In a steady-state (independent of time) :

\(\frac{\partial \rho}{\partial t} =0\;\;\) so \(\;\;div\vec{j} =0\).

Thus, it can be deduced that :

• The total intensity exiting a given closed surface \((S)\) is equal to zero in steady-state :

\(i = \oint_{(S)}\vec {j}.\vec{n} \ dS=\iiint_{(V)}div\vec{j}\ d\tau=0\)

• In steady-state, the intensity has the same value through any section of a given field tube :

Let us consider a closed surface \((S)\) formed by one field tube \((T)\) (called current tube) and two surfaces \((S_1)\) and \((S_2)\).

These two surfaces are based on two loops of \((T)\) and have the same orientation.

Let \(I_1\) and \(I_2\) be the intensities, or respectively the flux of the density of current \(\vec j\) through \((S_1)\) and \((S_2)\) :

\(\oint_S\vec {j}.\vec{n} \ dS=-I_1+I_2+\iint_S\vec {j}.\vec{n} \ dS=-I_1+I_2\)

(We can notive that the two vectors \(\vec j\) and \(\vec n\) are perpendicular).

So :

\(I_1=I_2\)

In steady-state, the intensity of the electric current has the same value through any section of a circuit branch.

We can easily deduce Kirchhoff's circuit laws (flux or current density vector conservation).