Why are they using a rectangle for Guass's Law?

[flyingpig]

Homework Statement

http://www.electron.rmutphysics.com...ers-Serway-Beichne 6edr-4/32 - Inductance.pdf

Please flip to page 11/30 you will see a coaxial cable.

Read to the end where they begin integrating using Guass's Law.

Now my question is, why a rectangle? The rectangle (surface) does NOT enclosed the B-field

 

[jhae2.718]

That is not Gauss's Law, but rather finding magnetic flux. (Gauss's Law for magnetism states that the flux through a closed surface is zero; i.e. that there are no magnetic monopoles.)

What the example does is find the B-field in a<r<b using Ampere's Law (w/o Maxwell's correction) and then integrates it w.r.t. dA=l dr. The rectange is just explaining what dA is. It's not like Gauss's Law for electric fields where you need a Gaussian surface enclosing charge.

 

[flyingpig]

Then why are they only concerning the rectangle penetrated by one rectangle?

 

[SammyS]

The authors state the reason for using this rectangle and what the circuit is that is involved.

Your question may be much deeper than it first appears to be. If that is the case, then explanation could be somewhat more involved. I suspect your confusion may stem from the fact that the conductors in most of the circuits dealing with magnetic fields have been thin, having negligible thickness, whereas in this example, the conductors have significant width, so the current is spread out -- but the example seams to only be concerned with the (magnetic) flux along a very narrow strip of the conductors.

 

[rwisz]

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I would like to clarify that the use of a rectangle in Gauss's Law is a common misconception. In fact, Gauss's Law is not limited to using a rectangle as the surface of integration. The choice of surface depends on the symmetry of the problem being solved. In the case of a coaxial cable, a cylindrical surface is used for integration because it follows the symmetry of the cable. This allows for a simpler calculation of the electric field.

Furthermore, the choice of surface for integration does not necessarily have to enclose the B-field. Gauss's Law states that the flux of the electric field through any closed surface is equal to the total charge enclosed by that surface divided by the permittivity of free space. This means that the surface chosen for integration does not need to enclose the B-field, but rather the charge distribution.

In conclusion, the use of a rectangle in the problem you mentioned is not a limitation of Gauss's Law, but rather a specific choice based on the symmetry of the problem. It is important to understand the fundamentals of Gauss's Law and its application in different scenarios, rather than relying on a specific surface for integration.