= Solution
The resistor's <Joule heating> power is $I^2R=(B^2\ell^2/R)v_0^2e^{-2t/\tau}$. Integrating up to time $t$ gives
$$
Q(t)=\frac{B^2\ell^2}{R}v_0^2\frac\tau2(1-e^{-2t/\tau})
=\frac12mv_0^2(1-e^{-2t/\tau}).
$$
The remaining <kinetic energy> is $K(t)=\tfrac12mv_0^2e^{-2t/\tau}$, so $\boxed{Q(t)+K(t)=\tfrac12mv_0^2}$ at every time. In particular $\boxed{Q(\infty)=\tfrac12mv_0^2}$: all initial <kinetic energy> becomes heat. The magnetic force's mechanical power is $-I^2R$, consistent with this energy balance.
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