Electric current density definition

Question

I'm just wondering why the current density $J$ is always defined as the amount of electric current traveling per unit cross-section area $J = \frac{I}{S}$, and not per volume unit $J = \frac{I}{V}$ so $\frac{A}{m^2}$ instead of $\frac{A}{m^3}$. Wouldn't it be useful to calculate electric current density per volume?

Answer 1

Wouldn't it be useful to calculate electric current density per volume?

No. That quantity would have very little use in physics because it is not physically meaningful.

The reason why we define current density as the way it is, is precisely because it is extremely natural to arrive at this particular definition, and that it is very useful in practice. The reason why it is natural, is because the physics tells us that it will be useful.

As for why the physics tells us it is useful, comes from two observations. One is Ohm's Law, namely, that, for quite a lot of materials at very low current flows through them and low voltage applied across them, the resistance $R$ is pretty much constant. $$\tag{Ohm's Law as usually taught}V=IR$$

The other observation, is that we live in 3D space (and 1D time). Imagine if you have a pure block of material of infinite size, and everywhere there is an applied (low) current passing through this material in a uniform and homogenous way. The amount of resistance of part of this block should then depend upon the perpendicular cross-sectional area $A$ and length along the current flow $\ell$ in the following way: $$\tag2R=\rho\frac\ell A$$ for some constant $\rho\in\mathbb R^+_0$ called the material's resistivity and pronounced rho. Combining this with the earlier equation means $$ \begin{align} \tag3V&=I\rho\frac\ell A\\ \tag4\frac V\ell&=\rho\frac IA\\ \tag5\vec E&=\rho\vec J\\ \tag{Ohm's Law in materials}\vec J&=\sigma\vec E \end {align}$$ with $\sigma=1/\rho$ being called the conductivity, pronounced sigma.

It should now be clear that we are actually doing physics. The above is one of a family of geometrical arguments, defining quantities like Young's Modulus. We use it all the time in physics and engineering.

Note that current flow always has a certain direction to it, and thus it is actually more sensible to define it over an area perpendicular to it, $|\vec J|=I/A$, rather than over a volume. The physical significance of the electric field $\vec E$, and that its magnitude is a voltage difference across an infinitesimal length, should be obvious.

The vast usefulness of the above is why we have chosen to define the things as they are.

There is no way for people to make tables of resistances, because the geometry of the devices would come into play. However, quantities like $\rho$ and $\sigma$ are basically only dependent upon the materials (assuming that infinitely large block of material is going to behave the same as the finite size piece of it we actually are considering), and thus we can measure their values for each particular material we have, and make tables of them for different materials and list that in a book or on a website. These will generally be rather good approximations, until, of course, we start getting so small that quantum corrections are needed.

Answer 2

I'm just wondering why the current density $J$ is always defined as the amount of electric current traveling per unit cross-section area $J = \frac{I}{S}$, and not per volume unit $J = \frac{I}{V}$ so $\frac{A}{m^2}$ instead of $\frac{A}{m^3}$. Wouldn't it be useful to calculate electric current density per volume?

If you are interested in keeping track of how much charge is in a given finite volume, you need to count up all the particles in the volume. This can be done by integrating the charge density $\rho$ over the given volume. Here, we see that the charge per unit volume is useful. If we want to see how this part of the accounting changes with time, we are interested in calculating: $$ \delta t \int_V d^3x \frac{\partial \rho}{\partial t}\tag{1}\;. $$

However, as time progresses, one also wants to keep track of how much charge moves into and out of that finite volume. This latter calculation is where the current density $\vec J$ is useful. Here, we are concerned with the boundaries of the volume, since we are concerned with flow into and out of the volume. In our three dimensional world, the boundaries of a volume are two-dimensional surfaces, which are measured in units of $m^2$.

At the boundaries, we take a surface element and multiply it by a velocity and then by the density to see how much charge moves across the surface in a given amount of time. The velocity has units of meters and so the surface element area times the velocity already has units of $m^3/s$. (The third factor of length comes from the numerator of the velocity factor of the incoming particles.) In equations, we are interested in: $$ \delta t\int_{\partial V}d^2\vec{A}\cdot\vec J = \delta t \int_{V}d^3x \vec \nabla \cdot \vec J\;,\tag{2} $$ where the notation $\partial V$ means the boundary of the volume $V$, which is an area.

Note, for a conserved quantity like charge the total charge is conserved, so we have to hae Eq. (1) plus Eq (2) equal to zero, which is the usual continuity equation of electromagnetism: $$ \frac{\partial \rho}{\partial t} + \vec \nabla\cdot \vec{J} = 0\;. $$