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

Voltage and flux graphs if a magnet is dropped. First graph shows that voltage drops from 0 to form a trough and increases rapidly towards a positive peak equal in magnitude. Then it decreases back to 0 again. In the second graph flux increases to a peak value before finally dropping to zero.

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?

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    $\begingroup$ 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 $\endgroup$
    – trula
    Mar 6, 2021 at 11:52

2 Answers 2

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The field (and flux) produced by a bar magnet is always in the same direction along the axis of the bar magnet.

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The closer the magnet gets to the coil, the bigger is the magnetic field going through the coil and thus the bigger the flux through 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 (through the surface $S$) $$\Phi=\int_S \vec B \cdot d\vec S.$$ Thus, when the magnet is inside the coil, the magnetic field going through $S$ 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 through the coil. Before or after reaching the coil, only some field lines go through 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.

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  • $\begingroup$ and we add the flux at the bottom and top of magnet to give us the total flux? $\endgroup$ Mar 6, 2021 at 13:12
  • $\begingroup$ @notsoanonperson notice that I extended my answer. $\endgroup$
    – Roger
    Mar 6, 2021 at 13:42

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