# Electromotive force is independent of path?

I am having some trouble understanding the EMF and how it provides a steady current (that is, the current isn't weaker further from the source). My present understanding is that if we have a battery with a wire connected to both ends, the current running through the wire is supplied by the battery, within which there is a force pushing the charges from one end to another against the natural flow, and this is what we call the EMF.

However, in Griffiths book he says that the EMF $$\mathcal{E}$$ is defined as $$\mathcal{E} \equiv \oint \textbf{f}_s \cdot d\textbf{l}$$ where $$\textbf{f}_s$$ is the source driving the current around the circuit and the integral is taken along the circuit. But this definition implies that the EMF is dependent on the length of the circuit (and so the wire length). How then do we speak of a battery having $$n$$ volts? Is the circuit he talks about referring to the electrodes within the battery and not the external circuit?

• Current has to be the same throughout the path because charge is not pooling in a spot or dissipating from a spot (without a capacitor for example). Conservation of charge Feb 11 at 6:59