Two long parallel smooth metal rails, 0.50 m apart, lie on an incline making 30^° with the horizontal, and are joined at the lower end by a resistor…

The rod moves down the incline with speed $v$, and the field is vertical, so only the component of $\vec B$ perpendicular to the plane of the rails, $B\cos\theta$, produces flux through the loop. The emf is $\varepsilon=B\cos\theta\,Lv$ and the current $I=\varepsilon/R$. The force on the rod is $BIL$ horizontal, and only its component along the incline, $BIL\cos\theta$, opposes the motion. At terminal speed $mg\sin\theta=BIL\cos\theta=\dfrac{B^2L^2v\cos^2\theta}{R}$. Hence $v=\dfrac{mgR\sin\theta}{B^2L^2\cos^2\theta}=\dfrac{0.2\times9.8\times0.4\times0.5}{0.64\times0.25\times0.75}=\dfrac{0.392}{0.12}=3.27\ \mathrm{m\,s^{-1}}$. Using the full field instead of its normal component would give a wrong factor of $\cos\theta$. With $B$ perpendicular to the plane (a common mistake) the terminal speed would instead be $mgR\sin\theta/(B^2L^2)$, about $2.45\ \mathrm{m\,s^{-1}}$ for these numbers, so the vertical field matters.