Правка разделов «Statement», «Answer»
en/12.1.3.md
+2 −3
| @@ -1,7 +1,6 @@ | |||
| ### Statement | |||
| − | $12.1.3.$ [Insert the problem statement] | ||
| − | |||
| + | $12.1.3.$ The figure illustrates the electric field of a plane sinusoidal wave at time $t = 0$. The arrow indicates the direction of wave propagation. How does the electric field intensity depend on the coordinate $z$ and time $t$? | ||
| ### Solution | |||
| For a plane sinusoidal wave, the general equation of the electromagnetic wave is given by: | |||
| \begin{equation} | |||
| E(z,t)=E_0\sin(At+Bz+\varphi_0) | |||
| \end{equation} | |||
| where $A$,$B$ and $\varphi_0$ are constants. | |||
| From the condition of the problem we know that: | |||
| \begin{equation} | |||
| E(z,0)=E_0\sin(\frac{2\pi}{\lambda}z) | |||
| \end{equation} | |||
| Using eq(1) we get: | |||
| \begin{equation} | |||
| E_0\sin{(Bz+\varphi_0)}=E_0\sin(\frac{2\pi}{\lambda}z) | |||
| \end{equation} | |||
| By comparing the phases we obtain that: | |||
| \begin{equation} | |||
| B=\frac{2\pi}{\lambda};\varphi_0=0 | |||
| \end{equation} | |||
| As the wave travels in the positive direction: | |||
| \begin{equation} | |||
| \frac{dz}{dt}=c | |||
| \end{equation} | |||
| Setting the phase($\varphi=At+Bz+\varphi_0$)to a constant value allows us to derive the phase velocity($c$): | |||
| \begin{equation} | |||
| A+B\frac{dz}{dt}=\frac{d\varphi}{dt}=0 | |||
| \end{equation} | |||
| So we get the constant $A$: | |||
| \begin{equation} | |||
| A=-\frac{2\pi c}{\lambda} | |||
| \end{equation} | |||
| Plugging all the constants into the eq(1) gives us: | |||
| \begin{equation} | |||
| E(z,t)=E_0\sin(\frac{2\pi}{\lambda}(z-ct)) | |||
| \end{equation} | |||
| @@ -73,4 +72,4 @@Solution | |||
| #### Answer | |||
| − | [Insert a concise answer or boxed result] | ||
| + | $E(z,t)=E_0\sin(\frac{2\pi}{\lambda}(z-ct))$ | ||
| @@ -1,7 +1,6 @@ | |||
| ### Statement | ### Statement | ||
| $12.1.3.$ [Insert the problem statement] | $12.1.3.$ The figure illustrates the electric field of a plane sinusoidal wave at time $t = 0$. The arrow indicates the direction of wave propagation. How does the electric field intensity depend on the coordinate $z$ and time $t$? | ||
| ### Solution | ### Solution | ||
| For a plane sinusoidal wave, the general equation of the electromagnetic wave is given by: | For a plane sinusoidal wave, the general equation of the electromagnetic wave is given by: | ||
| \begin{equation} | \begin{equation} | ||
| E(z,t)=E_0\sin(At+Bz+\varphi_0) | E(z,t)=E_0\sin(At+Bz+\varphi_0) | ||
| \end{equation} | \end{equation} | ||
| where $A$,$B$ and $\varphi_0$ are constants. | where $A$,$B$ and $\varphi_0$ are constants. | ||
| From the condition of the problem we know that: | From the condition of the problem we know that: | ||
| \begin{equation} | \begin{equation} | ||
| E(z,0)=E_0\sin(\frac{2\pi}{\lambda}z) | E(z,0)=E_0\sin(\frac{2\pi}{\lambda}z) | ||
| \end{equation} | \end{equation} | ||
| Using eq(1) we get: | Using eq(1) we get: | ||
| \begin{equation} | \begin{equation} | ||
| E_0\sin{(Bz+\varphi_0)}=E_0\sin(\frac{2\pi}{\lambda}z) | E_0\sin{(Bz+\varphi_0)}=E_0\sin(\frac{2\pi}{\lambda}z) | ||
| \end{equation} | \end{equation} | ||
| By comparing the phases we obtain that: | By comparing the phases we obtain that: | ||
| \begin{equation} | \begin{equation} | ||
| B=\frac{2\pi}{\lambda};\varphi_0=0 | B=\frac{2\pi}{\lambda};\varphi_0=0 | ||
| \end{equation} | \end{equation} | ||
| As the wave travels in the positive direction: | As the wave travels in the positive direction: | ||
| \begin{equation} | \begin{equation} | ||
| \frac{dz}{dt}=c | \frac{dz}{dt}=c | ||
| \end{equation} | \end{equation} | ||
| Setting the phase($\varphi=At+Bz+\varphi_0$)to a constant value allows us to derive the phase velocity($c$): | Setting the phase($\varphi=At+Bz+\varphi_0$)to a constant value allows us to derive the phase velocity($c$): | ||
| \begin{equation} | \begin{equation} | ||
| A+B\frac{dz}{dt}=\frac{d\varphi}{dt}=0 | A+B\frac{dz}{dt}=\frac{d\varphi}{dt}=0 | ||
| \end{equation} | \end{equation} | ||
| So we get the constant $A$: | So we get the constant $A$: | ||
| \begin{equation} | \begin{equation} | ||
| A=-\frac{2\pi c}{\lambda} | A=-\frac{2\pi c}{\lambda} | ||
| \end{equation} | \end{equation} | ||
| Plugging all the constants into the eq(1) gives us: | Plugging all the constants into the eq(1) gives us: | ||
| \begin{equation} | \begin{equation} | ||
| E(z,t)=E_0\sin(\frac{2\pi}{\lambda}(z-ct)) | E(z,t)=E_0\sin(\frac{2\pi}{\lambda}(z-ct)) | ||
| \end{equation} | \end{equation} | ||
| @@ -73,4 +72,4 @@Solution | |||
| #### Answer | #### Answer | ||
| [Insert a concise answer or boxed result] | $E(z,t)=E_0\sin(\frac{2\pi}{\lambda}(z-ct))$ | ||