A plane electromagnetic wave, with wavelength 3.0m, travels in vacuum in the positive direction of anxaxis. The electric field, of amplitude 300 V/m, oscillates parallel to theyaxis. What are the (a) frequency (b) angular frequency, and (c) angular wave number of the wave? (d) What is the amplitude of the magnetic field component? (e) Parallel to which axis does the magnetic field oscillate? (f) What is the time-averaged rate of energy flow in watts per square meter associated with this wave? The wave uniformly illuminates a surface of area 2.0 m². If the surface totally absorbs the wave, (g) What is the rate at which momentum is transferred to the surface? and (h) What is the radiation pressure on the surface?

The amplitude of the electric field,$E_m=300\text{V/m}$.

Wavelength is, $\lambda =3.0\text{m}$.

The surface area is, $A=2.0{\text{\hspace{0.17em}m}}^{\text{2}}$.

Find the frequency, angular frequency, and wave number using respective formulae. Amplitude can be found by using the relationship between electric and magnetic fields. The direction of propagation can be found by using the cross product of E and B. Intensity can be found by using the formula which relates the electric field with the velocity of light. The rate of change of momentum can be written in terms of intensity, velocity of light, and area. Finally, the rate of momentum transferred can be used to find the radiation pressure.

Formulae are as follows:

The rate of energy flow in watts per meter square is

$I=\frac{E_m^2}{2\times {\mu }_{0}c}$

The amplitude of the magnetic field (B_m):

$B_m=\frac{E_m}{c}$

Where µ₀ is the permeability of free space,B_m is the amplitude of the magnetic field, I is the rate of energy flow, and c is the speed of the light.

Frequency (f) can be calculated as,

$f=\frac{c}{\lambda }$

Substitute the values in the above expression, and we get,

$$\begin{gathered} f=\frac{3\times {10}^8}{3} \\ f=1.0\times {10}^8\text{Hz} \end{gathered}$$

Therefore, frequency is $f=1.0\times {10}^8\text{Hz}$.

Angular frequency$\left(\omega \right)$ can be calculated as,

$\omega =2\pi f$

Substitute the values in the above expression, and we get,

$$\begin{gathered} \omega =2\pi \times 1.0\times {10}^8 \\ \omega =6.28\times {10}^8\text{Hz} \end{gathered}$$

Therefore, the angular frequency is $\omega =6.28\times {10}^8\text{Hz}$.

Wave number (k) can be calculated as,

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

Substitute the values in the above expression, and we get,

$\begin{array}{l}k=\frac{2\pi }{6.28\times {10}^8}\\ k=2.09\text{\hspace{0.17em}rad/m}\end{array}$

Therefore, the wave number is$k=2.09\text{\hspace{0.17em}rad/m}$

The amplitude of the magnetic field (B_m) can be calculated as,

$B_m=\frac{E_m}{c}$

Substitute the values in the above expression, and we get,

$$\begin{gathered} B_{\text{m}}=\frac{300}{3\times {10}^8} \\ {B}_{\text{m}}=1.0\times {10}^{-6}\text{T} \end{gathered}$$

Therefore, the amplitude of the magnetic field is ${\text{B}}_{\text{m}}=1.0\times {10}^{-6}\text{T}$.

$\overrightarrow{E}$is in the positive y direction, and the direction of propagation $\overrightarrow{E}\times \overrightarrow{B}$is in the positive x direction, so $\overrightarrow{B}$should be in the positive z direction.

Thus, the magnetic field B oscillates in a positive z direction.

The rate of energy flow in watts per meter square can be calculated as,

$I=\frac{E_m^2}{2\times {\mu }_{0}c}$

Substitute the values in the above expression, and we get,

$\begin{array}{c}I=\frac{{300}^2}{2\times 4\pi \times 3\times {10}^8}\\ I=119.43\frac{\text{W}}{{\text{m}}^{\text{2}}}\\ \approx 120\frac{\text{W}}{{\text{m}}^{\text{2}}}\end{array}$

Therefore, the rate of energy flow in watts per meter square is $I=\text{120}\frac{\text{W}}{{\text{m}}^{\text{2}}}$.

The momentum transferred $\frac{dp}{dt}$can be calculated as,

$\frac{dp}{dt}=\frac{IA}{c}$

Substitute the values in the above expression, and we get,

$$\begin{gathered} \frac{dp}{dt}=\frac{119.43\times 2}{3\times {10}^8} \\ =79.62\times {\text{10}}^{\text{- 8}}\text{N} \\ ~\text{80}\times {\text{10}}^{\text{- 8}}\text{N} \end{gathered}$$

Therefore, the momentum transferred is $\frac{dp}{dt}=\text{80}\times {\text{10}}^{\text{- 8}}\text{N}$.

Radiation pressure (P) can be calculated as,

$P=\frac{dp/dt}{A}$

Substitute the values in the above expression, and we get,

$\begin{array}{c}P=\frac{79.62\times {10}^{-8}}{2}\\ P=3.981\times {10}^{-7}\text{\hspace{0.17em}Pa}\\ ~\text{4.0}\times {\text{10}}^{\text{-7}}\text{\hspace{0.17em}Pa}\end{array}$

Therefore, the radiation pressure is 4.0x10^-7 Pa.