# The Unapologetic Mathematician

## Ohm’s Law

When calculating the potential energy of the magnetic field, we calculated the power needed to run a certain current around a certain circuit. Let’s look into that a little more deeply.

We 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 faster. As a formula:

$\displaystyle V=IR$

The electromotive force — or “voltage” — on the left is equal to the current around the circuit times the “resistance”. What’s the resistance? Well, here it’s basically just a constant of proportionality, which we read as “how hard is it to push charge around this circuit?”

But let’s dig in a bit more. A current doesn’t really flow around an infinitely-thin wire; it flows around a wire with some thickness. The thicker the wire is — the bigger its cross-sectional area — the easier it should be to push charge around, while the longer the circuit is, the harder. We’ll write down our resistance in the form

$\displaystyle R=\eta\frac{l}{A}$

where $l$ is the length of the wire, $A$ is its cross-sectional area, and $\eta$ is a new proportionality constant we call “resistivity”. Putting this together with the first form of Ohm’s law we find

$\displaystyle V=\eta\frac{l}{A}I$

But look at this: the current is made up of a current density flowing along the wire, integrated across a cross-section. If the wire is running in the $z$ direction and the current density in that direction is constantly $J_z$, then $I=JA$. Further — at least to a first approximation — the electromotive force is the $z$-component of the electric field $E_z$ times the length $l$ traveled in that direction.

Thus we conclude that $E_z=\eta J_z$. But since there’s nothing really special about the $z$ direction, we actually find that

$\displaystyle E=\eta J$

which is Ohm’s law again, but now in terms of fields and current distributions.

But what about the power? We’ve got a battery pushing a current around a circuit and using power to do it; where does the energy go? Well, if we think about pushing little bits of charge around the wire, they’re going to hit parts of the wire and lose some energy in the process. The parts they hit get shaken up, and this appears as heat energy; the process is called “Ohmic” or “Joule” heating, the latter from Joule’s own experiments using a resistive wire to heat up a tub of water.

If we have a current $I$ made up of $N$ bits of charge $q$ per unit time, then each bit takes an energy of $qV$ to go around the circuit once. This happens $N$ times per unit time, so the total power expenditure is

$\displaystyle P=NqV=IV$

just as we said last time. But now we can do the same trick as above and write

$\displaystyle P=IV=(J\cdot E)Al$

or

$\displaystyle\frac{P}{Al}=E\cdot J$

which measures the power per unit volume dissipated through Joule heating in the circuit.

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## 1 Comment	»

1. [...] second term on the right is the energy density lost to Joule heating per unit time. The only thing left is this vector [...]

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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”).

I’m in the process of tweaking some aspects of the site to make it easier to refer back to older topics, so try to make the best of it for now.

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