# Why is current not considered to be a vector?

1. Dec 5, 2007

### PFStudent

1. The problem statement, all variables and given/known data

Currently, I am going through the magnetism section of my University Physics II course and I recognized that current has a magnitude and a direction. So, I was wondering why is that we do not treat current as a vector (Since, the direction current moves in can be represented using a unit vector.)?

2. Relevant equations

Biot-Savart Law

$${d{\vec{B}}} = {{\frac{{\mu}_{0}}{4{\pi}}}{\cdot}{\frac{Id{\vec{s}}{\times}{\vec{r}}}{{r}^{3}}}}$$

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

$$q = n_{e}e, {{.}}{{.}}{{.}}{n_{e}} = \pm1, \pm 2, \pm 3,...,$$

e $\equiv$ elementary charge

$${n_{e}} = {\pm}N, {{.}}{{.}}{{.}}{N} = 1, 2, 3,...,$$

$${N_{V}} = \frac{n_{e}}{V}$$

$${n_{e}} = {N_{V}}{V}$$

3. The attempt at a solution

I guess, we do not treat current as a vector because we recognize that any given current can go in two possible directions. Either forward or backward. So, because of the limit on the directions current can take we consider it a scalar? Is that right?

Thanks,

-PFStudent

Last edited: Dec 5, 2007
2. Dec 5, 2007

3. Dec 5, 2007

### PFStudent

Hey,

Thanks for the link dst. I understand now why it is difficult to consider current as a vector. In addition, I remembered now that current can be defined as follows,

$${I} = {\int_{}^{}}{\vec{J}}{\cdot}{d{\vec{A}}}$$

and

$${I} = {\frac{dq}{dt}}$$

However, what if you take,

$${\vec{J}} = {{N_{V}}{e}{\vec{v}_{d}}}$$

and rewrite as,

$${{J}{\hat{u}}} = {{N_{V}}{e}{{v}_{d}}{\hat{u}}}$$

Where, $${\hat{u}}$$ is some unit vector.

And rearranging as,

$${{I}{\hat{u}}} = {{n_{e}}{e}{{v}_{d}}{A}{\hat{u}}}$$

Why would the above be considered incorrect?

Thanks,

-PFStudent