# Boundary condition on $\mathbf{B}$ to describe resonant cavities from waveguides

In Jackson- Classical Electrodynamics when resonant cavities are discussed (8.6, page 252) (but also at page 7 here or at page 19 here) the explanation is made by saying that the solution is the same of a rectangular waveguide, which we can suppose that is for $E$

$$\mathbf{E}(x,y,z,t)=\mathbf{E_0}(x,y) e^{i(\alpha z-\omega t )}$$ $$\mathbf{B}(x,y,z,t)=\mathbf{B_0}(x,y) e^{i(\alpha z-\omega t )}$$

but instead of $e^{i(\alpha z )}$ there is a factor $H cos(wz)+J sin(wz)$, so $$\mathbf{E}(x,y,z,t)=\mathbf{E_0}(x,y) [H cos(wz)+J sin(wz)]e^{-i \omega t }$$ $$\mathbf{B}(x,y,z,t)=\mathbf{B_0}(x,y) [H' cos(wz)+J' sin(wz)]e^{-i \omega t }$$ And the boundary conditions imposed are teh following (supposing the lenght of cavity to be $d$ in $z$ direction) $$B_z(z=0)=B(z=d)=0 \,\,\,\,\,\,\,\,\,\,\, \forall x,y \tag{1}$$ $$E_x(z=0)=E_x(z=d)=E_y(z=0)=E_y(z=d)=0 \,\,\,\,\,\,\,\,\,\,\, \forall x,y \tag{2}$$

While $(2)$ is clear because it simply requires the electric field to be normal to the surface (which is the necessary condition to avoid dissipation), I do not see why $(1)$ is required.

It would be explained by the fact that $E$ is perpendicular to $B$ but that's a conclusion that one should get from the solution, not an assumption to impose the boundary conditions. Unfortunately I did not find an explanation for $(1)$ neither on Jackson neither on the references so why is $(1)$ a necessary boundary conditions to impose?

If you're not sure why the magnetic field is always zero inside a PEC, consider Faraday's law: $$\Bbb \nabla \times \mathbf E(\mathbf r,t) = -\frac \partial{\partial t} \mathbf B(\mathbf r,t)$$ Inside a PEC, the electric field is always zero, so $\mathbf E=0$ and we get: $$\frac \partial{\partial t} \mathbf B(\mathbf r,t)=0$$ This means that the magnetic field is constant in time; i.e. it has not, and will not change from the beginning of time to its end (if it has one !). Now since in all practical cases, the magnetic field is zero inside the PEC at some point in time (e.g. before one has turned on the circuit), the magnetic field is always identically zero, because it cannot change.