# How can flux have a unit?

1. May 17, 2015

### Brajesh kedia

Flux is the no. Of electric feild lines per unit area..how can it have a unit

2. May 17, 2015

### Staff: Mentor

Area always has a unit. Cm2, M2, Ft2, etc.

3. May 17, 2015

### rumborak

Also keep in mind that the flux is *proportional* to the number of field lines. It's not just a number.

4. May 17, 2015

### stedwards

Sure. Tell this to the mechanical engineers, students, professors, and text writers alike, attempting to explain the stress tensor and the pressure vector. Area, for these indocrinated, is unitless.

Last edited: May 17, 2015
5. May 29, 2015

### Brajesh kedia

Frankly speaking i have not got ur answers

6. May 29, 2015

### Brajesh kedia

Total Flux as explained is no of lines of electric feild through a surface

7. May 29, 2015

### Staff: Mentor

Then you need to explain more about what you don't understand.

8. May 29, 2015

### nasu

No, the flux is not defined in terms of number of lines. There is no way to count these lines, they don't have a physical existence. Not more than the longitude lines, for example. The reference to the number of lines (if it is used at all) is just to offer some sort of intuitive understanding of the flux. In this case it should be said that the flux is proportional to the number of field lines.

The real definition is for example here (for magnetic field):
http://en.wikipedia.org/wiki/Magnetic_flux

9. May 29, 2015

### sophiecentaur

As you say, nasu, there is no such thing as a 'field line'; it's just a way of visualising things. Magnetic flux Φ is just BA, where B is the field and A is the area at right angles to the field vector.

10. May 29, 2015

### stedwards

Of some historical interest: A "line" is an obsolete unit of magnetic flux density. 1 line = 1e-8 weber [Wb].

11. May 29, 2015

### HomogenousCow

The units are pretty clear from the definition.

12. May 29, 2015

### Staff: Mentor

Are you saying that engineers don't regard the stress tensor as having units of force per unit area?

Chet

13. May 29, 2015

### morrobay

I thought flux was quantified with the Poynting Vector, http://en.wikipedia.org/wiki/Poynting_vector . And regarding physical existence: Is it not the flux of the E and B fields that propagate at c ?

14. May 29, 2015

### stedwards

Oh, no. It's worse than that. The pressure tensor $\sigma$, or pressure matrix is a linear map. It takes a vector, normal to an area $a$ and spit-out a pressure $P$.

$P= \sigma \cdot a$

Pressure has units of pressure. That's easy enough. Area has units area. Properly speaking, the stress tensor has units of Force per Length^4.

Maybe you don't like the idea of the area normal having units of area, and want it to have units of length. Now the stress tensor has different units, but these units are not pressure.

Last edited: May 29, 2015
15. May 29, 2015

### Staff: Mentor

As an engineer with over 50 years of experience in solid mechanics and fluid mechanics, I can assure you that your recollection about this is not correct. The Cauchy stress relationship specifies dotting the stress tensor with a unit normal vector to an area to spit-out the traction vector (or stress vector) acting on the area. The pressure tensor is the isotropic part of the stress tensor. If the stress tensor happens to be isotropic (as in the case of a fluid at hydrostatic equilibrium, for example), the dot product of the stress tensor with the unit normal gives you the hydrostatic pressure p times the unit normal. So, since the stress tensor is being dotted with a unit normal, the units of the pressure vector are still force per unit area. I think what you are referring to is that if you dot the stress tensor with a unit normal times a differential element of area dA, you obtain the differential force acting on that area. In this case, the units of the differential force are those of force (since the area units cancel out).

Chet

16. May 30, 2015

### stedwards

As an engineer with a wee bit less than 50 years of experience, I invite you balance the units as well as the variables.
For the equation $P[F/A] = A[D^2] \sigma[F/A]$, the units do not balance. Units are in square brackets.

F =Force
P=Pressure
A=Distance^2

Everything you learned in school is wrong.

17. May 30, 2015

### morrobay

From my text : The units of the Poynting Vector, S = 1/μ0 E x B are in watts/meter2
From S = dU/dtA =EBAdx/μ0 c(dx/c)A
S in direction of EM wave propagation.

Last edited: May 30, 2015
18. May 30, 2015

### Staff: Mentor

The correct equation is not $\vec{P}=\vec{σ}\centerdot \vec{a}$. The correct equation is $\vec{F}=\vec{σ}\centerdot \vec{a}$, where vector $\vec{F}$ is the force acting on the area, not the pressure. Now, if you continue issuing this string of misinformation, you will be receiving infraction points, which can lead to a ban from Physics Forums.

Chet

19. May 30, 2015

### Brajesh kedia

Please note i just know about total electric flux(means flow) stating that no. Of electric feild lines which i myself believe cannot be counted and if counted cannot have unit..please explain?

20. May 30, 2015

Staff Emeritus
Repeating the question is not going to help, I'm afraid.