The rate at which sunlight and other electromagnetic waves transport energy is described by the wave intensity. For a plane wave traveling in the direction of the positive x-axis with the phase of the wave chosen so that the wave maximum is at the origin at \(t = 0\), the electric and magnetic fields obey the equations, \[E_y (x,t) = E_0 \, \cos \, (kx - \omega t)\], \[B_x (x,t) = B_0 \, \cos \, (kx - \omega t).\], The energy in any part of the electromagnetic wave is the sum of the energies of the electric and magnetic fields. We therefore need to take some time to develop the mathematics behind this effect. The amplitude of the magnetic field can be obtained from Equation 16.20: As an Amazon Associate we earn from qualifying purchases. 1 Answer Sorted by: 1 I could explain it here, but someone has already done it: http://physics.info/intensity/. This is possible only if the wave is propagating to the right in the diagram, in which case, the relative orientations show that \(\vec{S} = \frac{1}{\mu_0} \vec{E} \times \vec{B}\) is specifically in the direction of propagation of the electromagnetic wave. Solution From Equation 16.31, the intensity of the laser beam is Therefore the only source of phase difference is the separation of the sources, \(\Delta x\). Large ocean breakers churn up the shore more than small ones. Notice that even though the resultant wave looks very different from its "parents," the medium somehow "remembers" the original waves, and when they no longer coincide, they continue along as exactly the waves they were before the superposition. These fields can exert forces and move charges in the system and, thus, do work on them. As the waves propagate along, the values of \(x\) and \(t\) will change, but as the two waves are identical (traveling in the same direction with the same speed), the differences in \(x\) and \(t\) dont change for a given phase. The intensity is proportional to the square of the amplitude, so the intensity of this combined wave is: \[I\propto A_{new}^2 = 4A^2\cos^2\left(\dfrac{\Delta\Phi}{2}\right)\]. For a plane wave traveling in the direction of the positive x -axis with the phase of the wave chosen so that the wave maximum is at the origin at t = 0, the electric and magnetic fields obey the equations (16.4.1) E y ( x, t) = E 0 cos ( k x t) (16.4.2) B x ( x, t) = B 0 cos ( k x t). Consider a sinusoidal wave on a string that is produced by a string vibrator, as shown in Figure \(\PageIndex{2}\). The intensity (I) of a wave is defined as the rate at which it transfers energy divided by the area over which the energy is spread. For example, if we have two stereo speakers putting out \(1.00 \, W/m^2\) each, there will be places in the room where the intensity is \(4.00 \, W/m^2\), other places where the intensity is zero, and others in between. The intensity of a sound depends upon its pressure amplitude. When two or more waves of the same type in the same medium coexist in the same region of space, they combine to create a new wave. The potential energy of the mass element is equal to, \[\Delta U = \frac{1}{2} k_{s} x^{2} = \frac{1}{2} \Delta m \omega^{2} x^{2} \ldotp \nonumber \]. As the energy propagates along the string, each mass element of the string is driven up and down at the same frequency as the wave. It allows us to relate acoustic power and intensity to acoustic pressure and flow, and to calculate and the reflection and transmission at boundaries. a. The energy of the wave spreads around a larger circumference and the intensity decreases proportional to 1 r, 1 r, which is also the same in the case of a spherical wave, since intensity is proportional to the amplitude squared. We can use whichever of the three preceding equations is most convenient, because the three equations are really just different versions of the same result: The energy in a wave is related to amplitude squared. The wave energy is determined by the wave amplitude. What intensity would such sunlight have if concentrated by a magnifying glass onto an area 200 times smaller than its own? It has to equal 1 when when the phase difference is 0 (modulo \(2\pi\)), since this means the waves constructively interfere. If either the angular frequency or the amplitude of the wave were doubled, the power would increase by a factor of four. are licensed under a, Heat Transfer, Specific Heat, and Calorimetry, Heat Capacity and Equipartition of Energy, Statements of the Second Law of Thermodynamics, Conductors, Insulators, and Charging by Induction, Calculating Electric Fields of Charge Distributions, Electric Potential and Potential Difference, Motion of a Charged Particle in a Magnetic Field, Magnetic Force on a Current-Carrying Conductor, Applications of Magnetic Forces and Fields, Magnetic Field Due to a Thin Straight Wire, Magnetic Force between Two Parallel Currents, Applications of Electromagnetic Induction, Maxwells Equations and Electromagnetic Waves, Energy carried by a wave depends on its amplitude. It has to vanish when the phase difference equals \(\pi\) (modulo \(2\pi\)), since this means the waves totally destructively interfere. Note that Lenzs and Faradays laws imply that when the magnetic field shown is increasing in time, the electric field is greater at x than at x+xx+x. Power is the rate at which energy is transferred by the wave. We will see that the average rate of energy transfer in mechanical waves is proportional to both the square of the amplitude and the square of the frequency. Consider the example of the seagull and the water wave earlier in the chapter (Figure 16.2.2). Consider a mass element of the string with a mass \(\Delta\)m, as seen in Figure \(\PageIndex{2}\). The amplitude of the electric field is therefore. The potential energy associated with a wavelength of the wave is equal to the kinetic energy associated with a wavelength. Other times, it is subtle, such as the unfelt energy of gamma rays, which can destroy living cells. Both of these are measures of intensity. In equation form, intensity I is I = P A, 14.5 where P is the power through an area A. In the case of the two-dimensional circular wave, the wave moves out, increasing the circumference of the wave as the radius of the circle increases. From Equation \ref{16.31}, the intensity of the laser beam is, \[I = \frac{1}{2}c\epsilon_0 E_0^2. If two identical waves, each having an intensity of \(1.00 \, W/m^2\). Notice that this relationship between total intensity and phase difference exactly matches the three criteria we outlined above. The energy of some waves can be directly observed. It is also proportional to the square of the frequency.The variation with time t of displacement x of particles when two progressive waves Q and P pass separately through a medium, are shown on the graphs. We can use whichever of the three preceding equations is most convenient, because the three equations are really just different versions of the same result: The energy in a wave is related to amplitude squared. k = \frac { {2\pi }} {\lambda } k = 2. The intensity of light moving at speed \(c\)in vacuum is then found to be, \[I = S_{avg} = \frac{1}{2}c\epsilon_0 E_0^2 \label{16.31}\], in terms of the maximum electric field strength \(E_0\), which is also the electric field amplitude. Sometimes this energy is obvious, such as in the warmth of the summer Sun. This page titled 16.5: Energy and Power of a Wave is shared under a CC BY 4.0 license and was authored, remixed, and/or curated by OpenStax via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request. 4 Answers Sorted by: 6 A light wave is a traveling disturbance in the electric and magnetic fields. The most common is the decibel. \[S = \frac{\text{Energy passing area } A \text{ in time } \Delta t}{A \Delta t} = uc = \epsilon_0cE^2 = \frac{1}{\mu_0} EB.\]. The quantity \(I\) is the intensity of the wave as a function of the phase difference of the two (identical) parent waves. \nonumber\], The amplitude of the electric field is therefore, \[ \begin{align*} E_0 &= \sqrt{\frac{2}{c\epsilon_0}I} \\[4pt] &= \sqrt{\frac{2}{(3.00 \times 10^8 m/s)(8.85 \times 10^{-12} F/m)}\left(1.0 \times 10^{-3} W/m^2 \right)} \\[4pt] &= 0.87 \, V/m. \end{align*}\]. We will examine a great many examples of interference in physical phenomena in the sections to come. The energy per unit area per unit time passing through a plane perpendicular to the wave, called the energy flux and denoted by \(S\), can be calculated by dividing the energy by the area \(A\) and the time interval \(\Delta t\). Decreasing the area increases the intensity considerably. This kinetic energy can be integrated over the wavelength to find the energy associated with each wavelength of the wave: \[\begin{split} dK & = \frac{1}{2} (\mu\; dx)[A^{2} \omega^{2} \cos^{2}(kx - \omega t)] \\ \int_{0}^{K_{\lambda}} dK & = \int_{0}^{\lambda} \frac{1}{2} \mu A^{2} \omega^{2} \cos^{2}(kx - \omega t) dx = \frac{1}{2} \mu A^{2} \omega^{2} \int_{0}^{\lambda} \cos^{2} (kx) dx, \\ K_{lambda} & = \frac{1}{2} \mu A^{2} \omega^{2} \Big[ \frac{1}{2} x + \frac{1}{4k} \sin (2kx) \Big]_{0}^{\lambda} \\ & = \frac{1}{2} \mu A^{2} \omega^{2} \Big[ \frac{1}{2} \lambda + \frac{1}{4k} \sin (2k \lambda) - \frac{1}{4k} \sin(0) \Big] \\ & = \frac{1}{4} \mu A^{2} \omega^{2} \lambda \ldotp \end{split}\], There is also potential energy associated with the wave. The larger the displacement \(x\) the larger the force \(F = kx\) needed to create it. What is your question? 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\newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\) \(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)\(\newcommand{\AA}{\unicode[.8,0]{x212B}}\), source@https://openstax.org/details/books/university-physics-volume-2, Express the time-averaged energy density of electromagnetic waves in terms of their electric and magnetic field amplitudes, Calculate the Poynting vector and the energy intensity of electromagnetic waves, Explain how the energy of an electromagnetic wave depends on its amplitude, whereas the energy of a photon is proportional to its frequency. This gives: \[\dfrac{I'}{I} = 4.\], Calculate to find \(I'\): \[I' = 4I = 4.00 \, W/m^2.\]. This equation can be used to find the energy over a wavelength. \nonumber\], Substitute known values into the equation: \[E = (700 \, W/m^2)(0.500 \, m^2)[(4.00 \, h)(3600 \, s/h)]. Using the proportionality of the areas to the squares of the distances, and solving, we obtain from the diagram, \[ \begin{align*} \frac{r_2^2}{r_1^2} &= \frac{A_2}{A_1} = \frac{90 \, W}{60 \, W}, \\[4pt] r_2 &= \sqrt{\frac{90}{60}}(100 \, km) \\[4pt] &= 122 \, km. The intensity of a sound wave is a combination of its rate and density of energy transfer. The string vibrator is a device that vibrates a rod up and down. Sunlight, for example, can be focused to burn wood. \nonumber\], Substitute known quantities: \[ I' = 200 I = 200(700 \, W/m^2). We recommend using a The SI unit for intensity is watts per square meter (W/m2). A laser beam can burn away a malignancy. Have a look on wikipedia and apply it for a . The area over which the intensity is \(4.00 \, W/m^2\) is much less than the area covered by the two waves before they interfered. There are other intensity-related units in use, too. OpenStax is part of Rice University, which is a 501(c)(3) nonprofit. are not subject to the Creative Commons license and may not be reproduced without the prior and express written Intensity is dependent on the density of the medium, speed, frequency, and amplitude of the sound wave; for an electromagnetic wave (here- light), medium is not required but other factors remain the same. The amplitude of the combined wave is therefore: \[A_{combined}=2A\cos\left(\dfrac{\Delta\Phi}{2}\right) = 2\left(8.0cm\right)\cos\left(0.41\pi\right) = 4.46cm\nonumber\]. Changing the area the waves cover has important effects. However, there is energy in an electromagnetic wave itself, whether it is absorbed or not. The kinetic energy K = \(\frac{1}{2}\)mv2 of each mass element of the string of length \(\Delta\)x is \(\Delta\)K = \(\frac{1}{2}\)(\(\Delta\)m)vy2, as the mass element oscillates perpendicular to the direction of the motion of the wave. 1 Answer Sorted by: 1 Sound wave velocity: (1) v s = B ; where the mass density of medium, B is the boulk modulous defined as ratio of pressure change P with percentage change of volume ( V V ) B = P V / V = V d P d V. Assume the area of the cross-section of propagation is A, and V = A d x, the volume unit without deformation. This page titled 16.4: Energy Carried by Electromagnetic Waves is shared under a CC BY 4.0 license and was authored, remixed, and/or curated by OpenStax via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request. Except where otherwise noted, textbooks on this site Creative Commons Attribution License in this video let's derive the expression for intensity in the young's double slit experiment so we'll derive expression for intensity of the resulting wave and we'll do that as a function of its phase difference phase difference is usually denoted by phi now just a quick heads up this most of this video is going to be visual so a way for you to. Here is a simple example of two pulses "colliding" (the "sum" of the top two waves yields the bottom wave). Sometimes this energy is obvious, such as in the warmth of the summer Sun. Furthermore, because these equations are based on the assumption that the electromagnetic waves are sinusoidal, the peak intensity is twice the average intensity; that is, \(I_0 = 2I\). Calculate the amount of energy that falls on a solar collector having an area of \(0.500 \, m^2\) in \(4.00 \, h\). Here, the velocity is the velocity of the oscillations of the medium, and not the velocity of the sound wave. The LibreTexts libraries arePowered by NICE CXone Expertand are supported by the Department of Education Open Textbook Pilot Project, the UC Davis Office of the Provost, the UC Davis Library, the California State University Affordable Learning Solutions Program, and Merlot. then you must include on every physical page the following attribution: If you are redistributing all or part of this book in a digital format, Identify the knowns: \[A = 200 A',\] \[\dfrac{I'}{I} = 200. As a spherical wave moves out from a source, the surface area of the wave increases as the radius increases (A = 4\(\pi\)r2). Earthquakes can shake whole cities to the ground, performing the work of thousands of wrecking balls. Figure \(\PageIndex{2}\) shows what this interference might look like. A simpler approach would be to note that the intensity is proportional to the square of the amplitude, and since changing the position of the second source increases the amplitude from \(4.46cm\) to \(16cm\) (double the individual wave amplitude), which is an increase of a factor of 3.59, the intensity increases by a factor of \(3.59^2=12.9\), giving the same result (except for some rounding errors along the way). So intensity of light is basically the power transmitted through electric and . The speed of the wave on the string can be derived from the linear mass density and the tension. The total energy associated with a wavelength is the sum of the potential energy and the kinetic energy: \[\begin{split} E_{\lambda} & = U_{\lambda} +K_{\lambda} \\ & = \frac{1}{4} \mu A^{2} \omega^{2} \lambda + \frac{1}{4} \mu A^{2} \omega^{2} \lambda \\ & = \frac{1}{2} \mu A^{2} \omega^{2} \lambda \ldotp \end{split}\]. 1999-2023, Rice University. Calculate the intensity and the power of rays and waves. For example, a sound speaker mounted on a post above the ground may produce sound waves that move away from the source as a spherical wave. For a plane wave traveling in the direction of the positive x-axis with the phase of the wave chosen so that the wave maximum is at the origin at t=0t=0, the electric and magnetic fields obey the equations, The energy in any part of the electromagnetic wave is the sum of the energies of the electric and magnetic fields. What does happen is intriguing. The electric field in the electromagnetic wave is E= 6sin(2(10 6=310 14t)) i. The intensity goes up by a factor of 4 when the amplitude doubles. This answer is a little disquieting. The way they combine is a simple process known as superposition. We will do this within the same framework that we have been using that of harmonic waves. Solution: Given : f 2 f 1 = 1 / 4 1 = 1 4 A 2 A 1 = 2 1 I = 2 2 f 2 A 2 v I 2 I 1 = 2 2 f 2 2 A 2 2 v 2 2 f 1 2 A 1 2 v The time-averaged power of a sinusoidal mechanical wave, which is the average rate of energy transfer associated with a wave as it passes a point, can be found by taking the total energy associated with the wave divided by the time it takes to transfer the energy. In this section, we examine the quantitative expression of energy in waves. We know from Superposition and Interference that when two identical waves, which have equal amplitudes \(X\) interfere perfectly constructively, the resulting wave has an amplitude of \(2X\). Note that this is four times the intensity of each individual wave, since the constructive interference adds the amplitudes (which are equal the waves are identical) and the intensity is proportional to the square of the amplitude. This phase difference is related to the change in position through the wavelength: \[\Delta \Phi = \frac{2\pi}{\lambda} \Delta x \;\;\;\Rightarrow\;\;\; \Delta x = \Delta \Phi\left(\frac{\lambda}{2\pi}\right) =\left(1.18\pi\right)\frac{61cm}{2\pi}=36cm \nonumber\]. The intensity for a spherical wave is therefore, \[I = \frac{P}{4 \pi r^{2}} \ldotp \label{16.12}\]. The electric field is decreasing with increasing \(x\) at the given time and location. The SI unit for I is W/m 2. At what distance in the same direction would the signal have the same maximum field strength if the transmitters output power were increased to 90 kW? If you toss a pebble in a pond, the surface ripple moves out as a circular wave. b. Each mass element of the string can be modeled as a simple harmonic oscillator. The wavelength of the wave divided by the period is equal to the velocity of the wave, \[P_{ave} = \frac{E_{\lambda}}{T} = \frac{1}{2} \mu A^{2} \omega^{2} \frac{\lambda}{T} = \frac{1}{2} \mu A^{2} \omega^{2} v \ldotp \label{16.10}\]. . Two identical harmonically-oscillating devices attached to a taut string are turned on simultaneously and in phase, vibrating at a frequency of \(20Hz\). The intensity of electromagnetic waves is derived by considering the sinusoidal expression of wave. The definition of intensity is valid for any energy in transit, including that carried by waves. The energy of the wave depends on both the amplitude and the frequency. \end{align*}\]. The electric field is decreasing with increasing x at the given time and location. 1 I was going through the texts of my book and read that intensity of a wave is proportional to its amplitude squared. Work is done on the seagull by the wave as the seagull is moved up, changing its potential energy. Formula Intensity Formula Download PDF NCERT Solutions CBSE CBSE Study Material Textbook Solutions CBSE Notes Intensity Formulas - Definitions & Solved Example Questions Intensity is a measure of the energy transmitted by a wave. The potential energy of the mass element can be found by considering the linear restoring force of the string, In Oscillations, we saw that the potential energy stored in a spring with a linear restoring force is equal to U = \(\frac{1}{2}\)ksx2, where the equilibrium position is defined as x = 0.00 m. When a mass attached to the spring oscillates in simple harmonic motion, the angular frequency is equal to \(\omega = \frac{k_{s}}{m}\). Decibels will be discussed in some detail in a later chapter. They simply create a new wave while they occupy the same space in the medium, and when their individual motions carry them to different parts of the medium, they return to being the waves they were before. One source is located at the origin, and the other is positioned at \(x=25cm\). The larger the amplitude, the higher the seagull is lifted by the wave and the larger the change in potential energy. It uses some results from The wave equation for sound. Expressions for both field energy densities were discussed earlier (\(u_E\) in Capacitance and \(u_B\) in Inductance). Combining these the contributions, we obtain, \[u (x,t) = u_E + u_B = \frac{1}{2}\epsilon_0 E^2 + \frac{1}{2\mu_0} B^2.\], The expression \(E = cB = \frac{1}{\sqrt{\epsilon_0\mu_0}}B\) then shows that the magnetic energy density \(u_B\) and electric energy density \(u_E\) are equal, despite the fact that changing electric fields generally produce only small magnetic fields. Legal. Note that Lenzs and Faradays laws imply that when the magnetic field shown is increasing in time, the electric field is greater at \(x\) than at \(x + \Delta x\). Direction: Select the correct option using below statement. A 60-kW radio transmitter on Earth sends its signal to a satellite 100 km away (Figure \(\PageIndex{3}\)). Loudness is a perceptual response to the physical property of intensity. The intensity of the (out of phase) combined wave is therefore: \[I=I_o\cos^2\left(\dfrac{\Delta\Phi}{2}\right)\]. The energy falling on the solar collector in 4 h in part is enough to be usefulfor example, for heating a significant amount of water. Similar questions. The SI unit for intensity is watts per square meter . Substituting the fact that \(cB_0 = E_0\), the previous expression becomes, \[I = \frac{E_0B_0}{2\mu_0} \label{16.33}.\]. Power is the rate at which energy is transferred by the wave. The energy flux at any place also varies in time, as can be seen by substituting u from Equation 16.23 into Equation 16.27. Accessibility StatementFor more information contact us [email protected]. Loud sounds have high-pressure amplitudes and come from larger-amplitude source vibrations than soft sounds. For example, the longer deep-heat ultrasound is applied, the more energy it transfers. Recall that intensity is proportional to amplitude squared. Displacement - Intensity can be related to the amplitude, A, or displacement, of a wave through the equation {eq}I = 2 \pi^2 \rho f^2v A^2 {/eq} where {eq}\rho {/eq} stands for the density of the . Variation of Intensity with amplitude of wave Example: If the amplitude of sound is doubled and the frequency is reduced to one-fourth, then find the intensity of sound at the same point. The amount of energy in a wave is related to its amplitude. If some energy is later absorbed, the field strengths are diminished and anything left travels on. We need to calculate the linear density to find the wave speed: $$\mu = \frac{m_{s}}{L_{s}} = \frac{0.070\; kg}{2.00\; m} = 0.035\; kg/m \ldotp$$, The wave speed can be found using the linear mass density and the tension of the string: $$v = \sqrt{\frac{F_{T}}{\mu}} = \sqrt{\frac{90.00\; N}{0.035\; kg/m}} = 50.71\; m/s \ldotp$$, The angular frequency can be found from the frequency: $$\omega = 2 \pi f = 2 \pi (60\; s^{-1}) = 376.80\; s^{-1} \ldotp$$, Calculate the time-averaged power: $$P = \frac{1}{2} \mu A^{2} \omega^{2} v = \frac{1}{2} (0.035\; kg/m)(0.040\; m)^{2}(376.80\; s^{-1})^{2}(50.71\; m/s) = 201.5\; W \ldotp$$. The function needs to have the following properties: To find this function, we start with two wave functions that are identical except for their phases and superpose them: \[f_{tot}=f_1+f_2=A\cos\left(\Phi_1\right)+A\cos\left(\Phi_2\right)=A\left[\cos\left(\Phi_1\right)+\cos\left(\Phi_2\right)\right] \]. In this video I explain two more properties of waves: amplitude and intensity, for A Level Physics.This catches lots of people out. The energy density moves with the electric and magnetic fields in a similar manner to the waves themselves. As defined in physics, the intensity of a wave is proportional to the square of its amplitude ( A2 I ). Since the string has a constant linear density \(\mu = \frac{\Delta m}{\Delta x}\), each mass element of the string has the mass \(\Delta\)m = \(\mu \Delta\)x. Water waves chew up beaches. Its wavelength is the distance from crest to crest or from trough to trough. Algebraic manipulation produces the relationship The average intensity of sunlight on Earths surface is about \(700 \, W/m^2\). then you must include on every digital page view the following attribution: Use the information below to generate a citation. Clearly, the larger the strength of the electric and magnetic fields, the more work they can do and the greater the energy the electromagnetic wave carries. The LibreTexts libraries arePowered by NICE CXone Expertand are supported by the Department of Education Open Textbook Pilot Project, the UC Davis Office of the Provost, the UC Davis Library, the California State University Affordable Learning Solutions Program, and Merlot. One more expression for IavgIavg in terms of both electric and magnetic field strengths is useful. If the two waves happen to be in phase, then the combined wave's intensity is \(I_o\) when the two waves are in phase. In general, the energy of a mechanical wave and the power are proportional to the amplitude squared and to the angular frequency squared (and therefore the frequency squared). Legal. \nonumber\], Calculate to find \(I'\): \[I' = 1.40 \times 10^5 \, W/m^2. Example \(\PageIndex{2}\): Determine the combined intensity of two waves: Perfect constructive interference. A light bulb emits 5.00 W of power as visible light. Intensity Formula measures area in the plane perpendicular to the direction of the propagation of energy. By the end of this section, you will be able to: Anyone who has used a microwave oven knows there is energy in electromagnetic waves.

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