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Use Maxwell’s second equation (Faraday’s law) to prove the flux continuity law: ${\rm div} (B) = 0$, where $B$ is time-varying.

My approach would be to prove the flux continuity law using Gauss' Law for magnetic fields, but I'm unsure on how to prove it using Faraday's Law. Any hints?

Thanks

caverac
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Adam
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1 Answers1

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$\newcommand{\uvect}{\hat{\bf e}}$

Frist note that

$$ \nabla\cdot (\nabla\times{\bf E}) = 0 \tag{1} $$

You can show this is true by expanding each term (abusing a bit Einstein's notation):

$$ (\uvect_i\partial_i)\cdot(\uvect_k \epsilon_{klm}\partial_l E_m) = \delta_{ik}\epsilon_{klm}\partial_i\partial_l E_m = \epsilon_{klm}\partial_i\partial_l E_m = 0 $$

where I used the fact that $\epsilon_{klm} = -\epsilon_{lkm}$ whereas $\partial_k\partial_l E_m = \partial_l\partial_k E_m$. With this in mind note that

\begin{eqnarray} \nabla\times{\bf E} &=& \frac{\partial{\bf B}}{\partial t} \\ \nabla\cdot(\nabla\times{\bf E}) &=& \nabla\cdot\left(\frac{\partial{\bf B}}{\partial t}\right) \\ 0 &=& \frac{\partial}{\partial t}(\nabla \cdot {\bf B}) \tag{2} \end{eqnarray}

caverac
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