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I am a bit confused. When i move a magnet through a coil thats in a closed circuit, what does my kinetic energy convert to?

I assume I will create a magnetic field, and that magnetic field will create a current trough the circuit. Is it true that the kinetic energy will only convert into magnetic potential energy from my magnet to the coil and electrical energy from the movement of electrons in the circuit, or are there more forms of energy that I need to take into account?

Also what would the equations look like for the different forms of energy?

Thanks a lot!

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Well, you are mixing different forms of energy here. If you are moving magnet trough coil you are generally making work, and not using kinetic energy of the magnet.

The case, where you will convert the kinetic energy of magnet will be following:

On ice surface (no friction) you have coil, and you slide magnet trough coil. In that case, the magnet will decelerate due to eddy currents. Due to induction some voltage will be induced and if you would connect the edges of coil some current will flow. So kinetic energy will be transformed into kinetic energy of charge carriers or "energy of current", I really don't know better term. Bear in mind, that electrons generally have quite high kinetic energy due to thermal movement, but in this instance we don't have to take that into account, since this energy doesn't transform.

In the case, where you are moving magnet with hand you are experiencing some force, and therefore you make some work. In that instance work is transformed into energy of the current and than after current flows into thermal energy of the wires.

The magnetic energy of coil will come into account only in other case, where we would have some current, which flows thru the coil, and we would suddenly stop that current. Then the magnetic field would generate some current for short period of time.

But in the case of moving magnet we are only converting work into moving of electrons. If we would use current to move magnet we would convert energy of moving electrons into work.

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  • $\begingroup$ And if I drop a magnet from above trough the coil, will the energy IN the coil be the same as if i let the magnet slide on ice trough the coil? I know the energy because of the movement will be different, on ice you only have $\frac{mv^2}{2}$ on the start but when falling you will have mgh and $\frac{mv^2}{2}$. Also, is the energy from the moving current equal to $\frac{LI^2}{2}$, or is it something else, and is the energy from heating equal to $\int_0^T RI(t)dt$? Thanks anyways for answering! $\endgroup$
    – Fooourier
    Commented May 2, 2021 at 18:12
  • $\begingroup$ Then you have $W=mgh+mc^2/2$, but that really doesn't change anything. Coil, trough which no current flows, doesn't have any magnetic energy. You have to think about this experiment in a way, that all induced current is immediately converted into head. Otherwise it will induce some magnetic field inside the coil, and energy will be converted into magnetic energy, and than, when magnetic field in coil will collapse into a heat energy. Your integral is technically correct, but do u know $I(r)$. $LI^2/2$ is magnetic energy of the coil, trough which the current I flows, and not of the current- $\endgroup$
    – Vid
    Commented May 4, 2021 at 14:10
  • $\begingroup$ thanks a lot man! really helped me understand this :) $\endgroup$
    – Fooourier
    Commented May 4, 2021 at 21:04
  • $\begingroup$ For the proper solution you will need to express $I(r)$. But in's neither very easy, either very nice, although it's possible. $\endgroup$
    – Vid
    Commented May 4, 2021 at 21:06

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