Loop Wire Antenna

From the post "Energy Accounting With Fourier" dated 08 Jul 2016, extra energy is needed for $B$ in circular motion.  The extra energy needed reduces the total power in $B$.  When $B$ returns from the time dimension, where is this energy when a radiating wire is not in a circular loop ($E$ not circular).

Is the power $\cfrac{1}{4}P_{total}$ the radiated power?

Yes! (but no, read further...)  Because in a circular loop wire antenna, the gap in the loop at the connections is radiating power.  This gap is the part of the wire loop that deviates from a circle.  When $B$ is transformed back to $E$, $E$ is supposedly circular, a complex sinusoidal, $A(cos(i2\pi f t)+isin(i2\pi f t))$ because $B$ is a complex sinusoidal.  That part of $E$ that encounters a break in the loop is radiated into space.

When the antenna is a straight wire, one of the components, ($Ecos(i2\pi f t$ or $Esin(i2\pi f t)$) of the circular $E$ is radiated.  The actual power fraction is $P_f=\cfrac{1}{2}\left(\cfrac{1}{2}\right)^2=\cfrac{1}{8}$.

And no!  The initial power fraction of $P_f=\cfrac{1}{4}$ is loss bending the $E$ into a circular wave.  This part of the power is not recovered as long as $E$ remains circular.

In the case of the gap in a circular loop,

$P_{str}=\left(\cfrac{\theta_{ant}}{2\pi}\right)^2$ a guess

where $\theta_{ant}$ is the angle subtended by the gap at the center of the loop.  The power fraction $P_f$ that applies to a straight wire antenna does not appear here because the applied $E$ is already circular.  No energy is needed to made $B$ circular.