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.
