Timeline for How much current is induced by a free falling magnet?
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| Oct 17, 2020 at 20:48 | comment | added | George Robinson | In a shorted purely inductive coil the maximum induced current DOES NOT depend on the speed of the flux change, see the answer to this question physics.stackexchange.com/questions/587133/… also the motion of the magnet will look like this youtu.be/c3kxyqbsERI. If the coil is not purely inductive and has some resistance then the motion of the magnet will look like this youtu.be/wUaqXk6axOo and the maximum induced current will be smaller than in the purely inductive coil. Do you want me to write it up as an answer ? | |
| Apr 22, 2018 at 2:47 | history | edited | rob♦ |
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| Apr 22, 2018 at 2:42 | answer | added | rob♦ | timeline score: 3 | |
| Apr 21, 2018 at 23:09 | history | edited | Feynmanfan85 | CC BY-SA 3.0 |
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| Apr 21, 2018 at 8:04 | comment | added | ACuriousMind♦ | Hi there! We've noted that you have made a large amount of edits to this question. Please note that editing a question bumps it on the front page as "active", so try to make as few and as substantial edits as possible. If you are continually trying to improve your question - which in itself is a good thing! - maybe don't submit each increment directly, but accumulate a few changes before you submit them. | |
| Apr 20, 2018 at 20:38 | history | edited | Feynmanfan85 | CC BY-SA 3.0 |
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| Apr 20, 2018 at 20:33 | history | edited | Feynmanfan85 | CC BY-SA 3.0 |
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| Apr 20, 2018 at 20:23 | history | edited | Feynmanfan85 | CC BY-SA 3.0 |
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| Apr 20, 2018 at 19:44 | comment | added | probably_someone | Let us continue this discussion in chat. | |
| Apr 20, 2018 at 19:41 | comment | added | probably_someone | I'm not considering a lifting back up either. I'm assuming that the magnet drops through the coils. In order to drop through the coils, it first has to enter, then it has to leave, the region from which you are trying to extract current. | |
| Apr 20, 2018 at 19:40 | comment | added | Feynmanfan85 | Right, but the above considers only the drop down. The current generated from the lift back up is "extra" current. In the extreme case we could simply let that extra current burn off as heat. Thus, for purposes of this example, the current flows in one direction. | |
| Apr 20, 2018 at 19:37 | comment | added | probably_someone | The situation is symmetric. The current generated when the magnet is entering the coils is equal and opposite to the current generated when the magnet is leaving the coils (the boundary between these two intervals is when the magnet is completely inside the coils). In the former case, the flux is increasing through the coils; in the latter case, the flux is decreasing, at an equal rate, because the magnet drops at a constant velocity through the coils. | |
| Apr 20, 2018 at 19:30 | comment | added | Feynmanfan85 | The current flows one way: the magnet is assumed to be the same length as the stack of coils. | |
| Apr 20, 2018 at 19:27 | comment | added | probably_someone | I know that. But just because it generates electricity does not mean the electricity is recoverable as stored electric energy. If you were to try to charge a battery using this coil, it wouldn't work, because the current spends an equal amount of time flowing in each direction. The energy contained in the electric currents is entirely dissipated as heat. | |
| Apr 20, 2018 at 19:23 | history | edited | Feynmanfan85 | CC BY-SA 3.0 |
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| Apr 20, 2018 at 19:21 | comment | added | Feynmanfan85 | Electricity is generated by the coils. See the link to Faraday induction above. | |
| Apr 20, 2018 at 19:16 | history | edited | Feynmanfan85 | CC BY-SA 3.0 |
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| Apr 20, 2018 at 19:16 | comment | added | probably_someone | What does it mean to "generate energy"? When you drop the rock to the ground, you're turning gravitational potential energy into heat and sound. When you drop the magnet into the pipe, you're turning gravitational potential energy into heat in the coils. How specifically do you "recover" this heat? | |
| Apr 20, 2018 at 19:08 | history | edited | Feynmanfan85 | CC BY-SA 3.0 |
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| Apr 20, 2018 at 14:54 | history | edited | Feynmanfan85 | CC BY-SA 3.0 |
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| Apr 20, 2018 at 14:12 | history | edited | Feynmanfan85 | CC BY-SA 3.0 |
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| Apr 20, 2018 at 4:02 | history | tweeted | twitter.com/StackPhysics/status/987179689201618944 | ||
| Apr 20, 2018 at 4:00 | comment | added | dmckee --- ex-moderator kitten | This is a question and answer site, not a threaded conversation. You can edit to clarify your question, but edits are not responses in the sense of a debate. You can also comment on answers related to your question, but those comments should be intended to elicit clarifications or improvements to the answer, so again they are not part of a debate. | |
| Apr 20, 2018 at 3:58 | history | rollback | dmckee --- ex-moderator kitten |
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| Apr 20, 2018 at 3:24 | history | edited | Feynmanfan85 | CC BY-SA 3.0 |
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| Apr 20, 2018 at 3:10 | history | edited | user4552 | CC BY-SA 3.0 |
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| Apr 20, 2018 at 2:53 | answer | added | rob♦ | timeline score: 6 | |
| Apr 20, 2018 at 0:44 | history | edited | Feynmanfan85 | CC BY-SA 3.0 |
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| Apr 19, 2018 at 22:53 | history | edited | Feynmanfan85 | CC BY-SA 3.0 |
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| Apr 19, 2018 at 21:45 | comment | added | Feynmanfan85 | Dropping a rock doesn't generate energy; dropping a magnet through a coil does. The energy generated by dropping a rock is 0. The energy expended lifting the rock is E > 0. The net energy of your hypothetical is therefore less than 0, meaning you exerted energy that you did not recover. In contrast, the net energy of the hypothetical I present above is 0, meaning no energy is lost, which would, if possible, be an example of perpetual energy. | |
| Apr 19, 2018 at 21:45 | history | edited | Feynmanfan85 | CC BY-SA 3.0 |
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| Apr 19, 2018 at 21:42 | comment | added | probably_someone | If you let a rock drop to the ground, and then lift it up back to its starting position, then you have found a height from which you can drop that rock and generate an amount of energy that is equal to the amount of energy you put into lifting that rock. Does this imply perpetual motion is possible? | |
| Apr 19, 2018 at 21:25 | history | edited | Feynmanfan85 | CC BY-SA 3.0 |
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| Apr 19, 2018 at 17:43 | history | asked | Feynmanfan85 | CC BY-SA 3.0 |