What is the Planck scale magnetic field strength? Using the constants $\mu_0$ (or $\varepsilon_0$), $c$, $\hbar$, $e$ and $G$, it is possible to define two quantities with units of magnetic field :
\begin{align}
B_1 &= \sqrt{\frac{\mu_0 c^7}{\hbar G^2}} \equiv \sqrt{\frac{c^5}{\varepsilon_0 \hbar G^2}} \approx 8 \times 10^{53} \, \mathrm{T}, \tag{1} \\[12pt]
B_2 &= \frac{c^3}{G e} \approx 3 \times 10^{54} \, \mathrm{T}. \tag{2}
\end{align}
Which one is really the Planck magnetic field?
While $B_2$ is simpler, I suspect it should be $B_1$, because it doesn't use the electric charge unit.  $e$ is not exactly as universal as $\mu_0$.  $B_1$ uses the Planck constant, so it's consistent to call it a Planck "unit", while $B_2$ doesn't use that constant.  Also, because of the square root, $B_1$ is a bit more of the same shape as the Planck length :
\begin{equation}\tag{3}
L_{P} \equiv \sqrt{\frac{\hbar G}{c^3}}.
\end{equation}
The Planck units are presented on wikipedia: https://en.wikipedia.org/wiki/Planck_units but it doesn't tell anything about the magnetic field.
We could also argue that $B_1$ is the answer because we can find it by equating the magnetic field energy density with the Planck density (dropping all the dimensionless constants) :
\begin{equation}
\frac{B_1^2}{2 \mu_0} = \frac{M_P c^2}{L_P^3}.
\end{equation}
But then, we could also find $B_2$ by equating the Planck cyclotron angular frequency with the Planck energy :
\begin{equation}
\hbar \omega_{\text{cyclotron}} \equiv \hbar \, \frac{e B_2}{2 M_P} = M_P c^2.
\end{equation}
Both methods are arbitrary.
So what is the Planck magnetic strength?
 A: Planck units are found simply by multiplying together powers of certain constants; one does not consider specific physical laws to get them, which is equivalent to motivating specific multiplicative constants. (We don't do it this way because setting each Planck unit to $1$, the ultimate goal of having Planck units, would be impossible on a law-based approach.)
Coulomb's constant $k_C=\frac{1}{4\pi\varepsilon_0}=\frac{\mu_0 c^2}{4\pi}$, which appears in an inverse-square law the same way $G$ does, is a Planck unit just like $G$. Thus in Planck units $\frac{\mu_0}{4\pi}=1$, so the Planck unit you want is $\frac{B_1}{\sqrt{4\pi}}$. It definitely isn't $B_2=\frac{c^3}{G\sqrt{\alpha}q_P}$, with Planck charge $q_P=\sqrt{4\pi\varepsilon_0 c\hbar}=\sqrt{\frac{c\hbar}{k_C}}$.
A: A Planck magnetic field strength is not uniquely defined.
Actually you can construct many of them:
$$B= \frac{c^3}{Ge}\left(\frac{e^2}{4\pi\epsilon_0\hbar c} \right)^n$$
where $n$ is an arbitrary number, and the factor in brackets
is the dimension-less fine-structure constant $\alpha \approx \frac{1}{137}$.
