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Transcript
CHAPTER 31: Maxwell’s Equations and Electromagnetic Waves Solutions to Assigned Problems 1. The electric field between the plates is given by E E 2. V d dE dt dV d dt d , where d is the distance between the plates. V m 120 V s 1.1 105 s 0.0011m 1 The displacement current is shown in section 31-1 to be I D 0 A dE . dt dE V 2 8 ID 0 A 8.85 1012 C2 N m 2 0.058 m 2.0 106 6.0 10 A dt m s 8. V Use Eq. 31-11 with v c. E0 B0 c B0 E0 c 0.57 104 V m 3.00 108 m s 1.9 1013 T 11. (a) If we write the argument of the cosine function as kz + t = k(z + ct), we see that the wave is traveling in the – z direction, or kˆ . (b) E and B are perpendicular to each other and to the direction of propagation. At the origin, the electric field is pointing in the positive x direction. Since E B must point in the negative z direction, B must point in the the – y direction, or jˆ . The magnitude of the magnetic field is found from Eq. 31-11 as B0 E0 c . 18. The length of the pulse is d ct. Use this to find the number of wavelengths in a pulse. 8 12 ct 3.00 10 m s 38 10 s N 10734 11,000 wavelengths 1062 109 m If the pulse is to be only one wavelength long, then its time duration is the period of the wave, which is the reciprocal of the wavelength. 9 1 1062 10 m T 3.54 1015 s 8 f c 3.00 10 m s 23. The energy per unit area per unit time is given by the magnitude of the Poynting vector. Let U represent the energy that crosses area A in a time T . © 2009 Pearson Education, Inc., Upper Saddle River, NJ. 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. 303 Physics for Scientists & Engineers with Modern Physics, 4th Edition S 2 cBrms t 0 0 U 2 AcBrms U At Instructor Solutions Manual 4 10 T m A 335J m 3.00 10 m s 22.5 10 T 7 1.00 10 4 2 9 8 2 0.194 W m 2 2.77 107 s 321 days 27. From Eq. 31-16b, the instantaneous energy density is u 0 E 2 . From Eq. 31-17, we see that this instantaneous energy density is also given by S c . The time-averaged value is therefore S c . Multiply that times the volume to get the energy. S 1350W m2 U uV V 1.00m3 4.50 106 J 8 c 3.00 10 m s 29. The radiation from the Sun has the same intensity in all directions, so the rate at which it reaches the Earth is the rate at which it passes through a sphere centered at the Sun with a radius equal to the Earth’s orbit radius. The 1350W m2 is the intensity, or the magnitude of the Poynting vector. 2 P S P SA 4 R2 S 4 1.496 1011 m 1350W m2 3.80 1026 W A 32. (a) Example 31-1 refers back to Example 21-13 and d Figure 21-31. In that figure, and the figure included here, the electric field between the plates is to the right. The magnetic field is shown as counterclockwise r0 circles. Take any point between the capacitor plates, B I I and find the direction of E B. For instance, at the E top of the circle shown in Figure 31-4, E is toward the viewer, and B is to the left. The cross product E B points down, directly to the line connecting the center of the plates. Or take the rightmost point on the circle. E is again toward the viewer, and B is upwards. The cross product E B points to the left, again directly to the line connecting the center of the plates. In cylindrical coordinates, ˆ . The cross product kˆ φˆ rˆ. E E kˆ and B Bφ (b) We evaluate the Poynting vector, and then integrate it over the curved cylindrical surface dE between the capacitor plates. The magnetic field (from Example 31-1) is B 12 0 0 r0 , dt 1 dE evaluated at r r0 . E and B are perpendicular to each other, so S E B 12 0 r0 E , 0 dt inward. In calculating S dA for energy flow into the capacitor volume, note that both S and dA point inward, and that S is constant over the curved surface of the cylindrical volume. dE dE 2 S dA SdA S dA SA S 2 r0d 12 0r0 E dt 2 r0d 0d r0 E dt The amount of energy stored in the capacitor is the energy density times the volume of the © 2009 Pearson Education, Inc., Upper Saddle River, NJ. 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. 304 Maxwell’s Equations and Electromagnetic Waves Chapter 31 capacitor. The energy density is given by Eq. 24-6, u 12 0 E 2 , and the energy stored is the energy density times the volume of the capacitor. Take the derivative of the energy stored with respect to time. dU dE U u Volume 12 0 E 2 r02d 0 E r02d dt dt dU We see that S dA . dt 49. The light has the same intensity in all directions, so use a spherical geometry centered on the source to find the value of the Poynting vector. Then use Eq. 31-19a to find the magnitude of the electric field, and Eq. 31-11 with v c to find the magnitude of the magnetic field. P P S 0 0 2 12 c 0 E02 A 4 r E0 75W P0 33.53V m 2 2 2 r c 0 2 2.00 m 3.00 108 m s 8.85 10 12 C2 n m 2 34 V m B0 33.53V m 1.1 107 T E0 c 3.00 108 m s 63. (a) To show that E and B are perpendicular, calculate their dot product. E B E0 sin kx t ˆj E0 cos kx t kˆ B0 cos kx t ˆj B0 sin kx t kˆ E0 sin kx t B0 cos kx t E0 cos kx t B0 sin kx t 0 Since E B 0, E and B are perpendicular to each other at all times. (b) The wave moves in the direction of the Poynting vector, which is given by S 1 0 E B. ˆi ˆj kˆ S EB 0 E0 sin kx t E0 cos kx t 0 0 0 B0 cos kx t B0 sin kx t 1 1 1 1 ˆi E B sin 2 kx t E B cos2 kx t ˆj 0 kˆ 0 E B ˆi 0 0 0 0 0 0 0 0 We see that the Poynting vector points in the negative x direction, and so the wave moves in the negative x direction, which is perpendicular to both E and B . (c) We find the magnitude of the electric field vector and the magnetic field vector. E E E0 sin kx t E0 cos kx t 2 E02 sin 2 kx t E02 cos2 kx t 1/ 2 E0 B B B0 cos kx t B0 sin kx t 2 B02 cos2 kx t B02 sin 2 kx t 1/ 2 2 1/ 2 2 1/ 2 B0 © 2009 Pearson Education, Inc., Upper Saddle River, NJ. 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. 305 Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual (d) At x = 0 and t = 0, E E0kˆ and B B0ˆj. See the figure. The x axis is coming out of the page toward the reader. As time increases, the component of the electric field in the z direction electric field begins to get smaller and the component in the negative y direction begins to get larger. At the same time, the component of the magnetic field in the y direction begins to get smaller, and the component in the z direction begins to get larger. The net effect is that both vectors rotate counterclockwise. © 2009 Pearson Education, Inc., Upper Saddle River, NJ. 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. 306