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Let a conducting loop of area A rotate about z with normal $(\cos \omega t,\sin \omega t,0)$ in a fixed field $B\hat{\mathbf{x}}$. Neglect inductance so $I=(BA\omega /R)\sin \omega t$. Put $m_0=B{A}^2\omega /R$. Its vector magnetic moment isThe static component does not radiate. The remaining component rotates at twice the mechanical frequency and has $|\ddot{\mathbf{m}}{|}^2=4m_0^2{\omega }^4$. Thus $\overline{P}=2{\mu }_0{B}^2{A}^4{\omega }^6/(3\pi c^3{R}^2)$. Treating the same sinusoidal current as a fixed-axis moment underestimates this rotating geometry by a factor of eight. The energy source is mechanical rotation, with additional ohmic dissipation and magnetic braking.

For a fixed contour, Faraday's law of induction isThe second equation is the differential Maxwell equation, related to the first by Stokes theorem. For a moving wire, the relevant motional electromotive force is $\mathcal{E}=\oint (\mathbf{E}+{\mathbf{v}}_{\mathrm{wire}}\times \mathbf{B})\cdot d\mathbf{l}=-d{\Phi }_B/dt$. Thus a static magnetic field can induce current in a moving circuit even when $\nabla \times \mathbf{E}=0$.

For translational magnetic damping of a square wire loop, while the upper horizontal leg remains in the field, a downward speed $v$ reduces the flux at rate $Bav$. Hence $|\mathcal{E}|=Bav$, $|I|=Bav/R$, and the upward magnetic braking force is $B^2{a}^{2}v/R$. With negligible electrical self-inductance,Let the wire cross-sectional area be $S$, mass density ${\rho }_m$, and resistivity ${\rho }_e$. Since $m=4aS{\rho }_m$ and $R=4a{\rho }_e/S$,These use the printed copper values and $B=1\mathrm{T}$. The wire dimensions cancel. The terminal-speed approximation is restricted to the interval before the upper leg leaves the field, as assumed in the question.

For rotational magnetic damping of a square wire loop, consider rotation about the central $x$ axis in a uniform field, the flux is $B{a}^2\cos \theta$. The induced current produces torque $-B^2{a}^4\dot{\theta }{\sin }^2\theta /R$. A thin square loop has moment of inertia ${\mathcal{I}}_x=m{a}^2/6$: the two legs parallel to $x$ contribute $m{a}^2/8$, and the other two $m{a}^2/24$. Averaging ${\sin }^2\theta$ over a rotation givesIf “decay time” means rotational kinetic energy rather than angular speed, $K\propto {\omega }^2$ gives ${\tau }_{\mathrm{rot},\mathrm{K}}={\tau }_t/6=0.400512\mathrm{ms}$. This distinction, and the assumed separation between rotation period and slow decay, are essential. The electrical and mechanical time scales in magnetic braking must be separated: electrical self-inductance is neglected in both calculations.

In a close binary star, the rotating effective gravitational potential defines two Roche lobes meeting at the inner Lagrange point. Stellar expansion or orbital shrinkage can bring the donor into Roche-lobe overflow. Gas then flows through the inner Lagrange point; its angular momentum can form an accretion disk around the companion. wind mass transfer in a binary star is also possible without Roche-lobe filling. For a circular, synchronized binary, the Eggleton Roche-lobe radius approximation isThis describes the lobe geometry, not the donor's dynamical response.^[ref]

conservative binary mass transfer has fixed total mass and orbital angular momentum, neglecting stellar spin. Using $J=M_d{M}_a\sqrt{Ga/(M_d+M_a)}$ givesSince ${\dot{M}}_d<0$, transfer from the more massive donor initially shrinks the orbit; after binary mass ratio reversal it expands. Stability depends on both stars and their changing lobes. Let ${\zeta }_{\ast }=d\ln R_d/d\ln M_d$ and ${\zeta }_L=d\ln R_L/d\ln M_d$. A small mass loss reduces the relative overfill if ${\zeta }_{\ast }>{\zeta }_L$. Use the adiabatic donor response for dynamical stability, and its thermal response for slower adjustment; an equilibrium mass-radius law alone cannot decide both.

Stable transfer can proceed on nuclear or thermal timescales, while dynamically unstable overflow may create a common envelope, orbital spiral-in and either envelope ejection or a merger. Escaping gas, gravitational radiation and magnetic braking make evolution nonconservative by changing orbital angular momentum. accretion onto a compact stellar object releases power of order $G{M}_a\dot{M}/R_a$, producing bright X-ray binaries; radiation-driven outflows can limit retained mass. Eclipses, radial velocities and disk emission constrain the geometry and masses, but inclination, irradiation and assumptions about mass or angular-momentum loss affect the inference. The essential feedback is the donor's radius relative to its evolving Roche lobe.