I learned that $$\epsilon_1 = -M_{12}\frac{di_2}{dt}$$ $$\epsilon_2 = -M_{21}\frac{di_1}{dt}$$

And the book tells us directly that $M_{12} = M_{21}$ without a reason. Is there a mathematical proof for this?


An elegant and elementary derivation is given by Crawford:"Mutual inductance $M_{12}=M_{21}$", American Journal of Physics, vol 60 , February 1962, p186. The idea is the following: the total stored energy rate, "power" is $$\frac{du}{dt} = L_1i_1\frac{di_1}{dt} + M_{12}i_1\frac{di_2}{dt} + L_2i_2\frac{di_2}{dt} + M_{21}i_2\frac{di_1}{dt}$$ This can be written as the differential $${du} = L_1d(i_1^2/2) + M_{12}i_1di_2 + L_2d(i_2^2/2) + M_{21}i_2di_1$$ Now integrate from $i_1(t=0)=i_2(t=0)=0$ to $I_1,I_2$ and get $$U=\frac{1}{2}L_1I_1^2 +M_{12}I_1I_2+\frac{1}{2}L_2I_2^2+(M_{21}-M_{12})\int {i_2di_1}$$.

In general, the integral $\int {i_2di_1}$ can be anything and will depend on the history of the currents, and thus $U$ is not necessarily single valued function of the present currents $I_1,I_2$ unless $M_{21}=M_{12}$! Now you can see that for a simple medium we must have equality of the mutual inductances.

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