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Suppose I have two identical current loops superimposed upon one another, but with their currents in opposite directions. The coefficient of coupling $K=1$, and therefore the mutual inductance $M =\sqrt {L_1L_2} = L$, giving the total magnetic energy as

$$E_m = 1/2(L_1i_1^2 + L_1i_1^2) + Mi_1i_2 = 2Li^2$$

But since the total magnetic field is zero, the total magnetic energy is zero, so where has the apparent excess energy come from?

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  • $\begingroup$ If $i_1=-i_2$, then the cross term is negative, isn't it? $\endgroup$
    – Vladimir Kalitvianski
    Commented Mar 21, 2013 at 19:30
  • $\begingroup$ @VladimirKalitvianski if you think the mutual inductive energy is negative and correct, then you could give an answer $\endgroup$
    – Larry Harson
    Commented Mar 21, 2013 at 22:29
  • $\begingroup$ Well, the self-inductance of a wire decreases if it goes in the opposite direction just because of the opposite currents in such a configuration. Why not the same with two different wires with the opposite currents? $\endgroup$
    – Vladimir Kalitvianski
    Commented Mar 22, 2013 at 9:41

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Your magnetic case is similar to the electric case of two opposite charges (balls) of finite density and with no polarization effects (bound charges). At infinite distance you have two equal electrostatic energies $E_1=E_2=E$ (self-energy). At a finite distance $r>2R$ you add to it the energy of electrostatic interaction $U(r)=-Q^2/r$, the latter being negative. This is an external force work necessary to stop attracting charges when they move from infinity to each other (otherwise they will be moving due to attraction). Now you have a "dipole" electric field in the space, i.e., a configuration with a smaller electric energy and it is reduced to $2E+U(r)$.

When you merge the two charges completely, they do not produce any field, so the external force work was entirely spent on bringing the charges together in a smooth way (no acceleration, no fast motion with kinetic energies). If there is no external (braking) force, the sum of kinetic energies and the energy of interaction is conserved: $2E+U(r)+K=const$. Same for the charges of the same sign.

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  • $\begingroup$ Yes, I can see you're right. Unfortunately you talk about the electrostatic case, so I can't mark your answer as correct $\endgroup$
    – Larry Harson
    Commented Mar 22, 2013 at 19:04
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First of all the definition of Magnetic Energy is the ammount of energy spent to build the aforementioned geometry.

So, you brought the first loop from infinity, in which case you didn't spent any energy. Then you connected your loop with a PSU. Let's assume that this also for free!

Now the closer you bring the second loop(let's assume that you did the same process in another place at infinty) the two loops repell each other, which means that you have to spent some work to keep the two loops at as fixed position with each other.

Even if you do that, there is also something else missing. Due to inductance, the current running in the wires isn't constant. The PSU's will have to provide the wires with the exact current, which means that you have to provide extra energy there!

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