In a book, it is derived that, in cyclotron, the radius is:
$$r=\frac{mcv}{eH\sqrt{1-v^2/c^2}}$$
If I let $v\ll c$, then I will get
$$r=\frac{mcv}{eH}$$
However, in classical physics, we have
$$r=\frac{mv}{eB}$$
Then does it mean
$$B=\frac{H}{c}$$
But isn't that
$$B=\mu H\,?$$
Possibly by asking this question you betray your age as the formula is not in SI units but in centimetre-gramme second (cgs) units.
In the cgs system there were two sets of units:
electrostatic units - esu
electromagnetic units - emu
In the emu system $H$ was measured in oersteds $\left(1 {\rm Oe} = \dfrac{10^{4}}{4 \pi} {\rm Am^{-1}}\right)$
and $B$ was measured in gauss $\left(1 {\rm G} = 10^{-4} {\rm T}\right)$
Also the permeability of free space $\mu_o$ was set as equal to one in the emu system.
You will note from the conversion factors out comes a familiar constant $4 \pi \times 10^{-7}$ the permeability of free space in ${\rm Hm^{-1}}$.
And if that not complicated enough there was a conversion factor between 1 esu of charge and 1 emu of charge ( $1$ emu of charge = $\dfrac 1 c$ esu of charge where c is the speed of light in cm s$^{-1}$.)
So are you not glad that you are young?
