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For an inductor in a circuit without a power source, does the induced emf found by using the mutual inductance value include self inductance, or is the self induced emf considered separately?

From Faraday's law I know that the induced emf is found by considering the total flux, and that this is equal to the self induced emf for the case where the inductor is connected to an external power source. For the emf induced by mutual inductance, the flux through the inductor is considered to be the flux due to the external inductor, but not the total flux, which leads me to think that there should be an added term in order to calculate the total induced emf. However, I do realize this could lead to doubly counting the flux produced by the self induced emf since this flux also affects the current in the inductor producing the external flux, which is already included in the calculation. But wouldn't this violate Faraday's law?

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2 Answers 2

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(a) The total emf in one coil (2) is $$\mathscr E_2 = -\frac{d\phi_\text{linked with 2}}{dt}=-L_2\frac{dI_2}{dt}-M\frac{dI_1}{dt}.$$ We can, if we wish, regard this as the sum of two emfs.

(b) You are quite right that the changing current induced in 2 produces changing flux some of which is linked with coil 1 and affects the current in it, which in turn affects the emf in coil 2, according to the equation above. But this is not double-counting: we simply have to accept that $I_1$ and $I_2$ are not independent of each other.

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  • $\begingroup$ The part I am still not sure of is that the EMF is determined by the change in the total flux, and so why would it be wrong to equate the total change in flux to the self-induced EMF? The total EMF induced in the inductor will determine the current in the self-induced EMF equation anyway, so if there was no way to know there was mutual induction taking place, this would be acceptable. $\endgroup$
    – Piksiki
    Commented Aug 28, 2022 at 17:21
  • $\begingroup$ I think that your difficulty lies with the concept of a self-induced emf. We define the coil's self inductance, $L$, by $$\mathscr E =-L\frac{dI}{dt}$$ in which $\mathscr E$ is the induced emf in the coil when the coil is not near any other current-carrying coil. We can then use the same equation even if the coil is near another current-carrying coil. This other coil will no doubt influence $I$ in the original coil, but nonetheless, $\mathscr E$ given by the equation $\mathscr E =-L\frac{dI}{dt}$ is the self-induced emf. To get the total emf we need to add $-M\frac{dI_1}{dt}$. $\endgroup$
    – Philip Wood
    Commented Aug 30, 2022 at 9:10
  • $\begingroup$ $L$ is a constant for the coil (I'm neglecting non-linear effects due to a possible ferromagnetic core). $L$ is, by definition, not influenced by the proximity of another current-carrying coil. $\endgroup$
    – Philip Wood
    Commented Aug 30, 2022 at 9:26
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No, the emf is in general due to the self inductance and the mutual inductance. $$e_{1}=-L \frac{di_{1}}{dt} -M\frac{di_{2}}{dt}$$

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  • $\begingroup$ I'm so sorry: the edit was made by mistake. I hope I've put right any damage. $\endgroup$
    – Philip Wood
    Commented Aug 26, 2022 at 21:41

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