Homework 2 Solutions - Electromagnetism | PHYS 323, Assignments of Electromagnetism and Electromagnetic Fields Theory

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323 HW2 solutions

323 HW3 solutions Problem 9.18 (a) Equation 9.120 = r = e/a. Now € = eotp (Eq. 4.34), c & n? (Eq. 9.70), and for glass the index of refraction is typically around 1.5, so ¢ & (1.5)* x 8.85 x 10-!? = 2x 10-"' C?/Nm?, while o = 1/p = 107"? 0m (Table 7.1). Then + = (2 x 10-")/10-" = (But the resistivity of glass varies enormously from one type to another, so this answer could be off by a factor of 100 in either direction.) (b) For silver, p = 1.59 x 10~* (Table 7.1), and € ss €9, 80 we = 24 x 10! x 8.85 x 107"? = 0.56. Since 7 = 1/p = 6.25 x 10? > we, the skin depth (Eq. 9.128) is 1 2 2 « /=, = a —4 d ~YV wo Qe x 10!© x 6.25 x 107 x dar x 10-7 64x 107 m= 64 x 107 oom, Td plate silver to a depth of about there’s no point in making it any thicker, since the fields don’t penetrate much beyond this anyway. (c) For copper, Table 7.1 gives o = 1/(1.68 x 107°) = 6 x 10", weg = (2m x 10°) x (8.85 x 107%) = 6 x 1075. Since o > we, Eq. 9.126 > k ve so (Eq. 9.129) 2 2 (Gam) =? —= —— 4 =[04mm. | "Gone 'V On x10 x OX 10x ae x TOT Om iFrom Eq. 9.129, the propagation speed is v = 2 = 2a = Av = (4 x 107 4) x 10° = k c_ 3x 108 = 5 = 08 3x 10° m/s. | (But really, in a good conductor the skin depth is so small, compared to the wavelength, that the notions of “wavelength” and “propagation speed” lose their meaning.) Problem 9.19 (a) Use the binomial expansion for the square root in Eq. 9.126: ye ww fH fig (Z) i) =. f#ts- Rew #li+3 (2) 1 w 2 fie 2 2 fe So (Eq. 9.128) d= == =e. ged €= 9 =80.1leq (Table 4.2), For pure water, ¢ = wo(1-+ Xm) = po(1 ~ 9.0 * 10-*) & ig (Table 6.1), o =1/(2.5 x 10°) (Table 7.1). (80.1)(8.85 x 10-1?) = (2)(2.5 x 10°) exit =x 10¢m. (b) In this case (a /ew)? dominates, so (Eq. 9.126) & & x, and hence (Eqs. 9.128 and 9.129) Qn A= SD = and, ord = 5 18 =7 7 Meanwhile « 2 w4/E = jSoeeer) Yar x W")00) _ gag qe = t= ew & 8x10" 13x 10-* = (13mm. So the fields do not vpenctrate far into a metal—which is what accounts for their opacity. (c) Since k & «, as we found in (b), Eq. 9.134 says @ = tan7'(1) = 45°. qed 2E || For a typical metal, then, =? = Meanwhile, Eq. 9.137 says 2 = 10-73/m.| (In vacuum, the ratio is Ife = 1/(3 x 10°) = 3 x 10-®s/m, so the magnetic field is comparatively about 100 times larger in a metal.)