Timeline for Forms of energy in a closed circuit with a coil
Current License: CC BY-SA 4.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 4, 2021 at 21:06 | comment | added | Vid | For the proper solution you will need to express $I(r)$. But in's neither very easy, either very nice, although it's possible. | |
| May 4, 2021 at 21:04 | comment | added | Fooourier | thanks a lot man! really helped me understand this :) | |
| May 4, 2021 at 21:03 | vote | accept | Fooourier | ||
| May 4, 2021 at 14:10 | comment | added | Vid | 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- | |
| May 2, 2021 at 18:12 | comment | added | Fooourier | 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! | |
| May 2, 2021 at 14:39 | history | answered | Vid | CC BY-SA 4.0 |