Real LC circuits have some resistance,
which wastes some of the energy as
thermal radiation, and the cycling
eventually dies. I think they also
have some other non-idealities that
allow energy to escape as far field
electromagnetic radiation, correct?
What are these non-idealities? Are
they independent of the resistive
component?
Energy will also be lost due to electromagnetic radiation. The amount of energy loss depends on the size of the circuitry. In the limit of infinitely small inductors and capacitors the loss to radiation falls to zero. In general, you want the circuit to be smaller than the natural wavelength of the radiation it produces. For example, if the frequency is 1GHz the period is 1 nsec. With the speed of light $3\times 10^8$ m/s, the wavelength is $(3\times 10^8)/10^9 = 0.3$ meters so you want your circuitry to be much smaller than 0.3 meters.
In addition to far field radiation, all circuit components have many other non-idealities and some of these result in thermal losses. Mostly it's about inductors, capacitors, and wires having resistance as the other circuit elements do not create heat.
If another identical LC circuit is
brought near this oscillating circuit,
will energy transfer from one circuit
to the other due to the changing
magnetic fields?
Yes.
Will this happen even with ideal
components that don't have far field
radiation?
No. To get near field radiation you need to have an antenna and these are not included in ideal circuit components. In the limit of infinitely small circuit components the near field is infinitely small. (This is an instinct from many years of circuit design; I'll check it more carefully and remove this comment if I'm assured of the answer.)
Is there no energy transfer if the
coils are perpendicular and the axis
of one is coplanar with the other,
like this?: | —
Correct.
Will the energy distribute itself
evenly among the circuits, so they
reach an equilibrium where each is
circulating half of the original
energy? Or will the energy oscillate
back and forth between the two
circuits, with one alternately being
depleted or full?
This is a wonderful question! What you have is a pair of coupled oscillators. In physics we analyze these sorts of things by looking at what kind of activity would NOT alternate between the two circuits. These are called the "normal modes" and since there are two harmonic oscillators here, there will be two normal modes.
Let's look at the problem from a general point of view. Let $(1,a)$ and $(1,b)$ be the two normal modes where $a$ and $b$ are complex numbers. That is, I've normalized the normal modes to the first oscillator. These two normal modes each has an exponential (i.e. cosine plus sine) dependence. That is, the first would actually look like $(1,a)\cos(\omega_1 t)$ and the second with an unequal $\omega_2$.
To get all the energy into the second oscillator (at time t=0), requires taking equal and opposite amounts of the two modes. Thus the values would be
$$(\cos(\omega_1 t) - \cos(\omega_2 t), a\cos(\omega_1 t)-b\cos(\omega_2 t))$$.
To answer your question, we look at the 2nd part of the above. That is, I've arranged to match the phases and sum to zero for the first part at time t=0. At some later time will the second part also match phase and sum to zero?
The answer is that this can only happen if the magnitudes of $a$ and $b$ are the same. So in general, the answer to your question is that a long term oscillation which sees all the energy in one of the circuits will not, in general, eventually get to a point where all the energy transfers to the other circuit.
Hints:
If you figure out the equivalent resistance of $R_V$, $R_{VI}$ and $R_{VII}$, you can figure out the current through $R_{VII}$.
If you know the current through $R_{VII}$, you can figure out the voltage across $R_{VII}$.
If you know the voltage across $R_{VII}$, you can figure out the voltage across $R_V$ and $R_{VI}$.
If you know the voltage across $R_V$ and $R_{VI}$, calculating the current through each of those is easy.
Best Answer
By the passive sign convention, the reference direction for the current through a circuit element is into the positive labelled terminal of the circuit element:
Assuming the starred (dotted) terminal denote the positive labelled terminal of each inductor and assuming the passive sign convention
we have
$$v_1 = L_1 \frac{di_1}{dt} + M\frac{di_2}{dt}$$
$$v_2 = M \frac{di_1}{dt} + L_2\frac{di_2}{dt}$$
where
$$M = k\sqrt{L_1L_2} $$
Clearly, if you change the reference direction for $i_2$
there must be a sign change in the equations
$$v_1 = L_1 \frac{di_1}{dt} - M\frac{di_2}{dt}$$
$$v_2 = M \frac{di_1}{dt} - L_2\frac{di_2}{dt}$$