Two long smooth parallel rails, a distance l=0.50 m apart, are inclined at 30^° to the horizontal and joined at the bottom by a resistor R=1.0 Ω. A…

The rod moves along the incline while $\vec{B}$ is vertical, so the component of $\vec{B}$ perpendicular to the velocity (and to the rod) is $B\cos30^\circ$ and the emf is $\mathcal{E}=B\,l\,v\cos30^\circ$. The current is $\mathcal{E}/R$ and the magnetic force on the rod is horizontal, with its component along the incline equal to $BIl\cos30^\circ$. At terminal speed $mg\sin30^\circ=\dfrac{B^2l^2v\cos^2 30^\circ}{R}$, so $v_t=\dfrac{0.5}{0.25\times0.75}=2.67\ \mathrm{m\,s^{-1}}$; (B) is correct and (A) wrong. Power in $R$ equals the rate of loss of gravitational energy: $mgv_t\sin30^\circ=1.33$ W, so (C) is wrong. The emf is $0.5\times2.67\times0.866=1.15$ V, not $1.33$ V, so (D) is wrong. Answer: B only.