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I'm studying magnetic fields and I have a big problem. The Gauss theorem states that the magnetic flux through a closed surface is null since the line of flux enter and exit from it in the same amount. However, the Faraday law talks about the flux linkage with a surface which can be not null in case, for example, the magnetic field is not constant in time. Why is not null as well, since it is the flux through a surface?

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2 Answers 2

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Faraday's law talks about the flux piercing through an open surface, while Gauss' law for magnetic fields is for a closed surface.

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  • $\begingroup$ Please can you give me an example? If I am correct, broadly speaking, a closed surface is a surface in which you cannot go from a side to another without crossing it. So a circumference, in which some lines of flux go through is closed or opened? $\endgroup$
    – user116008
    Jul 5, 2016 at 14:41
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    $\begingroup$ Here's a Wikipedia article I think might help Just as an example, imagine a sphere--that's a closed surface. Now imagine a sphere with a pinprick-sized hole in it--that's an open surface. A circumference is open, so is this $\endgroup$
    – GeeJay
    Jul 5, 2016 at 14:45
  • $\begingroup$ Now the difference between closed and opened surfaces is very clear, but why the flux through this two types of surfaces is different? I have to clear up my mind: for sake of simplicity I consider the flux of a fluid, I know that the flux is a vectorial quantity so entering - exiting = 0 (closed surface), but means that there is no net effect or no flux at all? $\endgroup$
    – user116008
    Jul 5, 2016 at 15:05
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    $\begingroup$ Magnetic field lines are always in closed loops. If the surface is closed, a field line that comes in must also go out. An open surface does not carry such restrictions. In a closed surface, it does not mean that there is no flux at all, simply no net flux. $\endgroup$
    – GeeJay
    Jul 5, 2016 at 15:22
  • $\begingroup$ So should not be the derivative of the flux over time to be null and the flux constant, if the net flux is null? $\endgroup$
    – user116008
    Jul 5, 2016 at 15:41
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The two concerned equations

\begin{align}\mathbf \nabla \times \mathbf E &= -\frac{\partial{\mathbf B}}{\partial t}\tag 1 \\ \mathbf \nabla \cdot \mathbf B &= 0\tag 2\end{align}

are conveying totally different meanings.

While the former $(1)$ correlates the circulation of the induced electric field with the negative time rate of change of the associated magnetic field; the later viz. $(2)$ says that the divergence of magnetic field is essentially zero.

Let's come to the present context where OP seems to confused with the integral forms of the equation: The integral forms respectively are:

\begin{align} \int_\textrm C \mathbf E \cdot \mathrm d\mathbf s &= -\frac{\mathrm d}{\mathrm dt}\, \underbrace{\color{red}{\int _\mathrm S\mathbf B\cdot\mathrm d\mathbf a}}_{\Phi\equiv\, \textrm{magnetic flux}}\tag 1 \\ \underbrace{\int_\mathrm S\mathbf B\cdot \mathrm d\mathbf a}_{\textrm{flux through}\,\color{red}{\textrm{closed surface}}} &= 0\tag 2\end{align}

OP can see the distinct red marked term in the first equation; the later doesn't talk like that. The possible misinterpretation could be averted if OP transforms the differential form to integral form keeping in mind the Gauss' or Divergence theorem and the Stokes' Theorem.

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  • $\begingroup$ Very clear answer and the difference between the two quantities is clearer. I realized that my problem is on the physical meaning of flux, I read some definitions which were usefulness. Thinking of a fluid and considering a cylindrical tube, the flux is the flow rate per unit of surface; now without any internal generation or withdrawal for the Gauss theorem it should be null: why? there is still matter flowing in and out from it. $\endgroup$
    – user116008
    Jul 5, 2016 at 15:23

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