# Is magnetic flux always zero?

So I have a doubt regarding the equations of Magnetic flux and Gauss' Law for magnetic fields.

According to Gauss' Law for magnetic fields, integral form of div(B) = $$0$$ can be written as:

$$\oint \mathbf{B}.d\mathbf{a} = 0$$

And magnetic flux through a surface is also defined as $$\iint \mathbf{B}.d\mathbf{S} = \Phi _{B}$$

If we combine the two, does it mean that flux is always zero through the surface? I believe $$\iint$$ and $$\oint$$ are the same thing (correct me if I am wrong) ? What is the significance of $$\Phi _{B}$$ then?

• You can think of the Gauss Law flux as a count of the number field lines leaving a closed surface. Magnetic field lines always form closed loops. So each line that comes in ( - flux ) must also go out. – R.W. Bird Sep 16 '19 at 18:40

I believe ∬ and ∮ are the same thing (correct me if I am wrong)?

$$\oint$$ is a closed contour integral symbol while $$\iint$$ is a double integral symbol.

Gauss's Law for magnetism in integral form is

$$\unicode{x222F}_S\,\mathbf{B}\cdot\mathrm{d}\mathbf{A}=0$$

which states that the magnetic flux through a closed (no boundary) surface is zero (there are no magnetic monopoles). The symbol $$\unicode{x222F}_S$$ indicates integration over the closed surface $$S$$.

The magnetic flux through a surface bounded by a closed countour is not necessarily zero. For example, the Maxwell-Faraday equation is (for a surface unchanging with time)

$$\oint_{\partial \Sigma} \mathbf{E} \cdot \mathrm{d}\mathbf{l} = - \frac{\mathrm{d}}{\mathrm{d}t} \iint_{\Sigma} \mathbf{B} \cdot \mathrm{d}\mathbf{A}$$

where $$\Sigma$$ is the surface of integration bounded by the closed countour $$\partial\Sigma$$.

• Shouldn't the surface integral over $\Sigma$ in the last equation be an $\iint$ symbol, for consistency with the notation conventions established earlier in this post? – Emilio Pisanty Sep 16 '19 at 17:35