= Solution
<Faraday's law of induction> states that the oriented <electromotive force> around a closed material circuit is minus the rate of change of its linked <magnetic flux>:
$$
\mathcal E=\oint_{C(t)}(\mathbf E+\mathbf u\times\mathbf B)\cdot d\boldsymbol\ell
=-\frac{d}{dt}\int_{S(t)}\mathbf B\cdot d\mathbf S.
$$
The surface orientation and circuit direction obey the <right-hand rule>. The minus sign expresses <Lenz's law>: the induced <electric current>, when allowed to flow, produces a <magnetic field> opposing the change of linked <magnetic flux>. For a stationary circuit this reduces to $\oint\mathbf E\cdot d\boldsymbol\ell=-d\Phi_B/dt$.
The source's “Lorenz force” is the <Lorentz force>, $q(\mathbf E+\mathbf v\times\mathbf B)$. A wire element of cross-section $A_c$ contains $nA_c\,d\ell$ carriers of charge $q$ with drift <velocity> $\mathbf v_d$. Adding their magnetic <forces> and using $I=nqA_cv_d$ gives
$$
\boxed{d\mathbf F=I\,d\boldsymbol\ell\times\mathbf B,\qquad
\mathbf F=I\mathbf L\times\mathbf B}
$$
for a straight wire in a uniform <magnetic field>. The length <vector> follows conventional <electric current>, irrespective of the carrier sign.
For the unlabelled upward-motion request, define $\hat{\mathbf t}=(\cos\phi,0,-\sin\phi)$ down the slope, $\hat{\mathbf b}=(0,1,0)$ across the rails and $\hat{\mathbf n}=(\sin\phi,0,\cos\phi)$ out of their plane. Choose the vertical <magnetic field> as $\mathbf B=B\hat{\mathbf z}$. An <electric current> $I$ across the bar in direction $-\hat{\mathbf b}$ gives a horizontal <magnetic force> $-IBs\hat{\mathbf x}$. The other <forces> are weight $-mg\hat{\mathbf z}$ and the rail <normal force> $N\hat{\mathbf n}$. Uniform upward <velocity> requires
$$
mg\sin\phi-IBs\cos\phi=0,\qquad
N-mg\cos\phi-IBs\sin\phi=0.
$$
Consequently
$$
\boxed{I=\frac{mg\tan\phi}{Bs},\qquad
F_B=\frac{mg\sin\phi}{\cos\phi},\qquad N=\frac{mg}{\cos\phi}.}
$$
Each symmetric rail supplies half of $N$. The <electric current> must make $I\mathbf L\times\mathbf B$ point horizontally toward the uphill side; for the coordinates just defined it flows along $-\hat{\mathbf b}$. Reversing the vertical <magnetic field> reverses this current direction. The external supply must maintain that <electric current> against the motional <electromotive force> during ascent. For $\cos\phi=0$, no finite horizontal <magnetic force> can support uniform motion along the vertical rails.
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