Chapter 23.4, Problem 1aTH

A long, thin steel wire is cut in half, and each half is connected to a different terminal of a light bulb. An electromagnetic (EM) plane wave $\left(\stackrel{\to }{E}\left(x,y,z,t\right)=E_o\sin \left(kx-\text{ω}t\right)\hat{y},\stackrel{\to }{B}\left(x,y,z,t\right)=B_o\sin \left(kx-\text{ω}t\right)\hat{z}\right)$moves past the wire, as shown.

a. In what direction is the wave propagating? Explain your reasoning.

To determine

The direction of the propagation of the wave.

Answer to Problem 1aTH

The direction of the propagation of the wave is along positive x axis.

Explanation of Solution

Introduction:

An electromagnetic wave is shown graphically as below:

  

Figure 1: An electromagnetic wave

The given electric field is in the form of sinusoidal wave form is:

  $\overrightarrow{E}\left(x,y,z,t\right)=E_{o}Sin\left(kx-\omega t\right)\widehat{y}$

According to the above equation, the direction of oscillating electrical field is along y and the direction of the wave propagation is along x direction. The corresponding magnetic field is given as below:

  $\overrightarrow{B}\left(x,y,z,t\right)=B_{o}Sin\left(kx-\omega t\right)\widehat{z}$

The magnetic field (black dotted sine wave) and electric field (red dotted sine wave) are shown in Figure 1.

The direction of the electromagnetic wave can be determined by the argument of the function, here is the Sin function which is $kx-\omega t$ which is constant.

  $kx-\omega t=\text{constant}\dots .\left(1\right)$

  $k$ is the propagation constant, x is the direction of propagation, $\omega$ is the angular frequency.

Derivative of this equation 1

  $\begin{array}{l}k\frac{dx}{dt}-\omega =0\\ \frac{dx}{dt}=\frac{\omega }{k}=v=positive\end{array}$

The speed $\frac{\omega }{k}=v$ is positive, hence wave is propagating along positive x axis.

Conclusion:

The direction of the propagation of the wave is along positive x axis.