Plane Electromagnetic Waves in Non-conducting Isotropic Medium

Isotropic medium is a non-conducting medium in which has a same property in all directions.

The Maxwell’s equations are

\begin{align*}\left. \begin{matrix}\nabla . \vec {D} = \rho \\\nabla . \vec B = 0 \\ \nabla \times \vec E = -\frac {\partial \vec {B}}{\partial t}\\\nabla \times \vec H = \vec J + \frac {\partial \vec {D}}{\partial t}\\ \end{matrix}\right )\dots (1)\end{align*}

In an isotropic dielectric medium

The Maxwell’s equations are

\begin{align*}\left. \begin{matrix} \vec {D} = \varepsilon\vec E \\ \vec B = \mu \vec H \\ \vec J = \sigma \vec E = 0 \\ \rho = 0 \end{matrix}\right )\dots (2)\end{align*}

From equations (1) and (2)

\begin{align*} \vec {\nabla }.\vec {E}= 0,\: \vec {\nabla }. \vec H = 0 \\ \nabla \times \vec E &= -\mu \frac {\partial \vec H}{\partial t}\: \text {and}\: \nabla \times \vec H &= \varepsilon \frac {\partial \vec E}{\partial t} \dots (3)\\ \text {Now}\: \vec {\nabla } \times (\vec {\nabla } \times \vec E) &= -\mu \frac {\partial}{\partial t}(\vec {\nabla } \times \vec H) \\ &= -\mu \frac {\partial}{\partial t} \left (\varepsilon \frac {\partial \vec E}{\partial t}\right ) \\ &= - \mu \varepsilon \frac {\partial ^2\vec E}{\partial t^2}\\ \vec {\nabla }\times (\vec {\nabla} \times \vec H) &= - \mu \varepsilon \frac {\partial ^2\vec E}{\partial t^2}\end{align*}

Hence the wave equations are

\begin{align*} \nabla^2\vec E - \mu \varepsilon \frac {\partial ^2\vec E}{\partial t^2} &= 0\: \:(\because\vec {\nabla}\times (\vec {\nabla}\times E) = \nabla ^2 \vec E) \\ \nabla^2\vec H - \mu \varepsilon \frac {\partial ^2\vec H}{\partial t^2} &= 0\end{align*}

\ \begin{align*} \text {And}\: \vec K \times \vec E &= \mu \omega \vec H \\ \vec K \times \vec H &= -\varepsilon \omega \vec E \\ \text {or,}\: \vec H \times \vec K &= \epsilon \omega \vec E\end{align*}

Phases of \(\vec E\) nad \(\vec H\) and wave Impedance:
\begin{align*} \vec H &= \frac {1}{\mu\omega } (\vec k \times \vec E) = \frac {k}{\mu\omega }(\hat n \times \vec E)\: [\because\vec k = \hat n \: k] \\ &= \frac {1}{\mu v} (\hat n \times \vec E)= \sqrt {\frac {\varepsilon}{\mu}}(\hat n \times \vec E)\: [\because k = \frac {\omega }{v} ,\: v \frac {1}{\sqrt {\mu \varepsilon}}]\end{align*}

Now, the ratio of magnitude of E to the magnitude of H is symbolized by Z i.e.

\begin{align*} Z &= \frac {|E|}{|H|} = \frac {|E_0|}{|H_0|}\\ &=\sqrt {\frac {\mu _0}{\varepsilon_0}} = \sqrt {\frac {k_m \mu_0}{k_e\varepsilon_0}} = \text {real} \end{align*}

Where \(\mu = \frac {\mu _0}{K_e}\) and \(\varepsilon = \frac {\varepsilon_0}{K_m}\).

This shows that that the field vectors \(\vec E\) and \(\vec H\) are in the same phase. i.e. they have same relative magnitudes at all points at all time.

Since unit of \(\vec E\) ,\(\vec H\) and Z is volt/meter, amp-turn/meter and ohm \(\Omega\).

Hence, the value of Z is referred to as wave impedance of isotropic dielectric medium. The wave impedance of medium is related to that of free space by the relation.

\begin{align*} Z &= \sqrt {\frac {k_m\mu_0}{K_e\varepsilon_0}} = \sqrt {\frac {k_m}{k_e}}. Z_0,\end{align*}

where \(Z_0 = \sqrt {\frac {\mu_0}{\epsilon _0}} \) is called wave impedance of free space.

We know

\begin{align*} \vec k. \vec E &= 0 \rightarrow \:\vec {\nabla}.\vec E = 0 \dots (1)\\ \vec k. \vec B &= 0 \rightarrow \:\vec {\nabla}.\vec B = 0 \dots (2) \\ \vec K \times \vec E &= \omega \vec B \rightarrow \vec {\nabla } \times \vec E = - \frac {\partial \vec B}{\partial t} \dots (3) \\ \vec K \times \vec B &= \mu\varepsilon \omega \: \vec E \rightarrow \vec {\nabla } \times \vec B = \mu \varepsilon \frac {\partial \vec E}{\partial t} \dots (4) \end{align*}

If the medium is non magnetic, isotropic and non conducting.

Then \(\mu = \mu_0\)

$$ \varepsilon_r = \frac {\varepsilon}{\varepsilon_0} \rightarrow \varepsilon = \varepsilon_0\varepsilon_r\: \text {where}\: k = \frac {\omega}{c}$$

\begin{align*} \text {From equation}\:(4)\\ \vec K \times \vec B &= -\mu_0\varepsilon_0\omega \vec E = \frac{1}{c^2}\varepsilon _r\omega \vec E\\\text {Now},\:\vec k \times \vec K \times \vec B &=-\frac{1}{c^2}\varepsilon _r\omega \vec K \times \vec E \\\vec k.(\vec k.\vec B)- (\vec k.\vec k)\vec B &= -\frac{1}{c^2}\varepsilon _r\omega (\vec K \times \vec E )\\\vec k.(\vec k.\vec B)- (\vec k.\vec k)\vec B &= -\frac{1}{c^2}\varepsilon _r.\omega \vec B \end{align*}

\begin{align*} \text {or,}\: 0 - k^2\vec B &= -\frac {1}{c^2}\varepsilon_r.\omega \vec B \\ \text {or,}\: k^2\vec B &= \frac {1}{c^2}\varepsilon_r.\omega \vec B \\ \text {From equation }\: (3),\\\vec k\times\vec k\times \vec E &= \omega . \vec k\times \vec B \\ \vec k.(\vec k .\vec B)- k^2 \vec B &=-\omega^2\varepsilon_r\vec E \mu_o \varepsilon_0 \end{align*}

\begin{align*} \text {or,}\: -k^2 \vec B &=-\frac {\omega^2}{c^2}\varepsilon_r\vec E \:\:\:\:\:\:\text {where}\: \omega ^2\varepsilon_0 =\frac {1}{c^2} \\\vec B &=-\frac {\varepsilon_r\omega^2}{c^2}\vec E \:\:\:\:\:\:\text {where}\: k ^2 =\frac {\omega^2}{c^2}\varepsilon_r \\ \vec H &=-\frac {\varepsilon_r\omega^2}{c^2\mu_0}\vec E \:\:\:\:\:\:\text {where}\: n = \sqrt {\varepsilon_r}\: \text {or,}\: n = \frac {ck}{\omega}\end{align*}

Bibliography

P.B. Adhikari, Bhoj Raj Gautam, Lekha Nath Adhikari. A Textbook of Physics. kathmandu: Sukunda Pustak Bhawan, 2011.

Jha, V. K.; 'Lecture title'; Maxwell's Equation; St. Xavier's College, Kathmandu; 2016.