Electromagnetic induction (finding ΔΦ).

  • #1
KarlsShwarzschild
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  • #3
I am unsure if I understand what you are doing. It seems ##\frac{d}{\sqrt{2gh}}## is approximately the time that the magnet spends inside the coil. You are also using $$\mathcal{E} = -N\frac{d\Phi}{dt}$$I suppose the average rate of change of flux in a time ##\Delta t## is ##\frac{\Delta \Phi}{\Delta t}##, from which it follows that the average EMF ##\mathcal{E} = -N \frac{\Delta \Phi}{\Delta t}##. So your construction ##\mathcal E = -N \frac{\Delta \Phi}{d}\sqrt{2gh}## is the time-averaged EMF inside the coil whilst the magnet is inside the coil. However this is zero, because ##\frac{1}{T} \int_0^T -N\frac{d\Phi}{dt} dt = 0##, since the flux linked in the coil is the same when the magnet is at the top of the coil as when it leaves (I assume necessary symmetry here :wink:).

Plus, the magnet actually still accelerates inside the coil, so a graph of ##\mathcal{E}## vs ##t## has an uneven shape (but still a total integral of 0).
 
  • #4
jtbell
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