Two long parallel conducting rails are fixed on a smooth plane inclined at 30^° to the horizontal and are 0.50 m apart. Their upper ends are joined…

The field is vertical, so the component of $\vec B$ normal to the plane is $B\cos\theta$. The rod moving at speed $v$ along the incline has emf $\varepsilon=B\,L\,v\cos\theta$, since only the normal component of $\vec B$ contributes to the flux change. The current is $I=\varepsilon/R$. The force on the rod is $BIL$ in the horizontal direction (perpendicular to both $\vec B$ and the rod). Its component along the incline is $BIL\cos\theta$, directed up the slope. At terminal speed $mg\sin\theta=BIL\cos\theta=\dfrac{B^2L^2v\cos^2\theta}{R}$, so $v=\dfrac{mgR\sin\theta}{B^2L^2\cos^2\theta}=\dfrac{0.2\times10\times0.4\times0.5}{1\times0.25\times0.75}=2.13\ \mathrm{m\,s^{-1}}$. The value $1.60$ comes from omitting the $\cos\theta$ factors, $1.85$ from keeping only one power of $\cos\theta$, and $6.40$ from using $\sin\theta$ in place of $\cos\theta$.