= Solution
<Faraday's law> for a moving circuit is $\mathcal E=\oint(\mathbf E+\mathbf v\times\mathbf B)\cdot d\boldsymbol\ell=-d\Phi_B/dt$. 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 $K=B_0a$. While $0<x<b$, flux grows at rate $Kv$. Define positive current as the direction whose <magnetic force> is upward. With negligible <self-inductance>, $I=Kv/R$ and
$$
m\dot v=mg-KI=mg-\frac{K^2}R v,\qquad \tau=\frac{mR}{K^2}.
$$
The <resistive magnetic entry of a falling loop> gives
$$
\boxed{v=g\tau(1-e^{-t/\tau}),\qquad I=\frac{Kg\tau}R(1-e^{-t/\tau}),\qquad
x=g\tau[t-\tau(1-e^{-t/\tau})].}
$$
The initial acceleration is $g$ and the speed approaches $g\tau$ if the loop remains partly immersed long enough. Joule heating $I^2R$ 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. Therefore
$$
m\ddot x=mg-\frac{K^2}Lx,\qquad \omega=\frac K{\sqrt{mL}},
$$
and the <lossless limit of magnetic loop braking> yields
$$
\boxed{x=\frac g{\omega^2}(1-\cos\omega t),\qquad
v=\frac g\omega\sin\omega t,\qquad
I=\frac{mg}K(1-\cos\omega t).}
$$
Here energy is stored reversibly in $LI^2/2$, 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.
\Image[/past-exam-of-the-physics-b-course-of-the-university-of-cambridge/2010/paper-2a-loop-entry.png]
{title=Resistive relaxation and reversible inductive motion during loop entry}
{description=Original dimensionless entry solutions. Curves assume that the loop is long enough for the indicated stage; a fully immersed loop has different subsequent dynamics.}
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