Magnetic flux question


by Reshma
Tags: flux, magnetic
Reshma
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#1
Mar4-06, 01:53 AM
P: 777
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?
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Galileo
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#2
Mar4-06, 03:42 AM
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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')
Reshma
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#3
Mar5-06, 01:12 AM
P: 777
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?

Galileo
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Mar5-06, 02:32 AM
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Magnetic flux question


Yes..........


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