# Difference between current density and current

I'm a little confused, I can't understand intuitively what current density is. From the formula looks obvious:

$$j=\frac{I}{A}$$

it is the current per area. Very similar to the "everyday life density" $$d=\frac{m}{a}$$, but it somehow confuses me. I've seen multiple times people explaining current like this:

and when I see this I can't tell the difference between current density and current.

"An ampere is the flow of one coulomb through an area in one second."

Can i say

"An ampere per $$cm^2$$ is the flow of one coulomb through an area of $$cm^2$$ in one second"?

Could you please give me an "intuitive" example for current density?

I found on quora this example:

Imagine a one way road (A) next to a 5-way highway (B). Picture a line of cars passing at 100 km/h through road A, and five lines of cars passing at 20 km/h each on road B. Now you are between both roads and you start counting the total number of cars that cross an imaginary perpendicular line that cuts both roads. You will agree that over the same period of time, you will count the same number of cars crossing the line, even though road B has five lines of - slower- cars. Now replace cars for charges and the roads for wires. Both wires have the same current (total number of charges/cars crossing the imaginary line), but different current density (1 car vs 5 cars crossing the line at the same time).

but I still a little confused, and I cant still really tell the difference..

The example of the road and the cars can be linked to a better definition of density of current in my opinion: $$\mathbf j = \rho \mathbf v$$, where $$\rho$$ is the density of moving charges and $$\mathbf v$$ its velocity.
If a snapshot at any time shows the same num of cars per unit area in both roads,(equivalent to the density of charges), the density of current of road A is 5 times greater that road B, because the $$v_a = 5v_b$$.
When multiplying by the road width, we see that the current is the same. The current $$\mathbf I = \mathbf jS$$, where $$S$$ is the cross section of the conductor.
At a current density of 1 A m$$^{-2}$$, a long square wire 1 mm by 1mm (area $$10^{-3}{\rm m}\times 10^{-3} {\rm m}= 10^{-6} {\rm m}^2$$) carries a current of $$10^{-6}$$ amps. Does that help?