Convert formula from CGS to SI I'd like to convert this formula 
\begin{equation}
l^2 =\frac{c\hbar}{eH}
\end{equation}
where $l$ is a length, and $H$ is in oersted, to SI units.
I am pretty sure it uses CGS, since Oe is mentioned in the text, and its from a theory paper (Kawabata1980, eq.3).
 A: As a plasma physicist I use the NRL Plasma Formulary to convert between CGS and SI units. It can be downloaded from here.
On page 18 it gives you a prescription on how to convert any formula. Remember to convert both sides of the equation. For your problem I get
$$ l^2 = \frac{\varepsilon_0 c^2 \hbar}{eH} $$
Step by step instruction:


*

*Identify all the quantities in your equation (with $\alpha=10^2\mathrm{cm\;m}^{-1}$ and $\beta=10^7\mathrm{erg\;J}^{-1}$) 


*

*$l$ length, factor $\alpha$

*$c$ velocity, factor $\alpha$

*$\hbar$ action = energy $\times$ time, factor $\beta \times 1$

*$e$ charge, factor $(\alpha \beta / 4 \pi \varepsilon_0)^{1/2}$

*$H$ magnetic intensity, factor $(4 \pi \mu_0\beta/\alpha^3)^{1/2}$


*Replace all quantities in the equation
$$ \alpha^2 l^2 = \frac{\alpha c \; \beta \hbar}{(\alpha \beta / 4 \pi \varepsilon_0)^{1/2}e\;(4 \pi \mu_0\beta/\alpha^3)^{1/2}H} $$

*Simplify
$$ l^2 = \frac{c \; \hbar}{(1 / \varepsilon_0)^{1/2}e\;\mu_0^{1/2}H} =  \frac{\varepsilon_0 c \; \hbar}{(\varepsilon_0 \mu_0)^{1/2}eH} $$

*Use $c = 1/\sqrt{\varepsilon_0 \mu_0}$
$$ l^2 = \frac{\varepsilon_0 c^2 \hbar}{eH} =  \frac{\hbar}{e\mu_0 H}$$

A: It's not completely clear to me what's going on there, but the authors provide a useful foothold by stating (under equation 4) that the cyclotron frequency of the problem is
$$
\omega=\frac{eH}{mc},
$$
and this needs to correspond directly with the SI expression,
$$
\omega_\mathrm{SI}=\frac{eB}{m}.
$$
From here you can get the correspondence 
$$
\frac{eH}{c} \leftrightarrow (eB)_\mathrm{SI}.
$$
Taking this and running with it, you get the SI version
$$
l^2=\frac{\hbar}{eB}.
$$
This checks out dimensionally: $eB$ is a force per unit velocity, with dimensions $[F/v]=[M\,T^{-1}]$, which matches $[\hbar/l^2]$.
Moreover, this coincides with Holger's black magic (which is probably the way to go), which reduces to $l^2=\hbar/e\mu_0H$, and this coincides with $l^2=\hbar/eB$ in vacuum.
