CollegePhysics StudentSolutionsManual Chapter24

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freespace( µ 0andε 0 )areenteredinto1theequationc .µ0ε 0SolutionWeknowthatµ 0 4π 10 7 T m / A,andfromSection19.1,weknowthatε 0 8.85 10 -12 F/m,sothattheequationbecomes:c 1(4π 10 7)T m / A (8.8542 10-12F/m) 2.998 10 8 m/s 3.00 10 8 m/sTheunitsworkasfollows:1[c] T F/A A T FC/s (N s/C m ) C 2 /J()J m N s2(N m )mN s2 m2 thefrequencyofa11.12- ‐m- ‐wavelengthchannel?SolutionUsingtheequationc fλ htandaregiventhewavelength;f 17.cλ 2.998 10 8 m/s 2.696 10 7 s 1 26.96 MHz11.2 eSunis1.50 1011 maway?182

pter24d, td 1.50 1011 msuntotheearth,wecancalculatethetime:t 500 sc 3.00 108 m/sWeknowthatv (a)Whatisthefrequencyofthe193- velengthoflight?Solution(a)Usingtheequationc ndthewavelengthoftheradiation:f cλ 3.00 10 8 m/s 1.55 1015 s 1 1.55 1015 Hz-71.93 10 hat:λvisible 380 nm 1.97.λUV193 eakmagneticfieldstrengthof4.00 10 9 T .2UsingtheequationI ave (2I avecB 0weseethat:2µ 0)(cB3.00 10 8 m/s 4.00 10 9 T 0 2 µ02 4π 10 7 T m/A()2) 1.91 10 3 W/m 2Theunitsworkasfollows:[I ] (m/s)T2T m/A T A (N/A m)(A ) NJ/mW 2sss.m s m m183

vewithamaximumelectricfieldstrengthof1.00 1011 V / c)Whatenergydoesitdeliverona1.00 - mm2area?E c ionB0 E0 1.00 1011 N/C 333 T,recallingthat1V/m 1N/C.c3.00 108 m/s(b)UsingtheequationI ave resultfrompart(a):I cε 0 E02,wecancalculatetheintensitywithoutusingthe2cε 0 E 022(3.00 108)(m/s 8.85 10 122)(11C /N m 1.00 10 N/C22) 1.33 1019 W/m ensity:P IA,andfromtheequationE Pt heenergydeliveredtoa1.00 mm 2areaperpulse:E PΔt IAΔt2 1 m 9 1.328 10 W/m 1.00 mm 1.00 10 s 1000 mm 4 1.33 10 J 13.3 itanceisneededinserieswithan800 - 196m?Usingtheequationf 0 12π LC,wecanfindthecapacitanceintermsofthe184

ncy:C ation4π Lf 0222(λ2196 m ) 1.35 10 11 F 13.5 pFc fλgives:C 2 2 24π Lc 4π (8.00 10 4 H )(3.00 10 8 m/s)Theunitsworkasfollows:[C ] 44.m22H(m/s) s2s2ssA s C FH Ω s Ω ma5.00- ‐mWlaserisconcentratedona1.00 - mm2area.(a)WhatistheintensityinW/m2 ?(b)Supposea2.00- 00m/s,whatmaximummagneticforcecanitfeel? Solution P 5.00 10 3 WP(a)Fromtheequation I ,weknow:I 5.00 10 3 W/m 2 62AA 1.00 10 m2(b)UsingtheequationI ave cε0 E0,wecansolveforthemaximumelectricfield:22I2(5.00 10 3 W/m 2 )E0 1.94 10 3 N/C.8 1222cε 0(3.00 10 m/s)(8.85 10 C /N m )So,usingtheequationE (Fwecancalculatetheforceona2.00nCcharges:q)()F qE0 2.00 10 9 C 1.94 10 3 N/C 3.88 10 6 NE c nsF qvB sin θ andFB ,max()()qvE02.00 10 9 C (400 m/s) 1.94 10 3 N/C qvB0 5.18 10 12 N8c3.00 10 tudestrongerthanthemagneticforce.185

easonableResultsAnLCcircuitcontaininga1.00- atesata300- uationsf 0 f L cλ 12π LCλ224π Cc12π LCandc fλ ,wecansolvefortheinductance:, sothat 722(3.00 10 m) 4π (1.00 10 F)(3.00 10 m/s)2 lengthistoosmall.Chapter241862 2.53 10 20 H

College"Physics" Student"Solutions"Manual" Chapter"24" 183" " Solution" We"know"that , t d v and"since"we"know"the"speed"of"light and"the"distance"from"the" sun"to"the"earth,"we"can"calculate"the"time:" 500 s 3.00 108 m/s 1.50 1011 m c d t " 23." (a)&What is&the&frequency&of&the&193;nm ultraviolet radiation&used&inlase re ye&

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