I've been trying to relearn electronics and have run into a conceptual block. Everywhere I look defines current as something along the lines of
$$
i = \frac{dq}{dt}
$$
where $i$ is current, $q$ is charge and $t$ is time.
Describing it as the rate of charge passing through a surface. But the definition seems odd to me because both charge and time are directionless quantities, but current has a direction, at least conceptually.
The units too seem directionless, but I would expect to see a unit like Coulomb meter per second. This has a direction and also would have the property that if the charges pass through at a rate of once a second but in one case go twice as fast as in the other then the value of one would be twice as high as the other, whereas they'd be the same for the way current is defined above.
I suspect the answer hinges on the way the surface is defined, but in my searching I haven't found anything.
Best Answer
Current is not a vector; it doesn't have a direction in the same way that a velocity has a direction.
There is a good analogy between electric current in a wire and the flux or rate of flow of water through a pipe. The idea here is that we put an imaginary surface right across the interior of the pipe. It can be on the skew, nor need it be plane. The (volume) rate of flow of water in the pipe is the volume of liquid crossing this surface per unit time. The only direction ('sense' would be a better word) that we assign to the flow rate is X to Y or Y to X, in which X and Y are the two sides of the surface. It is the same with electric current, rate of flow of charge, in a wire.
By contrast, we can describe the flow using vectors. For the water we have the velocity, $\mathbf v$, of the individual chunks of water. For the electric charge we have the velocity of the charge carriers (charge $q$, say), or a special vector called 'current density, $\mathbf J$.
The relationship between the velocity of the water (which can vary from point to point) and the volume flow rate, $\Phi$, is $$\Phi = \int_S \mathbf v.d\mathbf S$$ in which the integral is evaluated over our imaginary surface, S.
In the same way, electric current, $I$, can be calculated from $$I=\int_S \nu q \mathbf v.d\mathbf S \ \ \ \text {or}\ \ \ I=\int_S \mathbf J.d\mathbf S.$$ [$\nu$ is the number of charge carriers per unit volume of the wire.]