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16.A: Electromagnetic Waves (Answer)

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    Check Your Understanding

    16.1. It is greatest immediately after the current is switched on. The displacement current and the magnetic field from it are proportional to the rate of change of electric field between the plates, which is greatest when the plates first begin to charge.

    16.2. No. The changing electric field according to the modified version of Ampère’s law would necessarily induce a changing magnetic field.

    16.3. (1) Faraday’s law, (2) the Ampère-Maxwell law

    16.4. a. The directions of wave propagation, of the E field, and of B field are all mutually perpendicular.

    b. The speed of the electromagnetic wave is the speed of light \(\displaystyle c=1/\sqrt{ε_0μ_0}\) independent of frequency.

    c. The ratio of electric and magnetic field amplitudes is \(\displaystyle E/B=c\).

    16.5. Its acceleration would decrease because the radiation force is proportional to the intensity of light from the Sun, which decreases with distance. Its speed, however, would not change except for the effects of gravity from the Sun and planets.

    16.6. They fall into different ranges of wavelength, and therefore also different corresponding ranges of frequency.

    Conceptual Questions

    1. The current into the capacitor to change the electric field between the plates is equal to the displacement current between the plates.

    3. The first demonstration requires simply observing the current produced in a wire that experiences a changing magnetic field. The second demonstration requires moving electric charge from one location to another, and therefore involves electric currents that generate a changing electric field. The magnetic fields from these currents are not easily separated from the magnetic field that the displacement current produces.

    5. in (a), because the electric field is parallel to the wire, accelerating the electrons

    7. A steady current in a dc circuit will not produce electromagnetic waves. If the magnitude of the current varies while remaining in the same direction, the wires will emit electromagnetic waves, for example, if the current is turned on or off.

    9. The amount of energy (about \(\displaystyle 100W/m^2\)) is can quickly produce a considerable change in temperature, but the light pressure (about \(\displaystyle 3.00×10^{−7}N/m^2\)) is much too small to notice.

    11. 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.

    13. The force on a surface acting over time \(\displaystyle Δt\) is the momentum that the force would impart to the object. The momentum change of the light is doubled if the light is reflected back compared with when it is absorbed, so the force acting on the object is twice as great.

    15. a. According to the right hand rule, the direction of energy propagation would reverse.

    b. This would leave the vector \(\displaystyle \vec{S}\), and therefore the propagation direction, the same.

    17. a. Radio waves are generally produced by alternating current in a wire or an oscillating electric field between two plates;

    b. Infrared radiation is commonly produced by heated bodies whose atoms and the charges in them vibrate at about the right frequency.

    19. a. blue;

    b. Light of longer wavelengths than blue passes through the air with less scattering, whereas more of the blue light is scattered in different directions in the sky to give it is blue color.

    21. A typical antenna has a stronger response when the wires forming it are orientated parallel to the electric field of the radio wave.

    23. No, it is very narrow and just a small portion of the overall electromagnetic spectrum.

    25. Visible light is typically produced by changes of energies of electrons in randomly oriented atoms and molecules. Radio waves are typically emitted by an ac current flowing along a wire, that has fixed orientation and produces electric fields pointed in particular directions.

    27. Radar can observe objects the size of an airplane and uses radio waves of about 0.5 cm in wavelength. Visible light can be used to view single biological cells and has wavelengths of about \(\displaystyle 10^{−7}m\).

    29. ELF radio waves

    31. The frequency of 2.45 GHz of a microwave oven is close to the specific frequencies in the 2.4 GHz band used for WiFi.


    33. \(\displaystyle B_{ind}=\frac{μ_0}{P2πr}I_{ind}=\frac{μ_0}{2πr}ε_0\frac{∂Φ_E}{∂t}=\frac{μ_0}{2πr}ε_0(A\frac{∂E}{∂t})=\frac{μ_0}{2πr}ε_0A(\frac{1}{d}\frac{dV(t)}{dt})=\frac{μ_0}{2πr}[\frac{ε_0A}{d}][\frac{1}{C}\frac{dQ(t)}{dt}]=\frac{μ_0}{2πr}\frac{dQ(t)}{dt}\) because \(\displaystyle C=\frac{ε_0A}{d}\)

    35. a. \(\displaystyle I_{res}=\frac{V_0sinωt}{R}\);

    b. \(\displaystyle I_d=CV_0ωcosωt\);

    c. \(\displaystyle I_{real}=_{Ires}+\frac{dQ}{dt}=\frac{V_0sinωt}{R}+CV_0\frac{d}{dt}sinωt=\frac{V_0sinωt}{R}+CV_0ωcosωt\); which is the sum of \(\displaystyle I_{res}\) and \(\displaystyle I_{real}\), consistent with how the displacement current maintaining the continuity of current.

    37. \(\displaystyle 1.77×10^{−3}A\)

    39. \(\displaystyle I_d=(7.97×10^{−10}A)sin(150t)\)

    41. 499 s

    43. 25 m

    45. a. 5.00 V/m;

    b. \(\displaystyle 9.55×10^8Hz\);

    c. 31.4 cm;

    d. toward the +x-axis;

    e. \(\displaystyle B=(1.67×10^{−8}T)cos[kx−(6×10^9s^{−1})t+0.40]\hat{k}\)

    47. \(\displaystyle I_d=πε_0ωR^2E_0sin(kx−ωt)\)

    49. The magnetic field is downward, and it has magnitude \(\displaystyle 2.00×10^{−8}T\).

    51. a. \(\displaystyle 6.45×10^{−3}V/m\);

    b. 394 m

    53. 11.5 m

    55. \(\displaystyle 5.97×10^{−3}W/m^2\)

    57. a.\(\displaystyle E_0=1027V/m, B_0=3.42×10^{−6}T\);

    b. \(\displaystyle 3.96×10^{26}W\)

    59. \(\displaystyle 20.8W/m^2\)

    61. a. \(\displaystyle 4.42×10^{‒6}W/m^2\);

    b. \(\displaystyle 5.77×10^{‒2}V/m\)

    63. a. \(\displaystyle 7.47×10^{−14}W/m^2\);

    b. \(\displaystyle 3.66×10−^{13}W\);

    c. 1.12 W

    65. \(\displaystyle 1.99×10^{−11}N/m^2\)

    67. \(\displaystyle F=ma=(p)(πr^2),p=\frac{ma}{πr^2}=\frac{ε_0}{2E^2_0}\)

    \(\displaystyle E_0=\sqrt{\frac{2ma}{ε_0πr^2}}=\sqrt{\frac{2(10^{−8}kg)(0.30m/s^2)}{(8.854×10^{−12}C^2/N⋅m^2)(π)(2×10^{−6}m)^2}}\)

    \(\displaystyle E_0=7.34×10^6V/m\)

    69. a. \(\displaystyle 4.50×10^{−6}N;\)

    b. it is reduced to half the pressure, \(\displaystyle 2.25×10^{−6}N\)

    71. a. \(\displaystyle W=\frac{1}{2}\frac{π^2r^4}{mc^2}I^2t^2\);

    b. \(\displaystyle E=πr^2It\)

    73. a. \(\displaystyle 1.5×10^{18}Hz\);

    b. X-rays

    75. a. The wavelength range is 187 m to 556 m.

    b. The wavelength range is 2.78 m to 3.41 m.

    77. \(\displaystyle P'=(\frac{12m}{30m})^2(100mW)=16mW\)

    79. time for 1 bit = \(\displaystyle 1.27×10^{−8}\) s, difference in travel time is \(\displaystyle 5.34×10^{−8}\) s

    81. a. \(\displaystyle 1.5×10^{−9}m\);

    b. \(\displaystyle 5.9×10^{−7}m\);

    c. \(\displaystyle 3.0×10^{−15}m\)

    83. \(\displaystyle 5.17×10^{−12}T\), the non-oscillating geomagnetic field of 25–65 \(\displaystyle μT\) is much larger

    85. a. \(\displaystyle 1.33×10^{−2}V/m\);

    b. \(\displaystyle 4.34×10^{−11}T\);

    c. \(\displaystyle 3.00×10^8m\)

    87. a. \(\displaystyle 5.00×10^6m\);

    b. radio wave;

    c. \(\displaystyle 4.33×10^{−5}T\)

    Additional Problems

    89. \(\displaystyle I_d=(10N/C)(8.845×10^{−12}C^2/N⋅m^2)π(0.03m)^2(5000)=1.25×10^{−5}mA\)

    91. \(\displaystyle 3.75×10^7km\), which is much greater than Earth’s circumference

    93. a. 564 W;

    b. \(\displaystyle 1.80×10^4W/m^2\);

    c. \(\displaystyle 3.68×10^3V/m\);

    d. \(\displaystyle 1.23×10^{−5}T\)

    95. a. \(\displaystyle 5.00×10^3W/m^2\);

    b. \(\displaystyle 3.88×10^{−6}N\);

    c. \(\displaystyle 5.18×10^{−12}N\)

    97. a. \(\displaystyle I=\frac{P}{A}=\frac{P}{4πr^2}∝\frac{1}{r^2}\);

    b. \(\displaystyle I∝E^2_0,B^2_0⇒E^2_0,B^2_0∝\frac{1}{r^2}⇒E_0,B_0∝\frac{1}{r}\)

    99. Power into the wire=\(\displaystyle ∫\vec{S}⋅d\vec{A}=(\frac{1}{μ_0}EB)(2πrL)=\frac{1}{μ_0}(\frac{V}{L})(\frac{μ_0i}{2πr})(2πrL)=iV=i^2R\)

    101. 0.431

    103. a. \(\displaystyle 1.5×10^{11}m\);

    b. \(\displaystyle 5.0×10^{−7}s\);

    c. 33 ns

    105. \(\displaystyle sound:λ_{sound}=\frac{v_s}{f}=\frac{343m/s}{20.0Hz}=17.2m\)

    \(\displaystyle radio:λ_{radio}=\frac{c}{f}=\frac{3.00×10^8m/s}{1030×10^3Hz}=291m; or 17.1 λ_{sound}\)

    Challenge Problems

    107. a. \(\displaystyle 0.29μm\);

    b. The radiation pressure is greater than the Sun’s gravity if the particle size is smaller, because the gravitational force varies as the radius cubed while the radiation pressure varies as the radius squared.

    c. The radiation force outward implies that particles smaller than this are less likely to be near the Sun than outside the range of the Sun’s radiation pressure.

    Contributors and Attributions

    Samuel J. Ling (Truman State University), Jeff Sanny (Loyola Marymount University), and Bill Moebs with many contributing authors. This work is licensed by OpenStax University Physics under a Creative Commons Attribution License (by 4.0).

    This page titled 16.A: Electromagnetic Waves (Answer) 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.