Two long smooth parallel rails are inclined at 30^° to the horizontal and are joined at the top by an ideal capacitor of 0.4 F. A conducting rod of…

The rod moves along the incline, so only the component $B\cos\theta$ of the vertical field normal to the plane of the rails contributes to the flux change, and the motional emf is $\varepsilon=B L v\cos\theta$. With no resistance the capacitor voltage equals the emf, $q=C B L v\cos\theta$, giving a current $i=CBL\cos\theta\,a$. The magnetic force on the rod is $B i L$, whose component along the incline is $BiL\cos\theta=CB^2L^2\cos^2\theta\,a$, opposing the motion. Newton's law: $mg\sin\theta-CB^2L^2\cos^2\theta\,a=ma$, so $a=\frac{g\sin\theta}{1+CB^2L^2\cos^2\theta/m}$. Numerically $CB^2L^2\cos^2\theta=0.4\times4\times0.25\times0.75=0.3$, divided by $0.2$ gives $1.5$, so $a=\frac{5}{2.5}=2.0\ \mathrm{m\,s^{-2}}$. The acceleration is constant.