When a loop A induces a current in an adjacent loop B does the change in current in the loop B affect the current in loop A? Let's say we take a Loop A connected to an AC voltage source and place it adjacent to a closed Loop B.

*

*we know that the change in the current in Loop A leads to a change in the magnetic flux
through loop B ($\Delta\phi_{\rm BA}$) and hence a current is induced in the loop B.


*If the induced current in the loop B changes won't it lead to  a change in the magnetic flux through loop A ($\Delta\phi_{\rm AB}$)?


*If yes, then won't the $\Delta\phi_{\rm AB}$ induce a current in the loop A in such a fashion that it increases the net current in the loop A at that instant to oppose $\Delta\phi_{\rm AB}$?


*Now it is once again loops back to step 1. This seems to be an infinite loop of the above three steps thus producing 'infinite' current in both the loops.
Can someone clarify where my understanding goes wrong and what actually happens in the process. It would be really helpful if I get an answer.
 A: technically a varying induced flux generates an electromotive force (emf). In order to relate it to a varying current, it will depend on the resistance, or more generally the impedance (it could be wired to a capacitor etc.) of the loop. Specifying the impedances, you therefore have a dynamical system, for the respective currents $i_A,i_B$ which you can solve depending on the varying degree of complexity.
For example, in your case, I think you are interested in the following model: loop A (resp. B) has resistance $R_A$ (resp. $R_B$) and applied AC voltage $u = u_0e^{i\omega t}$, and they have mutual inductance $M$. In this case, you get the system:
$$
R_Ai_A+M\frac{di_B}{dt}=u
$$
$$
R_Bi_B+M\frac{di_A}{dt}=0
$$
Since the model was simple, you can easily get a closed form setting $\tau = M/\sqrt{R_AR_B}$:
$$
i_A(t) = i_1\cosh(t/\tau)-i_2\sinh(t/\tau)+u\frac{R_B}{R_AR_B+\omega^2M^2}
$$
$$
i_B(t) = -i_1\sinh(t/\tau)+i_2\cosh(t/\tau)+u\frac{-i\omega M}{R_AR_B+\omega^2M^2}
$$
with $i_1,i_2$ fixed by the initial conditions. Actually, for typical initial values, the solution diverges from the stationary solution as you suspected. These pathologies arise because the model is not realistic, you neglected the self-inductances of the loops which help converge to a stationary solution. Writing their respective self-inductances $L_A,L_B$, the equations become:
$$
RI+L\frac{dI}{dt} = U
$$
where I wrote vectorially the system: $I = (i_A,i_B)$, $U=(u,0)$, $R = \begin{pmatrix} R_A & 0 \\ 0 & R_B\end{pmatrix}$ (resistance matrix) and $L = \begin{pmatrix} L_A & M \\ M & L_B\end{pmatrix}$ (inductance matrix). While before $-L^{-1}R$ had eigenvalues $\pm 1/\tau$ whose positive one resulted in the exponential instability, the new one with has only negative eigenvalues which you can check with  the determinant (positive so same sign) and the trace (negative so both negative). Physically, the self-inductance saves us thanks to Lenz's law which acts as a break to the amplification effect you noticed.
Hope this helps and please tell me if you find some mistakes.
A: I'm not sure that this answers your question, but the situations are similar. In an ideal transformer in which no current is being drawn from the secondary, the current in the primary is determined by the inductive reactance of the primary and is out of phase with the voltage, thus drawing no power.  The voltage in each coil is determined by the rate of change of the flux. If the input voltage is fixed, then the magnitude of the flux is fixed. If current flows in the secondary, its field causes a phase change in the flux. That causes the current in the primary to shift toward being in phase with the voltage.  The input power must match the output power. Both can be dependent on the phase difference between current and voltage. Any resistance in the coils will dissipate power and complicate the situation. (In many modern devices which use integrated circuits, the input 120  volt power is rectified to give a high voltage DC. That drives a high frequency oscillator which drives the step down transformer. At high frequency, the transformer can use fewer coils and have much less resistance. It becomes much smaller, lighter, and more efficient. After being stepped down, the high frequency low voltage can be filtered by a much smaller capacitor. Net effect: the devices are smaller with a much lower shipping weight.)
