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I don't see Faraday's law of induction and Maxwell-Ampere's law are totally symmetric in the sense that Maxwell-Ampere's law has a factor of $ϵ_0μ_0$: \begin{align} \nabla\times\mathbf E&=-\frac{\partial\mathbf B}{\partial t} \\ \nabla\times\mathbf B&=\color{blue}{\mu_0\varepsilon_0}\frac{\partial\mathbf E}{\partial t} \end{align} Can anyone explain me why.

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marked as duplicate by Rob Jeffries, Kyle Kanos, Jon Custer, John Rennie, Mike Nov 30 '17 at 17:04

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Faraday's law and Ampere's law are manifestly asymmetric, as Faraday's law is a homogeneous equation and Ampere's law is not. Therefore, I assume you mean in the absence of a current density, in which case the equations become (in SI units) $$\nabla \times \vec E = -\frac{\partial \vec B}{\partial t}$$ and $$\nabla \times \vec B = \epsilon_0 \mu_0 \frac{\partial \vec E}{\partial t}$$

There is a necessary asymmetry present because $\vec E$ and $\vec B$ have different dimensions, as do $\nabla$ and $\frac{\partial}{\partial t}$.

In the gaussian unit system, the electric field and magnetic field have the same units, so the equations become $$ \nabla \times \vec E = -\frac{1}{c} \frac{\partial \vec B}{\partial t}$$ $$ \nabla \times \vec B = \frac{1}{c} \frac{\partial \vec E}{\partial t}$$

We can do even better and use natural units where $c=1$, in which case we have $$ \nabla \times \vec E = -\frac{\partial \vec B}{\partial t}$$ $$ \nabla \times \vec B = \frac{\partial \vec E}{\partial t}$$


The lesson here is that it's important to be able to see through the outward appearance of an equation to its inner structure. Often times, constants can be "absorbed" by redefining certain quantities and/or changing unit systems. This is one aspect of a larger concept called nondimensionalization, and it's an important skill to develop as time goes on.

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  • $\begingroup$ @YassineSifeddine If you're satisfied with this answer, then you can mark it "accepted." Otherwise, if you wish for more clarification, then let me know. $\endgroup$ – J. Murray Nov 9 '17 at 3:25
  • $\begingroup$ To be honest i hope you to clarify me more please. $\endgroup$ – Yassine Sifeddine Nov 9 '17 at 3:28
  • $\begingroup$ Can you explain what is causing you trouble? $\endgroup$ – J. Murray Nov 9 '17 at 3:40
  • $\begingroup$ For exemple if we add magnetic monopoles gauss laws for electric field and magnetic field will be symmetric but that's not the case for amper and faraday laws. $\endgroup$ – Yassine Sifeddine Nov 9 '17 at 3:47
  • $\begingroup$ They are not precisely the same because of the minus sign, but why should they be? $\endgroup$ – J. Murray Nov 9 '17 at 3:48

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