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?
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$\begingroup$ The electric force formula, $\vec{F}=q\vec{E}$ also has no constant of proportionalty... $\endgroup$– jacob1729Commented Nov 6, 2020 at 11:44
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$\begingroup$ That's because the proportion is used to decide the electric field strength. $\endgroup$– Jeffrey SuenCommented Nov 6, 2020 at 12:53
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1$\begingroup$ Right, and the magnetic case differs from the electric case how? $\endgroup$– jacob1729Commented Nov 6, 2020 at 12:58
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$\begingroup$ So there is a vacuum permeability for the magnetic field strength... I feel dumb now $\endgroup$– Jeffrey SuenCommented Nov 6, 2020 at 13:28
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2 Answers
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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
$\vec{F_m}=k_m\frac{g_1g_2}{r^2}\hat{r}=g_1\vec{B}$
because there isn't a magnetic charge.
answered Nov 6, 2020 at 13:26
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1$\begingroup$ Yeah it hit me when I realize the proportionality is hidden in the magnetic field strength. $\endgroup$ Commented Nov 6, 2020 at 13:30
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This expression is about a charged particle moving through a medium with some speed. For the movement of the charged particle, any solid medium is out of question. Only the liquid and gaseous medium remain. If you choose to ignore the drag force, there is no change in speed due to the medium. Now there is only one factor that the medium can influence and that is the magnitude of external magnetic field. So, whatever the source of magnetic field (either a current carrying conductor or a permanent magnet) that is present in the given situation, its field will depend on the medium. And therefore the term of permeability will be in the expression for this magnetic field.
