**1 Answer**

written 11 months ago by |

**Solution:**

In a simple-free. Linear, i.wtropic. the homogeneous region, Maxwell's curl equations in phasor form is,

$\nabla \times \bar{E}=-j \omega \mu \bar{H}$,

$\nabla \times \bar{H}=j \omega \epsilon \bar{E}$.

and constitute two equations for the two unknowns, $\bar{E}$ and $\bar{H}$. As such. Lhey can be solved for either $\bar{E}$ or $\bar{H}$. Thus, taking the curl of gives,

$$ \nabla \times \nabla \times \bar{E}=-j \omega \mu \nabla \times \bar{H}=\omega^2 \mu \epsilon \bar{E}, $$

which is an equation for $\bar{E}$.

This result can be simplified through the use of vector identity (B.), $\nabla \times \nabla \times \bar{A}=\nabla(\nabla \cdot \bar{A})-\nabla^2 \bar{A}$, which is valid for the rectangular components of an arbitrary vector $\bar{A}$. Then:

$\nabla^2 \bar{E}+\omega^2 \mu \epsilon \bar{E}=0$

since $\nabla \cdot \bar{E}=0$ in a source-free region. The equation is the wave equation. or Helmholtz equation. for $\bar{E}$. An identical equation for $\bar{H}$ can be derived in the same manner:

$$ \nabla^2 \bar{H}+\omega^2 \mu \epsilon \bar{H}=0 $$

A constant $k=\omega \sqrt{\mu \epsilon}$ is defined and called the wavenumber. or propagation constant. of the medium; its units are $1 / \mathrm{m}$.

As a way of introducing wave behavior, we will next study the solutions to the above wave equations in their simplest forms, first for a lossless medium and then for a lossy (conducting) medium.