Formula: 4th Maxwell Equation in Integral Form Magnetic flux density (B-field) Electric current Electric field (E field)
$$\oint_{L} \class{violet}{\boldsymbol{B}} ~\cdot~ \text{d}\boldsymbol{l} ~=~ \mu_0 \, \class{red}{I} ~+~ \mu_0 \, \varepsilon_0 \, \int_{A} \frac{\partial \class{blue}{\boldsymbol{E}}}{\partial t} ~\cdot~ \text{d}\boldsymbol{a}$$
Magnetic field
$$ \class{violet}{B} $$ Unit $$ \mathrm{T} = \frac{\mathrm{kg}}{\mathrm{A} \, \mathrm{s}^2} $$
The magnetic flux density indicates how strong the magnetic field is at a certain location \((x,y,z)\) and in which direction it points.
Closed line
$$ L $$ Unit $$ \mathrm{m} $$
Line (e.g. a current-carrying conductor) over which you integrate. It is the edge of the surface \( A \) (for example the edge of a circle). The line must be closed, i.e. its beginning and its end must be connected.
Surface
$$ A $$ Unit $$ \mathrm{m}^2 $$
This surface is enclosed by the closed loop \(L\). This can be, for example, the area of a circle.
Electric current
$$ \class{red}{\boldsymbol I} $$ Unit $$ \mathrm{A} = \frac{ \mathrm C }{ \mathrm s } $$
Electric current running along the line \(L\).
Electric field
$$ \class{purple}{\boldsymbol E} $$ Unit $$ \frac{\mathrm{V}}{\mathrm{m}} = \frac{\mathrm{N}}{\mathrm{C}} = \frac{\mathrm{kg} \, \mathrm{m}}{\mathrm{A} \, \mathrm{s}^3} $$
Electric field tells what would be the electric force on a sample charge at a given location \((x,y,z)\) if that sample charge is placed at that location.
Here the E-field is differentiated with respect to time: \( \frac{\partial \boldsymbol{E}}{\partial t} \).
Vacuum permeability
$$ \mu_0 $$ Unit $$ \frac{\mathrm{Vs}}{\mathrm{Am}} = \frac{ \mathrm{kg} \, \mathrm{m} }{ \mathrm{A}^2 \, \mathrm{s}^2 } $$
The vacuum permeability is a physical constant and has the following experimentally determined value:
$$ \mu_0 ~=~ 1.256 \, 637 \, 062 \, 12 ~\cdot~ 10^{-6} \, \frac{\mathrm{Vs}}{\mathrm{Am}} $$