Electromagnetic induction (finding ΔΦ).

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The discussion focuses on understanding electromagnetic induction and calculating the average electromotive force (EMF) when a magnet moves through a coil. It highlights that the time spent by the magnet inside the coil can be approximated using the formula ##\frac{d}{\sqrt{2gh}}##. The average rate of change of magnetic flux is expressed as ##\frac{\Delta \Phi}{\Delta t}##, leading to the conclusion that the average EMF is zero due to symmetry in the flux linked when the magnet enters and exits the coil. Additionally, it notes that the magnet accelerates while inside the coil, resulting in a non-uniform graph of EMF over time, but the total integral remains zero. The conversation reflects on the complexities of accurately estimating these values in the context of electromagnetic induction.
KarlsShwarzschild
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What is your specific problem?
 
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).
 
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KarlsShwarzschild said:
I'm not too sure if I overestimated the prefix, guess we'll find out
:oldconfused:
 
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