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When finding the EMF induced in an inductor due to both self-inductance and mutual inductance, the EMFs due to both of these contributions are summed. However, I have doubts regarding their summation.

The self-induced EMF is equal to the negative of the change in the net magnetic flux passing through the inductor, which is proportional to the rate of change of the current in its windings. When there is mutual induction in addition to self-induction, then the flux through the inductor changes, which also changes the current in the inductor windings. Considering these facts, I do not see why the net flux through the inductor doesn't equal the self induced EMF.

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  • $\begingroup$ > "The self-induced EMF is equal to the negative of the change in the net magnetic flux passing through the inductor" No, not net magnetic flux. Self-induced EMF in an inductor is minus rate of change of that contribution to total magnetic flux that is due to that same inductor. The contribution due to other bodies (inductors) is not part of the self-induced EMF. That is another independent contribution to the total EMF. $\endgroup$
    – Ján Lalinský
    Commented Aug 29, 2022 at 20:00
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    $\begingroup$ This question seems very similar to your previous one – in which you seem to have lost interest! $\endgroup$
    – Philip Wood
    Commented Aug 29, 2022 at 22:50
  • $\begingroup$ @PhilipWood . Yes, but that question has a definite answer and I thought it would be messy to modify that question to accommodate these doubts after receiving answers. $\endgroup$
    – Piksiki
    Commented Aug 30, 2022 at 2:10
  • $\begingroup$ Try to have a look at basics.altervista.org/test/Physics/EM/… for the magnetic circuit in a ideal transformer and in a transformer with dispersed flux, and the relationship between the currents in the windings, the voltage drop across them, and the magnetic flux. Only equations, almost no word, but it can be something to start with $\endgroup$
    – basics
    Commented Aug 30, 2022 at 6:41
  • $\begingroup$ I wasn't sure whether or not you were satisfied with the answers to that other question. I've just replaced my last comment following my answer to that question, hoping to alleviate your unease. $\endgroup$
    – Philip Wood
    Commented Aug 30, 2022 at 9:21

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As suggested by Jan, your sentence about self-induction

The self-induced EMF is equal to the negative of the change in the net magnetic flux passing through the inductor

is wrong. A valid sentence about EMF and the net magnetic flux through the inductor is

  • The total EMF is equal to the negative of the change in the total net magnetic flux $\Psi_i = N_i \Phi_i$ passing through the inductor, being $N_1$ the number of windings of the inductor. Then, you can exploit linearity of the problem to distinguish the contributions of the self-induction and the mutual inductions

    $\Phi_{i} = \Phi_{ii} + \sum_{k \ne i} \Phi_{ik}$

    so that you can get the relationship between the EMF in the winding (or the opposite of voltage drop) and the total magnetic flux passing through it

    $v_i = \dfrac{d \Psi_i}{dt} = N_i \dfrac{d\Phi_{ii}}{dt} + N_i \sum_{k \ne i} \dfrac{d\Phi_{ik}}{dt}$

    Then, you can write each flux contribution as a function of currents in the set of inductors of your problem, as shown in https://basics.altervista.org/test/Physics/EM/magnetic_circuits_transformer.html.

A more detailed derivation below, including symmetry of the mutual-induction coefficients and energy balance. For more details of the magnetic circuit analogy, derived as a circuit approximation from the fundamental equations of the electromagnetic field (i.e. Maxwell's equations), you can find a reference here https://basics.altervista.org/test/Physics/EM/magnetic_circuits.html.

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  • $\begingroup$ Based on the above description, the net flux can be divided into components: the flux through the inductor 1 due to inductor 1, and the flux through inductor 1 due to inductor 2. So the flux causing the mutually induced EMF of the second inductor is due to the flux through inductor 1 due to inductor 1. The same reasoning applies for the first inductor. However, this doesn't lead to an equality between mutual inductances. $\endgroup$
    – Piksiki
    Commented Aug 30, 2022 at 15:21
  • $\begingroup$ You get symmetry of mutual inductances with arguments about the conservative nature of the process. I'll attach you a sketch of what I mean, if you need it $\endgroup$
    – basics
    Commented Aug 30, 2022 at 15:49
  • $\begingroup$ I would appreciate it if you provided a sketch $\endgroup$
    – Piksiki
    Commented Aug 30, 2022 at 15:50
  • $\begingroup$ I edited my answered and left some more references about the topic. If you have other doubts, I'd suggest you to accept the answer, while opening other questions, otherwise we'll do a whole course on Electromagnetism in this thread $\endgroup$
    – basics
    Commented Aug 30, 2022 at 17:25
  • $\begingroup$ I'm having some difficulty understanding the diagram. Does Φbb=Φab ? $\endgroup$
    – Piksiki
    Commented Aug 30, 2022 at 21:08

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