• Current is a scalar $I$ with units of $\mathrm{[J/s]}$. It is defined as $I=\frac{\mathrm{d}Q}{\mathrm{d}t}$.
  • Current density is a vector $\vec{J}$ (with magnitude $J$) with units of $\mathrm{[J/s/m^2]}$. It is current per unit cross-sectional area, and is defined as $\vec{J}=nq\vec{v_d}$ (where $n$ is the number of moving $q$-charges with drift velocity $v_d$).

Why is $I$ defined to not have a direction? Current density $\vec{J}$ is defined as a vector, so why is current $I$ not?

There are many questions about current vs. current density

... like this, this and this, but none answers my question about one being vector and the other being scalar. Is it simply just a definition? It just seems so obvious to define current as a vector too.

Another but equivalent definition of current density is $I = \int \vec{J} \cdot d\vec{A}$. Mathematically, the dot product gives a scalar. But, for me this doesn't give much explanation still, as we could just as well have mathematically defined current as a vector and then used the area in scalar form in an equivalent formula like: $\vec I = \int \vec{J} \cdot dA$.

Is it just a definition without further reason, or is there a point in keeping $I$ in scalar form?

  • 5
    $\begingroup$ If I'm studying a circuit and I want to know how much charge is flowing through a wire I don't need a vector. The wire is effectively one-dimensional, so the + and - sign of $I$ tells me all the direction information I need to know. That's really it. $\endgroup$
    – DanielSank
    Commented Jan 2, 2015 at 18:31
  • $\begingroup$ The current is a scalar because the charge Q is a scalar - like mass for example. So you're asking how much of this scalar quantity is flowing through a wire per time. Since there is no vector quantity in the question, the answer, i.e. the current I, is also a scalar quantity. (hmm essentially I said the same as danielsank but with different wording ... maybe you get it now. ^^ ) $\endgroup$
    – user42076
    Commented Jan 2, 2015 at 19:00
  • 2
    $\begingroup$ @Steeven - Current does have a magnitude and direction. However, it is usually just expressed as a scalar due to the 1D nature of simple "wire problems," as explained by Daniel. Current is a type of flux and thus, satisfies the continuity equation whether you are using linear current I, surface current density $\boldsymbol{\kappa}$, or volume current density $\mathbf{J}$. $\endgroup$ Commented Jan 2, 2015 at 20:18
  • 1
    $\begingroup$ Current is a measure of the total charge passing through a surface, in a direction normal to that surface at every point on the surface. It has no direction vector for the same reason basketball scores do not have a direction vector, you're just counting stuff up. $\endgroup$
    – crobar
    Commented Apr 24, 2018 at 8:45

6 Answers 6


According to my understanding, indeed you could define a physical quantity like $$\vec{I} = I\;\vec{n_d}$$ where $\vec{n_d}$ is the unitary drift direction. There is no problem with that. But what is the most important is to understand the harmony between different quantities. I mean that there is some little subtleties between $I$ and $\vec{j}$.

The current is defined according to a surface $A_t$ (and is a local quantity: the position of the surface). Since the surface may be tilted (not perpendicular to $\vec{n_d}$), then in general, we should write $$I=n\;q \; \vec{A_t}.\vec{v_d}=n\;q\;A_t \; \vec{n_{A_t}}.\vec{v_d}=n\;q\;A_t \; cos(\theta).v_d$$ where $\theta$ is the angle between $\vec{A_t}$ and $\vec{v_d}$ (If the charge $q$ is negative, $\vec{I}$ and $\vec{n_d}$ would have opposite orientations, which fits with the usual convention of the electric current). Of course, if we have considered a surface $A$ which is perpendicular to the drift, the current would be the same but we would write $$I=n\;q \; \vec{A}.\vec{v_d}=n\;q\;A \; v_d$$

That is, let's talk about the densities. We have $$\mathrm{d}I=n\;q\;v_d\;\mathrm{d}A=n\;q\;v_d\;cos(\theta)\;\mathrm{d}A_t$$ The scalar current density is given by $$j=\frac{\mathrm{d}I}{\mathrm{d}A}=n\;q\;v_d=\frac{1}{cos(\theta)}\frac{\mathrm{d}I}{\mathrm{d}A_t}$$ Yet, you see that you have to be careful about the differential area that you put in the denominator. The current could then be calculated as $$I=\int_A j\;\mathrm{d}A=\int_{A_t} j\;cos(\theta)\;\mathrm{d}A_t$$ Here also, we see a possible source of confusion. This can be fixed if we define a vector current density as $$\vec{j}=j\;\vec{n_d}=n\;q\;v_d\;\vec{n_d}$$ The direction of $\vec{j}$ is resolved too: it's that of the local drift in the considered material position. That direction may be different than that of the average drift of the whole current $\vec{I}$. That is, the expression of the current can be written simply as $$I=\int_A \vec{j}\;\mathrm{d}\vec{A}=\int_{A_t} \vec{j}\;\mathrm{d}\vec{A_t}$$ Or, you could use your own convention and write $$\vec{I}=\bigg(\int_A \vec{j}\;\mathrm{d}\vec{A}\bigg)\;\vec{n_d}=\bigg(\int_{A_t} \vec{j}\;\mathrm{d}\vec{A_t}\bigg)\;\vec{n_d}$$


As you already mentioned $$ I = \int \vec{J} \cdot d\vec{A} $$ current is the charge-flow through a given surface. So If one talks about currents, the surface (and therefore its normal direction) is to be understood beforehand.

You could assign a direction like this: $$ \frac{\vec{I}}{I} = \frac{\int d\vec{A}}{A} $$ which is the mean normal direction of the surface but this makes no sense for curved surfaces (e.g $\vec{I} = 0$ for a ball, regardless of the magnitude $I$).


This is basically just rephrasing some of the existing answers, but:

Current is a scalar $I$ with units of [J/s]. It is defined as $I = dQ / dt$.

Not quite - $dQ/dt$ isn't precisely defined. What is $Q$ the charge of? Current is actually defined to be the charge per time passing through some surface $S$. In terms of current density it can be expressed as $I := \iint_S {\bf J} \cdot d{\bf S}$. The fact that it is only defined with respect to some surface $S$ means that it is an inherently global quantity, unlike current density ${\bf J}$, which can be unambiguously defined at a single point. (Current is also in general not a very physical quantity, unlike current density, because in principle $S$ can be any crazy Gaussian surface. It's exactly analagous to the difference between electric field vs. electric flux.) As others have mentioned, if the surface $S$ is curved then your proposal to integrate with respect to $dS$ instead of $d{\bf S}$ doesn't work.

The problem is that we almost always consider the current going through thin wires, in which case none of these subtleties come up. For the mathematical intuition, it's better to think of current going through a bulk conductor (possibly including eddy currents, etc.) instead.


An old question, maybe to help anyone else with this issue. According to Griffths Introduction to Electrodynamics current is a vector, defined by

$\mathbf{I} = \lambda \mathbf{v}$

Where $\mathbf{v}$ is the velocity of the charges.

The only reason we don't do that is because where the electrons are going is completely dictated by the wire. If a train were running on a track, you would talk about its speed, rather than its velocity.


The current is defined as the rate at which charge flows, therefore, it does not make much sense the add a direction to something that expresses a rate. On the other hand current density involves an area, how much electric current is flowing through a given cross section something physical with a direction, so it makes sense to define it as a vector.

  • $\begingroup$ Rates can have direction. Consider velocity $\vec v \equiv d\vec x/dt$ or net force $\vec F = d\vec p/dt$. $\endgroup$
    – BMS
    Commented Jan 2, 2015 at 20:10
  • $\begingroup$ @AlanZ2223 - Current, whether density or simple current, is a type of flux. It satisfies a continuity equation and so does have a direction associated with it. The lack of use of a vector direction is typically due to laziness and the 1D nature of a simple wire problem. $\endgroup$ Commented Jan 2, 2015 at 20:14
  • $\begingroup$ That is because the components of those measurements have vectors, and rightly so. A rate of change in displacement makes sense to have it as a vector. Rate of change of momentum all makes sense. $\endgroup$
    – AlanZ2223
    Commented Jan 2, 2015 at 20:36
  • $\begingroup$ @honeste_vivere, I am probably missing your point but, how can a flow have a direction? The "direction" of the flow is given by the surface you have chosen. $\endgroup$
    – Gilgamesh
    Commented Jun 18, 2020 at 4:07
  • $\begingroup$ @MarcoCiafa - Flow always has a direction, which is why it's described by the 3-vector fluid velocity of a particular system. I am not sure what you mean by "surface you have chosen" though. $\endgroup$ Commented Jun 18, 2020 at 12:34

What I think is the best way to look at it is that current is completely local. What I mean is that actually current is defined a distinct region of space. We care about how much charge there is in a region of space and then how much charge is in that same region without worrying to what happened to the first amount of charge. On the other hand, the current density $\vec{J}$ at a point $\vec{r}$ has to do to where is the charge at the point going. That's something that is usually forgotten when teaching circuits.


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