How concentrated energy is in an electric field. Lorentz force moves charged particles to positions of higher potential energy.
Divide the energy by the amount of charge and you get electromotive force.

To argue with a "potential" is misleading in this case since the Lorentz force is not a potential force but reads
$$\vec{F}=q \left (\vec{E}+\frac{\vec{v}}{c} \times \vec{B} \right),$$
and thus the correct electromotive force in Faraday's Law of induction in integral form is
$$\mathcal{E}(t)=\int_{C} \mathrm{d} \vec{r} \cdot \left [\vec{E}(t,\vec{x})+\frac{\vec{v}(t,\vec{x})}{c} \times \vec{B}(t,\vec{x}) \right ]=-\frac{1}{c} \dot{\Phi}_B,$$
where $$\vec{v}(t,\vec{x})$$ is the velocity field of the path ##C##.