= Solution
Substituting the electron Lorentz factor in the supplied energy-loss formula gives
$$
\boxed{U_0=\frac{e^2\gamma_e^4}{3\epsilon_0\rho}\simeq5.53\,\mathrm{MeV\ per\ turn}.}
$$
Without replenishment, the electron's momentum decreases. The reference dipoles then bend it more tightly, producing a displaced orbit. In linear <beam optics>, $x=D_x\delta$ and $y=D_y\delta$, where $\delta=\Delta p/p\simeq-U_0/E\simeq-1.11\times10^{-3}$ per uncompensated turn. In an ideal planar ring the horizontal dispersion is nonzero and the orbit shifts horizontally, inward where $D_x>0$. There is no corresponding vertical closed-orbit shift when $D_y=0$. Real coupling or vertical dispersion can change that idealization.
With current $I=0.500$ A, the number of circulating particles is $N=IT_{\rm rev}/e$, so the radiated power is
$$
\boxed{P=\frac{NU_0}{T_{\rm rev}}=\frac{IU_0}{e}\simeq2.76\,\mathrm{MW}.}
$$
The $2\,\mu\mathrm s$ revolution time cancels when using beam current. Radio-frequency cavities restore the lost energy longitudinally, with synchronous gain $eV_{\rm RF}\sin\varphi_s=U_0$ in the sine-phase convention. Their RF power system must supply at least this beam power, in addition to its own losses. Radiation damping and quantum excitation determine beam emittances; they should not be confused with the dispersive closed-orbit shift from a momentum change.
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