Both charge carriers participate in a current, so $dq = e n v_n S \, dt + e p v_p S \, dt$, where $v_n = \mathcal{E} \mu_n$ and $v_p = \mathcal{E} \mu_p$. So, on the one hand
\[ dq = \mathcal{E}eS dt (n \mu_n + p \mu_p)\]
and on the other hand
\[ \frac{dq}{dt} = I = \frac{U}{R} = \frac{\mathcal{E} S}{\rho}\]
Let's describe both cases. The second Newton's law for the particle is
\[ q\vec{\mathcal{E}}-\frac{|q|}{\mu}\vec{v}+q\vec{\mathcal{E}}_\perp + q \vec{v} \times \vec{B} = 0\]
Thus, the case of the stationary motion is described by $\vec{\mathcal{E}}_\perp + \vec{v} \times \vec{B} = 0$.
Since $V_H$ is positive the direction of the electric field is from the bottom to the top. This is the case of negative charge carriers: electrons.
\[\frac{V_H}{W} = \mathcal{E}_\perp, \quad \mathcal{E}_\perp = v_nB = \mu_n \mathcal{E}B= \mu_n \rho_n j B = \frac{IB}{en A}\]
We have the electronetruality condition $n-N_A = p$, which has the form
\[ n_i \sqrt{\frac{T}{T_0}} e^{\frac{\Delta E_F}{k_\text{B}T}} - N_A = n_i \sqrt{\frac{T}{T_0}} e^{-\frac{\Delta E_F}{k_\text{B}T}}.\]
Also $n \gg n_i$, then $p \simeq 0$ and $n \simeq N_A$. It means that $V_H$ doesn't depend on the temperature in this case.