# Magnetic Dipole from 'Current' in LC Circuit

We can work out the 'current' in an LC (series) circuit by creating and solving an ODE:

$$I\frac{1}{LC}+\ddot{I}=0$$ This provides us with a sinusoidal resonant response to any initial current.

However, as there isn't any current flowing through the capacitor, and there are no junctions, which means that, given a Kirchhoffian understanding of current, we get a contradiction depending on whether we look at current within the wire or between the plates of a capacitor. The solution to this is to recognise that the current varies throughout the wire, but that leaves me with the question as to what, if any, physical meaning was there in the 'current' we just calculated in the above equation. Is it something like average current?

Secondly, would it be possible, then, to use this 'current', which we calculate above, in an equation for the magnetic moment induced by a planar loop:

$$\mathbf{m}=I\mathbf{S}$$

There is a displacement current $\frac{\partial {\bf D}}{\partial t}$ between the plates of the capcitor. The total flux of this current is equal to the $I$ current in the wire, so that Kirchhoff's circuit law is obeyed with $I$ being the same everywhere in the circuit. The displacement current was introduced by Maxwell specificly to make the laws of electromagnetism internally consistent. Without the displacement current there would be no electromagnetic waves. In answwr to your question $\frac{\partial {\bf D}}{\partial t}$ does contribute tho the dipole moment of the loop in the LC circuit.