= Solution
For $\mathcal E_0=0$ and $B_0\ne0$, $v=v_0e^{-t/\tau}$ and $I=-(B_0L/R)v$. The instantaneous <Joule heating> is
$$
RI^2=\frac{B_0^2L^2}{R}v_0^2e^{-2t/\tau}
=\frac{mv_0^2}{\tau}e^{-2t/\tau}.
$$
Hence the energy dissipated up to time $t$ and its infinite-time value are
$$
\boxed{Q(t)=\frac12mv_0^2(1-e^{-2t/\tau}),\qquad
Q(\infty)=\frac12mv_0^2.}
$$
The remaining <kinetic energy> is $\tfrac12mv_0^2e^{-2t/\tau}$, so \b[kinetic energy plus dissipated heat equals the initial kinetic energy at every time]. The magnetic field redirects charge motion, and the circuit transfers mechanical energy to heat through the electric forces; there is no battery input in this case. If $B_0=0$, no current is induced, no energy is dissipated, and all the initial kinetic energy remains.
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