Update textbook.fr.md

12.temporary_ins/96.electromagnetism-in-media/20.reflexion-refraction-at-interfaces/20.metallic-waveguides/10.main/textbook.fr.md

@@ -284,8 +284,8 @@ and
 $`\overrightarrow{B}_{\perp}=
 \left(\begin{array}{l}
 0\\
-\dfrac{E_0}{c}\dfrac{k_z}{k}\sin\big(\frac{n\pi}{b}\,y\big)\sin\,(k_z\,z -\omega\,t)\\
-\dfrac{E_0}{c}\dfrac{n\pi}{b\,k}\cos\big(\frac{n\pi}{b}\,y\big)\sin\,(k_z\,z -\omega\,t)
+\dfrac{E_0}{c}\dfrac{k_z}{k}\sin\big(\dfrac{n\pi}{b}\,y\big)\sin\,(k_z\,z -\omega\,t)\\
+\dfrac{E_0}{c}\dfrac{n\pi}{b\,k}\cos\big(\dfrac{n\pi}{b}\,y\big)\sin\,(k_z\,z -\omega\,t)
 \end{array}\right)`$
 $`\quad(eq. 4.12)`$  
 
@@ -295,17 +295,28 @@ $`\langle\overrightarrow{P}\rangle=
 \dfrac{1}{2}\,\dfrac{E_0^{\;2}}{c\mu_0}\,\dfrac{k_z}{z}\,\sin^2\Big(\dfrac{n\pi}{b}\,y\Big)\,\overrightarrow{e_z}`$
 $`\quad(eq. 4.13)`$  
 
-The power transmitted by the guide (units \[*W* \]) can be found by
+The power transmitted by the guide (units $`W`$) can be found by
 integrating the previous results over the cross-section of the
 waveguide
 
-
-
+$`\displaystyle\mathscr{P}=\int_0^a\int_0^b\langle P\rangle\,dx\,dy=\dfrac{1}{4}\,\dfrac{E_0^{\;2}}{c\mu_0}\,\dfrac{k_z}{k}`$
+$`\quad(eq. 4.14)`$  
 
 i.e. the transmitted power is proportional to the cross-sectional area
 of the waveguide. The practical limit of transmittable power is set by
 the dielectric breakdown of the dielectric filling the waveguide. In
-case of dry air, this limit is of about 3 MV/m.
+case of dry air, this limit is of about $`3\;MV/m`$.
+
+!!!!! *Exercice 4.2 : Transmitted power*   
+!!!!! 
+!!!!! 1) Estimate the maximum power transmittable by a rectangular waveguide
+!!!!! of dimensions $`b=2.5\,cm\;,\; a=1b=2.5, a=1\,cm`$ at $`f=9.24\,GHz`$ filled with dry air
+!!!!! for the mode $`TE_{0,1}`$.   
+!!!!!
+!!!!! 2) Rewrite equation 4.14 in case the guide is filled with a perfect dielectric
+!!!!! with refractive index $`n`$.
+
+
 
 # Exercices solved during tutorials