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What is the reason for some writing Faraday's Induction Law as $$ \nabla \times E= -\frac{1}{c}\frac{\partial B}{\partial t} $$ versus $$ \nabla \times E= -\frac{\partial B}{\partial t} ?$$

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The difference has to do with the units in which $\vec{B}$ is measured in.

In SI units Faraday's Law reads as,

$$ \nabla \times \vec{E} = - \frac{\partial \vec{B}}{\partial t} $$

In Gaussian and Heaviside-Lorentz Units it reads as,

$$ \nabla \times \vec{E} = - \frac{1}{c}\frac{\partial \vec{B}}{\partial t} $$


Basically this amounts to redifining $\vec{B}$.

$$ c\vec{B}_\text{(SI)} \equiv \vec{B}_\text{(Gaussian)}$$

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  • $\begingroup$ Would the latter units be more formerly referred to as CGS units by chance? The particular text I have that makes reference of the latter equation typically operates in CGS units. $\endgroup$ Apr 11 '14 at 4:17
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Argue by dimensional analysis. The force on a charged particle is usually taken to be $$\mathbf F = q(\mathbf E + \mathbf v \times \mathbf B)$$ and this defines the $\mathbf E$ and $\mathbf B$ fields. With this definition $[\mathbf E] = [c][\mathbf B]$. However you could take as definition $$\mathbf F = q(\mathbf E + \frac{\mathbf v}{c} \times \mathbf B)$$ and then since $\mathbf v/c$ is dimensionless the fields have the same dimension.

In the induction law we have a space derivative on one side and a time derivative on the other side. If the dimensions of $\mathbf E$ and $\mathbf E$ already differ by $[c] = L/T$ this is not a problem, but if you take them to have the same dimension you have to correct for it with a factor $1/c$.

Of course the adult way is to take $\mathbf E$ and $\mathbf B$ to have the same dimension and $c = 1$.

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