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?

  • $\begingroup$ 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 $\endgroup$
    – RC_23
    Feb 11 at 6:59

1 Answer 1


Griffiths is correct, this is the general concept of electromotive force acting on current in a closed path. It applies to any kind of electromotive force due to any source. In general, an EMF can act at any point of the circuit where mobile charges are present.

For example, the definition applies to net induced electromotive force due to induced electric field that is present when magnetic field changes in time.

Indeed this means that the larger the circuit, or the more coils it has, the larger the induced EMF can be.

The definition also applies to electrochemical EMF due to a battery. However, as you have realized, battery acts with its EMF on current only inside the battery. Thus the part of the closed path outside the battery does not contribute. In this special case, the integration path can be reduced to an unclosed path that goes from the minus terminal through insides of the battery to the plus terminal.

  • $\begingroup$ Now I finally understand, thank you! $\endgroup$
    – CBBAM
    Feb 11 at 3:55

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