# Current flows which way? [closed]

Current flows through a resistor (with no direction since it is not a vector/can flow from any potential)?

I thinks it is with no direction since it is not a vector. Is that right?

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## closed as unclear what you're asking by Chris White, Nathaniel, Waffle's Crazy Peanut, Emilio Pisanty, akhmeteliJul 5 '13 at 22:57

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Note that the current density is a vector. – Qmechanic Jul 4 '13 at 0:12

A current is like the flow of a fluid, so it's a vector field. We tend to think of resistors as one dimensional when we draw circuit diagrams, but of course they aren't.

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Should we be careful with the use of current vs current density? The current density is a vector field defined proportional to the electric field, whereas the current is defined as the integral of the component of the current density through some particular cross-sectional area. – Will Jul 3 '13 at 18:46
Oops, yes, good catch. – John Rennie Jul 4 '13 at 9:24
@JohnRennie Isn't current a scalar, though? It doesn't follow the Vector Law of Addition, neither can you divide it into components like vectors. – mikhailcazi Jul 4 '13 at 16:12
@mikhailcazi How does it not follow vector addition? Show me. See my solution. – Will Jul 4 '13 at 16:19
@Will Well if you take wires arranged in a closed polygon, vector sum says the resultant current should be zero, but that's not true - a circuit is a closed polygon. And current does flow. – mikhailcazi Jul 4 '13 at 16:24

I feel really bad for the OP getting these downvotes. The question (if worded better) is a really good question. There is obviously a lot of confusion over this - given the range of solutions posted. I think the main confusion comes from two places: (1) using current density in place of current (whether consciously or not), (2) Not realizing that a 1-D vector is a scalar.

Current density:

For simplicity, assume we have a conductor with a constant conductance, $\sigma$. Also for simplicity, let's assume that the electric field is time-independent, but could vary in space. The electric field, $\vec{E}(\vec{x})$, is a vector field, that is, at each point in space, we associate a vector space. The relationship between the electric field and the current density in this simplified model is $$\vec{J}(\vec{x}) = \sigma~\vec{E}(\vec{x})$$ From this equation, we see that $\vec{J}(\vec{x})$, the current density, is also a vector field.

Current:

The current is defined with reference to some area as $$I_A = \int_A \vec{J}(\vec{x})\cdot d\vec{A}$$ In circuits, the area we care about is the cross-sectional area of the wire. With a DC source the electric field is constant throughout the wire (if AC, still constant spatially but has time dependence), that is $\vec{E}(\vec{x}) = E_0~\hat{z}$ (assuming the wire is along the $z$-axis). This means that the current density is $$\vec{J} = \sigma~E_0~\hat{z}$$ and so the current is (area of interest is the cross-section of the wire, with $d\vec{A} = dA\hat{z}$) $$I = \int dA~\sigma E_0~\hat{z}\cdot\hat{z} = A\sigma ~ E_0$$ In this idealized situation, we see that $I$ can be treated as a 1-D vector, so far as the vector space axioms are concerned. We also note that a 1-D vector is a scalar! Note that I am not saying scalar field or vector field.

Let's see what happens when we aren't confined to a wire and have spatial varying electric fields.

Take some area through a region in a conductor (with constant $\sigma$ as before) where the electric field has spatial dependence $\vec{E}(\vec{x})$. The current associated with this electric field and area is $$I_A = \sigma\int_A \vec{E}(\vec{x})\cdot d\vec{A}$$ With a little thought, one can see that this doesn't change the argument in our wire case. The current defined this way still produces a 1-D vector, whose direction indicates which way the total flow of positive charges (by convention) is flowing, through the prescribed surface. And again, a 1-D vector IS a scalar.

Note: in this situation, adding currents is physically identical to adding electric fields (as $\sigma$ is assumed constant).

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Electric current has a direction and that is the direction in which positive charge flows.

In circuit theory, electric current enters the positive terminal of a resistor and exits the negative terminal.

If electric current did not have a direction, how would we determine the polarity of a magnetic dipole due to a current loop?

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