#set document(title: "16.3 Energy Carried by Electromagnetic Waves", author: "OpenStax / XYZ Homework") #set page(width: 8.5in, height: auto, margin: 1in) #import "@preview/cetz:0.5.2" #set text(font: ("STIX Two Text", "Libertinus Serif", "New Computer Modern"), size: 10.5pt, lang: "en") #show math.equation: set text(font: ("STIX Two Math", "New Computer Modern Math")) #set par(justify: true, leading: 0.62em, spacing: 0.9em) #set enum(spacing: 1.1em) // room between list items so tall inline fractions don't collide #set list(spacing: 1.1em) #set table(stroke: 0.5pt + rgb("#c7ccd3")) #let BLUE = rgb("#183B6F") // brand navy — section bars + example/solution labels (white on navy 11.09:1) #let ORANGE = rgb("#A94509") // brand primary-700 — AA-safe deep orange for TEXT (5.93:1 on white; raw brand #F37021 is 2.94:1 and must never carry text) #let RED = rgb("#DC2626") // brand error-600 #let GREEN = rgb("#059669") // brand success-600 (decoration only; small green text uses green-text #007942) #show heading.where(level: 1): it => block(width: 100%, above: 0pt, below: 16pt, fill: gradient.linear(BLUE, rgb("#2C5AA0")), inset: (x: 14pt, y: 12pt), radius: 3pt, text(fill: white, weight: "bold", size: 19pt, it.body)) #show heading.where(level: 2): it => block(width: 100%, above: 18pt, below: 10pt, fill: BLUE, inset: (x: 10pt, y: 6pt), radius: 2pt, text(fill: white, weight: "bold", size: 12pt, it.body)) #show heading.where(level: 3): it => text(fill: ORANGE, weight: "bold", size: 12.5pt, it.body) #show heading.where(level: 4): it => text(fill: BLUE, weight: "bold", size: 10.5pt, it.body) #let examplebox(label, title, body) = block(width: 100%, breakable: true, fill: rgb("#EFF1F5"), stroke: 0.5pt + rgb("#CFDDF0"), radius: 4pt, inset: 10pt, above: 12pt, below: 12pt)[ #block(below: 6pt)[#box(fill: BLUE, inset: (x: 6pt, y: 2pt), radius: 2pt, text(fill: white, weight: "bold", size: 8.5pt, label)) #h(0.4em) #strong[#title]] #body] // rail = decorative left rule (raw brand token); labelcolor = AA-safe label text shade #let notebox(label, rail, labelcolor, tint, body) = block(width: 100%, breakable: true, fill: tint, stroke: (left: 3pt + rail), inset: (left: 10pt, rest: 8pt), radius: (right: 4pt), above: 11pt, below: 11pt)[ #text(fill: labelcolor, weight: "bold", size: 7.5pt, tracking: 0.5pt)[#upper(label)] #linebreak() #body] #let solutionbox(body) = block(above: 4pt, below: 8pt)[ #text(fill: BLUE, weight: "bold", size: 8.5pt)[Solution] #linebreak() #body] #let figph(msg) = block(width: 100%, height: 60pt, fill: rgb("#f6f7f9"), stroke: (paint: rgb("#c7ccd3"), dash: "dashed"), radius: 4pt, inset: 10pt)[ #align(center + horizon, text(fill: rgb("#889"), style: "italic", size: 9pt, msg))] // Standardize inlined figure sizes: measure the natural CeTZ canvas, then scale to a // consistent envelope (aspect-aware; see build_typst.py FIG_* constants). Unlike the // print preamble, dimensions are FLOORED: in an editor a user can trim a figure to a // degenerate 1-D shape (a bare line), and w/h or tw/w would then divide by zero. #let _STD_W = 3.5 #let _WIDE_W = 5.6 #let _MAX_H = 3.4 #let _ASPECT_WIDE = 2.2 #let _UPSCALE_MAX = 1.15 #let stdfig(body) = context { let m = measure(body) let w = calc.max(m.width / 1in, 0.01) let h = calc.max(m.height / 1in, 0.01) let tw = if w / h > _ASPECT_WIDE { _WIDE_W } else { _STD_W } let s = calc.min(tw / w, _MAX_H / h, _UPSCALE_MAX) align(center, box(scale(x: s * 100%, y: s * 100%, reflow: true, body))) } #show figure: set block(breakable: false) #set figure(gap: 8pt) #show figure.caption: set text(size: 8.5pt, fill: rgb("#555")) == 16.3#h(0.6em)Energy Carried by Electromagnetic Waves Anyone who has used a microwave oven knows there is energy in electromagnetic waves. Sometimes this energy is obvious, such as in the warmth of the summer Sun. Other times, it is subtle, such as the unfelt energy of gamma rays, which can destroy living cells. Electromagnetic waves bring energy into a system by virtue of their electric and magnetic fields. These fields can exert forces and move charges in the system and, thus, do work on them. However, there is energy in an electromagnetic wave itself, whether it is absorbed or not. Once created, the fields carry energy away from a source. If some energy is later absorbed, the field strengths are diminished and anything left travels on. 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. In electromagnetic waves, the amplitude is the maximum field strength of the electric and magnetic fields . The wave energy is determined by the wave amplitude. #figure(figph[Figure on the left shows an electromagnetic wave with electric field E and magnetic field B. It is labeled u. Figure on the right shows an electromagnetic wave with electric field 2E and magnetic field 2B. Here, the amplitudes of the sine waves are doubled. The wave is labeled 4u.], alt: "Figure on the left shows an electromagnetic wave with electric field E and magnetic field B. It is labeled u. Figure on the right shows an electromagnetic wave with electric field 2E and magnetic field 2B. Here, the amplitudes of the sine waves are doubled. The wave is labeled 4u.", caption: [Energy carried by a wave depends on its amplitude. With electromagnetic waves, doubling the #emph[E] fields and #emph[B] fields quadruples the energy density #emph[u] and the energy flux #emph[uc].]) For a plane wave traveling in the direction of the positive #emph[x]-axis with the phase of the wave chosen so that the wave maximum is at the origin at #math.equation(block: false, alt: "t equals 0")[$t = 0$], the electric and magnetic fields obey the equations #math.equation(block: true, alt: "E sub y open parenthesis x , t close parenthesis equals E sub 0 cos open parenthesis k x minus ω t close parenthesis; B sub z open parenthesis x , t close parenthesis equals B sub 0 cos open parenthesis k x minus ω t close parenthesis .")[$E_(y) ( x , t ) = E_(0) #h(0.2em) "cos" #h(0.2em) ( k x − ω t ) \ B_(z) ( x , t ) = B_(0) #h(0.2em) "cos" #h(0.2em) ( k x − ω t ) .$] The energy in any part of the electromagnetic wave is the sum of the energies of the electric and magnetic fields. This energy per unit volume, or #strong[energy density] #emph[u], is the sum of the energy density from the electric field and the energy density from the magnetic field. Expressions for both field energy densities were discussed earlier (#math.equation(block: false, alt: "u sub E")[$u_(E)$] in Capacitance and #math.equation(block: false, alt: "u sub B")[$u_(B)$] in Inductance). Combining these the contributions, we obtain #math.equation(block: true, alt: "u open parenthesis x , t close parenthesis equals u sub E plus u sub B equals the fraction 1 over 2 ε sub 0 E squared plus the fraction 1 over 2 μ sub 0 B squared .")[$u ( x , t ) = u_(E) + u_(B) = frac(1, 2) ε_(0) E^(2) + frac(1, 2 μ_(0)) B^(2) .$] The expression #math.equation(block: false, alt: "E equals c B equals the fraction 1 over the square root of ε sub 0 μ sub 0 B")[$E = c B = frac(1, sqrt(ε_(0) μ_(0))) B$] then shows that the magnetic energy density #math.equation(block: false, alt: "u sub B")[$u_(B)$] and electric energy density #math.equation(block: false, alt: "u sub E")[$u_(E)$] are equal, despite the fact that changing electric fields generally produce only small magnetic fields. The equality of the electric and magnetic energy densities leads to #math.equation(block: true, alt: "u open parenthesis x , t close parenthesis equals ε sub 0 E squared equals the fraction B squared over μ sub 0 .")[$u ( x , t ) = ε_(0) E^(2) = frac(B^(2), μ_(0)) .$] The energy density moves with the electric and magnetic fields in a similar manner to the waves themselves. We can find the rate of transport of energy by considering a small time interval #math.equation(block: false, alt: "Δ t")[$"Δ" t$]. As shown in , the energy contained in a cylinder of length #math.equation(block: false, alt: "c Δ t")[$c "Δ" t$] and cross-sectional area #emph[A] passes through the cross-sectional plane in the interval #math.equation(block: false, alt: "Δ t .")[$"Δ" t .$] #figure(figph[Figure shows a cylinder of length c delta t and cross sectional area A. Arrows indicate the direction of a wave to be along the length of the cylinder. A plane is shown perpendicular to the direction of wave.], alt: "Figure shows a cylinder of length c delta t and cross sectional area A. Arrows indicate the direction of a wave to be along the length of the cylinder. A plane is shown perpendicular to the direction of wave.", caption: [The energy #math.equation(block: false, alt: "u A c Δ t")[$u A c "Δ" t$] contained in the electric and magnetic fields of the electromagnetic wave in the volume #math.equation(block: false, alt: "A c Δ t")[$A c "Δ" t$] passes through the area #emph[A] in time #math.equation(block: false, alt: "Δ t")[$"Δ" t$].]) The energy passing through area #emph[A] in time #math.equation(block: false, alt: "Δ t")[$"Δ" t$] is #math.equation(block: true, alt: "u times volume equals u A c Δ t .")[$u #h(0.2em) × #h(0.2em) "volume" = u A c "Δ" t .$] The energy per unit area per unit time passing through a plane perpendicular to the wave, called the #strong[energy flux] and denoted by #emph[S], can be calculated by dividing the energy by the area #emph[A] and the time interval #math.equation(block: false, alt: "Δ t")[$"Δ" t$]. #math.equation(block: true, alt: "S equals the fraction Energy passing area A in time Δ t over A Δ t equals u c equals ε sub 0 c E squared equals the fraction 1 over μ sub 0 E B .")[$S = frac("Energy passing area" #h(0.2em) A #h(0.2em) "in time" #h(0.2em) "Δ" t, A "Δ" t) = u c = ε_(0) c E^(2) = frac(1, μ_(0)) E B .$] More generally, the flux of energy through any surface also depends on the orientation of the surface. To take the direction into account, we introduce a vector #math.equation(block: false, alt: "S →")[$arrow(S)$], called the #strong[Poynting vector], with the following definition: #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #math.equation(block: true, alt: "S → equals the fraction 1 over μ sub 0 E → times B → .")[$arrow(S) = frac(1, μ_(0)) arrow(E) #h(0.2em) × #h(0.2em) arrow(B) .$] ] The cross-product of #math.equation(block: false, alt: "E →")[$arrow(E)$] and #math.equation(block: false, alt: "B →")[$arrow(B)$] points in the direction perpendicular to both vectors. To confirm that the direction of #math.equation(block: false, alt: "S →")[$arrow(S)$] is that of wave propagation, and not its negative, return to . Note that Lenz’s and Faraday’s laws imply that when the magnetic field shown is increasing in time, the electric field is greater at #emph[x] than at #math.equation(block: false, alt: "x plus Δ x")[$x + "Δ" x$]. The electric field is decreasing with increasing #emph[x] at the given time and location. The proportionality between electric and magnetic fields requires the electric field to increase in time along with the magnetic field. This is possible only if the wave is propagating to the right in the diagram, in which case, the relative orientations show that #math.equation(block: false, alt: "S → equals the fraction 1 over μ sub 0 E → times B →")[$arrow(S) = frac(1, μ_(0)) arrow(E) #h(0.2em) × #h(0.2em) arrow(B)$] is specifically in the direction of propagation of the electromagnetic wave. The energy flux at any place also varies in time, as can be seen by substituting #emph[u] from into . #math.equation(block: true, alt: "S open parenthesis x , t close parenthesis equals c ε sub 0 E 0 2 cos squared open parenthesis k x minus ω t close parenthesis")[$S ( x , t ) = c ε_(0) E_(0)^(2) "cos"^(2) ( k x − ω t )$] Because the frequency of visible light is very high, of the order of #math.equation(block: false, alt: "10 to the power 14 Hz,")[$10^(14) #h(0.2em) "Hz,"$] the energy flux for visible light through any area is an extremely rapidly varying quantity. Most measuring devices, including our eyes, detect only an average over many cycles. The time average of the energy flux is the #strong[intensity] #emph[I] of the electromagnetic wave and is the power per unit area. It can be expressed by averaging the cosine function in over one complete cycle, which is the same as time-averaging over many cycles (here, #emph[T] is one period): #math.equation(block: true, alt: "I equals S sub avg equals c ε sub 0 E 0 2 the fraction 1 over T ∫ 0 T cos squared open parenthesis 2 π the fraction t over T close parenthesis d t .")[$I = S_("avg") = c ε_(0) E_(0)^(2) frac(1, T) display(limits(∫)^(T)_(0) "cos"^(2) ( 2 π frac(t, T) )) d t .$] We can either evaluate the integral, or else note that because the sine and cosine differ merely in phase, the average over a complete cycle for #math.equation(block: false, alt: "cos squared open parenthesis ξ close parenthesis")[$"cos"^(2) ( ξ )$] is the same as for #math.equation(block: false, alt: "sin squared open parenthesis ξ close parenthesis")[$"sin"^(2) ( ξ )$], to obtain #math.equation(block: true, alt: "〈 cos squared ξ 〉 equals the fraction 1 over 2 [ 〈 cos squared ξ 〉 plus 〈 sin squared ξ 〉 ] equals the fraction 1 over 2 〈 1 〉 equals the fraction 1 over 2 .")[$〈 "cos"^(2) ξ 〉 = frac(1, 2) #h(0.2em) [ 〈 "cos"^(2) ξ 〉 + 〈 "sin"^(2) ξ 〉 ] = frac(1, 2) 〈 1 〉 = frac(1, 2) .$] where the angle brackets #math.equation(block: false, alt: "〈 ⋯ 〉")[$〈 "⋯" 〉$] stand for the time-averaging operation. The intensity of light moving at speed #emph[c] in vacuum is then found to be #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #math.equation(block: true, alt: "I equals S sub avg equals the fraction 1 over 2 c ε sub 0 E 0 2")[$I = S_("avg") = frac(1, 2) c ε_(0) E_(0)^(2)$] ] in terms of the maximum electric field strength #math.equation(block: false, alt: "E sub 0 ,")[$E_(0) ,$] which is also the electric field amplitude. Algebraic manipulation produces the relationship #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #math.equation(block: true, alt: "I equals the fraction c B 0 2 over 2 μ sub 0")[$I = frac(c B_(0)^(2), 2 μ_(0))$] ] where #sub[#math.equation(block: false, alt: "B sub 0")[$B_(0)$]] is the magnetic field amplitude, which is the same as the maximum magnetic field strength. One more expression for #math.equation(block: false, alt: "I sub avg")[$I_("avg")$] in terms of both electric and magnetic field strengths is useful. Substituting the fact that #math.equation(block: false, alt: "c B sub 0 equals E sub 0 ,")[$c B_(0) = E_(0) ,$] the previous expression becomes #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #math.equation(block: true, alt: "I equals the fraction E sub 0 B sub 0 over 2 μ sub 0 .")[$I = frac(E_(0) B_(0), 2 μ_(0)) .$] ] 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. 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, #math.equation(block: false, alt: "I sub 0 equals 2 I .")[$I_(0) = 2 I .$] #examplebox("Example 1")[A Laser Beam][ The beam from a small laboratory laser typically has an intensity of about #math.equation(block: false, alt: "1.0 times 10 to the power −3 W/m squared")[$1.0 #h(0.2em) × #h(0.2em) 10^(−3) attach(#h(0.2em) "W/m", t: 2)$]. Assuming that the beam is composed of plane waves, calculate the amplitudes of the electric and magnetic fields in the beam. Strategy Use the equation expressing intensity in terms of electric field to calculate the electric field from the intensity. Solution From , the intensity of the laser beam is #math.equation(block: true, alt: "I equals the fraction 1 over 2 c ε sub 0 E 0 2 .")[$I = frac(1, 2) c ε_(0) E_(0)^(2) .$] The amplitude of the electric field is therefore #math.equation(block: true, alt: "E sub 0 equals the square root of the fraction 2 over c ε sub 0 I equals the square root of the fraction 2 over open parenthesis 3.00 times 10 to the power 8 m/s close parenthesis open parenthesis 8.85 times 10 to the power −12 F/m close parenthesis open parenthesis 1.0 times 10 to the power −3 W/m squared close parenthesis equals 0.87 V/m .")[$E_(0) = sqrt(frac(2, c ε_(0)) I) = sqrt(frac(2, ( 3.00 #h(0.2em) × #h(0.2em) 10^(8) #h(0.2em) "m/s" ) ( 8.85 #h(0.2em) × #h(0.2em) 10^(−12) #h(0.2em) "F/m" )) ( 1.0 #h(0.2em) × #h(0.2em) 10^(−3) attach(#h(0.2em) "W/m", t: 2) )) = 0.87 #h(0.2em) "V/m" .$] The amplitude of the magnetic field can be obtained from : #math.equation(block: true, alt: "B sub 0 equals the fraction E sub 0 over c equals 2.9 times 10 to the power −9 T .")[$B_(0) = frac(E_(0), c) = 2.9 #h(0.2em) × #h(0.2em) 10^(−9) #h(0.2em) "T" .$] ] #examplebox("Example 2")[Light Bulb Fields][ A light bulb emits 5.00 W of power as visible light. What are the average electric and magnetic fields from the light at a distance of 3.0 m? Strategy Assume the bulb’s power output #emph[P] is distributed uniformly over a sphere of radius 3.0 m to calculate the intensity, and from it, the electric field. #figure(figph[Figure shows a light bulb in the centre illuminating a circular area around it. This area has a radius of 3 m.], alt: "Figure shows a light bulb in the centre illuminating a circular area around it. This area has a radius of 3 m.", caption: none) Solution The power radiated as visible light is then #math.equation(block: true, alt: "I, equals, the fraction P over 4 π r squared equals the fraction c ε sub 0 E 0 2 over 2 ,; E sub 0, equals, the square root of 2 the fraction P over 4 π r squared c ε sub 0 equals the square root of 2 the fraction 5.00 W over 4 π open parenthesis 3.0 m close parenthesis squared open parenthesis 3.00 times 10 to the power 8 m/s close parenthesis open parenthesis 8.85 times 10 to the power −12 C squared /N times m squared close parenthesis equals 5.77 N/C,; B sub 0, equals, E sub 0 / c equals 1.92 times 10 to the power −8 T .")[$I & = & frac(P, 4 π r^(2)) = frac(c ε_(0) E_(0)^(2), 2) , \ E_(0) & = & sqrt(2 frac(P, 4 π r^(2) c ε_(0))) = sqrt(2 frac(5.00 #h(0.2em) "W", 4 π attach(( 3.0 #h(0.2em) "m" ), t: 2) ( 3.00 #h(0.2em) × #h(0.2em) 10^(8) #h(0.2em) "m/s" ) ( 8.85 #h(0.2em) × #h(0.2em) 10^(−12) attach(#h(0.2em) "C", t: 2) "/N" · "m"^(2) ))) = 5.77 #h(0.2em) "N/C," \ B_(0) & = & E_(0) "/" c = 1.92 #h(0.2em) × #h(0.2em) 10^(−8) #h(0.2em) "T" .$] Significance The intensity #emph[I] falls off as the distance squared if the radiation is dispersed uniformly in all directions. ] #examplebox("Example 3")[Radio Range][ A 60-kW radio transmitter on Earth sends its signal to a satellite 100 km away . At what distance in the same direction would the signal have the same maximum field strength if the transmitter’s output power were increased to 90 kW? #figure(figph[A point is labeled radio source. A small square labeled A1 is in the path of the lines radiating from the radio source. The lines continue from the corners of A1 and reach A2, a slightly bigger square. A1 is at a distance r1 from the source and A2 is at a distance R2.], alt: "A point is labeled radio source. A small square labeled A1 is in the path of the lines radiating from the radio source. The lines continue from the corners of A1 and reach A2, a slightly bigger square. A1 is at a distance r1 from the source and A2 is at a distance R2.", caption: [In three dimensions, a signal spreads over a solid angle as it travels outward from its source.]) Strategy The area over which the power in a particular direction is dispersed increases as distance squared, as illustrated in the figure. Change the power output #emph[P] by a factor of (90 kW/60 kW) and change the area by the same factor to keep #math.equation(block: false, alt: "I equals the fraction P over A equals the fraction c ε sub 0 E 0 2 over 2")[$I = frac(P, A) = frac(c ε_(0) E_(0)^(2), 2)$] the same. Then use the proportion of area #emph[A] in the diagram to distance squared to find the distance that produces the calculated change in area. Solution Using the proportionality of the areas to the squares of the distances, and solving, we obtain from the diagram #math.equation(block: true, alt: "the fraction r 2 2 over r 1 2, equals, the fraction A sub 2 over A sub 1 equals the fraction 90 W over 60 W ,; r sub 2, equals, the square root of the fraction 90 over 60 open parenthesis 100 km close parenthesis equals 122 km .")[$frac(r_(2)^(2), r_(1)^(2)) & = & frac(A_(2), A_(1)) = frac(90 #h(0.2em) "W", 60 #h(0.2em) "W") , \ r_(2) & = & sqrt(frac(90, 60)) ( 100 #h(0.2em) "km" ) = 122 #h(0.2em) "km" .$] Significance The range of a radio signal is the maximum distance between the transmitter and receiver that allows for normal operation. In the absence of complications such as reflections from obstacles, the intensity follows an inverse square law, and doubling the range would require multiplying the power by four. ] === Summary - The energy carried by any wave is proportional to its amplitude squared. For electromagnetic waves, this means intensity can be expressed as #math.equation(block: true, alt: "I equals the fraction c ε sub 0 E 0 2 over 2")[$I = frac(c ε_(0) E_(0)^(2), 2)$] where #emph[I] is the average intensity in #math.equation(block: false, alt: "W/m squared")[$"W/m"^(2)$] and #math.equation(block: false, alt: "E sub 0")[$E_(0)$] is the maximum electric field strength of a continuous sinusoidal wave. This can also be expressed in terms of the maximum magnetic field strength #math.equation(block: false, alt: "B sub 0")[$B_(0)$] as #math.equation(block: true, alt: "I equals the fraction c B 0 2 over 2 μ sub 0")[$I = frac(c B_(0)^(2), 2 μ_(0))$] and in terms of both electric and magnetic fields as #math.equation(block: true, alt: "I equals the fraction E sub 0 B sub 0 over 2 μ sub 0 .")[$I = frac(E_(0) B_(0), 2 μ_(0)) .$] The three expressions for #math.equation(block: false, alt: "I sub avg")[$I_("avg")$] are all equivalent. === Conceptual Questions When you stand outdoors in the sunlight, why can you feel the energy that the sunlight carries, but not the momentum it carries? #solutionbox[ The amount of energy (about #math.equation(block: false, alt: "100 W/m squared")[$attach(100 #h(0.2em) "W/m", t: 2)$]) is can quickly produce a considerable change in temperature, but the light pressure (about #math.equation(block: false, alt: "3.00 times 10 to the power −7 N/m squared")[$3.00 #h(0.2em) × #h(0.2em) 10^(−7) "N/m"^(2)$]) is much too small to notice. ] How does the intensity of an electromagnetic wave depend on its electric field? How does it depend on its magnetic field? What is the physical significance of the Poynting vector? #solutionbox[ It has the magnitude of the energy flux and points in the direction of wave propagation. It gives the direction of energy flow and the amount of energy per area transported per second. ] A 2.0-mW helium-neon laser transmits a continuous beam of red light of cross-sectional area #math.equation(block: false, alt: "0.25 cm squared")[$0.25 #h(0.2em) "cm"^(2)$]. If the beam does not diverge appreciably, how would its rms electric field vary with distance from the laser? Explain. === Problems While outdoors on a sunny day, a student holds a large convex lens of radius 4.0 cm above a sheet of paper to produce a bright spot on the paper that is 1.0 cm in radius, rather than a sharp focus. By what factor is the electric field in the bright spot of light related to the electric field of sunlight leaving the side of the lens facing the paper? A plane electromagnetic wave travels northward. At one instant, its electric field has a magnitude of 6.0 V/m and points eastward. What are the magnitude and direction of the magnetic field at this instant? #solutionbox[ The magnetic field is downward, and it has magnitude #math.equation(block: false, alt: "2.00 times 10 to the power −8 T")[$2.00 #h(0.2em) × #h(0.2em) 10^(−8) "T"$]. ] The electric field of an electromagnetic wave is given by #emph[E] = #linebreak() #math.equation(block: false, alt: "open parenthesis 6.0 times 10 to the power −3 V/m close parenthesis sin [ 2 π open parenthesis the fraction x over 18 m minus the fraction t over 6.0 times 10 to the power −8 s close parenthesis ] j ^ .")[$( 6.0 #h(0.2em) × #h(0.2em) 10^(−3) #h(0.2em) "V/m" ) #h(0.2em) "sin" #h(0.2em) [ 2 π ( frac(x, 18 #h(0.2em) "m") − frac(t, 6.0 #h(0.2em) × #h(0.2em) 10^(−8) #h(0.2em) "s") ) ] hat(j) .$] #linebreak() Write the equations for the associated magnetic field and Poynting vector. A radio station broadcasts at a frequency of 760 kHz. At a receiver some distance from the antenna, the maximum magnetic field of the electromagnetic wave detected is #math.equation(block: false, alt: "2.15 times 10 to the power −11 T")[$2.15 #h(0.2em) × #h(0.2em) 10^(−11) "T"$]. #linebreak() (a) What is the maximum electric field? (b) What is the wavelength of the electromagnetic wave? #solutionbox[ a. #math.equation(block: false, alt: "6.45 times 10 to the power −3 V/m;")[$6.45 #h(0.2em) × #h(0.2em) 10^(−3) #h(0.2em) "V/m;"$] b. 394 m ] The filament in a clear incandescent light bulb radiates visible light at a power of 5.00 W. Model the glass part of the bulb as a sphere of radius #math.equation(block: false, alt: "r sub 0 equals 3.00 cm")[$r_(0) = 3.00 #h(0.2em) "cm"$] and calculate the amount of electromagnetic energy from visible light inside the bulb. At what distance does a 100-W lightbulb produce the same intensity of light as a 75-W lightbulb produces 10 m away? (Assume both have the same efficiency for converting electrical energy in the circuit into emitted electromagnetic energy.) #solutionbox[ 11.5 m ] An incandescent light bulb emits only 2.6 W of its power as visible light. What is the rms electric field of the emitted light at a distance of 3.0 m from the bulb? A 150-W lightbulb emits 5% of its energy as electromagnetic radiation. What is the magnitude of the average Poynting vector 10 m from the bulb? #solutionbox[ #math.equation(block: true, alt: "5.97 times 10 to the power −3 W/m squared")[$5.97 #h(0.2em) × #h(0.2em) 10^(−3) attach(#h(0.2em) "W/m", t: 2)$] ] A small helium-neon laser has a power output of 2.5 mW. What is the electromagnetic energy in a 1.0-m length of the beam? At the top of Earth’s atmosphere, the time-averaged Poynting vector associated with sunlight has a magnitude of about #math.equation(block: false, alt: "1.4 kW/m squared .")[$1.4 #h(0.2em) "kW/m"^(2) .$] #linebreak() (a) What are the maximum values of the electric and magnetic fields for a wave of this intensity? (b) What is the total power radiated by the sun? Assume that the Earth is #math.equation(block: false, alt: "1.5 times 10 to the power 11 m")[$1.5 #h(0.2em) × #h(0.2em) 10^(11) "m"$] from the Sun and that sunlight is composed of electromagnetic plane waves. #solutionbox[ #math.equation(block: true, alt: "a. E sub 0 equals 1027 V/m , B sub 0 equals 3.42 times 10 to the power −6 T ; b. 3.96 times 10 to the power 26 W")[$"a." #h(0.2em) E_(0) = 1027 #h(0.2em) "V/m" , B_(0) = 3.42 #h(0.2em) × #h(0.2em) 10^(−6) "T" ; #h(0.2em) "b." #h(0.2em) 3.96 #h(0.2em) × #h(0.2em) 10^(26) #h(0.2em) "W"$] ] The magnetic field of a plane electromagnetic wave moving along the #emph[z] axis is given by #math.equation(block: false, alt: "B → equals B sub 0 cos open parenthesis k z plus ω t close parenthesis j ^")[$arrow(B) = B_(0) "cos" #h(0.2em) ( k z + ω t ) hat(j)$], where #math.equation(block: false, alt: "B sub 0 equals 5.00 times 10 to the power −10 T")[$B_(0) = 5.00 #h(0.2em) × #h(0.2em) 10^(−10) #h(0.2em) "T"$] and #math.equation(block: false, alt: "k equals 3.14 times 10 to the power −2 m to the power −1 .")[$k = 3.14 #h(0.2em) × #h(0.2em) 10^(−2) attach(#h(0.2em) "m", t: −1) .$] #linebreak() (a) Write an expression for the electric field associated with the wave. (b) What are the frequency and the wavelength of the wave? (c) What is its average Poynting vector? What is the intensity of an electromagnetic wave with a peak electric field strength of 125 V/m? #solutionbox[ #math.equation(block: true, alt: "20.8 W/m squared")[$20.8 #h(0.2em) "W/m"^(2)$] ] Assume the helium-neon lasers commonly used in student physics laboratories have power outputs of 0.500 mW. (a) If such a laser beam is projected onto a circular spot 1.00 mm in diameter, what is its intensity? (b) Find the peak magnetic field strength. (c) Find the peak electric field strength. An AM radio transmitter broadcasts 50.0 kW of power uniformly in all directions. (a) Assuming all of the radio waves that strike the ground are completely absorbed, and that there is no absorption by the atmosphere or other objects, what is the intensity 30.0 km away? (#emph[Hint:] Half the power will be spread over the area of a hemisphere.) (b) What is the maximum electric field strength at this distance? #solutionbox[ a. #math.equation(block: false, alt: "4.42 times 10 to the power ‒6 W/m squared")[$4.42 #h(0.2em) × #h(0.2em) 10^(‒6) attach(#h(0.2em) "W/m", t: 2)$]; b. #math.equation(block: false, alt: "5.77 times 10 to the power ‒2 V/m")[$5.77 #h(0.2em) × #h(0.2em) 10^(‒2) #h(0.2em) "V/m"$] ] Suppose the maximum safe intensity of microwaves for human exposure is taken to be #math.equation(block: false, alt: "1.00 W/m squared")[$1.00 attach(#h(0.2em) "W/m", t: 2)$]. (a) If a radar unit leaks 10.0 W of microwaves (other than those sent by its antenna) uniformly in all directions, how far away must you be to be exposed to an intensity considered to be safe? Assume that the power spreads uniformly over the area of a sphere with no complications from absorption or reflection. (b) What is the maximum electric field strength at the safe intensity? (Note that early radar units leaked more than modern ones do. This caused identifiable health problems, such as cataracts, for people who worked near them.) A 2.50-m-diameter university communications satellite dish receives TV signals that have a maximum electric field strength (for one channel) of #math.equation(block: false, alt: "7.50 μV/m")[$7.50 #h(0.2em) "μV/m"$] (see below). (a) What is the intensity of this wave? (b) What is the power received by the antenna? (c) If the orbiting satellite broadcasts uniformly over an area of #math.equation(block: false, alt: "1.50 times 10 to the power 13 m squared")[$1.50 #h(0.2em) × #h(0.2em) 10^(13) attach(#h(0.2em) "m", t: 2)$] (a large fraction of North America), how much power does it radiate? #figure(figph[Figure shows waves incident on a dish antenna.], alt: "Figure shows waves incident on a dish antenna.", caption: none) #solutionbox[ a. #math.equation(block: false, alt: "7.47 times 10 to the power −14 W/m squared")[$7.47 #h(0.2em) × #h(0.2em) 10^(−14) attach(#h(0.2em) "W/m", t: 2)$]; b. #math.equation(block: false, alt: "3.66 times 10 to the power − 13 W")[$3.66 #h(0.2em) × #h(0.2em) 10^("−" 13) #h(0.2em) "W"$]; c. 1.12 W ] Lasers can be constructed that produce an extremely high intensity electromagnetic wave for a brief time—called pulsed lasers. They are used to initiate nuclear fusion, for example. Such a laser may produce an electromagnetic wave with a maximum electric field strength of #math.equation(block: false, alt: "1.00 times 10 to the power 11 V / m")[$1.00 #h(0.2em) × #h(0.2em) 10^(11) #h(0.2em) "V" "/" "m"$] for a time of 1.00 ns. (a) What is the maximum magnetic field strength in the wave? (b) What is the intensity of the beam? (c) What energy does it deliver on an #math.equation(block: false, alt: "1.00 -mm squared")[$1.00 "-mm"^(2)$] area?