In this wikipedia article there is a description of the difference between the Maxwell-Faraday equation $$ \vec\nabla\times\vec E = -\frac{\partial\vec B}{\partial t} \tag{1}\label{uno} $$ and the Faraday law of induction $$ \mathcal E = \frac d{dt}\Phi_\Sigma(\vec B), \tag{2}\label{due} $$ where $\mathcal E$ is the electromotive force. The two are equivalent as long as $\partial\Sigma$ is a fixed circuit, in which case $\mathcal E = \int_{\partial\Sigma} \vec E\cdot d\vec l = \int_\Sigma (\vec\nabla\times\vec E)\cdot d\vec\Sigma$. However, if the circuit is moving and/or changing shape, \eqref{uno} and \eqref{due} are no longer equivalent since in the general case (sometimes this is given as a definition) $$ \mathcal E = \int_{\partial\Sigma} (\vec E + \vec v \times \vec B)\cdot d\vec l. \tag{3}\label{tre} $$ Now, in the article linked above it is stated that \eqref{tre} is the work (per unit charge) done by the Lorentz force, but I can't see how this can be, since 1. the magnetic part of the Lorentz force does no work, so \eqref{tre} would be equal to the circuitation of the electric field, and 2. the circuit on which the integral is computed isn't the path of any charge, since the charges are moving with the circuit while the integral is taken at a fixed time.
Question: What does \eqref{tre} actually represent? i.e. what actually is the physical meaning of the electromotive force? Is it linked to the work done to move/deform the circuit? If so, can this be shown mathematically?