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The vacuum permeability, or the capability of the vacuum to permit magnetic field lines, contains the value of $\pi$. Why? What does this have to do with the ratio of a circle's circumference to its diameter?

Qmechanic
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Joel DeWitt
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    Note that $\pi$ only appears in $\epsilon$ in some sets of units. See https://en.wikipedia.org/wiki/Centimetre%E2%80%93gram%E2%80%93second_system_of_units#Alternate_derivations_of_CGS_units_in_electromagnetism for instance. – dmckee --- ex-moderator kitten Nov 18 '15 at 15:48
  • Possible duplicates: http://physics.stackexchange.com/q/1673/2451 , http://physics.stackexchange.com/q/28673/2451 and links therein. – Qmechanic Nov 18 '15 at 17:36

1 Answers1

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This is nothing but a choice of units.

Let me make that (hopefully) more clear by explaining more about how choosing units in electromagnetism works:

  • Coulomb's Law is $\vec F = k \frac{q_1 q_2}{r^2} \hat{\vec r}$ for the force between two charges $q_1$ and $q_2$. $k$ is different in the various systems of units - essentially, it depends on how the unit of charge is defined.

  • Ampere's Law is $\vec F = k' I_1 I_2 \oint_{C_1} \oint_{C_2} \frac{d\vec r_1 \times (d\vec r_2 \times \vec r)}{|r|^3}$ for the force between two currents along $C_1$ and $C_2$. $k'$ is different in the various systems of units - essentially, it depends on how the unit of current is defined.

In electrodynamics, we find out that $$ k / k' = c^2 $$ otherwise, we are free to choose. In CGS, we define $k = 1$, and in SI $k' = 10^{-7} \frac{Vs}{Am}$.

Now we introduce constants $$ \mu_0 = 4\pi\, k', \quad \varepsilon_0 = \frac{1}{4\pi k} $$ That is nothing more than a definition - the factors of $4\pi$ simplify some formulas later on, e.g. when fields are integrated over the surface of a sphere.

Noiralef
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