= Solution
For a moving loop, <Faraday's law of induction> is $\mathcal E=\oint(\mathbf E+\mathbf u\times\mathbf B)\cdot d\mathbf l=-d\Phi_B/dt$, where $\mathbf u$ is the local wire <velocity>, $\Phi_B=\int\mathbf B\cdot d\mathbf S$, and the oriented boundary and surface follow the right-hand rule. For a fixed loop the motional term vanishes. <Lenz's law> is the minus sign: the induced response opposes the change of linked flux, not necessarily the applied field itself.
Let $s$ increase downhill and choose positive current $I$ to produce an uphill magnetic <force>. The normal component of the imposed field is $B\cos\theta$, so the changing external flux is $Bl\cos\theta\,s$. With initially zero current and no resistance, conservation of <flux linkage> gives $LI=Bl\cos\theta\,s$. The rod feels gravity vertically downward, a rail reaction normal to the slope, and magnetic <force> of magnitude $IlB$ horizontally uphill. Its downhill component is $-IlB\cos\theta$; its normal component presses the rod into the rails, giving $N=mg\cos\theta+IlB\sin\theta$.
The <inductive oscillation of a sliding rod on inclined rails> follows from
$$
m\ddot s=mg\sin\theta-\frac{B^2l^2\cos^2\theta}{L}s,
\qquad\boxed{\kappa=\frac{l^2\cos^2\theta}{L}.}
$$
The magnetic <force> is not itself parallel to the rails. Defining $\omega_B=Bl\cos\theta/\sqrt{mL}$ and $s_e=mg\sin\theta/(m\omega_B^2)$, release from rest gives
$$
s(t)=s_e(1-\cos\omega_Bt),\qquad
\boxed{s_{\rm first\ stop}=2s_e=\frac{2mgL\sin\theta}{B^2l^2\cos^2\theta}.}
$$
The first stop occurs at $t=\pi/\omega_B$; the rod then returns and oscillates between $0$ and $2s_e$. At $s_e$ the <force> is zero but the <speed> is maximal. Mechanical <potential energy> is reversibly exchanged with <kinetic energy> and $LI^2/2$; a perfectly conducting circuit provides no damping.
For the <resistive long-time drift of an inductive sliding rod>, with loop resistance $R>0$, use
$$
L\dot I+RI=Bl\cos\theta\,\dot s,\qquad
m\ddot s=mg\sin\theta-Bl\cos\theta\,I.
$$
<Joule heating> damps the transient oscillation. On sufficiently long rails the asymptotic motion is steady sliding, not permanent rest:
$$
\boxed{I_\infty=\frac{mg\sin\theta}{Bl\cos\theta},\qquad
v_\infty=\frac{Rmg\sin\theta}{B^2l^2\cos^2\theta}.}
$$
The zero-resistance model has persistent oscillations; its late-time limit cannot be interchanged with taking $R\to0$. These results assume $B\cos\theta\ne0$, smooth contacts, negligible rod self-inductance and constant geometry.
\Image[/past-exam-of-the-physics-b-course-of-the-university-of-cambridge/2013/paper-1-inductive-rails.png]
{title=Original rod force resolution and calculated ideal inductive oscillation along the slope}
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