SKEDSOFT

Physics For Engineers - 2
To see that propagation is really a wave disturbance, take y-component of Eqn. (3) and x-component of Eqn. (4)
To get the wave equation for $E_x$, take the derivative of eqn. (5) with respect to $z$ and substitute in eqn. (6) and interchange the space and time derivatives,
\begin{displaymath}\frac{\partial^2 E_x}{\partial z^2} = \mu_0\epsilon_0 \frac{... ...z}\right)= \mu_0\epsilon_0 \frac{\partial^2 E_x}{\partial t^2}\end{displaymath}
Similarly, we can show, We get
\begin{displaymath} \frac{\partial^2 B_y}{\partial z^2} = \mu_0\epsilon_0 \frac{\partial^2 B_y}{\partial t^2} \end{displaymath}
Each of the above equations represents a wave disturbance propagating in the z-direction with a speed
\begin{displaymath}c = \frac{1}{\sqrt{\mu_0\epsilon_0}}\end{displaymath}
On substituting numerical values, the speed of electromagnetic waves in vacuum is $3\times 10^{8}$ m/sec.
Consider plane harmonic waves of angular frequency $\omega$ and wavlength $\lambda=2\pi/k$. We can express the waves as
\begin{eqnarray*} E_x &=& E_0\sin(kz-\omega t)\\ B_y &=& B_0\sin(kz-\omega t) \end{eqnarray*}
The amplitudes $E_0$ an $B_0$ are not independent as they must satisfy eqns. (5) and (6) :
\begin{eqnarray*} \frac{\partial E_x}{\partial z} &=& E_0 k \cos(kz-\omega t)\\ \frac{\partial B_y}{\partial t} &=& -B_0 \omega \cos(kz-\omega t) \end{eqnarray*}
Using Eqn. (5) we get
\begin{displaymath}E_0k = B_0\omega\end{displaymath}
The ratio of the electric field amplitude to the magnetic field amplitude is given by
\begin{displaymath}\frac{E_0}{B_0}=\frac{\omega}{k} = c\end{displaymath}
Fields $\vec E$ and $\vec B$ are in phase, reaching their maximum and minimum values at the same time. The electric field oscillates in the x-z plane and the magnetic field oscillates in the y-z plane. This corresponds to a polarized wave . Conventionally, the plane in which the electric field oscillates is defined as the plane of polarization. In this case it is x-z plane. The figure shows a harmonic plane wave propagating in the z-direction. Note that $\vec E, \vec B$ and the direction of propagation $\hat k$ form a right handed triad.