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

  • Thread starter Thread starter flyingpig
  • Start date Start date
  • Tags Tags
    Law Rectangle
AI Thread Summary
The discussion centers on the use of a rectangle in the application of Gauss's Law for a coaxial cable example, questioning its appropriateness since it does not enclose the B-field. It clarifies that the rectangle is utilized to illustrate the concept of magnetic flux rather than to apply Gauss's Law directly, which requires a closed surface. The integration performed relates to finding the B-field using Ampere's Law, with the rectangle serving to define the differential area element dA. The confusion arises from the significant width of the conductors in the example, which contrasts with typical cases involving thin conductors. Overall, the rectangle's role is to simplify the explanation of magnetic flux in the context of the example presented.
flyingpig
Messages
2,574
Reaction score
1
Physics news on Phys.org
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.
 
Then why are they only concerning the rectangle penetrated by one rectangle?
 
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.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top