Timeline for Faraday's law in circuits with multiple loops and different magnetic fields
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 2, 2016 at 8:49 | vote | accept | Sørën | ||
| Nov 2, 2016 at 6:35 | comment | added | GeeJay | @Sørën I see no reason why this extra emf should be taken into account. | |
| S Nov 2, 2016 at 5:29 | history | edited | user36790 | CC BY-SA 3.0 |
added 104 characters in body
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| Nov 2, 2016 at 5:26 | review | Suggested edits | |||
| S Nov 2, 2016 at 5:29 | |||||
| Nov 1, 2016 at 14:15 | comment | added | Sørën | That is (calling the left solenoid $1$ and the right solenoid $2$, and considering that the magnetic field $B$ is the same in both solenoid but varying in time) $$\begin{cases} i_1 R_1+ (i_1+i_2) R_2=\frac{dB}{dt} \pi r_1^2 + \frac{1}{4}\frac{dB}{dt} \pi r_2^2 \\ i_2 R_2+ (i_1+i_2) R_2=\frac{dB}{dt} \pi r_2^2 + \frac{1}{4}\frac{dB}{dt} \pi r_1^2 \end{cases}$$ I can't explain why this $1/4$ of the emf should be taken into account, besides the emf caused by enclosed changing magnetic flux. Could you give me any suggestions about this? | |
| Nov 1, 2016 at 14:15 | comment | added | Sørën | Thanks so much for this clear and complete answer! If I may, on my textbook the exercise from which the picture is taken, is solved in a way that is a bit in contrast with 1. (and that's why I asked). Consider situation A: on textbook mesh current method is used and the equation are the same ones you proposed, besides the fact that the emf considered in each loop is (minus) time derivative of magnetic flux enclosed by that loop plus $1/4$ of the time derivative of the flux in the other loop, and this is justified by saying "that's because of the common central branch". | |
| Nov 1, 2016 at 4:20 | history | answered | GeeJay | CC BY-SA 3.0 |