When graphing the induced current in a coil while a magnet is dropped through it why is the total area equal to 0? The area represents the charge in the coil but why must the resultant flow of charge be 0?
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$\begingroup$ Short answer: current goes one way as the magnet enters, and the other way as it leaves. Never is the coil charged; the area represents the charge that has passed through any point (cross-section) of the wire. Only with a capacitor does this translate to collected charge. $\endgroup$– Blackbody BlacklightCommented May 24, 2014 at 15:27
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The EMF induced in the coil is given by:
$$ \varepsilon=-N\frac{d\Phi_B}{dt} $$
where $d\Phi_B$ is the magnetic flux through the coil and $N$ is the number of windings. The current through the coil is given by Ohm's law:
$$ I=\frac{\varepsilon}{R}=-\frac{N}{R}\frac{d\Phi_B}{dt} $$
where $R$ is the resistance of the coil. The total charge $C$ having gone through a conductor over a period of time is the time integral of the current:
$$ C=\int^{t_2}_{t_1}Idt=-\frac{N}{R}\int^{t_2}_{t_1}\frac{d\Phi_B}{dt}dt=-\frac{N}{R}\left(\Phi_B(t_2)-\Phi_B(t_1)\right) $$
Assuming $\Phi_B$ has roughly the same value when the magnet has gone through the coil, $t_2$, as when it was dropped, $t_1$, this will be zero.
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$\begingroup$ Are those two times, $t_1$ and $t_2$ referring to when the magnet is far above the coil, and far below the coil respectively? $\endgroup$– kηivesCommented Jan 15, 2013 at 23:54
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$\begingroup$ @kηives The formulas are valid for arbitrary times $t_1$ and $t_2$, but I guess they need to be when the magnet is far above and far below for the question to make sense. $\endgroup$– jkejCommented Jan 16, 2013 at 0:00