# The Unapologetic Mathematician

## Electromotive Force

When we think of electricity, we don’t usually think of charged particles pushing and pulling on each other, mediated by vector fields. Usually we think of electrons flowing along wires. But what makes them flow?

The answer is summed up in something that is named — very confusingly — “electromotive force”. The word “force” is just a word here, so try to keep from thinking of it as a force. In fact, it’s more analogous to work, in the same way the electric field is analogous to force.

We calculate the work done by a force $F$ in moving a particle along a path $C$ is given by the line integral

$\displaystyle W=\int\limits_CF\cdot dr$

If $F$ is conservative, this amounts to the difference in “potential energy” between the start and end of the path. We often interpret a work integral as an energy difference even in a more general setting. Colloquially, we sometimes say that particles “want to move” from high-energy states to low-energy ones.

Similarly, we define the electromotive force along a path to be the line integral

$\displaystyle\mathcal{E}=-\int\limits_CE\cdot dr$

If the electric field $E$ is conservative — the gradient of some potential function $\phi$ — then the electromotive force over a path is the difference between the potential at the end and at the start of the path. But, we may ask, why the negative sign? Well, it’s conventional, but I like to think of it in terms of electrons, which have negative charge; given electric field “pushes” an electron in the opposite direction, thus we should take the negative to point the other way.

In any event, just like the electric field measures force per unit of charge, electromotive force measures work per unit of charge, and is measured in units of energy per unit of charge. In the SI system, there is is a unit called a volt — with symbol $\mathrm{V}$ — which is given by

$\displaystyle\mathrm{V}=\frac{\mathrm{kg}\cdot\mathrm{m}^2}{\mathrm{C}\cdot\mathrm{s}^2}$

And often electromotive force is called “voltage”. For example, a battery — through chemical processes — induces a certain difference in the electric potential between its two terminals. In a nine-volt battery this difference is, predictably enough, $9\mathrm{V}$, and the same difference is induced along the path of a wire leading from one terminal, through some electric appliance, and back to the other terminal. Just like the potential energy difference “pushes” particles along from high-energy states to low-energy ones, so the voltage difference “pushes” charged particles along the wire.

## 4 Comments	»

1. [...] wire following a closed path . There’s no battery or anything that might normally induce an electromotive force around the circuit by chemical or other means. And, as we saw when discussing Gauss’ law, [...]

Pingback by | January 14, 2012 | Reply

2. [...] But we don’t work explicitly with force as much as we do with the fields. We do have an analogue for work, though — electromotive force: [...]

Pingback by | February 1, 2012 | Reply

3. [...] Faraday’s law tells us about the electromotive force induced on the [...]

Pingback by | February 14, 2012 | Reply

4. [...] start with Ohm’s law, which basically says that — as a first approximation — the electromotive force around a circuit is proportional to the current around it; push harder and you’ll move charge [...]

Pingback by | February 16, 2012 | Reply

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## About this weblog

This is mainly an expository blath, with occasional high-level excursions, humorous observations, rants, and musings. The main-line exposition should be accessible to the “Generally Interested Lay Audience”, as long as you trace the links back towards the basics. Check the sidebar for specific topics (under “Categories”).

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