 Solution

The resistor's Joule heating power is $I^{2}R=(B^2{\ell }^2/R)v_0^2{e}^{-2t/\tau }$. Integrating up to time $t$ givesThe remaining kinetic energy is $K(t)=\frac{1}{2}m{v}_0^2{e}^{-2t/\tau }$, so $\boxed{Q(t)+K(t)=\frac{1}{2}m{v}_0^2}$ at every time. In particular $\boxed{Q(\infty )=\frac{1}{2}m{v}_0^2}$: all initial kinetic energy becomes heat. The magnetic force's mechanical power is $-I^{2}R$, consistent with this energy balance.