Why is current not considered to be a vector?

  • Thread starter Thread starter PFStudent
  • Start date Start date
  • Tags Tags
    Current Vector
Click For Summary
SUMMARY

The discussion centers on the classification of electric current as a scalar rather than a vector despite its directional properties. Participants reference the Biot-Savart Law and the definition of current through the equation I = ∫ J · dA. The consensus is that current is treated as a scalar because it can flow in two opposite directions, which limits its vectorial representation. The conversation highlights the distinction between current density (J) as a vector and current (I) as a scalar quantity.

PREREQUISITES
  • Understanding of electric current and its properties
  • Familiarity with vector calculus and scalar quantities
  • Knowledge of the Biot-Savart Law
  • Basic concepts of current density (J) and its relation to electric current (I)
NEXT STEPS
  • Study the Biot-Savart Law in detail to understand its applications in magnetism
  • Learn about current density (J) and its vector nature
  • Research the mathematical treatment of scalars versus vectors in physics
  • Explore advanced topics in electromagnetism, particularly the implications of treating current as a scalar
USEFUL FOR

Students in physics, particularly those studying electromagnetism, educators explaining the concepts of current and current density, and anyone interested in the mathematical foundations of vector and scalar quantities in physics.

PFStudent
Messages
169
Reaction score
0

Homework Statement



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.)?

Homework Equations



Biot-Savart Law

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

<br /> {J} = {\frac{I}{A}}<br />

<br /> q = n_{e}e, {\textcolor[rgb]{1.00,1.00,1.00}{.}}{\textcolor[rgb]{1.00,1.00,1.00}{.}}{\textcolor[rgb]{1.00,1.00,1.00}{.}}{n_{e}} = \pm1, \pm 2, \pm 3,...,<br />

e \equiv elementary charge

<br /> {n_{e}} = {\pm}N, {\textcolor[rgb]{1.00,1.00,1.00}{.}}{\textcolor[rgb]{1.00,1.00,1.00}{.}}{\textcolor[rgb]{1.00,1.00,1.00}{.}}{N} = 1, 2, 3,...,<br />

<br /> {N_{V}} = \frac{n_{e}}{V}<br />

<br /> {n_{e}} = {N_{V}}{V}<br />

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:
Physics news on Phys.org
Hey,

dst said:

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,

<br /> {I} = {\int_{}^{}}{\vec{J}}{\cdot}{d{\vec{A}}}<br />

and

<br /> {I} = {\frac{dq}{dt}}<br />

However, what if you take,

<br /> {\vec{J}} = {{N_{V}}{e}{\vec{v}_{d}}}<br />

and rewrite as,

<br /> {{J}{\hat{u}}} = {{N_{V}}{e}{{v}_{d}}{\hat{u}}}<br />

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

And rearranging as,

<br /> {{I}{\hat{u}}} = {{n_{e}}{e}{{v}_{d}}{A}{\hat{u}}}<br />

Why would the above be considered incorrect?

Thanks,

-PFStudent
 

Similar threads

Replies
8
Views
14K
  • · Replies 3 ·
Replies
3
Views
5K
Replies
2
Views
5K
  • · Replies 5 ·
Replies
5
Views
15K
Replies
7
Views
9K
  • · Replies 9 ·
Replies
9
Views
10K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 1 ·
Replies
1
Views
6K
Replies
12
Views
2K
  • · Replies 5 ·
Replies
5
Views
1K