# Magnetic flux question

by Reshma
Tags: flux, magnetic
 P: 777 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?
 HW Helper Sci Advisor P: 2,004 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 $\vec F=\vec \nabla \times \vec A$ 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')
 P: 777 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?
HW Helper