Curiosity about current density

[STEMucator]

Why does the current across the cross section of an element, with perpendicular area vector $d \vec{A}$, have a normal integral symbol in its definition:

$i = \int \vec{J} \cdot d \vec{A}$​

Why is it not a closed surface integral?

$i = \oint \vec{J} \cdot d \vec{A}$​

Usually we take the current density $\vec{J}$ to be parallel with $d \vec{A}$ (current density uniform) as to derive this equation:

$i = \int \vec{J} \cdot d \vec{A} = J \int \cos (0°) dA = J \int dA = JA$

$i = JA$

 

[Hardik Batra]

\overline{}\overline{}

Usually we take the current density $\vec{J}$ to be parallel with $d \vec{A}$ (current density uniform) as to derive this equation:

$i = \int \vec{J} \cdot d \vec{A} = J \int \cos (0°) dA = J \int dA = JA$

$i = JA$

i = JA
This equation is used when current density is uniform over the cross section Area.

It the current density is non uniform then
i=∫J⃗ ⋅dA⃗
this equation will be used.

 

[STEMucator]

\overline{}\overline{}
i = JA
This equation is used when current density is uniform over the cross section Area.

It the current density is non uniform then
i=∫J⃗ ⋅dA⃗
this equation will be used.

Yes, I understand this. My concern was as to why a closed surface integral is not being used.

 

[AlephZero]

My concern was as to why a closed surface integral is not being used.

Because it is not necessarily a closed surface.

For example, if a conducting wire is in an electromagnetic field, you might want to find the current in the wire by taking the integral over the cross section of the the wire.

 

[STEMucator]

Because it is not necessarily a closed surface.

For example, if a conducting wire is in an electromagnetic field, you might want to find the current in the wire by taking the integral over the cross section of the the wire.

This confused me a little.

If we slice the wire, is the circular area we integrate over not a closed surface?

For example, suppose the radius of the cross section is $R = 2.0 m$. Take $\vec J (r) = ar^4$ where $a$ is a constant and $r$ is the varying radius.

The current that flows closer to the outer surface and the current that flows closer to the center of the surface can be calculated:

$I_{Surface} = \oint \vec J \cdot d \vec A = a \int_{1.8 m}^{2.0 m} r^4 2 \pi r dr = 31.40a$

$I_{Inside} = \oint \vec J \cdot d \vec A = a \int_{0 m}^{1.8 m} r^4 2 \pi r dr = 35.62a$

This of course validates that most of the current density will be near the surface if we used $\vec J (r) = ar^6$ perhaps.

EDIT: Indeed $r^6$ is enough to show this:

Surface:

http://www.wolframalpha.com/input/?i=2+\pi+*+%28integrate+r^7++dr+from+1.8+to+2.0%29&dataset=&equal=Submit

Inside:

http://www.wolframalpha.com/input/?i=2+\pi+*+%28integrate+r^7++dr+from+0+to+1.8%29

 

[AlephZero]

I think a "closed surface" means a surface that encloses a volume of space, and does not have a boundary. http://en.wikipedia.org/wiki/Closed_surface#Closed_surfaces

I think the $\oint$ notation is only used for closed surfaces or closed curves (which do not have end points).

Your integrals are fine, but they are not closed surfaces in that sense.

 

[STEMucator]

I think a "closed surface" means a surface that encloses a volume of space, and does not have a boundary. http://en.wikipedia.org/wiki/Closed_surface#Closed_surfaces

I think the $\oint$ notation is only used for closed surfaces or closed curves (which do not have end points).

Your integrals are fine, but they are not closed surfaces in that sense.

Are the surfaces integrated over not compact? Here's a diagram I drew to help visualize:

http://gyazo.com/230a42591eb0c596955e236bcbab7746

EDIT:

Also, I can think of other examples such as this, where it's easy to see why $\oint$ is used:

$V = - \oint \vec E \cdot d \vec s = - \int_i^f \vec E \cdot d \vec s$

Where we calculate $V$ from the electric field. The electric force is conservative, the path integral is independent.

 

[ehild]

Also, I can think of other examples such as this, where it's easy to see why $\oint$ is used:

$V = - \oint \vec E \cdot d \vec s = - \int_i^f \vec E \cdot d \vec s$

Where we calculate $V$ from the electric field. The electric force is conservative, the path integral is independent.

The potential difference ΔV is calculated from the electric field strength as the line integral

$V_f−V_i=−\int_i^f\vec E \cdot \vec{ds}$

The contour integral along a closed path produces zero, as $\oint \vec E \cdot \vec{ds}$ means that you integrate from i to f along a path and then from f to i along an other path. Changing the boundaries in the second integral changes the sign, so you have the difference of integrals between the same points along two different lines, which are the same if the electric field is conservative. Therefore $\oint \vec E \cdot \vec{ds}=0$

The same with the integral of the current density along a closed surface. The closed surface embeds a volume V. $\vec J \cdot \vec{d A }$ means the charge leaving the volume through the surface element dA in unit time when it is positive, and it means the charge entering the volume in unit time if the product is negative. The total integral over the closed surface is equal to the rate of change of the charge in the volume.

Check what you learned in the school about the definition of current: The current in a wire/conductor is the charge flowing through the cross section of a conductor in unit time. It is the integral of the current density for the cross section and not the closed surface integral for the whole piece of wire, which is zero.

A surface is enclosed by its boundary line. Integrating over the surface is not a closed-surface integral and also is not the same as the line integral over the closed boundary line.
ehild