Why is flux of magnetic field always zero for closed surfaces?

Question

I'm having troubles accepting that the magnetic flux through a closed surface is always zero. I understand the fact that no magnetic monopoles exist , therefore all the 'lines' of magnetic flux that exit the surface must enter back, but if the surface is irregular, the angle the field creates with the normal to the surface would be different in the 2 cases, therefore making their dot product different and the flux is not going to be zero, also the value of the magnetic field that exits the surface could differ from the value of the magnetic field that enters.

What is the factor that makes the flux zero? Am I missing some details?

Answer 1

If it holds, the local form of the of the "non-existence" of the magnetic monopole is the Gauss' law for the magnetic field,

$$\nabla \cdot \mathbf{b}(\mathbf{r},t) = 0 \qquad \forall \, \mathbf{r},t .$$

Since it holds for every point in space,

$$\int_{\mathbf{r}\in V} \nabla \cdot \mathbf{b}(\mathbf{r},t) = 0 \ , \qquad \forall \, V $$

If divergence theorem holds, it's possible to transform the volume integral over $V$ of the divergence as a flux integral across the closed boundary $\partial V$,

$$0 = \int_{\mathbf{r} \in V} \nabla \cdot \mathbf{b}(\mathbf{r},t) = \oint_{\partial V} \mathbf{b}(\mathbf{r},t) \cdot \mathbf{\hat{n}} = \Phi_{\partial V}(\mathbf{b}) \ , \qquad \forall \, V$$

i.e. the flux of the magnetic flux across any closed surface $\partial V$, whatever is its shape, is identically zero.

Answer 2

All these things are certainly true, but that is exactly the beauty of working in the differential notation of $\nabla \cdot \overrightarrow{B} = 0$. With the divergence theorem (https://en.wikipedia.org/wiki/Divergence_theorem), you know that the surface integral over any surface is zero. In your representation, it is indeed the case that for your single field line in green, the two contributions at the edges are different, and would not give zero. It is only when considering the whole space of all field lines that these would indeed average to zero, independent of the surface.