second Graph.
By Faraday's Induction Law $$\vec \nabla\times \vec E=-\partial_t \vec B,$$ the inductive voltage is defined as $$U_{\rm ind}=-\partial_t \Phi.$$ Thus the inductive voltage is proportional to the derivative of the Flux $$\Phi$$ with respect to time. This explains the first Graph.