# Current is a scalar quantity, but how does it have a definite direction? [duplicate]

A scalar quantity is one which has magnitude but no direction. A vector quantity is one which has both magnitude and direction. The direction of current is opposite to the flow of electrons.

Electric current has both magnitude and direction so ideally it 'should' be a vector. But all sources say it's a scalar just because it does not follow vector addition.

Is it enough to ignore the direction of current, which also is an important factor in determining whether a quantity is scalar or vector?

• Current is a vector. Mar 22 '16 at 9:12
• Duplicate of Why is current a scalar quantity? This is a stupid terminology issue where people say that current is "not a vector" just because you don't get "total" current by adding individual currents, but any current density, in contrast, is a perfectly fine vector. Mar 22 '16 at 10:19

Electric current has a magnitude given by $$\left|\int_\Sigma \mathbf{J}\cdot\mathbf{N}_e\,\mathrm d\sigma\right|$$where, in the surface integral, $\Sigma$ is the orientable surface through which the current flows, $\mathbf{J}$ is the current density and $\mathbf{N}_e$ is the external normal unit vector to $\Sigma$. You see that, by changing the orientation of the surface, the normal vector changes its sign and therefore the integral$$\int_\Sigma \mathbf{J}\cdot\mathbf{N}_e\,\mathrm d\sigma$$also changes its sign.
You can arbitrarily choose one of the two sides, having external normal $\mathbf{N}_{e,1}$, of the orientable surface $\Sigma$ in a way such that, at a particular instant, the current is $i\ge 0$ when $$\int_\Sigma \mathbf{J}\cdot\mathbf{N}_{e,1}\,\mathrm d\sigma\ge 0.$$Of course, if we call $\mathbf{N}_{e,2}$ the external normal on the other side of $\Sigma$, we have that $$\int_\Sigma \mathbf{J}\cdot\mathbf{N}_{e,1}\,\mathrm d\sigma\ge 0\iff \int_\Sigma \mathbf{J}\cdot\mathbf{N}_{e,2}\,\mathrm d\sigma\le 0.$$ In that sense, the direction of the current is represented by the orientation of the surface, considered to calculate it, that makes the integral $\int_\Sigma \mathbf{J}\cdot\mathbf{N}_{e,1}\,\mathrm d\sigma\ge 0$ positive.
So you have that current is not at all a vector belonging to $\mathbb{R}^3$ - while the current density $\mathbf{J}$, of course, is - because its direction can only be mono-dimensional: in or out of a side of the surface. If you wish, any real number can be seen as an element of a mono-dimensional vector space, in fact a structure of vector space is quite trivially defined on $\mathbb{R}$, but that is not, as far as I know, what is usually meant by vector quantity in elementary physics, which usually is an element of a vector of a space of dimension greater than $1$.
• @drvrm If, by current of water, you mean the mass flow rate, then it's a scalar for the same reason electric current is and all I've said about electric current apply to it with $\mathbf{J}$ = mass flux. Mar 22 '16 at 11:46