Current density and units

Question

I have a question about current densities. My understanding is that there are 3, volumic $(\vec{J})$ , surface $(\vec{K})$ and linear $(\vec{\lambda})$. The first question is, that if J is volumetric, why is it used in the equation $I=\int_s \vec{J} d\vec{S}$. The second is, if J has units A/m^2 and K A/m, does lambda have units of A? how does it differ then with Current Intensity?

Answer 1

Volumetric

Current density is typically the amount of current per area flowing through a cross-section. Since current is a word for charge-per-time, current density is charge-per-time per area. In SI units that would be amperes-per-square-metre, so coulombs-per-second-per-square-metre:

$$[\mathrm{A/m^2}]=[\mathrm{C/s/m^2}]=\left[\mathrm{\frac{C}{s\cdot m^2}}\right].$$

If you keep the current constant through a wire, then a thicker wire results in a smaller current density. Think of a water park's water tube; a larger tube will not change the amount of water-particles-per-second flowing through (current), but it will result in fewer water-particles-per-second flowing through each square metre of a cross-section (current density).

The reason that this can be considered a volumetric (3-dimensional) measure, is that the direction of the current is perpendicular to the 2-dimensional surface of the cross-section, so spanning into the third dimension. In other words, current density is not a meaningful measure in less than 3 dimensions.

Surface

Now, redo this thought process with not a cross-sectional area but a "cross-sectional" length. For instance, imagine that you take your wire, which is cylindrical (round) and then you hammer it flat. Very flat. So it becomes a flat sheet, so thin that its thickness is negligible for your purpose. Then you can consider the "cross-section" to be a straight line with a length, rather than a cross-sectional surface with an area. Your current in amperes can now be considered per-metre of this line rather than per-square-metre, so

$$[\mathrm{A/m}]=\left[\mathrm{\frac{C}{s\cdot m}}\right].$$

The line is 1-dimensional, but the current still flows in a perpendicular direction to it, so spanning a second dimension. Thus, this is often considered a 2-dimensional, or surface, measure of current density.

Linear

Now hammer your thinned wire from the other side as well. You are now flattening it so much that its cross-section essentially, to all practical purposes, fills just a single point. Imagine, for example, than only a single electron of charge can pass through at a time. Then you are essentially working with a 0-dimensional "cross-section". There is no need to talk about how much that flows through per square metre or per metre. All you need to know is how much that flows through, because there can be no influence by any size-measure to take into account here. So, the answer is just the current itself, with units of ampere $[\mathrm A].$

This current spans 1 dimension, so this is concidered a linear concept of current density.

To your last question, sure, you can say that linear current density $\lambda$ is a property with the same numerical value as current (intensity) $I$. But, if you have no further information about the wire, then it does make a difference whether I tell you that "We have measured the linear current density to be $x$" vs. "We have measured the current to be $x$". In the former case you have more information.

Answer 2

These are the physical dimensions, since a current density is defined as the product of a charge density and the average velocity of the charges, namely

$$\mathbf{j} = \rho \mathbf{v} \qquad , \qquad \mathbf{k} = \sigma \mathbf{v} \qquad , \qquad \boldsymbol{\lambda} = \mu \mathbf{v} \qquad , \qquad$$

being here $\rho$, $\sigma$ and $\mu$ the volume, surface and line densities respectively.

Volume current density, $\mathbf{j} = \rho \mathbf{v}$, can be interpreted as the elementary charge flux across a surface. This interpretation comes from the integral balance equation for electric charge, that reads

$$0 = \dfrac{d}{dt}\int_{V} \rho + \oint_{\partial V} \rho\mathbf{v} \cdot \hat{\mathbf{n}} = \dfrac{d}{dt}\int_{V} \rho + \oint_{\partial V} \mathbf{j} \cdot \hat{\mathbf{n}}$$

for a volume $V$ fixed in space, stating electric charge conservation (you can't destroy or create out of nothing, as you can't destroy or create mass in classical mechanics).

Thus, dimensional analysis gives

$$[\rho] = \frac{[\text{charge}]}{[\text{length}]^3}$$ $$[\mathbf{j}] = [\rho] [\mathbf{v}] =\frac{[\text{charge}]}{[\text{length}]^3} \frac{[\text{length}]}{[\text{time}]} = \frac{[\text{el. current}]}{[\text{length}]^2}$$

while surface and linear density comes from one or two integration in space, and thus

$$[\mathbf{k}] = \frac{[\text{el. current}]}{[\text{length}]} \qquad , \qquad [\boldsymbol{\lambda}] = [\text{el. current}]$$

Answer 3

According to your comment, $\vec{\lambda}$ has simply the dimension of current. It is a linear current produced by a line charge with a line charge density, not a "linear current density". You need a 2-D delta function to describe it as the limit of a current described by the integral of a current density in space normal to an area where the area goes to zero and the current density to infinity with the current staying finite. You have an analogous 3-D situation with charge in the case of a point charge.