Supongamos que el medio en el que se propaga la onda electromagnética es isótropo, lineal y homogéneo, no conductor. La onda electromagnética se propaga a lo largo del eje Z. En coordenadas cilíndricas

${\displaystyle \begin{array}{l}\overrightarrow{E}=E_{\rho }\widehat{\rho }+E_{\phi }\widehat{\phi }+E_z\widehat{k},\{\begin{array}{l}{E}_{\rho }=E_{0\rho }\exp \left(i\left(kz-\omega t\right)\right)\\ E_{\phi }=E_{0\phi }\exp \left(i\left(kz-\omega t\right)\right)\\ E_z=E_{0z}\exp \left(i\left(kz-\omega t\right)\right)\end{array}\\ \overrightarrow{B}=B_{\rho }\widehat{\rho }+B_{\phi }\widehat{\phi }+B_z\widehat{k},\{\begin{array}{l}{B}_{\rho }=B_{0\rho }\exp \left(i\left(kz-\omega t\right)\right)\\ B_{\phi }=B_{0\phi }\exp \left(i\left(kz-\omega t\right)\right)\\ B_z=B_{0z}\exp \left(i\left(kz-\omega t\right)\right)\end{array}\end{array}}$

Las amplitudes E_0ρ, E_0φ, E_0z, B_0ρ, B_0φ, B_0z son funciones de ρ e φ, que por ahora no conocemos, la dependencia con z y t aparece en la función exponencial. La onda electromagnética no es transversal ya que tiene una compoenete E_z a lo largo de la dirección de propagación

Apicamos la ley de Faraday en forma diferencial

${\displaystyle \begin{array}{l}\overrightarrow{\nabla }\times \overrightarrow{E}=-\frac{\partial \overrightarrow{B}}{\partial t}\\ \left(\frac{1}{\rho }\frac{\partial E_z}{\partial \phi }-\frac{\partial E_{\phi }}{\partial z}\right)\widehat{\rho }+\left(\frac{\partial E_{\rho }}{\partial z}-\frac{\partial E_z}{\partial \rho }\right)\widehat{\phi }+\frac{1}{\rho }\left(\frac{\partial }{\partial \rho }\left(\rho E_{\phi }\right)-\frac{\partial E_{\rho }}{\partial \phi }\right)\widehat{k}=-\frac{\partial }{\partial t}\left(B_{\rho }\widehat{\rho }+B_{\phi }\widehat{\phi }+B_z\widehat{k}\right)\\ \{\begin{array}{l}\frac{1}{\rho }\frac{\partial E_{0z}}{\partial \phi }{e}^{i(kz-\omega t)}-ik{E}_{0\phi }{e}^{i(kz-\omega t)}=i\omega B_{0\rho }{e}^{i(kz-\omega t)}\\ ik{E}_{0\rho }{e}^{i(kz-\omega t)}-\frac{\partial E_{0z}}{\partial \rho }{e}^{i(kz-\omega t)}=i\omega B_{0\phi }{e}^{i(kz-\omega t)}\\ \frac{1}{\rho }\left(\frac{\partial \left(\rho E_{0\phi }\right)}{\partial \rho }{e}^{i(kz-\omega t)}-\frac{\partial E_{0\rho }}{\partial \phi }{e}^{i(kz-\omega t)}\right)=i\omega B_{0z}{e}^{i(kz-\omega t)}\end{array}\\ \{\begin{array}{l}\frac{1}{\rho }\frac{\partial E_{0z}}{\partial \phi }-ik{E}_{0\phi }=i\omega B_{0\rho }\\ ik{E}_{0\rho }-\frac{\partial E_{0z}}{\partial \rho }=i\omega B_{0\phi }\\ \frac{1}{\rho }\left(\frac{\partial \left(\rho E_{0\phi }\right)}{\partial \rho }-\frac{\partial E_{0\rho }}{\partial \phi }\right)=i\omega B_{0z}\end{array}\end{array}}$

La ley de Ampère-Maxwell, sin corrientes

${\displaystyle \begin{array}{l}\overrightarrow{\nabla }\times \overrightarrow{B}-\frac{1}{c^2}\frac{\partial \overrightarrow{E}}{\partial t}=0\\ \left(\frac{1}{\rho }\frac{\partial B_z}{\partial \phi }-\frac{\partial B_{\phi }}{\partial z}\right)\widehat{\rho }+\left(\frac{\partial B_{\rho }}{\partial z}-\frac{\partial B_z}{\partial \rho }\right)\widehat{\phi }+\frac{1}{\rho }\left(\frac{\partial }{\partial \rho }\left(\rho B_{\phi }\right)-\frac{\partial B_{\rho }}{\partial \phi }\right)\widehat{k}-\frac{1}{c^2}\frac{\partial }{\partial t}\left(E_{\rho }\widehat{\rho }+E_{\phi }\widehat{\phi }+E_z\widehat{k}\right)=0\\ \{\begin{array}{l}\frac{1}{\rho }\frac{\partial B_{0z}}{\partial \phi }-ik{B}_{0\phi }+\frac{1}{c^2}i\omega E_{0\rho }=0\\ ik{B}_{0\rho }-\frac{\partial B_{0z}}{\partial \rho }+\frac{1}{c^2}i\omega E_{0\phi }=0\\ \frac{1}{\rho }\left(\frac{\partial }{\partial x}\left(\rho B_{0\phi }\right)-\frac{\partial B_{0\rho }}{\partial \phi }\right)+\frac{1}{c^2}i\omega E_{0z}=0\end{array}\end{array}}$

Las cuatro componentes transversales E_0ρ, E_0φ, B_0ρ, B_0φ se pueden deducir de las componentes longitudinales, E_0z, B_0z del siguiente modo

A partir de las ecuaciones

${\displaystyle \{\begin{array}{l}\frac{1}{\rho }\frac{\partial B_{0z}}{\partial \phi }-ik{B}_{0\phi }+\frac{1}{c^2}i\omega E_{0\rho }=0\\ ik{E}_{0\rho }-\frac{\partial E_{0z}}{\partial \rho }=i\omega B_{0\phi }\end{array}}$

Despejamos E_0ρ y B_0φ

${\displaystyle \begin{array}{l}{E}_{0\rho }=\frac{i}{{\gamma }^2}\left(\frac{\omega }{\rho }\frac{\partial B_{0z}}{\partial \phi }+k\frac{\partial E_{0z}}{\partial \rho }\right),{\gamma }^2=\frac{{\omega }^2}{c^2}-k^2\\ B_{0\phi }=\frac{i}{{\gamma }^2}\left(\frac{k}{\rho }\frac{\partial B_{0z}}{\partial \phi }+\frac{\omega }{c^2}\frac{\partial E_{0z}}{\partial \rho }\right)\end{array}}$

A partir de las ecuaciones

${\displaystyle \{\begin{array}{l}\frac{1}{\rho }\frac{\partial E_{0z}}{\partial \phi }-ik{E}_{0\phi }=i\omega B_{0\rho }\\ ik{B}_{0\rho }-\frac{\partial B_{0z}}{\partial \rho }+\frac{1}{c^2}i\omega E_{0\phi }=0\end{array}}$

Despejamos B_0ρ y E_0φ

${\displaystyle \begin{array}{l}{E}_{0\phi }=\frac{i}{{\gamma }^2}\left(\frac{k}{\rho }\frac{\partial E_{0z}}{\partial \phi }-\omega \frac{\partial B_{0z}}{\partial \rho }\right)\\ B_{0\rho }=\frac{i}{{\gamma }^2}\left(k\frac{\partial B_{0z}}{\partial \rho }-\frac{\omega }{c^2}\frac{1}{\rho }\frac{\partial E_{0z}}{\partial \phi }\right)\end{array}}$

Las ecuaciones de propagación de las ondas electromagnéticas son

${\displaystyle \{\begin{array}{l}{\nabla }^2\overrightarrow{E}-{\epsilon }_0{\mu }_0\frac{{\partial }^2\overrightarrow{E}}{\partial t^2}=0\\ {\nabla }^2\overrightarrow{B}-{\epsilon }_0{\mu }_0\frac{{\partial }^2\overrightarrow{B}}{\partial t^2}=0\end{array}}$

La onda electromagnética se propaga a lo largo del eje Z

${\displaystyle \begin{array}{l}\left(\frac{1}{\rho }\frac{\partial }{\partial \rho }\left(\rho \frac{\partial E_{0z}}{\partial \rho }\right)+\frac{1}{{\rho }^2}\frac{{\partial }^2{E}_{0z}}{\partial {\phi }^2}\right)e^{i\left(kz-\omega t\right)}-k^2{E}_{0z}{e}^{i\left(kz-\omega t\right)}+\frac{{\omega }^2}{c^2}{E}_{0z}{e}^{i\left(kz-\omega t\right)}=0\\ \frac{1}{\rho }\frac{\partial }{\partial \rho }\left(\rho \frac{\partial E_{0z}}{\partial \rho }\right)+\frac{1}{{\rho }^2}\frac{{\partial }^2{E}_{0z}}{\partial {\phi }^2}+\left(\frac{{\omega }^2}{c^2}-k^2\right)E_{0z}=0\\ \frac{1}{\rho }\frac{\partial }{\partial \rho }\left(\rho \frac{\partial E_{0z}}{\partial \rho }\right)+\frac{1}{{\rho }^2}\frac{{\partial }^2{E}_{0z}}{\partial {\phi }^2}+{\gamma }^2{E}_{0z}=0\end{array}}$

De modo similar

${\displaystyle \frac{1}{\rho }\frac{\partial }{\partial \rho }\left(\rho \frac{\partial B_{0z}}{\partial \rho }\right)+\frac{1}{{\rho }^2}\frac{{\partial }^2{B}_{0z}}{\partial {\phi }^2}+{\gamma }^2{B}_{0z}=0}$

Los valores del número de onda k son discretos y vienen determinados por la frecuencia ω de la onda electromagnética y por la geometría y características del medio en el que se propaga ε, μ (no conductor)

Resolvemos la ecuación diferencial utilizando el procedimiento de variables separadas

${\displaystyle B_z\left(x,y,t\right)=R(\rho )\Psi (\phi )\exp \left(i\left(kz-\omega t\right)\right)}$

Introducimos en la ecuación diferencial

${\displaystyle \begin{array}{l}\Psi \frac{1}{\rho }\frac{\partial }{\partial \rho }\left(\rho \frac{\partial R}{\partial \rho }\right)+R\frac{1}{{\rho }^2}\frac{{\partial }^2\Psi }{\partial {\phi }^2}+{\gamma }^{2}R\Psi =0\\ \frac{\rho }{R}\frac{\partial }{\partial \rho }\left(\rho \frac{\partial R}{\partial \rho }\right)+\frac{1}{\Psi }\frac{{\partial }^2\Psi }{\partial {\phi }^2}+{\gamma }^2{\rho }^2=0\end{array}}$

El segundo solamente función de φ. Convertimos una ecuación diferencial en un sistema de dos ecuaciones diferenciales cuya solución es sencilla

${\displaystyle \{\begin{array}{l}\frac{1}{\Psi }\frac{d^2\Psi }{d{\phi }^2}=-m^2\\ \rho \frac{d}{d\rho }\left(\rho \frac{dR}{d\rho }\right)+\left({\gamma }^2{\rho }^2-m^2\right)R=0\end{array}}$

La solución de la primera ecuación diferencial es

${\displaystyle \Psi \left(\phi \right)=\{\begin{array}{l}{C}_1\sin \left(m\phi \right)+C_2\cos \left(m\phi \right),m\ne 0\\ C_1+C_2\phi ,m=0\end{array}}$

Como la solución es periódica

${\displaystyle \begin{array}{l}\Psi \left(\phi \right)=\Psi \left(2\pi +\phi \right)\\ C_1\sin \left(m\phi \right)+C_2\cos \left(m\phi \right)=C_1\sin \left(m\phi +2\pi m\right)+C_2\cos \left(m\phi +2\pi m\right)=\\ C_1\sin \left(m\phi \right)+C_2\cos \left(m\phi \right)=C_1\left\{\sin \left(m\phi \right)\cos \left(2\pi m\right)+\cos \left(m\phi \right)\sin \left(2\pi m\right)\right\}+C_2\left\{\cos \left(m\phi \right)\cos \left(2\pi m\right)-\sin \left(m\phi \right)\sin \left(2\pi m\right)\right\}\end{array}}$

Esta igualdad es posible si m es entero

Resolvemos la segunda ecuación difeencial

${\displaystyle {\rho }^2\frac{d^{2}R}{d{\rho }^2}+\rho \frac{dR}{d\rho }+\left({\gamma }^2{\rho }^2-m^2\right)R=0}$

Sustituyendo r=ργ

${\displaystyle r^2\frac{d^{2}R}{d{r}^2}+r\frac{dR}{dr}+\left(r^2-m^2\right)R=0}$

La solución se expresa en términos de las funciones de Bessel

${\displaystyle \begin{array}{l}R(r)=A{J}_m(r)+B{Y}_m(r)\\ R(\rho )=A{J}_m\left(\gamma \rho \right)+B{Y}_m\left(\gamma \rho \right)\end{array}}$

Se descarta el segundo término ya que tiende a infinito cuando ρ→0

La expresión de la amplitud B_0z del campo magnético es

${\displaystyle B_{0z}=\{\begin{array}{l}A{J}_m\left(\gamma \rho \right)\left(C_1\sin \left(m\phi \right)+C_2\cos \left(m\phi \right)\right),m\ne 0\\ A{J}_m\left(\gamma \rho \right)\left(C_1+C_2\phi \right),m=0\end{array}}$

Eligiendo un origen apropiado de ángulos podemos hacer que C₁=0. Denominamos B₀ al producto A·C₂

Estudiamos el caso m≠0

${\displaystyle B_{0z}=B_0{J}_m\left(\gamma \rho \right)\cos \left(m\phi \right)}$

La componente B_z

${\displaystyle B_z=\mathrm{Re}\left\{B_{0z}\exp \left(i\left(kz-\omega t\right)\right)\right\}=B_0{J}_m\left(\gamma \rho \right)\cos \left(m\phi \right)\cos \left(kz-\omega t\right)}$

Derivamos B_0z respecto de ρ e φ

${\displaystyle \begin{array}{l}\frac{\partial B_{0z}}{\partial \rho }=B_0\frac{d{J}_m\left(\gamma \rho \right)}{d\rho }\cos \left(m\phi \right)\\ \frac{\partial B_{0z}}{\partial \phi }=-m{B}_0{J}_m\left(\gamma \rho \right)\sin \left(m\phi \right)\end{array}}$

TE, ondas transversales eléctricas

Las ondas transversales eléctricas están caracterizadas por que E_z=0

Sea una guía circular de radio a que se extiende a lo largo del eje Z

La condición de contorno es${\frac{\partial B_z}{\partial n}|}_{\text{pared}}=0$, donde el símbolo n indica derivada en la dirección normal a la superficie metálica evaluada en la posición de dicha superficie de la guía. Esta condición es equivalente a la siguiente

Sabiendo que E_0z=0, de la relación

${\displaystyle \begin{array}{l}{E}_{0\phi }=\frac{i}{{\gamma }^2}\left(\frac{k}{\rho }\frac{\partial E_{0z}}{\partial \phi }-\omega \frac{\partial B_{0z}}{\partial \rho }\right)=-\frac{i}{{\gamma }^2}\omega \frac{\partial B_{0z}}{\partial \rho }\\ {E_{0\phi }|}_{\rho =a}={-\frac{i}{{\gamma }^2}\omega \frac{\partial B_{0z}}{\partial \rho }|}_{\rho =a}=0\end{array}}$

El campo eléctrico es perpendicular a la superficie cilíndrica conductora

La condición de contorno en ρ=a

${\displaystyle \begin{array}{l}{\frac{\partial B_z}{\partial \rho }|}_{\rho =a}=0\\ {\frac{d{J}_m\left(\gamma \rho \right)}{d\rho }|}_{\rho =a}=0\end{array}}$

La derivada de las funciones J_m(x) de Bessel es

${\displaystyle \begin{array}{l}\frac{d{J}_m(x)}{dk}=\frac{m}{x}{J}_m(x)-J_{m+1}(x)\hfill \\ \frac{d{J}_m\left(\gamma \rho \right)}{d\rho }=\frac{m}{\rho }{J}_m\left(\gamma \rho \right)-\gamma J_{m+1}\left(\gamma \rho \right)\hfill \end{array}}$

Denominamos t_m,n a las raíces de la la ecuación transcendente

${\displaystyle \begin{array}{l}\begin{array}{l}\frac{m}{a}{J}_m\left(\gamma a\right)-\gamma J_{m+1}\left(\gamma a\right)=0\hfill \\ {\gamma }_{m,n}=\frac{t_{m,n}}{a}\hfill \end{array}\\ {\omega }^2=c^2\left(\frac{t_{m,n}^2}{a^2}+k^2\right)\end{array}}$

Para m=0, calculamos las raíces de la ecuación J₁(x)=0

function guia_1
    x=linspace(0,10,10);
    f=@(x) besselj(1,x);
    rr=raices(f,x); 
    disp('dJ_0(x)/dx=0')
    disp( rr)
    
    f=@(x) besselj(1,x)./x-besselj(2,x);
     rr=raices(f,x); 
    disp('dJ_1(x)/dx=0')
    disp (rr)

    f=@(x) 2*besselj(2,x)./x-besselj(3,x);
    rr=raices(f,x); 
    disp('dJ_2(x)/dx=0')
    disp(rr)

    function r = raices(f, x)
        y=f(x);
        indices=find(y(1:end-1).*y(2:end)<0);
        r=zeros(1,length(indices));
        for k=1:length(indices)
            r(k)=fzero(f, [x(indices(k)), x(indices(k)+1)]);
        end
    end
end

dJ_0(x)/dx=0,     3.8317    7.0156
dJ_1(x)/dx=0,     1.8412    5.3314    8.5363
dJ_2(x)/dx=0,     3.0542    6.7061    9.9695

Amplitudes

Conocido B_0z y E_0z=0, determinamos las otras amplitudes de las componentes del campo eléctrico y magnético

• Componente E_ρ

${\displaystyle \begin{array}{l}{E}_{0\rho }=\frac{i}{{\gamma }^2}\left(\frac{\omega }{\rho }\frac{\partial B_{0z}}{\partial \phi }+k\frac{\partial E_{0z}}{\partial \rho }\right)=\frac{i}{{\gamma }^2}\frac{\omega }{\rho }\frac{\partial B_{0z}}{\partial \phi }\\ E_{0\rho }=-\frac{i}{{\gamma }^2}\frac{\omega }{\rho }m{B}_0{J}_m\left(\gamma \rho \right)\sin \left(m\phi \right)\\ E_{\rho }=\mathrm{Re}\left\{E_{0\rho }\exp \left(i\left(kz-\omega t\right)\right)\right\}=\frac{\omega B_0}{{\gamma }^2}\frac{m}{\rho }{J}_m\left(\gamma \rho \right)\sin \left(m\phi \right)\sin \left(kz-\omega t\right)\end{array}}$

• Componente E_φ

${\displaystyle \begin{array}{l}{E}_{0\phi }=\frac{i}{{\gamma }^2}\left(\frac{k}{\rho }\frac{\partial E_{0z}}{\partial \phi }-\omega \frac{\partial B_{0z}}{\partial \rho }\right)=-\frac{i}{{\gamma }^2}\omega \frac{\partial B_{0z}}{\partial \rho }\\ E_{0\phi }=-\frac{i}{{\gamma }^2}\omega B_0\frac{d{J}_m\left(\gamma \rho \right)}{d\rho }\cos \left(m\phi \right)\\ E_{\phi }=\mathrm{Re}\left\{E_{0\phi }\exp \left(i\left(kz-\omega t\right)\right)\right\}=\frac{\omega B_0}{{\gamma }^2}\frac{d{J}_m\left(\gamma \rho \right)}{d\rho }\cos \left(m\phi \right)\sin \left(kz-\omega t\right)\end{array}}$

• Componente B_ρ

${\displaystyle \begin{array}{l}{B}_{0\rho }=\frac{i}{{\gamma }^2}\left(k\frac{\partial B_{0z}}{\partial \rho }-\frac{\omega }{c^2}\frac{1}{\rho }\frac{\partial E_{0z}}{\partial \phi }\right)=\frac{i}{{\gamma }^2}k\frac{\partial B_{0z}}{\partial \rho }\\ B_{0\rho }=\frac{i}{{\gamma }^2}k{B}_0\frac{d{J}_m\left(\gamma \rho \right)}{d\rho }\cos \left(m\phi \right)\\ B_{\rho }=\mathrm{Re}\left\{B_{0\rho }\exp \left(i\left(kz-\omega t\right)\right)\right\}=-\frac{k{B}_0}{{\gamma }^2}\frac{d{J}_m\left(\gamma \rho \right)}{d\rho }\cos \left(m\phi \right)\sin \left(kz-\omega t\right)\end{array}}$

• Componente B_φ

${\displaystyle \begin{array}{l}{B}_{0\phi }=\frac{i}{{\gamma }^2}\left(\frac{k}{\rho }\frac{\partial B_{0z}}{\partial \phi }+\frac{\omega }{c^2}\frac{\partial E_{0z}}{\partial \rho }\right)=\frac{i}{{\gamma }^2}\frac{k}{\rho }\frac{\partial B_{0z}}{\partial \phi }\\ B_{0\phi }=-\frac{i}{{\gamma }^2}\frac{k}{\rho }m{B}_0{J}_m\left(\gamma \rho \right)\sin \left(m\phi \right)\\ B_{\phi }=\mathrm{Re}\left\{B_{0\phi }\exp \left(i\left(kz-\omega t\right)\right)\right\}=\frac{k{B}_0}{{\gamma }^2}\frac{m}{\rho }{J}_m\left(\gamma \rho \right)\sin \left(m\phi \right)\sin \left(kz-\omega t\right)\end{array}}$

Resumiendo, las componentes del campo eléctrico $\overrightarrow{E}$ y del campo magbético $\overrightarrow{B}$, son

${\displaystyle \begin{array}{l}\overrightarrow{E}=\{\begin{array}{l}{E}_{\rho }=\frac{\omega B_0}{{\gamma }^2}\frac{m}{\rho }{J}_m\left(\gamma \rho \right)\sin \left(m\phi \right)\sin \left(kz-\omega t\right)\\ E_{\phi }=\frac{\omega B_0}{{\gamma }^2}\frac{d{J}_m\left(\gamma \rho \right)}{d\rho }\cos \left(m\phi \right)\sin \left(kz-\omega t\right)\\ E_z=0\end{array}\\ \overrightarrow{B}=\{\begin{array}{l}{B}_{\rho }=-\frac{k{B}_0}{{\gamma }^2}\frac{d{J}_m\left(\gamma \rho \right)}{d\rho }\cos \left(m\phi \right)\sin \left(kz-\omega t\right)\\ B_{\phi }=\frac{k{B}_0}{{\gamma }^2}\frac{m}{\rho }{J}_m\left(\gamma \rho \right)\sin \left(m\phi \right)\sin \left(kz-\omega t\right)\\ B_z=B_0{J}_m\left(\gamma \rho \right)\cos \left(m\phi \right)\cos \left(kz-\omega t\right)\end{array}\end{array}}$

TM, ondas transversales magnéticas

Las ondas transversales magnéticas están caracterizadas por que B_z=0.

La solución de la ecuación diferencial

${\displaystyle \frac{1}{\rho }\frac{\partial }{\partial \rho }\left(\rho \frac{\partial E_{0z}}{\partial \rho }\right)+\frac{1}{{\rho }^2}\frac{{\partial }^2{E}_{0z}}{\partial {\phi }^2}+{\gamma }^2{E}_{0z}=0}$

es, de modo análogo al primer apartado

${\displaystyle E_{0z}\left(\rho ,\phi ,t\right)=\{\begin{array}{l}A{J}_m\left(\gamma \rho \right)\left(C_1\sin \left(m\phi \right)+C_2\cos \left(m\phi \right)\right),m\ne 0\\ A{J}_m\left(\gamma \rho \right)\left(C_1+C_2\phi \right),m=0\end{array}}$

Eligiendo un origen apropiado de ángulos podemos hacer que C₁=0. Denominamos E₀ al producto A·C₂

${\displaystyle E_{0z}=E_0{J}_m\left(\gamma \rho \right)\cos \left(m\phi \right)}$

La condición de contorno E_z(a)=0 para ρ=a implica

${\displaystyle \begin{array}{l}\begin{array}{l}{J}_m\left(\gamma a\right)=0\hfill \\ {\gamma }_{m,n}=\frac{s_{m,n}}{a}\hfill \end{array}\\ {\omega }^2=c^2\left(\frac{s_{m,n}^2}{a^2}+k^2\right)\end{array}}$

s_m,n son las raíces de la ecuación transcendente

function guia
    x=linspace(0,10,10);
    f=@(x) besselj(0,x);
    rr=raices(f,x); 
    disp('J_0(x)=0')
    disp( rr)
    
    f=@(x) besselj(1,x);
    rr=raices(f,x); 
    disp('J_1(x)=0')
    disp (rr)

    f=@(x) besselj(2,x);
    rr=raices(f,x); 
    disp('J_2(x)=0')
    disp(rr)

    function r = raices(f, x)
        y=f(x);
        indices=find(y(1:end-1).*y(2:end)<0);
        r=zeros(1,length(indices));
        for k=1:length(indices)
            r(k)=fzero(f, [x(indices(k)), x(indices(k)+1)]);
        end
    end
end

J_0(x)=0,    2.4048    5.5201    8.6537
J_1(x)=0,    3.8317    7.0156
J_2(x)=0,    5.1356    8.4172

Amplitudes

La componente E_0z es

${\displaystyle \begin{array}{l}{E}_{0z}=E_0{J}_m\left(\gamma \rho \right)\cos \left(m\phi \right)\\ E_z=\mathrm{Re}\left\{E_{0z}\exp \left(i\left(kz-\omega t\right)\right)\right\}=E_0{J}_m\left(\gamma \rho \right)\cos \left(m\phi \right)\cos \left(kz-\omega t\right)\end{array}}$

Conocido E_0z y B_0z=0, determinamos las otras amplitudes de las componentes del campo eléctrico y magnético

• Componente E_ρ

${\displaystyle \begin{array}{l}{E}_{0\rho }=\frac{i}{{\gamma }^2}\left(\frac{\omega }{\rho }\frac{\partial B_{0z}}{\partial \phi }+k\frac{\partial E_{0z}}{\partial \rho }\right)=\frac{i}{{\gamma }^2}k\frac{\partial E_{0z}}{\partial \rho }\\ E_{0\rho }=\frac{i}{{\gamma }^2}k{E}_0\frac{d{J}_m\left(\gamma \rho \right)}{d\rho }\cos \left(m\phi \right)\\ E_{\rho }=\mathrm{Re}\left\{E_{0\rho }\exp \left(i\left(kz-\omega t\right)\right)\right\}=-\frac{k{E}_0}{{\gamma }^2}\frac{d{J}_m\left(\gamma \rho \right)}{d\rho }\cos \left(m\phi \right)\sin \left(kz-\omega t\right)\end{array}}$

• Componente E_φ

${\displaystyle \begin{array}{l}{E}_{0\phi }=\frac{i}{{\gamma }^2}\left(\frac{k}{\rho }\frac{\partial E_{0z}}{\partial \phi }-\omega \frac{\partial B_{0z}}{\partial \rho }\right)=\frac{i}{{\gamma }^2}\frac{k}{\rho }\frac{\partial E_{0z}}{\partial \phi }\\ E_{0\phi }=-\frac{i}{{\gamma }^2}\frac{k}{\rho }{E}_0{J}_m\left(\gamma \rho \right)m\sin \left(m\phi \right)\\ E_{\phi }=\mathrm{Re}\left\{E_{0\phi }\exp \left(i\left(kz-\omega t\right)\right)\right\}=\frac{k{E}_0}{{\gamma }^2}\frac{m}{\rho }{J}_m\left(\gamma \rho \right)\sin \left(m\phi \right)\sin \left(kz-\omega t\right)\end{array}}$

Para ρ=a, J_m(γa)=0, por lo que E_φ=0, el campo eléctrico tiene dirección radial perpendicular a la superficie cilíndrica conductora

• Componente B_ρ

${\displaystyle \begin{array}{l}{B}_{0\rho }=\frac{i}{{\gamma }^2}\left(k\frac{\partial B_{0z}}{\partial \rho }-\frac{\omega }{c^2}\frac{1}{\rho }\frac{\partial E_{0z}}{\partial \phi }\right)=-\frac{i}{{\gamma }^2}\frac{\omega }{c^2}\frac{1}{\rho }\frac{\partial E_{0z}}{\partial \phi }\\ B_{0\rho }=\frac{i}{{\gamma }^2}\frac{\omega }{c^2}\frac{1}{\rho }{E}_{0}m{J}_m\left(\gamma \rho \right)\sin \left(m\phi \right)\\ B_{\rho }=\mathrm{Re}\left\{B_{0\rho }\exp \left(i\left(kz-\omega t\right)\right)\right\}=-\frac{\omega E_0}{c^2{\gamma }^2}\frac{m}{\rho }{J}_m\left(\gamma \rho \right)\sin \left(m\phi \right)\sin \left(kz-\omega t\right)\end{array}}$

• Componente B_φ

${\displaystyle \begin{array}{l}{B}_{0\phi }=\frac{i}{{\gamma }^2}\left(\frac{k}{\rho }\frac{\partial B_{0z}}{\partial \phi }+\frac{\omega }{c^2}\frac{\partial E_{0z}}{\partial \rho }\right)=\frac{i}{{\gamma }^2}\frac{\omega }{c^2}\frac{\partial E_{0z}}{\partial \rho }\\ B_{0\phi }=\frac{i}{{\gamma }^2}\frac{\omega }{c^2}{E}_0\frac{d{J}_m\left(\gamma \rho \right)}{d\rho }\cos \left(m\phi \right)\\ B_{\phi }=\mathrm{Re}\left\{B_{0\phi }\exp \left(i\left(kz-\omega t\right)\right)\right\}=-\frac{\omega E_0}{c^2{\gamma }^2}\frac{d{J}_m\left(\gamma \rho \right)}{d\rho }\cos \left(m\phi \right)\sin \left(kz-\omega t\right)\end{array}}$

Resumiendo, las componentes del campo eléctrico $\overrightarrow{E}$ y del campo magnético $\overrightarrow{B}$, son

${\displaystyle \begin{array}{l}\overrightarrow{E}=\{\begin{array}{l}{E}_{\rho }=-\frac{k{E}_0}{{\gamma }^2}\frac{d{J}_m\left(\gamma \rho \right)}{d\rho }\cos \left(m\phi \right)\sin \left(kz-\omega t\right)\\ E_{\phi }=\frac{k{E}_0}{{\gamma }^2}\frac{m}{\rho }{J}_m\left(\gamma \rho \right)\sin \left(m\phi \right)\sin \left(kz-\omega t\right)\\ E_z=E_0{J}_m\left(\gamma \rho \right)\cos \left(m\phi \right)\cos \left(kz-\omega t\right)\end{array}\\ \overrightarrow{B}=\{\begin{array}{l}{B}_{\rho }=-\frac{\omega E_0}{c^2{\gamma }^2}\frac{m}{\rho }{J}_m\left(\gamma \rho \right)\sin \left(m\phi \right)\sin \left(kz-\omega t\right)\\ B_{\phi }=-\frac{\omega E_0}{c^2{\gamma }^2}\frac{d{J}_m\left(\gamma \rho \right)}{d\rho }\cos \left(m\phi \right)\sin \left(kz-\omega t\right)\\ B_z=0\end{array}\end{array}}$

Nota: las expresiones de B_ρ y B_φ difieren de las proporcionadas en la referencia, véase la página 372

Referencias

J. Pierrus. Solved Problems in Classical Electromagnetism. Analytical and numerical solutions with comments. Oxford University Press (2018). Question 7.16, pp. 369-372