##### Coaxial Lines
emdm.emwm.wg.c

We now consider a cylindrical geometry, formed by two concentric conducting cylinders. We let \(a\) (resp. \(b\)) represent the radius of the inner (resp. outer) cylinder.

In this case, we can find nontrivial solutions to Maxwell's equations for \(E_z = 0 = B_z\). Simple inspection gives \(k = \frac{\omega}{c}\) and

\begin{align} c B_y &= E_x, &\hspace{10mm} \frac{\partial E_x}{\partial x} + \frac{\partial E_y}{\partial y} &= 0, &\hspace{10mm} \frac{\partial E_y}{\partial x} - \frac{\partial E_x}{\partial y} &= 0, \nonumber \\ c B_x &= -E_y, & \frac{\partial B_x}{\partial x} + \frac{\partial B_y}{\partial y} &= 0, & \frac{\partial B_y}{\partial x} - \frac{\partial B_x}{\partial y} &= 0 \end{align}

The derivative equations are precisely the electrostatic and magnetic equations for empty space with cylindrical symmetry. The solution is thus that of an infinite line charge and an infinite straight current: \[ {\boldsymbol E}_0 (s, \varphi) = \frac{A}{s} \hat{\boldsymbol s}, \hspace{10mm} {\boldsymbol B}_0 (s, \varphi) = \frac{A}{cs} \hat{\boldsymbol \varphi} \] where \(A\) is a constant amplitude. Substituting and taking the real part, \[ {\boldsymbol E} (s, \varphi, z, t) = \frac{A}{s} \cos (kz - \omega t) \hat{\boldsymbol s}, \hspace{10mm} {\boldsymbol B} (s, \varphi, z, t) = \frac{A}{cs} \cos (kz - \omega t) \hat{\boldsymbol \varphi}. \]

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Author: Jean-Sébastien Caux

Created: 2022-03-24 Thu 08:42