Magnetic Flux Q: How Is Flux Independent of Surface?

Reshma
Messages
749
Reaction score
6
This question might seem rather naive.
We define the magentic flux through a loop by \Phi = \int \vec B \cdot d\vec a. But an infinite number of different surfaces can be fitted to a given boundary line...so how is the flux independent of the nature of the surface used?
 
Physics news on Phys.org
In general, flux isn't defined through a loop (to my knowlegde), it is always defined through a surface. (It's a surface integral). Different surfaces bounding the same loop will in general give different answers.

Exception: If the divergence of the field F is zero everywhere:\vec \nabla \cdot \vec F =0, then we can write \vec F=\vec \nabla \times <br /> \vec A for some field A. Now you can use Stokes' theorem to prove that for a given boundary line, the flux is independent of the surface bounded by that line. Since div B=0 always and everywhere, you can unambigously talk about the magnetic flux through a loop (although I would still never say 'flux through a loop')
 
Thanks for the reply.
So, that means \vec \nabla \cdot \vec B = 0 guarantees that \int \vec B \cdot d\vec a is the same for all surfaces within a given boundary?
 
Yes...
 
Thread 'Help with Time-Independent Perturbation Theory "Good" States Proof'
(Disclaimer: this is not a HW question. I am self-studying, and this felt like the type of question I've seen in this forum. If there is somewhere better for me to share this doubt, please let me know and I'll transfer it right away.) I am currently reviewing Chapter 7 of Introduction to QM by Griffiths. I have been stuck for an hour or so trying to understand the last paragraph of this proof (pls check the attached file). It claims that we can express Ψ_{γ}(0) as a linear combination of...
Back
Top