$\lambda = A q = \pi R^2 \rho_{charge} e =$ {{charge_per_unit_length_in_C_per_m = charge_density_in_C_per_m3 * wire_cross_section_area_in_m2}} $\frac{C}{m}$ = charge density per unit meter along wire
Mean velocity of moving electrons within wire:
$v = \frac{I}{\lambda}
= \frac{V}{\rho_{charge} e \cdot \rho_{res} \cdot l}
=$ {{mean_electron_velocity_in_m_per_s = current_in_A / (wire_cross_section_area_in_m2 * charge_density_in_C_per_m3)}} $\frac{m}{s}$
Electric field: $E(x) = \int \frac{q(x')}{|x' - x|^2} \cdot \frac{(x' - x)}{|x' - x|} dx'$
Electric field around a wire:
TODO explain the origin of the $\frac{v^2}{c^2}$ term... from the sources, is this a Taylor-expanded $\gamma - 1$?
$E(r) = 2 k_e \lambda \frac{1}{r} \frac{v^2}{c^2}
= \frac{\lambda}{2 \pi \epsilon_0 r} \frac{v^2}{c^2}
= 2 k_e \pi R^2 \rho_{charge} e \frac{1}{r} \frac{v^2}{c^2}
=$ {{electric_field_in_kg_m2_per_C_s2_r
= 2 * Coulomb_constant_in_kg_m3_per_C2_s2 * charge_per_unit_length_in_C_per_m * mean_electron_velocity_in_m_per_s * mean_electron_velocity_in_m_per_s / (speed_of_light_in_m_per_s * speed_of_light_in_m_per_s) }}
$\cdot \frac{1}{r} \cdot \frac{kg \cdot m^2}{C \cdot s^2}$
$E_i$ has units $\frac{N}{C} = \frac{kg \cdot m}{C \cdot s^2}$.
$B_i$ has units $\frac{N \cdot s}{C \cdot m} = \frac{kg}{C \cdot s}$
$F_{ab}$ has units of $\frac{kg}{C \cdot s}$
In terms of electromagnetic four-potential:
$F_{ab} = 2 \partial_{[a} A_{b]}$
$F_{ab} = \downarrow a \overset{\rightarrow b}{\left[ \begin{matrix}
0 & \frac{1}{c} \partial_t \vec{A} - \vec{\nabla} A_t \\
\vec{\nabla} A_t - \frac{1}{c} \partial_t \vec{A} & (\vec{\nabla} \times \vec{A}) \times
\end{matrix}\right]}$
$A_t$ is in units of $V = \frac{kg \cdot m^2}{C \cdot s^2}$
$A_i$ is in units of $\frac{V \cdot s}{m} = \frac{kg \cdot m}{C \cdot s}$
For a resting electromagnetic field of magnitude $E$ around a wire oriented in the x direction:
$F_{ab} = \downarrow a \overset{\rightarrow b}{\left[ \begin{matrix}
0 & 0 & -\frac{1}{c} E \frac{y}{r} & -\frac{1}{c} E \frac{z}{r} \\
0 & 0 & 0 & 0 \\
\frac{1}{c} E \frac{y}{r} & 0 & 0 & 0 \\
\frac{1}{c} E \frac{z}{r} & 0 & 0 & 0
\end{matrix}\right]}$
$F'_{ab} = \downarrow a \overset{\rightarrow b}{\left[ \begin{matrix}
0 &
0 &
-\gamma \frac{1}{c} E \frac{y}{r} &
-\gamma \frac{1}{c} E \frac{z}{r} \\
0 &
0 &
\beta \gamma \frac{1}{c} E \frac{y}{r} &
\beta \gamma \frac{1}{c} E \frac{z}{r} \\
\gamma \frac{1}{c} E \frac{y}{r} & -\beta \gamma \frac{1}{c} E \frac{y}{r} & 0 & 0 \\
\gamma \frac{1}{c} E \frac{z}{r} & -\beta \gamma \frac{1}{c} E \frac{z}{r} & 0 & 0
\end{matrix}\right]}$
One possible pre-boosted 4-potential:
$A_i = 0 = $ magnetic vector potential.
$r_0 = $ = electromagnetic potential reference length.
$A_t = 2 \lambda k_e ln( \frac{r}{r_0} ) = $ {{
2 * charge_per_unit_length_in_C_per_m * Coulomb_constant_in_kg_m3_per_C2_s2 * Math.log(field_measure_distance_in_m / electric_potential_reference_dist_in_m)
}} $ \frac{ kg \cdot m^2 }{ C \cdot s^2} $ = electric potential.
So the electric field in the direction perpendicular to the boost is scaled up by $\gamma \ge 1$.
And a magnetic field is created equal to:
$\vec{B} =
\beta \gamma \frac{1}{c} E \cdot
\left[\begin{matrix}
0 \\
-\frac{z}{r} \\
\frac{y}{r}
\end{matrix}\right]$
$B = \frac{1}{c} \beta \gamma E = \frac{2 \cdot \gamma \cdot V \cdot \pi \cdot R^2 \cdot k_e}{c^2 \cdot \rho_{res} \cdot l \cdot r}$ = magnetic field magnitude
Measuring the magnetic field at a distance of {{field_measure_distance_in_m}} $ m$ from the wire gives us:
$E =$ {{electric_field_in_kg_m_per_C_s2 = electric_field_in_kg_m2_per_C_s2_r / field_measure_distance_in_m}} $\frac{kg \cdot m}{C \cdot s^2} =$ {{electric_field_in_kg_m_per_C_s2}} $\frac{V}{m}$
$B =$ {{magnetic_field_in_kg_per_C_s = magnetic_field_in_kg_m_per_C_s_r / field_measure_distance_in_m}} $\frac{kg}{C \cdot s} =$ {{magnetic_field_in_kg_per_C_s}} T