 Solution

For ${\mathcal{E}}_0=0$ and $B_0\ne 0$, $v=v_0{e}^{-t/\tau }$ and $I=-(B_{0}L/R)v$. The instantaneous Joule heating isHence the energy dissipated up to time $t$ and its infinite-time value areThe remaining kinetic energy is $\frac{1}{2}m{v}_0^2{e}^{-2t/\tau }$, so 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.