= Solution
Choose the guide axis as $z$, with $0<x<a$ and $0<y<b$. The <electric field> is along $y$, parallel to the walls' $b$ dimension. Take a homogeneous lossless nonmagnetic <dielectric> with <relative permittivity> $\epsilon$, ideal conducting walls, and time dependence $e^{-i\omega t}$. The lowest <transverse electric mode of a rectangular waveguide> is
$$
\widetilde{\mathbf E}=\widehat{\mathbf y}E_0\sin\!\left(\frac{\pi x}a\right)e^{i\beta z},\qquad
\beta^2=\frac{\epsilon\omega^2}{c^2}-\frac{\pi^2}{a^2},\qquad \omega=2\pi f.
$$
At $x=0,a$, this tangential <electric field> vanishes. At $y=0,b$ the <electric field> is normal to the wall; a nonzero value is supported by surface <electric charge> and violates no conducting boundary condition. The field has zero <divergence>. The associated normal <magnetic field> also vanishes at each wall, as verified below. The fundamental sine across the wider dimension minimizes the nonzero transverse wave number; $\mathrm{TE}_{00}$ is not a physical propagating mode of a hollow simply connected guide.
Below the <waveguide cutoff frequency> $f_c=c/(2a\sqrt\epsilon)$, put $\beta=i\kappa$ with $\kappa>0$ to select decay for $z>0$. The <evanescent transverse electric mode in a rectangular waveguide> then has
$$
\widetilde E_y=E_0\sin(\pi x/a)e^{-\kappa z},\qquad
\boxed{\ell^{-2}=\kappa^2=\pi^2\!\left(\frac1{a^2}-\frac{4\epsilon f^2}{c^2}\right).}
$$
For the fixed <frequency> in the question,
$$
\boxed{\frac\ell a=\frac1{\pi\sqrt{1-\epsilon/4}}\quad(0<\epsilon<4).}
$$
The decay length increases monotonically and diverges as $\epsilon\to4^-$; its formal limit at $\epsilon\to0^+$ is $a/\pi$. At $\epsilon=4$ the mode is at cutoff, with zero axial wave number. For $\epsilon>4$, $\beta=(\pi/a)\sqrt{\epsilon/4-1}$ is real and the mode propagates without axial attenuation in this ideal lossless model. There is no real finite evanescent decay length to continue beyond cutoff. The paper's graph notation $l$ denotes the same $\ell$ as its preceding formula.
\Image[/past-exam-of-the-physics-b-course-of-the-university-of-cambridge/2015/paper-1-waveguide-cutoff.png]
{title=Evanescent decay length and propagating axial wave number on opposite sides of cutoff}
From <Faraday's law of induction>, $\nabla\times\widetilde{\mathbf E}=i\omega\widetilde{\mathbf B}$. Writing $q=\pi/a$ gives
$$
\boxed{\widetilde B_x=-\frac{i\kappa}{\omega}E_0\sin(qx)e^{-\kappa z},\quad
\widetilde B_y=0,\quad
\widetilde B_z=-\frac{iq}{\omega}E_0\cos(qx)e^{-\kappa z}.}
$$
Thus $B_x$ vanishes on the $x$ walls and $B_y$ vanishes on the $y$ walls, meeting the normal magnetic boundary conditions. The fields also satisfy $\nabla\cdot\mathbf B=0$. For real $E_0$ the physical fields are $E_y=E_0\sin(qx)e^{-\kappa z}\cos\omega t$, $B_x=-(\kappa/\omega)E_0\sin(qx)e^{-\kappa z}\sin\omega t$, and $B_z=-(q/\omega)E_0\cos(qx)e^{-\kappa z}\sin\omega t$.
The time-averaged <Poynting vector> is
$$
\langle\mathbf S\rangle=\frac1{2\mu_0}\operatorname{Re}
(\widetilde{\mathbf E}\times\widetilde{\mathbf B}^{\,*})=\mathbf0,
\qquad \boxed{\langle P_z\rangle=0.}
$$
All products are imaginary because the <electric field> and <magnetic field> are in quadrature. A single decaying <evanescent wave> stores reactive <energy> without transmitting average power into a semi-infinite lossless guide. A finite below-cutoff section can transmit power when both independent evanescent solutions are present; that is a different boundary-value problem.
The <skin effect> also gives exponential field decay, but in a good metal $k$ has both real and imaginary parts and there is a real inward mean <Poynting vector>. That flux supplies <Joule heating> $\sigma|\widetilde E|^2/2$. Here attenuation comes from geometric cutoff in a lossless region, with no conductivity and no dissipative power sink; exponential decay alone does not establish absorption.
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