Why is electromotive force in magnetohydrodynamics a vector quantity? In the mean-field dynamo theory in magnetohydrodynamics, I frequently came across a quantity;
$\langle v'\times B' \rangle$, which is termed as the mean electromotive force. I want to know that why is it termed as electromotive force, if it is a vector.
Everywhere else I have seen  emf is just the potential difference and hence a scalar. Is this emf different than the emf used in mean-field dynamo theory?
 A: $\left\langle \mathbf{v}' \times \mathbf{B}' \right\rangle$ has dimensions of electric field, rather than potential. Therefore, it is different from the standard definition of electromotive force. In a highly conductive fluid it would be equal to $-\left\langle \mathbf{E}' \right\rangle$ (by Ohm's law). It could be considered the electromotive force per unit length in the direction parallel to the vector resulting from the motion of the fluid.
A: The usual EMF in circuits refers to a closed oriented path: it is the integral of net motional force per unit charge acting on current. Physical unit of this EMF is Volt. Sometimes EMF for a non-closed path is discussed, which is based on the same idea, only the integration path is not closed but has starting point and ending point.
The MHD electromotive force $\mathbf E^*$ is clearly a different concept, but still related: it is the motional electromotive force (due to motion in magnetic field) acting on current in medium, per unit charge. So the actual force on current in volume element $\Delta V$ would be
$$
\Delta \mathbf F  = \rho_m \Delta V\mathbf E^* 
$$
where $\rho_m$ is density of mobile charge.
A: Strictly speaking, you are right. However, in magnetohydrodynamics, it happens that people refer to the force per unit charge $ \frac{{\bf J}}{e}\times {\bf B}$ as emf, with the implicit assumption that its line integral over a path provides the real emf.
