Electromagnetism – How Induced Current in Adjacent Loops Affects Each Other

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Let's say we take a Loop A connected to an AC voltage source and place it adjacent to a closed Loop B.

  1. 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.

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

  3. 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 $\Delta\phi_{\rm AB}$?

  4. 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.

Best Answer

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.

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