Is All Magnetic Field Really Zero According to Maxwell's Equations?

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Maxwell's equations indicate that the divergence of the magnetic field B is zero, leading to the conclusion that B can be expressed as the curl of another vector function A. This results in the application of Stokes' theorem, which suggests that the line integral of A around a closed path is also zero, implying that A is path independent. Consequently, it follows that A can be represented as the gradient of a scalar function ψ, leading to the conclusion that the magnetic field B must be zero. However, there is a concern regarding the application of Stokes' theorem in this context, as the region may not be simply connected. This raises questions about the validity of the proof and the assumptions made regarding the magnetic field.
typhoonss821
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One of Maxwell's equations says that <br /> \nabla\cdot\vec{B}{=0}<br /> where B is any magnetic field.
Then using the divergence theore, we find
<br /> \int\int_S \vec{B}\cdot\hat{n}dS=\int\int\int_V \nabla\cdot\vec{B}dV=0<br />.

Because B has zero divergence, there must exist a vector function, say A, such that
<br /> \vec{B}=\nabla\times\vec{A}<br /> .
Combining these two equations, we get
<br /> \int\int_S \hat{n}\cdot\nabla\times\vec{A}dS=0<br /> .
Next we apply Stoke's theorem and the preceding result to find
<br /> \oint_C\vec{A}\cdot\hat{t}ds=\int\int_S\hat{n}\cdot\nabla\times\vec{A}dS=0<br /> .
Thus A is path independent. It follows that we can write
<br /> \vec{A}=\nabla\psi <br /> , where ψ is some scalar fution.

Since the curl of the gradient of a function is zero, we arrive at the remarkable fact that
<br /> \vec{B}=\nabla\times\nabla\psi=0<br />
that is, all magnetic field is zero!

Wow, there must be something wrong...

My thought is that we can not apply Stoke's theorem in this case because the region we discuss is not simply connected.

But I'm not sure if I am right, please help me check the proof^^
 
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The surface you use for the divergence theorem is a closed surface that encloses a volume. The surface is from the Stokes theorem is a surface that has the contour C as edge.
 
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