Question

The electric component of a beam of polarized light is

${\mathit{E}}_{\mathbf{y}}\mathbf{=}\left(5.00\text{V/m}\right)\mathbf{}\mathit{s}\mathit{i}\mathit{n}\left[\left(1.00\times {10}^6{\text{m}}^{\text{- 1}}\right)z+\omega t\right]$.

(a) Write an expression for the magnetic field component of the wave, including a value for $\mathit{\omega }$.

What are the (b) wavelength, (c) period, and (d) intensity of this light?

(e) Parallel to which axis does the magnetic field oscillate?

(f) In which region of the electromagnetic spectrum is this wave?

Short answer

a. $B_x=\left(1.67\times {10}^{-8}\text{T}\right)\sin \left[\left(1.00\times {10}^6{\text{m}}^{-1}\right)z+\left(3.00\times {10}^{14}\text{rad/s}\right)t\right]$

b. $\lambda =6.28\times {10}^{-6}\text{m}$

c. $T=2.09\times {10}^{-14}s$

d. $I=0.0332\frac{\text{W}}{{\text{m}}^{\text{2}}}$

e. The magnetic field oscillates along thex axis.

f. The region of the electromagnetic spectrum of the wave is an infrared region.

Step-by-step solution

Given Information

The expression of the electric field is$E_y=\left(5.00\text{V/m}\right)\sin \left[\left(1.00\times {10}^6{\text{m}}^{\text{- 1}}\right)z+\omega t\right]$

Understanding the concept

We can use the concept of the electric field and magnetic field assinusoidal functions of positionxand timet. By comparing the given equation with this expression, we get the values of the wave number and maximum electric field. Use the expression of the amplitude of the magnetic field and find its value. We can use the expression of the period, and intensity of the light. By using the electromagnetic spectrum concept, we can find the region of the oscillation of the electromagnetic spectrum.

Formula:

$E_y=E_m\sin \left(kx-\omega t\right)$

$\frac{E}{B}=c$

$\lambda =\frac{2\pi }{k}$

$T=\frac{2\pi }{\omega }$

$I=\frac{1}{c{\mu }_0}{\left(\frac{E}{\sqrt{2}}\right)}^2$

a. Find an expression for the magnetic field component of the wave including a value for  ω

An expression for the magnetic field component:

The expression of the electric field as sinusoidal functions of position $x$ and time $t$ is

$E_y=E_m\sin \left(kx-\omega t\right)$

The given expression as

$E_y=\left(5.00\text{V/m}\right)\sin \left[\left(1.00\times {10}^6{\text{m}}^{\text{- 1}}\right)z+\omega t\right]$

By comparing the given equations, we get

$E_m=5.00\text{V/m}$and

$k=1.00\times {10}^6{\text{m}}^{\text{- 1}}$

The expression of the angular velocity is

$$\begin{gathered} \omega =kc \\ \omega =1.00\times {10}^6{\text{m}}^{-1}\times 3\times {10}^8\text{m/s} \\ \omega =3.00\times {10}^{14}\text{rad/s} \end{gathered}$$

The expression of amplitude of magnetic field is

$$\begin{gathered} \frac{E}{B}=c \\ B=\frac{E}{c} \\ B=\frac{5.00V/m}{3\times {10}^{8}m/s} \\ B=1.67\times {10}^{-8}T \end{gathered}$$

From the given wave, the electromagnetic wave is propagating along thez-axis axis and the electric field along the y-axis.

$\overrightarrow{E}=E_y\hat{j}$

Hence, the magnetic wave propagates along the x-axis axis. Then the x component of the magnetic field is $B_x$ as$B_x=\left(1.67\times {10}^{-8}\text{T}\right)\sin \left[\left(1.00\times {10}^6{\text{m}}^{-1}\right)z+\left(3.00\times {10}^{14}\text{rad/s}\right)t\right]$

b. Calculating the wavelength 

The expression of the wavelength is

$$\begin{gathered} \lambda =\frac{2\pi }{k} \\ \lambda =\frac{2\times 3.14}{1.00\times {10}^6{\text{m}}^{-1}} \\ \lambda =6.28\times {10}^{-6}\text{m} \end{gathered}$$

Hence, the wavelength is $\lambda =6.28\times {10}^{-6}\text{m}$.

c. Calculating the period

The period:

The expression of the period is

$$\begin{gathered} T=\frac{2\pi }{\omega } \\ T=\frac{2\times 3.14}{3.00\times {10}^{14}\text{rad/s}} \\ T=2.09\times {10}^{-14}\text{s} \end{gathered}$$

Hence, the period is $T=2.09\times {10}^{-14}s$.

d. Calculating the intensity

The expression of the intensity is

$$\begin{gathered} I=\frac{1}{c{\mu }_0}{\left(\frac{E}{\sqrt{2}}\right)}^2 \\ I=\frac{1}{3\times {10}^{8}m/s\times 4\times 3.14}{\left(\frac{5.00\text{V/m}}{\sqrt{2}}\right)}^2 \\ I=0.0332{\text{W/m}}^{\text{2}} \end{gathered}$$

Hence, the intensity of the light is $I=0.0332\frac{\text{W}}{{\text{m}}^{\text{2}}}$.

e. Calculating the parallel to which axis the magnetic field oscillates 

The only non-zero component of $\overrightarrow{B}$ is $B_x$. Hence the magnetic field oscillates along the x-axis perpendicular to the electric field and electromagnetic wave.

f. Calculating the region of electromagnetic spectrum

According to the value of wavelength we get,

$\lambda =6.28\times {10}^{-6}\text{m}$

According to the electromagnetic spectrum, the region of the infrared is ranging from ${10}^{-6}\text{m}$ to ${10}^{-3}\text{m}.$

Hence, it indicates that, the region of the electromagnetic spectrum is the infrared region of the spectrum.