Value of Vacuum Permeability/Proof 2

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 $1 \cdotp 5 \times 10^{-10}$.

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 $

$=$

$\ds \dfrac 1 {8 \cdotp 85418 \, 78128 (13) \times 10^{-12} \times \paren {299 \, 792 \, 458}^2}$

$\ds \dfrac 1 {\frac {\mathrm s^4 \, \mathrm A^2} {\mathrm {kg}^{-1} \, \mathrm m^{-3} } \times \paren {\mathrm m \, \mathrm s^{-1} }^2}$

$\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 $

$=$

$\ds 1 \cdotp 25663 \, 70621 \, 2 (19) \times 10^{-6}$

$\ds \dfrac {\mathrm H} {\mathrm m}$

$\blacksquare$