A straight conducting rod of mass 0.10 kg and length 0.50 m slides without friction while remaining horizontal and perpendicular to two parallel…

Rod speed $v$ is along the incline. The flux through the circuit changes through the component of $\vec B$ normal to the incline, $B\cos\theta$, so emf $=Blv\cos\theta$ and current $I=Blv\cos\theta/R$. The force from the field on the rod is horizontal with magnitude $BIl$, and its component along the incline is $BIl\cos\theta$, which opposes motion: $F=B^2l^2v\cos^2\theta/R$. At terminal speed $mg\sin\theta=B^2l^2v\cos^2\theta/R$, giving $v_t=\dfrac{0.1\times10\times0.5\times0.5}{1\times0.25\times0.75}=\dfrac43\ \mathrm{m\,s^{-1}}$: (A) correct. Power dissipated $=mgv_t\sin\theta=\tfrac23\ \mathrm{W}$, not $\tfrac43$ W: (B) false (that value omits $\sin\theta$). Time constant $\tau=\dfrac{mR}{B^2l^2\cos^2\theta}=0.267\ \mathrm{s}$ and $v=v_t(1-e^{-t/\tau})$, so half of $v_t$ is reached at $\tau\ln2=0.185\ \mathrm{s}$: (C) false (0.27 s is $\tau$). Current at terminal speed $=\dfrac{1\times0.5\times\frac43\times0.866}{0.5}=1.155\ \mathrm{A}$, not $\tfrac43$ A: (D) false.