1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Magnetic flux question

  1. Mar 4, 2006 #1
    This question might seem rather naive.
    We define the magentic flux through a loop by [itex]\Phi = \int \vec B \cdot d\vec a[/itex]. 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?
  2. jcsd
  3. Mar 4, 2006 #2


    User Avatar
    Science Advisor
    Homework Helper

    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:[itex]\vec \nabla \cdot \vec F =0[/itex], then we can write [itex]\vec F=\vec \nabla \times
    \vec A[/itex] 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')
  4. Mar 5, 2006 #3
    Thanks for the reply.
    So, that means [itex]\vec \nabla \cdot \vec B = 0[/itex] guarantees that [itex]\int \vec B \cdot d\vec a[/itex] is the same for all surfaces within a given boundary?
  5. Mar 5, 2006 #4


    User Avatar
    Science Advisor
    Homework Helper

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Magnetic flux question
  1. Magnetic flux (Replies: 1)

  2. Magnetic flux (Replies: 7)

  3. Magnetic flux (Replies: 2)