To test the understanding of the lesson
Question
Write the electric current density vector in an environment that contains \(n\) carriers of charge by volume unit, which move at the velocity \(\vec v\).
Solution
The electric current density vector is :
\(\vec j=nq\vec v=\rho_{mob}\vec v\)
Question
How can we define the intensity in the presence of a volume current ? And of a surface current ?
Solution
Volume current : the intensity of current is the flux of the vector \(\vec j\) through a surface S :
\(I=\iint_{(S)}\vec j.\vec n\; dS\)
If the vector \(\vec j\) is uniform and perpendicular to the surface, then :
\(I=jS\)
Surface current : the integral becomes linear :
\(I=\int_{(AB)}\vec j_S.d\vec \ell\)
If the vector \(\vec j_S\) is uniform and perpendicular to the segment \(AB\) of length \(\ell\) :
\(I=j_S\ell\)
Question
Give the local expression of the charge conservation principle.
Solution
It is a classical conservation equation :
\(div \vec j +\frac{\partial\rho}{\partial t}=0\)
\(\rho\) is the total density of charges, and not only the density of the mobile charges \(\rho_{mob}\) which appears in the definition of \(\vec j=\rho_{mob}\vec v\).
Question
Demonstrate Kirchhoff's law in steady-state.
Solution
In steady state, \(div\vec j = 0\). The flux of \(\vec j\) is thus conserved :
\(\oint_{(S)} \vec j.\vec n \;dS = 0\)
Or, equivalently, on a field tube (or current tube) at a nodal point :
\(\iint_{(S_1)}\vec j.\vec n\; dS=\iint_{(S_2)}\vec j.\vec n\; dS+\iint_{(S_3)}\vec j.\vec n\; dS\)
Hence we have demonstrated Kirchhof's law :
\(I_1=I_2+I_3\)
Question
Write all four equations of Maxwell in the general case.
How can they be simplified in a conductor ?
Solution
Maxwell's equations are :
\(\mathrm{div} \vec{E} =\frac{\rho}{\epsilon_0}\) (Gauss' law for electricity, MG)
\(\mathrm{div} \vec{B} =0\) (Gauss' law for magnetism)
\(\overrightarrow{\mathrm{rot}} \vec E = -\frac{\partial \vec B}{\partial t}\) (Faraday's law of induction, MF)
\(\overrightarrow{\mathrm{rot}} \vec B = \mu_0 \vec{\jmath} + \epsilon_0 \mu_0 \frac{\partial \vec E}{\partial t}\) (Ampère's law, MA)
In a metallic conductor :
\(\begin{array}{l}div\;\vec B = 0 \\div\;\vec E = 0 \\\overrightarrow {rot} \;\vec E = - \frac{{\partial \vec B}}{{\partial t}} \\\;\overrightarrow {rot} \;\vec B = \mu _0 \vec j = \mu _0 \sigma \;\vec E \\\end{array}\)
Question
Prove the integral form of Gauss' law of electricity
Solution
Let us apply the divergence theorem :
\(\iiint_{(V)} div\vec E\; d\tau =\oint_{(S)} \vec E.\vec n\; dS=\frac{1}{\varepsilon_0}\iiint_{(V)}\rho (M)\; d\tau\)
So :
\(\oint_{(S)} \vec E.\vec n\; dS=\frac{1}{\varepsilon_0}Q_{int}\)
Question
Prove the integral form of Ampere's circuital law.
Solution
Let us apply Stoke's theorem :
\(\iint_{(S)} \vec {rot}\vec B.\vec n\; dS=\oint_{(C)}\vec B.d\vec \ell=\mu_0 \iint_{(S)} \vec j.\vec n\; dS\)
So :
\(\oint_{(C)}\vec B.d\vec \ell=\mu_0 \iint_{(S)} \vec j.\vec n \; dS=\mu_0 I_{enl}\)
Question
Why is it said that the flux of the magnetic field is conserved ?
Solution
The Gauss' law of magnetism, \(div\vec B=0\), leads to :
\(\oint_{(S)}\vec B.\vec n \; dS=0\)
Which means the flux of the magnetic field is conserved.
Question
Give the density of electric energy of an electromagnetic field.
Give the density of magnetic energy of an electromagnetic field.
Solution
Density of electric energy of an electromagnetic field :
\(e_E=\frac{1}{2}\varepsilon_0 E^2\)
Density of magnetic energy of an electromagnetic field :
\(e_B=\frac{1}{2}\frac{B^2}{\mu_0}\)
\(e_{EM}=e_E+e_B\) is the total density of energy of the electromagnetic field.
Question
Define Poynting's vector in electromagnetism, writen \(\vec \Pi\).
Give the density of power received by the matter from an electromagnetic field. What is it in the case of a metal conductor ?
In the presence of volume current \(\vec j\), write the local then global electromagnetic energy of the electromagnetic field \((\vec E,\vec B)\).
Solution
Poynting's vector in electromagnetism :
\(\vec \Pi = \frac{{\vec E \wedge \vec B}}{{{\mu _0}}}\)
Density of power received by the matter from an electromagnetic field :
\(p_V=\vec j.\vec E\)
For a metal conductor (for which Ohm's local law is verified, \(\vec j =\gamma \vec E\)) :
\(p_V=\gamma E^2\)
Local conservation of electromagnetic energy :
\(\frac{{\partial e_{em} }}{{\partial t}} = - div\;\vec \Pi - \vec j.\vec E\)
The integral form of electromagnetic energy conservation is :
\(\frac{d}{dt}(\iiint_{(V)} e_{EM}\; d\tau)=- \oint_{(S)}\vec {\vec \Pi}.\vec{n} \ dS - \iiint_{(V)} \vec j .\vec E \; d\tau\)
Question
Give the magnetic energy instantly stored in a coil of inductance \(L\), travelled by the intensity \(i(t)\).
A system is made of two circuits which have respectively the self-inductances \(L_1\) and \(L_2\) and a mutual inductance \(M\).
Both circuits are crossed respectively by two currents \(I_1\) and \(I_2\).
What is the magnetic energy of the system ?
Solution
The magnetic energy instantly stored in a coil of inductance \(L\), crossed by the intensity \(i(t)\) is :
\(E_m=\frac{1}{2}Li(t)^2\)
The magnetic energy of the system is :
\(E_m=\frac{1}{2}L_1I_1^2+\frac{1}{2}L_2I_2^2+MI_1I_2\)
Question
What is the energy stored by a capacitor ?
Define the capacitance of a capacitor. Define it in the case of a plane capacitor.
Solution
Electric energy instantly stored in a capacitor of capacitance \(C\) under the voltage \(u(t)\) :
\(E_C=\frac{1}{2}Cu(t)^2\)
The capacitance of a capacitor is defined by :
\(u(t)=\frac{q}{C}\)
For a plane capacitor :
\(C=\frac{\varepsilon_0 S}{e}\)
Where \(S\) is the surface of the frames and \(e\) the distance between them.
If a dielectric of relative permittivity \(\varepsilon_r\) is put between the two frames, the capacitance becomes :
\(C=\frac{\varepsilon_0 \varepsilon_rS}{e}\)