M32 YOUN7066 13 ISM c32 tủ tài liệu training

ELECTROMAGNETIC WAVES 32.1 32 IDENTIFY: Since the speed is constant, distance x = ct SET UP: The speed of light is c = 3.00 × 108 m/s y = 3.156 × 107 s EXECUTE: (a) t = x 3.84 × 108 m = = 1.28 s c 3.00 × 108 m/s (b) x = ct = (3.00 × 108 m/s)(8.61 y)(3.156 × 107 s/y) = 8.15 × 1016 m = 8.15 × 1013 km 32.2 EVALUATE: The speed of light is very great The distance between stars is very large compared to terrestrial distances IDENTIFY: Find the direction of propagation of an electromagnetic wave if we know the directions of the electric and magnetic fields G G SET UP: The direction of propagation of an electromagnetic wave is in the direction of E × B , which is G G related to the directions of E and B according to the right-hand rule for the cross product The directions G G of E and B in each case are shown in Figures 32.2a-d Figure 32.2 © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 32-1 32-2 Chapter 32 EXECUTE: (a) The wave is propagating in the + z direction (b) + z direction (c) – y direction (d) – x direction 32.3 G G EVALUATE: In each case, the direction of propagation is perpendicular to the plane of E and B G G IDENTIFY: Emax = cBmax E × B is in the direction of propagation SET UP: c = 3.00 × 108 m/s Emax = 4.00 V/m 32.4 G G G EXECUTE: Bmax = Emax /c = 1.33 × 10−8 T For E in the + x-direction, E × B is in the + z -direction G when B is in the + y -direction G G EVALUATE: E , B and the direction of propagation are all mutually perpendicular G G IDENTIFY and SET UP: The direction of propagation is given by E × B EXECUTE: (a) Sˆ = iˆ × ( − ˆj ) = − kˆ (b) Sˆ = ˆj × iˆ = − kˆ (c) Sˆ = (− kˆ ) × (− iˆ) = ˆj (d) Sˆ = iˆ × (− kˆ ) = ˆj 32.5 G G EVALUATE: In each case the directions of E , B and the direction of propagation are all mutually perpendicular IDENTIFY: Knowing the wavelength and speed of x rays, find their frequency, period, and wave number All electromagnetic waves travel through vacuum at the speed of light 2π SET UP: c = 3.00 × 108 m/s c = f λ T = k = λ f f = EXECUTE: c λ = 3.0 × 108 m/s 0.10 × 10−9 m = 3.0 × 1018 Hz, 1 2π 2π = = 3.3 × 10−19 s, k = = = 6.3 × 1010 m −1 f 3.0 × 1018 Hz λ 0.10 × 10−9 m EVALUATE: The frequency of the x rays is much higher than the frequency of visible light, so their period is much shorter 2π IDENTIFY: c = f λ and k = T= 32.6 λ SET UP: c = 3.00 × 108 m/s c EXECUTE: (a) f = UVA: 7.50 × 1014 Hz to 9.38 × 1014 Hz UVB: 9.38 × 1014 Hz to 1.07 × 1015 Hz (b) k = 32.7 2π λ λ UVA: 1.57 × 107 rad/m to 1.96 × 107 rad/m UVB: 1.96 × 107 rad/m to 2.24 × 107 rad/m EVALUATE: Larger λ corresponds to smaller f and k IDENTIFY: c = f λ Emax = cBmax k = 2π /λ ω = 2π f SET UP: Since the wave is traveling in empty space, its wave speed is c = 3.00 × 108 m/s EXECUTE: (a) f = c λ = 3.00 × 108 m/s 432 × 10−9 m = 6.94 × 1014 Hz (b) Emax = cBmax = (3.00 × 108 m/s)(1.25 × 10−6 T) = 375 V/m (c) k = 2π λ = 2π rad 432 × 10−9 m = 1.45 × 107 rad/m ω = (2π rad)(6.94 × 1014 Hz) = 4.36 × 1015 rad/s E = Emax cos( kx − ωt ) = (375 V/m)cos([1.45 × 107 rad/m]x − [4.36 × 1015 rad/s]t ) B = Bmax cos( kx − ωt ) = (1.25 × 10−6 T)cos([1.45 × 107 rad/m]x [ 4.36 ì 1015 rad/s]t ) â Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher Electromagnetic Waves 32-3 EVALUATE: The cos(kx − ωt ) factor is common to both the electric and magnetic field expressions, since 32.8 these two fields are in phase IDENTIFY: c = f λ Emax = cBmax Apply Eqs (32.17) and (32.19) SET UP: The speed of the wave is c = 3.00 × 108 m/s EXECUTE: (a) f = (b) Bmax 32.9 c λ = 435 × 10 −9 m = 6.90 × 1014 Hz E 2.70 × 10−3 V/m = max = = 9.00 × 10−12 T c 3.00 × 108 m/s G 2π (c) k = = 1.44 × 107 rad/m ω = 2π f = 4.34 × 1015 rad/s If E (z , t ) = iˆEmax cos( kz + ωt ), then λ G G G B (z , t ) = − ˆjBmax cos(kz + ωt ), so that E × B will be in the − kˆ direction G ˆ 2.70 × 10−3 V/m)cos([1.44 × 107 rad/m)z + [ 4.34 × 1015 rad/s]t ) and E (z , t ) = i( G B (z , t ) = − ˆj( 9.00 × 10−12 T)cos([1.44 × 107 rad/m)z + [4.34 × 1015 rad/s]t ) G G EVALUATE: The directions of E and B and of the propagation of the wave are all mutually perpendicular The argument of the cosine is kz + ωt since the wave is traveling in the − z -direction Waves for visible light have very high frequencies IDENTIFY: Electromagnetic waves propagate through air at essentially the speed of light Therefore, if we know their wavelength, we can calculate their frequency or vice versa SET UP: The wave speed is c = 3.00 × 108 m/s c = f λ c EXECUTE: (a) (i) f = (ii) f = 3.00 × 108 m/s 5.0 × 10−6 m (iii) f = λ = 3.00 × 108 m/s 5.0 × 103 m = 6.0 × 104 Hz = 6.0 × 1013 Hz 3.00 × 108 m/s 5.0 × 10−9 m (b) (i) λ = = 6.0 × 1016 Hz c 3.00 × 108 m/s = = 4.62 × 10−14 m = 4.62 × 10−5 nm f 6.50 × 1021 Hz 3.00 × 108 m/s = 508 m = 5.08 × 1011 nm 590 × 103 Hz EVALUATE: Electromagnetic waves cover a huge range in frequency and wavelength IDENTIFY: For an electromagnetic wave propagating in the negative x direction, E = Emax cos(kx + ωt ) (ii) λ = 32.10 3.00 × 108 m/s ω = 2π f and k = 2π λ T= Emax = cBmax f SET UP: Emax = 375 V/m, k = 1.99 × 107 rad/m and ω = 5.97 × 1015 rad/s Emax = 1.25 μ T c ω 2π (b) f = = 9.50 × 1014 Hz λ = = 3.16 × 10−7 m = 316 nm T = = 1.05 × 10−15 s This wavelength 2π k f is too short to be visible (c) c = f λ = (9.50 × 1014 Hz)(3.16 × 10−7 m) = 3.00 × 108 m/s This is what the wave speed should be for an EXECUTE: (a) Bmax = electromagnetic wave propagating in vacuum ⎛ ω ⎞⎛ 2π ⎞ ω EVALUATE: c = f λ = ⎜ is an alternative expression for the wave speed ⎟⎜ ⎟= ⎝ 2π ⎠⎝ k ⎠ k © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 32-4 32.11 Chapter 32 G IDENTIFY and SET UP: Compare the E (y, t ) given in the problem to the general form given by G G Eq (32.17) Use the direction of propagation and of E to find the direction of B (a) EXECUTE: The equation for the electric field contains the factor cos(ky − ωt ) so the wave is traveling in the + y -direction G (b) E (y, t ) = (3.10 × 105 V/m)kˆ cos[ky − (12.65 × 1012 rad/s)t ] Comparing to Eq (32.17) gives ω = 12.65 × 1012 rad/s 2π c ω = 2π f = λ so λ = 2π c ω = 2π (2.998 × 108 m/s) (12.65 × 1012 rad/s) = 1.49 × 10−4 m (c) G G E × B must be in the + y -direction (the direction in which the wave is traveling) G G When E is in the + z -direction then B must be in the + x -direction, as shown in Figure 32.11 Figure 32.11 k= 2π λ = ω c = 12.65 × 1012 rad/s 2.998 × 108 m/s = 4.22 × 104 rad/m Emax = 3.10 × 105 V/m Emax 3.10 × 105 V/m = = 1.03 × 10−3 T c 2.998 × 108 m/s G G Using Eq (32.17) and the fact that B is in the + iˆ direction when E is in the + kˆ direction, G B = +(1.03 × 10−3 T)iˆ cos[(4.22 × 104 rad/m) y − (12.65 × 1012 rad/s)t ] G G EVALUATE: E and B are perpendicular and oscillate in phase IDENTIFY: Apply Eqs (32.17) and (32.19) f = c /λ and k = 2π /λ Then Bmax = 32.12 SET UP: B y ( x, t ) = − Bmax cos(kx + ωt ) EXECUTE: (a) The phase of the wave is given by kx + ωt , so the wave is traveling in the − x direction 2π f kc (1.38 × 104 rad/m)(3.0 × 108 m/s) f = = = 6.59 × 1011 Hz λ c 2π 2π (c) Since the magnetic field is in the − y -direction, and the wave is propagating in the − x-direction, then G G the electric field is in the –z-direction so that E × B will be in the − x-direction G E (x, t ) = + cB( x, t )kˆ = − cBmax cos(kx + ω t ) kˆ G E (x, t ) = −(c (8.25 × 10−9 T))cos((1.38 × 104 rad/m) x + (4.14 × 1012 rad/s)t ) kˆ G E (x, t ) = −(2.48 V/m)cos((1.38 × 104 rad/m) x + (4.14 × 1012 rad/s)t ) kˆ G G EVALUATE: E and B have the same phase and are in perpendicular directions (b) k = 2π = © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher Electromagnetic Waves 32.13 32-5 IDENTIFY and SET UP: c = f λ allows calculation of λ k = 2π /λ and ω = 2π f Eq (32.18) relates the electric and magnetic field amplitudes c 2.998 × 108 m/s EXECUTE: (a) c = f λ so λ = = = 361 m f 830 × 103 Hz (b) k = 2π = 2π rad = 0.0174 rad/m 361 m λ (c) ω = 2π f = (2π )(830 × 103 Hz) = 5.22 × 106 rad/s (d) Eq (32.18): Emax = cBmax = (2.998 × 108 m/s)(4.82 × 10−11 T) = 0.0144 V/m 32.14 EVALUATE: This wave has a very long wavelength; its frequency is in the AM radio braodcast band The electric and magnetic fields in the wave are very weak IDENTIFY: Apply Eq (32.21) Emax = cBmax v = f λ SET UP: K = 3.64 K m = 5.18 EXECUTE: (a) v = (b) λ = c (3.00 × 108 m/s) = = 6.91 × 107 m/s KK m (3.64)(5.18) v 6.91 × 107 m/s = = 1.06 × 106 m f 65.0 Hz Emax 7.20 × 10−3 V/m = = 1.04 × 10−10 T v 6.91 × 107 m/s EVALUATE: The wave travels slower in this material than in air IDENTIFY and SET UP: v = f λ relates frequency and wavelength to the speed of the wave Use Eq (32.22) to calculate n and K v 2.17 × 108 m/s EXECUTE: (a) λ = = = 3.81 × 10−7 m f 5.70 × 1014 Hz (c) Bmax = 32.15 (b) λ = (c) n = c 2.998 × 108 m/s = = 5.26 × 10−7 m f 5.70 × 1014 Hz c 2.998 × 108 m/s = = 1.38 v 2.17 × 108 m/s (d) n = KK m ≈ K so K = n = (1.38) = 1.90 32.16 EVALUATE: In the material v < c and f is the same, so λ is less in the material than in air v < c always, so n is always greater than unity IDENTIFY: We want to find the amount of energy given to each receptor cell and the amplitude of the magnetic field at the cell SET UP: Intensity is average power per unit area and power is energy per unit time I = 12 ⑀0cEmax , I = P /A, and Emax = cBmax EXECUTE: (a) For the beam, the energy is U = Pt = (2.0 × 1012 W)(4.0 × 10−9 s) = 8.0 × 103 J = 8.0 kJ This energy is spread uniformly over 100 cells, so the energy given to each cell is 80 J (b) The cross-sectional area of each cell is A = π r , with r = 2.5 × 10−6 m I= P 2.0 × 1012 W = = 1.0 × 1021 W/m A (100)π (2.5 × 10−6 m) (c) Emax = 2I 2(1.0 × 1021 W/m ) = = 8.7 × 1011 V/m − ⑀ 0c (8.85 × 10 12 C2 /N ⋅ m )(3.00 × 108 m/s) Emax = 2.9 × 103 T c EVALUATE: Both the electric field and magnetic field are very strong compared to ordinary fields Bmax = © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 32-6 Chapter 32 32.17 IDENTIFY: I = P/A I = 12 ⑀0cEmax Emax = cBmax SET UP: The surface area of a sphere of radius r is A = 4π r ⋅ ⑀0 = 8.85 × 10−12 C2 /N ⋅ m EXECUTE: (a) I = (b) Emax = P (0.05)(75 W) = = 330 W/m A 4π (3.0 × 10−2 m)2 2I 2(330 W/m ) = = 500 V/m 12 − ⑀ 0c (8.85 × 10 C2 /N ⋅ m )(3.00 × 108 m/s) Emax = 1.7 × 10−6 T = 1.7 μ T c EVALUATE: At the surface of the bulb the power radiated by the filament is spread over the surface of the bulb Our calculation approximates the filament as a point source that radiates uniformly in all directions 2 IDENTIFY: The intensity of the electromagnetic wave is given by Eq (32.29): I = 12 ⑀0cEmax = ⑀0cErms Bmax = 32.18 The total energy passing through a window of area A during a time t is IAt SET UP: ⑀ = 8.85 × 10−12 F/m EXECUTE: Energy = ⑀0cErms At = (8.85 × 10−12 F/m)(3.00 × 108 m/s)(0.0200 V/m)2 (0.500 m2 )(30.0 s) = 15.9 μJ 32.19 EVALUATE: The intensity is proportional to the square of the electric field amplitude IDENTIFY and SET UP: Use Eq (32.29) to calculate I, Eq (32.18) to calculate Bmax , and use I = Pav /4π r to calculate Pav (a) EXECUTE: I = 12 ⑀0cEmax ; Emax = 0.090 V/m, so I = 1.1 × 10−5 W/m (b) Emax = cBmax so Bmax = Emax /c = 3.0 × 10−10 T (c) Pav = I (4π r ) = (1.075 × 10−5 W/m )(4π )(2.5 × 103 m) = 840 W 32.20 (d) EVALUATE: The calculation in part (c) assumes that the transmitter emits uniformly in all directions IDENTIFY and SET UP: I = Pav /A and I = ⑀ 0cErms EXECUTE: (a) The average power from the beam is Pav = IA = (0.800 W/m2 )(3.0 ×10−4 m2 ) = 2.4 ×10−4 W (b) Erms = 32.21 I ⑀ 0c = 0.800 W/m (8.85 × 10−12 F/m)(3.00 × 108 m/s) =17.4 V/m EVALUATE: The laser emits radiation only in the direction of the beam IDENTIFY: I = Pav /A SET UP: At a distance r from the star, the radiation from the star is spread over a spherical surface of area A = 4π r EXECUTE: Pav = I (4π r ) = (5.0 × 103 W/m )(4π )(2.0 × 1010 m) = 2.5 × 1025 W 32.22 EVALUATE: The intensity decreases with distance from the star as 1/r IDENTIFY and SET UP: c = f λ , Emax = cBmax and I = Emax Bmax /2 μ0 EXECUTE: (a) f = (b) Bmax c λ = 3.00 × 108 m/s = 8.47 × 108 Hz 0.354 m E 0.0540 V/m = max = = 1.80 × 10−10 T c 3.00 × 108 m/s (c) I = Sav = Emax Bmax (0.0540 V/m)(1.80 × 10−10 T) = = 3.87 × 10−6 W/m μ0 μ0 EVALUATE: Alternatively, I = 12 ⑀0cEmax © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher Electromagnetic Waves 32.23 32-7 IDENTIFY: Pav = IA and I = Emax /2μ0c SET UP: The surface area of a sphere is A = 4π r 32.24 ⎛ E2 ⎞ P cμ (60.0 W)(3.00 × 108 m/s) μ0 = 12.0 V/m EXECUTE: Pav = Sav A = ⎜ max ⎟ (4π r ) Emax = av 20 = ⎜ 2cμ ⎟ 2π r 2π (5.00 m) 0⎠ ⎝ E 12.0 V/m = 4.00 × 10−8 T Bmax = max = c 3.00 × 108 m/s EVALUATE: Emax and Bmax are both inversely proportional to the distance from the source G G G IDENTIFY: The Poynting vector is S = E × B SET UP: The electric field is in the − y -direction, and the magnetic field is in the + z -direction cos φ = 12 (1 + cos 2φ ) EXECUTE: (a) Sˆ = Eˆ × Bˆ = ( − ˆj ) × kˆ = − iˆ The Poynting vector is in the – x-direction, which is the direction of propagation of the wave E ( x, t ) B ( x, t ) Emax Bmax E B cos ( kx + ωt ) = max max (1 + cos(2(ωt + kx ))) But over one (b) S ( x, t ) = = μ0 μ0 μ0 period, the cosine function averages to zero, so we have Sav = 32.25 Emax Bmax This is Eq (32.29) μ0 EVALUATE: We can also show that these two results also apply to the wave represented by Eq (32.17) IDENTIFY: Use the radiation pressure to find the intensity, and then Pav = I (4π r ) I SET UP: For a perfectly absorbing surface, prad = c EXECUTE: prad = I/c so I = cprad = 2.70 × 103 W/m Then Pav = I (4π r ) = (2.70 × 103 W/m )(4π )(5.0 m) = 8.5 × 105 W 32.26 EVALUATE: Even though the source is very intense the radiation pressure 5.0 m from the surface is very small IDENTIFY: The intensity and the energy density of an electromagnetic wave depends on the amplitudes of the electric and magnetic fields SET UP: Intensity is I = Pav /A, and the average radiation pressure is Pav = 2I/c, where I = 12 ⑀0cEmax The energy density is u = ⑀0 E EXECUTE: (a) I = Pav /A = 316,000 W 2π (5000 m) = 0.00201 W/m prad = 2I/c = 2(0.00201W/m2 ) 3.00 ×108 m/s = 1.34 ×10−11 Pa (b) I = 12 ⑀0cEmax gives Emax = 2I 2(0.00201 W/m ) = = 1.23 N/C − 12 ⑀ 0c (8.85 × 10 C2 /N ⋅ m )(3.00 × 108 m/s) Bmax = Emax /c = (1.23 N/C)/(3.00 × 108 m/s) = 4.10 × 10−9 T (c) u = ⑀0 E , so uav = ⑀0 ( Erms )2 and Erms = Emax , so (8.85 × 10−12 C2 /N ⋅ m )(1.23 N/C) = 6.69 × 10−12 J/m3 2 (d) As was shown in Section 32.4, the energy density is the same for the electric and magnetic fields, so each one has 50% of the energy density EVALUATE: Compared to most laboratory fields, the electric and magnetic fields in ordinary radiowaves are extremely weak and carry very little energy uav = ⑀0 Emax = © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 32-8 Chapter 32 32.27 IDENTIFY: We know the greatest intensity that the eye can safely receive P SET UP: I = I = 12 ⑀0cEmax Emax = cBmax A EXECUTE: (a) P = IA = (1.0 × 102 W/m )π (0.75 × 10−3 m) = 1.8 × 10−4 W = 0.18 mW (b) E = 2I 2(1.0 × 102 W/m ) E = = 274 V/m Bmax = max = 9.13 × 10−7 T − 12 2 ⑀ 0c c (8.85 × 10 C /N ⋅ m )(3.00 × 10 m/s) (c) P = 0.18 mW = 0.18 mJ/s 32.28 ⎛ 1m ⎞ (d) I = (1.0 × 102 W/m ) ⎜ ⎟ = 0.010 W/cm ⎝ 10 cm ⎠ EVALUATE: Both the electric and magnetic fields are quite weak compared to normal laboratory fields IDENTIFY: Apply Eqs (32.32) and (32.33) The average momentum density is given by Eq (32.30), with S replaced by Sav = I SET UP: atm = 1.013 × 105 Pa EXECUTE: (a) Absorbed light: prad = prad = 8.33 × 10−6 Pa 1.013 × 10 Pa/atm = 8.23 × 10−11 atm (b) Reflecting light: prad = prad = 1.67 × 10−5 Pa 1.013 × 10 Pa/atm 2500 W/m I = = 8.33 × 10−6 Pa Then c 3.0 × 108 m/s I 2(2500 W/m ) = = 1.67 × 10−5 Pa Then c 3.0 × 108 m/s = 1.65 × 10−10 atm (c) The momentum density is dp Sav 2500 W/m = = = 2.78 × 10−14 kg/m ⋅ s dV c (3.0 × 108 m/s) EVALUATE: The factor of in prad for the reflecting surface arises because the momentum vector totally 32.29 reverses direction upon reflection Thus the change in momentum is twice the original momentum IDENTIFY: We know the wavelength and power of the laser beam, as well as the area over which it acts SET UP: P = IA A = π r Emax = cBmax The intensity I = Sav is related to the maximum electric field by I = 12 ⑀0cEmax The average energy density uav is related to the intensity I by I = uav c EXECUTE: (a) I = (b) Emax = 0.500 × 10−3 W P = = 637 W/m A π (0.500 × 10−3 m) 2I 2(637 W/m ) E = = 693 V/m Bmax = max = 2.31 μ T 12 − ⑀ 0c c (8.85 × 10 C2 /N ⋅ m )(3.00 × 108 m/s) I 637 W/m = = 2.12 × 10−6 J/m3 c 3.00 × 108 m/s EVALUATE: The fields are very weak, so a cubic meter of space contains only about µJ of energy (c) uav = 32.30 IDENTIFY: We know the intensity of the solar light and the area over which it acts We can use the light intensity to find the force the light exerts on the sail, and then use the sail’s density to find its mass Newton’s second law will then give the acceleration of the sail 2I Pressure is force per unit area, and Fnet = ma The SET UP: For a reflecting surface the pressure is c mass of the sail is its volume V times its density ρ The area of the sail is π r 2, with r = 4.5 m Its volume is π r 2t , where t = 7.5 ì 106 m is its thickness â Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher Electromagnetic Waves 32-9 2(1400 W/m ) ⎛ 2I ⎞ EXECUTE: (a) F = ⎜ ⎟ A = π (4.5 m) = 5.9 × 10−4 N 3.00 × 108 m/s ⎝ c ⎠ (b) m = ρV = (1.74 × 103 kg/m3 )π (4.5 m) (7.5 × 10−6 m) = 0.83 kg a= F 5.9 × 10−4 N = = 7.1 × 10−4 m/s m 0.83 kg (c) With this acceleration it would take the sail 1.4 × 106 s = 16 days to reach a speed of km/s This 32.31 would be useful only in specialized applications The acceleration could be increased by decreasing the mass of the sail, either by reducing its density or its thickness EVALUATE: The calculation assumed the only force on the sail is that due to the radiation pressure The sun would also exert a gravitational force on the sail, which could be significant IDENTIFY: The nodal and antinodal planes are each spaced one-half wavelength apart SET UP: 12 wavelengths fit in the oven, so 12 λ = L, and the frequency of these waves obeys the ( ) equation f λ = c ( ) EXECUTE: (a) Since 12 λ = L, we have L = (5/2)(12.2 cm) = 30.5 cm (b) Solving for the frequency gives f = c/λ = (3.00 × 108 m/s)/(0.122 m) = 2.46 × 109 Hz ( ) (c) L = 35.5 cm in this case 12 λ = L, so λ = 2L/5 = 2(35.5 cm)/5 = 14.2 cm f = c /λ = (3.00 × 108 m/s)/(0.142 m) = 2.11× 109 Hz 32.32 EVALUATE: Since microwaves have a reasonably large wavelength, microwave ovens can have a convenient size for household kitchens Ovens using radiowaves would need to be far too large, while ovens using visible light would have to be microscopic IDENTIFY: The electric field at the nodes is zero, so there is no force on a point charge placed at a node SET UP: The location of the nodes is given by Eq (32.36), where x is the distance from one of the planes λ = c /f c 3.00 × 108 m/s = = 0.200 m = 20.0 cm There must be nodes at the planes, 2 f 2(7.50 × 108 Hz) which are 80.0 cm apart, and there are two nodes between the planes, each 20.0 cm from a plane It is at 20 cm, 40 cm, and 60 cm from one plane that a point charge will remain at rest, since the electric fields there are zero EVALUATE: The magnetic field amplitude at these points isn’t zero, but the magnetic field doesn’t exert a force on a stationary charge IDENTIFY and SET UP: Apply Eqs (32.36) and (32.37) G EXECUTE: (a) By Eq (32.37) we see that the nodal planes of the B field are a distance λ /2 apart, so λ /2 = 3.55 mm and λ = 7.10 mm G (b) By Eq (32.36) we see that the nodal planes of the E field are also a distance λ /2 = 3.55 mm apart EXECUTE: Δxnodes = 32.33 32.34 λ = (c) v = f λ = (2.20 × 1010 Hz)(7.10 × 10−3 m) = 1.56 × 108 m/s G G EVALUATE: The spacing between the nodes of E is the same as the spacing between the nodes of B Note that v < c, as it must G G IDENTIFY: The nodal planes of E and B are located by Eqs (32.26) and (32.27) c 3.00 × 108 m/s = 4.00 m SET UP: λ = = f 75.0 × 106 Hz EXECUTE: (a) Δx = λ = 2.00 m (b) The distance between the electric and magnetic nodal planes is one-quarter of a wavelength, so is λ Δx 2.00 m = = =1.00 m 2 © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 32-10 32.35 Chapter 32 G EVALUATE: The nodal planes of B are separated by a distance λ /2 and are midway between the nodal G planes of E G (a) IDENTIFY and SET UP: The distance between adjacent nodal planes of B is λ /2 There is an antinodal G plane of B midway between any two adjacent nodal planes, so the distance between a nodal plane and an adjacent antinodal plane is λ /4 Use v = f λ to calculate λ EXECUTE: λ = λ 32.36 v 2.10 × 108 m/s = = 0.0175 m f 1.20 × 1010 Hz 0.0175 m = 4.38 × 10−3 m = 4.38 mm 4 G (b) IDENTIFY and SET UP: The nodal planes of E are at x = 0, λ /2, λ , 3λ /2, , so the antinodal G G planes of E are at x = λ /4, 3λ /4, 5λ /4, The nodal planes of B are at x = λ /4, 3λ /4, 5λ /4, , so G the antinodal planes of B are at λ /2, λ , 3λ /2, G G EXECUTE: The distance between adjacent antinodal planes of E and antinodal planes of B is therefore λ /4 = 4.38 mm G G (c) From Eqs (32.36) and (32.37) the distance between adjacent nodal planes of E and B is λ /4 = 4.38 mm G G G EVALUATE: The nodes of E coincide with the antinodes of B and conversely The nodes of B and the G nodes of E are equally spaced IDENTIFY: Evaluate the derivatives of the expressions for E y ( x, t ) and Bz ( x, t ) that are given in = Eqs (32.34) and (32.35) ∂ ∂ ∂ ∂ SET UP: sin kx = k cos kx, sin ωt = ω cos ωt cos kx = − k sin kx, cos ωt = −ω sin ωt ∂t ∂t ∂x ∂x EXECUTE: (a) ∂ E y ( x, t ) ∂x Similarly: ∂ Bz ( x , t ) ∂ E y ( x, t ) ∂x = ∂2 ∂x (−2 Emax sin kx sin ωt ) = = 2k Emax sin kx sin ωt = ∂ Bz ( x , t ) ∂x = ∂2 ∂x c2 Emax sin kx sin ωt = ⑀0 μ0 ( −2 Bmax cos kx cos ωt ) = = 2k Bmax cos kx cos ωt = ∂x ∂E y ( x, t ) ω2 ω2 c ∂ ( −2kEmax cos kx sin ωt ) and ∂x ∂ E y ( x, t ) ∂t ∂ ( +2kBmax sin kx cos ωt ) and ∂x Bmax cos kx cos ωt = ⑀ μ0 ∂ Bz ( x , t ) ∂t ∂ ( −2 Emax sin kx sin ωt ) = −2kEmax cos kx sin ωt ∂x ω E = − Emax cos kx sin ωt = −ω max cos kx sin ωt = −ω Bmax cos kx sin ωt ∂x c c ∂E y ( x, t ) ∂ ∂Bz ( x, t ) = + (2 Bmax cos kx cos ωt ) = − ∂x ∂t ∂t ∂B ( x, t ) ∂ Similarly: − z = ( +2 Bmax cos kx cos ωt ) = −2kBmax sin kx cos ωt ∂x ∂x ∂Bz ( x, t ) ω ω − = − Bmax sin kx cos ωt = − 2cBmax sin kx cos ωt ∂x c c ∂E y ( x, t ) ∂Bz ( x, t ) ∂ − = −⑀0 μ0ω Emax sin kx cos ωt = ⑀0 μ0 (−2 Emax sin kx sin ωt ) = ⑀ μ0 ∂x ∂t ∂t EVALUATE: The standing waves are linear superpositions of two traveling waves of the same k and ω (b) ∂x ∂E y ( x, t ) = © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher Electromagnetic Waves 32.37 32-11 IDENTIFY: We know the wavelength and power of a laser beam as well as the area over which it acts and the duration of a pulse I P SET UP: The energy is U = Pt For absorption the radiation pressure is , where I = The c A λ0 wavelength in the eye is λ = I = ⑀0cEmax and Emax = cBmax n EXECUTE: (a) U = Pt = (250 × 10−3 W)(1.50 × 10−3 s) = 3.75 × 10−4 J = 0.375 mJ (b) I = P 250 × 10−3 W = = 1.22 × 106 W/m The average pressure is A π (255 × 10−6 m) I 1.22 × 106 W/m = = 4.08 × 10−3 Pa c 3.00 × 108 m/s 810 nm v c 3.00 × 108 m/s = = 3.70 × 1014 Hz; f is the same in the air and = 604 nm f = = n 1.34 λ λ0 810 × 10−9 m in the vitreous humor (c) λ = λ0 (d) Emax = = 2I 2(1.22 × 106 W/m ) = = 3.03 × 104 V/m ⑀ 0c (8.85 × 10−12 C2 /N ⋅ m )(3.00 × 108 m/s) Emax = 1.01 × 10−4 T c EVALUATE: The intensity of the beam is high, as it must be to weld tissue, but the pressure it exerts on the retina is only around 10−8 that of atmospheric pressure The magnetic field in the beam is about twice that of the earth’s magnetic field IDENTIFY: Evaluate the partial derivatives of the expressions for E y ( x, t ) and Bz ( x, t ) Bmax = 32.38 ∂ ∂ cos(kx − ωt ) = − k sin( kx − ωt ), cos( kx − ωt ) = ω sin( kx − ωt ) ⋅ ∂x ∂t ∂ ∂ sin(kx − ωt ) = k cos(kx − ωt ), sin(kx − ωt ) = −ω cos( kx − ωt ) ∂x ∂t G G EXECUTE: Assume E = Emax ˆjcos(kx − ωt ) and B = Bmax kˆ cos(kx − ωt + φ ), with − π < φ < π Eq (32.12) SET UP: ∂E y ∂Bz This gives kEmax sin(kx − ωt ) = +ω Bmax sin(kx − ωt + φ ), so φ = 0, and kEmax = ω Bmax , ∂x ∂t ∂E y ∂B ω 2π f Bmax = f λ Bmax = cBmax Similarly for Eq (32.14), − z = ⑀0 μ0 gives so Emax = Bmax = ∂x ∂t 2π /λ k kBmax sin( kx − ωt + φ ) = ⑀0 μ0ω Emax sin(kx − ωt ), so φ = and kBmax = ⑀ 0μ0ω Emax , so is 2π f fλ Emax = Emax = Emax = Emax k c c 2π /λ c G G EVALUATE: The E and B fields must oscillate in phase IDENTIFY: The light exerts pressure on the paper, which produces an upward force This force must balance the weight of the paper I SET UP: The weight of the paper is mg For a totally absorbing surface the radiation pressure is and for c 2I P a totally reflecting surface it is The force is F = PA, and the intensity is I = c A ⎛I⎞ EXECUTE: (a) The radiation force must equal the weight of the paper, so ⎜ ⎟ A = mg ⎝c⎠ Bmax = 32.39 =− I= ⑀0 μ0ω mgc (1.50 × 10−3 kg)(9.80 m/s )(3.00 × 108 m/s) = = 7.16 × 107 W/m (0.220 m)(0.280 m) A © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 32-12 Chapter 32 (b) I = 12 ⑀0cEmax Solving for Emax gives Emax = 2I 2(7.16 × 107 W/m ) = = 2.32 × 105 V/m ⑀ 0c (8.85 × 10−12 C2 /N ⋅ m )(3.00 × 108 m/s) Emax 2.32 × 105 V/m = = 7.74 × 10−4 T c 3.00 × 108 m/s 2I mgc ⎛ 2I ⎞ , so ⎜ ⎟ A = mg I = = 3.58 × 107 W/m (c) The pressure is c 2A ⎝ c ⎠ Bmax = (d) I = 32.40 P 0.500 × 10−3 W = = 637 W/m A π (0.500 × 10−3 m) EVALUATE: The intensity of this laser is much less than what is needed to support a sheet of paper And to support the paper, not only must the intensity be large, it also must be over a large area IDENTIFY: The average energy density in the electric field is uE ,av = 12 ⑀0 ( E )av and the average energy density in the magnetic field is u B ,av = ( B )av μ0 SET UP: (cos ( kx − ωt ))av = 12 2 EXECUTE: E y ( x, t ) = Emax cos( kx − ωt ) u E = 12 ⑀0 E y2 = 12 ⑀0 Emax cos (kx − ωt ) and u E , av = 14 ⑀0 Emax Bz ( x, t ) = Bmax cos(kx − ωt ), so u B = μ0 Bz2 = Emax = cBmax , so uE , av = 14 ⑀0c Bmax c= μ0 ⑀ μ0 cos (kx − ωt ) and u B ,av = Bmax , so u E ,av = 2μ0 μ0 , which equals u B,av Bmax EVALUATE: Our result allows us to write uav = 2u E ,av = 12 ⑀0 Emax and uav = 2u B ,av = 32.41 Bmax Bmax μ0 IDENTIFY: The intensity of an electromagnetic wave depends on the amplitude of the electric and magnetic fields Such a wave exerts a force because it carries energy SET UP: The intensity of the wave is I = Pav /A = 12 ⑀0cEmax , and the force is F = prad A where prad = I /c EXECUTE: (a) I = Pav /A = (25,000 W)/[4π (575 m) ] = 0.00602 W/m 2 (b) I = 12 ⑀0cEmax , so Emax = 2I 2(0.00602 W/m ) = = 2.13 N/C − 12 ⑀ 0c (8.85 × 10 C2 /N ⋅ m )(3.00 × 108 m/s) Bmax = Emax /c = (2.13 N/C)/(3.00 × 108 m/s) = 7.10 × 10−9 T (c) F = prad A = (I/c)A = (0.00602 W/m )(0.150 m)(0.400 m)/(3.00 × 108 m/s) = 1.20 ×10−12 N 32.42 EVALUATE: The fields are very weak compared to ordinary laboratory fields, and the force is hardly worth worrying about! IDENTIFY: c = f λ Emax = cBmax I = 12 ⑀0cEmax For a totally absorbing surface the radiation pressure is I c SET UP: The wave speed in air is c = 3.00 × 108 m/s EXECUTE: (a) f = c = 3.00 × 108 m/s = 7.81 × 109 Hz λ 3.84 × 10−2 m E 1.35 V/m (b) Bmax = max = = 4.50 × 10−9 T c 3.00 × 108 m/s (c) I = 12 ⑀0cEmax = 12 (8.854 × 10−12 C2 /N ⋅ m )(3.00 × 108 m/s)(1.35 V/m) = 2.42 ì 103 W/m â Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher Electromagnetic Waves 32-13 IA (2.42 × 10−3 W/m )(0.240 m ) = = 1.94 × 10−12 N c 3.00 × 108 m/s EVALUATE: The intensity depends only on the amplitudes of the electric and magnetic fields and is independent of the wavelength of the light (a) IDENTIFY and SET UP: Calculate I and then use Eq (32.29) to calculate Emax and Eq (32.18) to (d) F = (pressure) A = 32.43 calculate Bmax EXECUTE: The intensity is power per unit area: I = I= P 4.60 × 10−3 W = = 937 W/m A π (1.25 × 10−3 m) 2 Emax , so Emax = μ0cI Emax = 2(4π × 10−7 T ⋅ m/A)(2.998 × 108 m/s)(937 W/m ) = 840 V/m μ 0c Emax 840 V/m = = 2.80 × 10−6 T c 2.998 × 108 m/s EVALUATE: The magnetic field amplitude is quite small compared to laboratory fields (b) IDENTIFY and SET UP: Eqs (24.11) and (30.10) give the energy density in terms of the electric and magnetic field values at any time For sinusoidal fields average over E and B to get the average energy densities EXECUTE: The energy density in the electric field is u E = 12 ⑀0 E E = Emax cos(kx − ωt ) and the average Bmax = value of cos (kx − ωt ) is 12 The average energy density in the electric field then is u E ,av = 14 ⑀0 Emax = 14 (8.854 × 10−12 C2 /N ⋅ m )(840 V/m)2 = 1.56 × 10−6 J/m3 The energy density in the B2 B2 (2.80 × 10−6 T) = 1.56 × 10−6 J/m3 The average value is u B ,av = max = 4μ0 4(4π × 10−7 T ⋅ m/A) μ0 EVALUATE: Our result agrees with the statement in Section 32.4 that the average energy density for the electric field is the same as the average energy density for the magnetic field (c) IDENTIFY and SET UP: The total energy in this length of beam is the average energy density (uav = u E ,av + u B ,av = 3.12 × 10−6 J/m3 ) times the volume of this part of the beam magnetic field is u B = EXECUTE: U = uav LA = (3.12 × 10−6 J/m3 )(1.00 m)π (1.25 × 10−3 m) = 1.53 × 10−11 J 32.44 EVALUATE: This quantity can also be calculated as the power output times the time it takes the light to 1.00 m ⎛L⎞ ⎛ ⎞ −11 travel L = 1.00 m: U = P ⎜ ⎟ = (4.60 × 10−3 W) ⎜ ⎟ = 1.53 × 10 J, which checks c ⎝ ⎠ ⎝ 2.998 × 10 m/s ⎠ IDENTIFY: We know the electric field in the plastic ω 2π , SET UP: The general wave function for the electric field is E = Emax cos( kx − ωt ) f = , λ= 2π k c v = f λ and v = n EXECUTE: (a) By comparing the equation for E to the general form, we have ω = 3.02 × 1015 rad/s and ω 2π k = 1.39 × 107 rad/m f = = 4.81 × 1014 Hz λ = = 4.52 × 10−7 m = 452 nm k 2π v = f λ = 2.17 × 108 m/s (b) n = c 3.00 × 108 m/s = = 1.38 v 2.17 × 108 m/s (c) In air, ω = 3.02 × 1015 rad/s, the same as in the plastic λ0 = λ n = (4.52 × 10−7 m)(1.38) = 6.24 × 10−7 m, so k = 2π λ = 1.01 × 107 rad/m The equation for E in air is E = (535 V/m)cos ⎡(1.01× 107 rad/m) x − (3.02 ì 1015 rad/s)t â Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 32-14 32.45 Chapter 32 EVALUATE: In the plastic, k and λ are different from their values in air, but f and ω are the same in both media I IDENTIFY: I = Pav /A For an absorbing surface, the radiation pressure is prad = c SET UP: Assume the electromagnetic waves are formed at the center of the sun, so at a distance r from the center of the sun I = Pav /(4π r ) EXECUTE: (a) At the sun’s surface: I = Pav 4π R = 3.9 × 1026 W 4π (6.96 × 108 m) = 6.4 × 107 W/m and I 6.4 × 107 W/m = = 0.21 Pa c 3.00 × 108 m/s Halfway out from the sun’s center, the intensity is times more intense, and so is the radiation pressure: I = 2.6 × 108 W/m and prad = 0.85 Pa At the top of the earth’s atmosphere, the measured sunlight prad = intensity is 1400 W/m and prad = × 10−6 Pa, which is about 100,000 times less than the values above 32.46 EVALUATE: (b) The gas pressure at the sun’s surface is 50,000 times greater than the radiation pressure, and halfway out of the sun the gas pressure is believed to be about × 1013 times greater than the radiation pressure Therefore it is reasonable to ignore radiation pressure when modeling the sun’s interior structure IDENTIFY: The intensity of the wave, not the electric field strength, obeys an inverse-square distance law SET UP: The intensity is inversely proportional to the distance from the source, and it depends on the amplitude of the electric field by I = Sav = 12 ⑀0cEmax , Emax ∝ I A point at 20.0 cm (0.200 m) from the source is 50 times EXECUTE: Since I = ⑀ 0cEmax closer to the source than a point that is 10.0 m from it Since I ∝ 1/r and (0.200 m)/(10.0 m) = 1/50, we have I 0.20 = 502 I10 Since Emax ∝ I , we have E0.20 = 50 E10 = (50)(1.50 N/C) = 75.0 N/C 32.47 EVALUATE: While the intensity increases by a factor of 502 = 2500, the amplitude of the wave only increases by a factor of 50 Recall that the intensity of any wave is proportional to the square of its amplitude IDENTIFY: The same intensity light falls on both reflectors, but the force on the reflecting surface will be twice as great as the force on the absorbing surface Therefore there will be a net torque about the rotation axis SET UP: For a totally absorbing surface, F = prad A = ( I /c) A, while for a totally reflecting surface the force will be twice as great The intensity of the wave is I = 12 ⑀0cEmax Once we have the torque, we can use the rotational form of Newton’s second law, τ net = Iα , to find the angular acceleration EXECUTE: The force on the absorbing reflector is FAbs = prad A = ( I/c) A = ⑀ cE A max c = 12 ⑀0 AEmax The net torque is For a totally reflecting surface, the force will be twice as great, which is ⑀ 0cEmax therefore τ net = FRefl ( L/2) − FAbs ( L/2) = ⑀0 AEmax L/4 Newton’s second law for rotation gives τ net = Iα ⑀0 AEmax L/4 = 2m( L/2) 2α Solving for α gives /(2mL) = α = ⑀0 AEmax (8.85 × 10−12 C2 /N ⋅ m )(0.0150 m) (1.25 N/C)2 = 3.89 × 10−13 rad/s (2)(0.00400 kg)(1.00 m) EVALUATE: This is an extremely small angular acceleration To achieve a larger value, we would have to greatly increase the intensity of the light wave or decrease the mass of the reflectors © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher Electromagnetic Waves 32.48 32-15 IDENTIFY: The changing magnetic field of the electromagnetic wave produces a changing flux through the wire loop, which induces an emf in the loop SET UP: Φ B = Bπ r = π r Bmax cos(kx − ωt ), taking x for the direction of propagation of the wave Faraday’s law says E = EXECUTE: f = c λ = Bmax = E = dΦB E B c c Bmax , and f = The intensity of the wave is I = max max = μ0 μ0 λ dt dΦB = ω Bmax sin( kx − ωt )π r E max = 2π fBmaxπ r dt 3.00 × 108 m/s E B c Bmax for Bmax gives = 4.348 × 107 Hz Solving I = max max = μ0 μ0 6.90 m μ0 I 2(4π × 10−7 T ⋅ m/A)(0.0195 W/m ) = = 1.278 × 10−8 T c 3.00 × 108 m/s E max = 2π (4.348 × 107 Hz)(1.278 × 10−8 T)π (0.075 m) = 6.17 × 10−2 V = 61.7 mV 32.49 EVALUATE: This voltage is quite small compared to everyday voltages, so it normally would not be noticed But in very delicate laboratory work, it could be large enough to take into consideration IDENTIFY and SET UP: In the wire the electric field is related to the current density by Eq (25.7) Use G Ampere’s law to calculate B The Poynting vector is given by Eq (32.28) and the equation that follows it G relates the energy flow through a surface to S G EXECUTE: (a) The direction of E is parallel to the axis of the cylinder, in the direction of the current From Eq (25.7), E = ρ J = ρ I /π a (E is uniform across the cross section of the conductor.) (b) A cross-sectional view of the conductor is given in Figure 32.49a; take the current to be coming out of the page Apply Ampere’s law to a circle of radius a G G v∫ B ⋅ dl = B(2π a) I encl = I Figure 32.49a G G μ I gives B (2π a ) = μ0 I and B = 2π a G The direction of B is counterclockwise around the circle G G (c) The directions of E and B are shown in Figure 32.49b v∫ B ⋅ dl = μ0 Iencl G G G The direction of S = E×B μ0 is radially inward 1 ⎛ ρ I ⎞⎛ μ0 I ⎞ S= EB = ⎟ μ0 μ0 ⎜⎝ π a ⎟⎜ ⎠⎝ 2π a ⎠ S= ρI 2π 2a Figure 32.49b © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 32-16 32.50 Chapter 32 (d) EVALUATE: Since S is constant over the surface of the conductor, the rate of energy flow P is given ρI ρ lI But by S times the surface of a length l of the conductor: P = SA = S (2π al ) = (2π al ) = π a2 2π a ρl R = , so the result from the Poynting vector is P = RI This agrees with PR = I R, the rate at which πa G electrical energy is being dissipated by the resistance of the wire Since S is radially inward at the surface of the wire and has magnitude equal to the rate at which electrical energy is being dissipated in the wire, this energy can be thought of as entering through the cylindrical sides of the conductor IDENTIFY: The nodal planes are one-half wavelength apart SET UP: The nodal planes of B are at x = λ /4, 3λ /4, 5λ /4, …, which are λ /2 apart EXECUTE: (a) The wavelength is λ = c /f = (3.000 × 108 m/s)/(110.0 × 106 Hz) = 2.727 m So the nodal planes are at (2.727 m)/2 = 1.364 m apart (b) For the nodal planes of E, we have λn = 2L/n, so L = nλ /2 = (8)(2.727 m)/2 = 10.91 m 32.51 EVALUATE: Because radiowaves have long wavelengths, the distances involved are easily measurable using ordinary metersticks IDENTIFY and SET UP: Find the force on you due to the momentum carried off by the light Express this force in terms of the radiated power of the flashlight Use this force to calculate your acceleration and use a constant acceleration equation to find the time (a) EXECUTE: prad = I /c and F = prad A gives F = IA/c = Pav /c ax = F/m = Pav /(mc) = (200 W)/[(150 kg)(3.00 × 108 m/s)] = 4.44 × 10−9 m/s Then x − x0 = v0 xt + 12 axt gives t = 2( x − x0 )/a x = 2(16.0 m)/(4.44 × 10−9 m/s ) = 8.49 × 104 s = 23.6 h 32.52 EVALUATE: The radiation force is very small In the calculation we have ignored any other forces on you (b) You could throw the flashlight in the direction away from the ship By conservation of linear momentum you would move toward the ship with the same magnitude of momentum as you gave the flashlight IDENTIFY: Pav = IA and I = 12 ⑀ 0cEmax Emax = cBmax SET UP: The power carried by the current i is P = Vi P EXECUTE: I = av = 12 ⑀0cEmax and A Emax = Bmax = Pav = A⑀0c 2Vi 2(5.00 × 105 V)(1000 A) = = 6.14 × 104 V/m A⑀0c (100 m )⑀0 (3.00 × 10 m/s) Emax 6.14 × 104 V/m = = 2.05 × 10−4 T c 3.00 × 10 m/s EVALUATE: I = Vi /A = 32.53 (5.00 × 105 V)(1000 A) 100 m = 5.00 × 106 W/m This is a very intense beam spread over a large area IDENTIFY: The orbiting satellite obeys Newton’s second law of motion The intensity of the electromagnetic waves it transmits obeys the inverse-square distance law, and the intensity of the waves depends on the amplitude of the electric and magnetic fields SET UP: Newton’s second law applied to the satellite gives mv /r = GmM/r , where M is the mass of the earth and m is the mass of the satellite The intensity I of the wave is I = Sav = 12 ⑀0cEmax , and by definition, I = Pav /A © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher Electromagnetic Waves 32-17 EXECUTE: (a) The period of the orbit is 12 hr Applying Newton’s second law to the satellite gives m(2π r /T ) GmM mv /r = GmM/r , which gives = Solving for r, we get r r2 1/3 ⎛ GMT ⎞ r =⎜ ⎜ 4π ⎟⎟ ⎝ ⎠ 1/3 ⎡ (6.67 × 10−11 N ⋅ m /kg )(5.97 × 1024 kg)(12 × 3600 s)2 ⎤ =⎢ ⎥ 4π ⎣⎢ ⎦⎥ = 2.66 × 107 m The height above the surface is h = 2.66 × 107 m – 6.38 × 106 m = 2.02 × 107 m The satellite only radiates its energy to the lower hemisphere, so the area is 1/2 that of a sphere Thus, from the definition of intensity, the intensity at the ground is I = Pav /A = Pav /(2π h ) = (25.0 W)/[2π (2.02 × 107 m) ] = 9.75 × 10−15 W/m 2 (b) I = Sav = 12 ⑀ 0cEmax , so Emax = 2I 2(9.75 × 10−15 W/m ) = = 2.71× 10−6 N/C ⑀ 0c (8.85 × 10−12 C2 /N ⋅ m )(3.00 × 108 m/s) Bmax = Emax /c = (2.71 × 10−6 N/C)/(3.00 × 108 m/s) = 9.03 × 10−15 T t = d /c = (2.02 × 107 m)/(3.00 × 108 m/s) = 0.0673 s (c) prad = I/c = (9.75 × 10 –15 W/m )/(3.00 × 108 m/s) = 3.25 × 10 –23 Pa (d) λ = c/f = (3.00 × 108 m/s)/(1575.42 × 106 Hz) = 0.190 m 32.54 EVALUATE: The fields and pressures due to these waves are very small compared to typical laboratory quantities 2I Find the force due to this pressure IDENTIFY: For a totally reflective surface the radiation pressure is c and express the force in terms of the power output P of the sun The gravitational force of the sun is mM sun Fg = G r2 SET UP: The mass of the sun is M sun = 1.99 × 1030 kg G = 6.67 × 10−11 N ⋅ m /kg EXECUTE: (a) The sail should be reflective, to produce the maximum radiation pressure P ⎛ 2I ⎞ (b) Frad = ⎜ ⎟ A, where A is the area of the sail I = , where r is the distance of the sail from the 4π r ⎝ c ⎠ PA mM sun PA ⎛ A ⎞⎛ P ⎞ sun Frad = ⎜ =G ⋅ Frad = Fg so ⎟⎜ ⎟= 2 2π r c r2 ⎝ c ⎠⎝ 4π r ⎠ 2π r c A= 2π cGmM sun 2π (3.00 × 108 m/s)(6.67 × 10−11 N ⋅ m /kg )(10,000 kg)(1.99 × 1030 kg) = P 3.9 × 1026 W A = 6.42 × 106 m = 6.42 km (c) Both the gravitational force and the radiation pressure are inversely proportional to the square of the distance from the sun, so this distance divides out when we set Frad = Fg 32.55 EVALUATE: A very large sail is needed, just to overcome the gravitational pull of the sun IDENTIFY and SET UP: The gravitational force is given by Eq (13.2) Express the mass of the particle in terms of its density and volume The radiation pressure is given by Eq (32.32); relate the power output L of the sun to the intensity at a distance r The radiation force is the pressure times the cross-sectional area of the particle mM EXECUTE: (a) The gravitational force is Fg = G The mass of the dust particle is m = ρV = ρ 43 π R3 r Thus Fg = ρ Gπ MR3 3r © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 32-18 Chapter 32 I (b) For a totally absorbing surface prad = If L is the power output of the sun, the intensity of the solar c L L Thus prad = The force Frad that corresponds to radiation a distance r from the sun is I = 4π r 4π cr prad is in the direction of propagation of the radiation, so Frad = prad A⊥ , where A⊥ = π R is the component of area of the particle perpendicular to the radiation direction Thus LR ⎛ L ⎞ Frad = ⎜ R ( π ) = ⎟ 4cr ⎝ 4π cr ⎠ (c) Fg = Frad ρ Gπ MR3 3r ⎛ ρ Gπ M ⎜ ⎝ R= = LR 4cr L 3L ⎞ and R = ⎟R = c 16 c Gπ M ρ ⎠ 3(3.9 × 1026 W) 16(2.998 × 108 m/s)(3000 kg/m3 )(6.673 × 10−11 N ⋅ m / kg )π (1.99 × 1030 kg) R = 1.9 × 10−7 m = 0.19 μ m EVALUATE: The gravitational force and the radiation force both have a r −2 dependence on the distance from the sun, so this distance divides out in the calculation of R ⎛ LR ⎞⎛ ⎞ F 3r 3L Frad is proportional to R and Fg is proportional to R3 , (d) rad = ⎜ ⎟⎜ ⎟⎟ = Fg ⎜⎝ 4cr ⎟⎜ c G MR 16 ρ π G mR ρ π ⎠⎝ ⎠ so this ratio is proportional to 1/R If R < 0.20 μ m then Frad > Fg and the radiation force will drive the 32.56 particles out of the solar system IDENTIFY and SET UP: Follow the steps specified in the problem EXECUTE: (a) E y ( x, t ) = Emax e− kC x cos (kC x − ωt ) ∂E y ∂x = Emax ( −kC )e− kC x cos(kC x − ωt ) + Emax (− kC )e− kC x sin(kC x − ωt ) ∂2Ey ∂x = Emax ( + kC2 )e− kC x cos(kC x − ωt ) + Emax (+ kC2 )e− kC x sin(kC x − ωt ) + Emax (+ kC2 )e− kC x sin(kC x − ωt ) + Emax ( −kC2 )e− kC x cos(kC x − ωt ) ∂2Ey ∂x = −2 Emax kC2e− kC x cos(kC x − ωt ) Setting true if ∂2Ey ∂x 2kC2 ω = = ∂E y ∂t = − Emax e− kC x ω sin(kC x − ωt ) μ∂E y gives Emax kC2e− kC x sin( kC x − ωt ) = μ /pEmax e− kC x ω sin(kC x − ωt ) This will only be ρ∂t μ ωμ , or kC = ρ 2ρ (b) The energy in the wave is dissipated by the i R heating of the conductor (c) E y = Ey0 e ⇒ kC x = 1, x = 2ρ 2(1.72 × 10−8 Ω ⋅ m) = = = 6.60 × 10−5 m ωμ kC 2π (1.0 × 106 Hz) μ0 EVALUATE: The lower the frequency of the waves, the greater is the distance they can penetrate into a conductor A dielectric (insulator) has a much larger resistivity and these waves can penetrate a greater distance in these materials © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher Electromagnetic Waves 32.57 IDENTIFY: The orbiting particle has acceleration a = 32-19 v2 R SET UP: K = 12 mv An electron has mass me = 9.11 × 10−31 kg and a proton has mass mp = 1.67 × 10−27 kg ⎡ q 2a ⎤ C2 (m/s ) N⋅m J ⎡ dE ⎤ EXECUTE: (a) ⎢ = = = = W = ⎢ ⎥ 3⎥ s s ⎣ dt ⎦ ⎢⎣ 6π ⑀0c ⎥⎦ (C /N ⋅ m )(m/s) (b) For a proton moving in a circle, the acceleration is a= 2(6.00 × 106 eV)(1.6 × 10−19 J/eV) v 12 mv = = = 1.53 × 1015 m/s The rate at which it emits energy −27 R mR (1.67 × 10 kg)(0.75 m) because of its acceleration is dE q 2a (1.6 × 10−19 C) (1.53 × 1015 m/s )2 = = = 1.33 × 10−23 J/s = 8.32 × 10−5 eV/s dt 6π ⑀0c 6π ⑀0 (3.0 × 108 m/s)3 Therefore, the fraction of its energy that it radiates every second is (dE /dt )(1 s) 8.32 × 10−5 eV = = 1.39 × 10−11 E 6.00 × 106 eV (c) Carry out the same calculations as in part (b), but now for an electron at the same speed and radius That means the electron’s acceleration is the same as the proton, and thus so is the rate at which it emits energy, since they also have the same charge However, the electron’s initial energy differs from the m (9.11× 10−31 kg) = 3273 eV Therefore, proton’s by the ratio of their masses: Ee = Ep e = (6.00 × 106 eV) mp (1.67 × 10−27 kg) the fraction of its energy that it radiates every second is EVALUATE: The proton has speed v = 32.58 (dE /dt )(1 s) 8.32 × 10−5 eV = = 2.54 × 10−8 3273 eV E 2E 2(6.0 × 106 eV)(1.60 × 10−19 J/eV) = = 3.39 × 107 m/s The mp 1.67 × 10−27 kg electron has the same speed and kinetic energy 3.27 keV The particles in the accelerator radiate at a much smaller rate than the electron in Problem 32.58 does, because in the accelerator the orbit radius is very much larger than in the atom, so the acceleration is much less v2 IDENTIFY: The electron has acceleration a = R SET UP: eV = 1.60 × 10−19 C An electron has q = e = 1.60 × 10−19 C EXECUTE: For the electron in the classical hydrogen atom, its acceleration is a= v 12 mv 2(13.6 eV)(1.60 × 10−19 J/eV) = = = 9.03 × 1022 m/s Then using the formula for the rate −31 −11 R mR (9 11 × 10 kg)(5 29 × 10 m) of energy emission given in Problem 32.57: dE q 2a (1.60 × 10−19 C) (9.03 × 1022 m/s ) = = = 4.64 × 10−8 J/s = 2.89 × 1011 eV/s This large value of dt 6π ⑀0c3 6π ⑀0 (3.00 × 108 m/s)3 dE would mean that the electron would almost immediately lose all its energy! dt EVALUATE: The classical physics result in Problem 32.57 must not apply to electrons in atoms © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher ... SET UP: atm = 1. 013 × 105 Pa EXECUTE: (a) Absorbed light: prad = prad = 8.33 × 10−6 Pa 1. 013 × 10 Pa/atm = 8.23 × 10−11 atm (b) Reflecting light: prad = prad = 1.67 × 10−5 Pa 1. 013 × 10 Pa/atm... 2 (b) I = 12 ⑀0cEmax , so Emax = 2I 2(0.00602 W/m ) = = 2 .13 N/C − 12 ⑀ 0c (8.85 × 10 C2 /N ⋅ m )(3.00 × 108 m/s) Bmax = Emax /c = (2 .13 N/C)/(3.00 × 108 m/s) = 7.10 × 10−9 T (c) F = prad A =... (0.75 × 10−3 m) = 1.8 × 10−4 W = 0.18 mW (b) E = 2I 2(1.0 × 102 W/m ) E = = 274 V/m Bmax = max = 9 .13 × 10−7 T − 12 2 ⑀ 0c c (8.85 × 10 C /N ⋅ m )(3.00 × 10 m/s) (c) P = 0.18 mW = 0.18 mJ/s 32.28