# Kinetic Energy and Inductor energy violates conservation of energy?

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!

• The drift kinetic energy of electron in a current flow is negligiably small. – ytlu Apr 16 at 11:23

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).
• 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? – SalahTheGoat Apr 16 at 13:20