# Why doesn't the magnetic force formula F=QvB have any constant proportionality?

Forces like the electric force has a Coulombs constant $$k_e$$ from the Coulomb's law $$F=k_e\frac{q_1q_2}{r^2}$$ , which is based on the vacuum permittivity $$\epsilon_0$$. However, the magnetic force can be directly calculated from the scales of charge, velocity and magnetic field strength in SI units by Lorentz law. $$\mathbf F_{mag}=Q(\mathbf v\times\mathbf B)$$ Is it because the unit for magnetic field is derived from the Lorentz law in SI units? If so, why can't the unit of charge or electric field be derived from SI units of distance and time just like magnetic field?

• The electric force formula, $\vec{F}=q\vec{E}$ also has no constant of proportionalty... – jacob1729 Nov 6 '20 at 11:44
• That's because the proportion is used to decide the electric field strength. – Jeffrey Suen Nov 6 '20 at 12:53
• Right, and the magnetic case differs from the electric case how? – jacob1729 Nov 6 '20 at 12:58
• So there is a vacuum permeability for the magnetic field strength... I feel dumb now – Jeffrey Suen Nov 6 '20 at 13:28

The answer to your question was correctly highlighted by jacob1729: you don't see a constant in $$F_{m}$$ because you are writing it in terms of the field $$B$$, and you're seeing it for $$F_e$$ because you're writing it using the charges. If you rewrite
$$\vec{F_e}=k_e\frac{q_1q_2}{r^2}\hat{r}=q_1\vec{E}$$
you absorbed $$k_e$$ (therefore $$\epsilon_0$$) into $$E$$.
But one may ask: can I do the opposite? Can I rewrite $$F_m$$ to make $$\mu_0$$ appear? Yes, but not in the same way. There are different ways to write $$B$$ depending on the setup, for example Biot-Savart law for solenoids, but as far as we know you can't find a formula like
$$\vec{F_m}=k_m\frac{g_1g_2}{r^2}\hat{r}=g_1\vec{B}$$