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Suppose we have an ideal LC circuit (no resistance) and an open switch where the capacitor has an initial voltage $V_o$. Initially, the energy stored in the capacitor at $t=0$ is $\frac{1}{2}CV_o^2$ and the energy in the magnetic field of the inductor is zero because no current is flowing. Now at time $t=0+dt$ we close the switch and current slowly begins to build up. When the current is a maximum, the energy stored in the magnetic field of the inductor is $\frac{1}{2}LI^2$ but now the energy stored in the capacitor is zero. Thus we must have that $\frac{1}{2}LI^2=\frac{1}{2}CV_o^2$ because no energy is dissipated since there is no resistance.

But there seems to be something very wrong here at a fundamental level. The charge (the electrons) traveling through the inductor at the instant that the current is a maximum have a non-zero kinetic energy (denote this kinetic energy $K_{charge}$). They have to have non-zero kinetic energy since they constitute a current. But if they do posses this energy in addition to the magnetic field energy $\frac{1}{2}LI^2$, then the total energy at the moment the current is a maximum will equal $E_{tot}=\frac{1}{2}LI^2+K_{charge} >E_{initial}=1/2CV_o^2$. So its seems we have created energy in this process?

The only way I can work around this issue is to assume that the kinetic energy is already somehow factored into the magnetic field energy but I am not sure.

Any help on this issue would be most appreciated!

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  • $\begingroup$ The drift kinetic energy of electron in a current flow is negligiably small. $\endgroup$
    – ytlu
    Apr 16, 2021 at 11:23

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Kinetic energy of electrons due to electric current $I$ in an inductor is much smaller than magnetic energy $\frac{1}{2}LI^2$ (provided the inductor has large enough $L$, which is usually the case).

So yes, strictly speaking total energy stored in the capacitor is transformed into magnetic energy and kinetic energy of current-carrying charges, but the latter energy is so small compared to magnetic energy it is customary to ignore it.

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  • $\begingroup$ Thanks for the response! Okay so would it be fair to say in my example, when the current through the inductor is maximum, we actually have that $E_{tot}=\frac{1}{2}LI^2+K_{charge} =E_{initial}=1/2CV_o^2$ but $K_{charge}<<< E_{mag}= \frac{1}{2}LI^2$ so in effect we have that $E_{mag}=E_{initial}$ where the equality can be used with impunity because the difference is completely negligible for all reasonable values of voltage, current and inductance? $\endgroup$
    – SalahTheGoat
    Apr 16, 2021 at 13:20
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    $\begingroup$ There is another reason energy conservation is not so simple here - losses during the discharge. It is not easy to discharge capacitor quasi-statically (without losses of energy). If we just move a piece of conductor (switch) and complete the circuit with the inductor, there will be very fast increase of current, which means the charges accelerate. Such accelerated charges produce EM radiation. So part of the energy that is initially in the capacitor goes to kinetic energy of charges, part goes away through EM radiation, and the major part is converted into magnetic energy. $\endgroup$
    – Ján Lalinský
    Apr 16, 2021 at 14:03

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