Value of Vacuum Permeability

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Theorem

The value of the vacuum permeability is calculated as:

$\mu_0 = 1 \cdotp 25663 \, 70621 \, 2 (19) \times 10^{-6} \, \mathrm H \, \mathrm m^{-1}$ (henries per metre)

with a relative uncertainty of $1 \cdotp 5 \times 10^{-10}$.


Proof 1

The vacuum permeability is the physical constant denoted $\mu_0$ defined as:

$\mu_0:= \dfrac {2 \alpha h} {e^2 c}$

where:

$e$ is the elementary charge
$\alpha$ is the fine-structure constant
$h$ is Planck's constant
$c$ is the speed of light defined in $\mathrm m \, \mathrm s^{-1}$


$e$ is defined precisely as:

$e = 1 \cdotp 60217 \, 6634 \times 10^{-19} \, \mathrm C$ (coulombs)

$\alpha$ has the value determined experimentally as:

$\alpha \approx 0 \cdotp 00729 \, 73525 \, 693 (11)$

$h$ is defined precisely as:

$h = 6 \cdotp 62607 \, 015 \times 10^{-34} \, \mathrm J \, \mathrm s$ (joule seconds)

$c$ is defined precisely as:

$c = 299 \, 792 \, 458 \, \mathrm m \, \mathrm s^{-1}$ (metres per second)


Hence $\mu_0$ can be calculated as:

\(\ds \mu_0\) \(=\) \(\ds \dfrac {2 \alpha h} {e^2 c}\) \(\ds \dfrac {\mathrm J \, s} {\mathrm C^2 \times \mathrm m \, \mathrm s^{-1} }\)
\(\ds \) \(=\) \(\ds \dfrac {0 \cdotp 00729 \, 73525 \, 693 (11) \times 6 \cdotp 62607 \, 015 \times 10^{-34} } {\paren {1 \cdotp 60217 \, 6634 \times 10^{-19} }^2 \times 299 \, 792 \, 458}\) \(\ds \dfrac {\paren {\frac {\mathrm {kg} \times \mathrm m^2} {\mathrm s} } } {\paren {\mathrm A \, \mathrm s}^2 \times \mathrm s \times \mathrm m \, \mathrm s^{-1} }\) Base Units of Coulomb, Base Units of Joule
\(\ds \) \(=\) \(\ds 1 \cdotp 25663 \, 70621 \, 2 (19) \times 10^{-6}\) \(\ds \dfrac {\mathrm {kg} \times \mathrm m} {\mathrm A^2 \times \mathrm s^4}\)
\(\ds \) \(=\) \(\ds 1 \cdotp 25663 \, 70621 \, 2 (19) \times 10^{-6}\) \(\ds \dfrac {\mathrm H} {\mathrm m}\) Base Units of Henry

$\blacksquare$


Proof 2

The vacuum permeability is the physical constant denoted $\mu_0$ defined as:

$\mu_0 := \dfrac 1 {\varepsilon_0c^2}$

where:

$\varepsilon_0$ is the vacuum permittivity defined in $\mathrm F \, \mathrm m^{-1}$ (farads per metre)
$c$ is the speed of light defined in $\mathrm m \, \mathrm s^{-1}$


$\varepsilon_0$ has the value determined as:

$\varepsilon_0 \approx 8 \cdotp 85418 \, 78128 (13) \times 10^{-12} \, \mathrm F \, \mathrm m^{-1}$

$c$ is defined precisely as:

$c = 299 \, 792 \, 458 \, \mathrm m \, \mathrm s^{-1}$


Hence $\mu_0$ can be calculated as:

\(\ds \mu_0\) \(=\) \(\ds \dfrac 1 {\varepsilon_0 c^2}\) \(\ds \dfrac 1 {\mathrm F \, \mathrm m^{-1} \times \paren {\mathrm m \, \mathrm s^{-1} }^2}\)
\(\ds \) \(=\) \(\ds \dfrac 1 {8 \cdotp 85418 \, 78128 (13) \times 10^{-12} \times \paren {299 \, 792 \, 458}^2}\) \(\ds \dfrac 1 {\frac {\mathrm s^4 \, \mathrm A^2} {\mathrm {kg}^{-1} \, \mathrm m^{-3} } \times \paren {\mathrm m \, \mathrm s^{-1} }^2}\) Base Units of Farad
\(\ds \) \(=\) \(\ds 1 \cdotp 25663 \, 70621 \, 2 (19) \times 10^{-6}\) \(\ds \dfrac {\mathrm {kg} \times \mathrm m} {\mathrm A^2 \times \mathrm s^2}\)
\(\ds \) \(=\) \(\ds 1 \cdotp 25663 \, 70621 \, 2 (19) \times 10^{-6}\) \(\ds \dfrac {\mathrm H} {\mathrm m}\) Base Units of Henry

$\blacksquare$


Also see


Interconnection between Vacuum Permittivity and Vacuum Permeability

Right from the start we see the equation $c^2 = \dfrac 1 {\varepsilon_0 \mu_0}$ and we understand immediately that these parameters, vacuum permittivity and vacuum permeability, are tightly wound up in the structure of the electromagnetic spectrum.

One of them defines the strength of the electric attraction.

The other defines the strength of the magnetic force.


Hence it is apparent that the units of measurement we use are to a certain extent arbitrary, as the concepts being discussed are more abstract than tangible physical measurements.

But in order to keep the relationships within the grasp of the tangible, we need to make sure the units we use are accessible enough to interface neatly with the existing measurement systems we already have.


Prevalent is the SI system, in which we can set up units of considerable convenience.

We notice that if we define $\mu_0$ as $4 \pi \times 10^{-7}$ in henries per metre, which evaluates to $\dfrac {\mathrm {kg} \times \mathrm m} {\mathrm A^2 \times \mathrm s^4}$, then many of the equations work out with manageable numbers all approximately the same size.

From this convenient number, we define the Ampere, and work out what conditions we require in order to produce it.

We use the Biot-Savart Law to work out what to expect, and then measure the force and work out the rest of the numbers from there.


Hence having chosen $\mu_0$ in that way, we must then define $\varepsilon_0$ as $\dfrac 1 {\mu_0 c^2}$.


Or instead, we could declare what value $\varepsilon_0$ should be based on accurately and reliably reproducible standards, that is $\dfrac {e^2} {2 \alpha h c}$, and then force $\mu_0$ to be $\dfrac 1 {\varepsilon_0 c^2}$.

The latter is the direction it currently lies, after the $2019$ standards redefinition.