 Solution

Faraday's law for a moving circuit is $\mathcal{E}=\oint (\mathbf{E}+\mathbf{v}\times \mathbf{B})\cdot d\mathbf{\ell }=-d{\Phi }_B/dt$. The magnetic flux ${\Phi }_B$ is through a surface bounded by the instantaneous loop, with consistent orientation. For the stationary external field here, the motion supplies the motional electromotive force.

Let $x$ be downward penetration of the bottom segment, $v=\dot{x}$, and $K=B_{0}a$. While $0<x<b$, flux grows at rate $Kv$. Define positive current as the direction whose magnetic force is upward. With negligible self-inductance, $I=Kv/R$ andThe resistive magnetic entry of a falling loop givesThe initial acceleration is $g$ and the speed approaches $g\tau$ if the loop remains partly immersed long enough. Joule heating $I^{2}R$ removes gravitational energy. At the first $x=b$, both horizontal segments are inside the uniform field: the net flux becomes constant, the net magnetic force vanishes, and this zero-inductance model has $I=0$ thereafter. The subsequent speed increases with acceleration $g$.

For negligible electrical resistance, conservation of total linked flux instead gives $LI=Kx$ in the braking-current convention. Thereforeand the lossless limit of magnetic loop braking yieldsHere energy is stored reversibly in $L{I}^2/2$, not dissipated. If $b>2g/{\omega }^2$, the loop reverses before complete entry and executes the ideal oscillatory motion while straddling the boundary. If $b<2g/{\omega }^2$, it reaches complete entry on its first descent; thereafter the current is constant, the forces on its two horizontal segments cancel, and it falls with acceleration $g$. The exact grazing case needs care at the ideal discontinuous field boundary. An entry-only sinusoid must not be extrapolated through full immersion.