工程电磁场与电磁波答案丁君

工程电磁场与电磁波习题解答（试用本）主编丁君第一章11解（1）ZYXAAABAVVVVV12593323ZYXAAAVVV712BAVV3194491441（2）266COS26216COSCOSCOSQQQQCBCBCBCBCVVVVVVVVVBR方向的单位矢量为26BBBBVVVVCR在BS方向的分矢量为33COS341313XYZCBBAAAQVVVVVV（3）191321COSCOSCOSQQQBABABABAVVVVVVVV113ARCCOS219Q（4）ZYXAAACBVVVVV553CBVV的单位矢量为ZYXZYXAAACBAAAVVVVVVVV59559559355312证明欲证明三矢量A、B、C能构成一个三角形，则须证出三个线性无关的非零矢量位于同一平面上。则有0CBA即0ZYXZYXZYXCCCBBBAAA代入得0000431110624431213ZYXZYXZYXCCCBBBAAA即得证13解（1）4321FFFFF合代入数据得XYF2AAVV合（2）514合F（3）合力方向与单位矢量YXAAVV5152方向相同，与X轴成1ARCSIN514证明设矢量RR的终点在ABC构成的平面上，则,RARBRCVVVVVV在此平面上，则必有不为0的实数,MNP满足0MRANRBPRCVVVVVV所以得PNMCPBNAMRVVVV，PNM,为实数15解设A点的坐标为,11YX，B点坐标为,22YX则AV,11YX，BV,22YX有题意得121211XXYYXXYY则过A,11YX，B,22YX点的方程为111212YXXXXYYY16解欲使,ABVV互相垂直，则有0ABVV则有8220ABAVV得3A17解矢量AV与坐标轴的夹角分别为72COS76COS7343693COSAAAAAAZYXGBA其中GBA,分别为矢量AV与三个坐标轴方向夹角。181证明11112222ABBCCACABAVVVVVVVVVV1111110222222ABBCCABCCAABRRVVVVVVVVVV所以四个面的面积之和为0（2）可以推广到任何闭曲面19解（1）2739XYZABAAAVVVVV（2）273942111XYZXYZABCAAAAAAVVVVVVVVV（3）111ABCABCVVVVVV（4）ABVV是垂直于AV、BV所在平面的矢量，有2739XYZABAAAVVVVV于是垂直于AV和BV所在平面的单位矢量为913919191NXYZABAAAAABVVVVM110证明2CBABAACCBABACBAACCB利用公式BACCABCBA可得CBAACBACBAACBAAACBBACACBABAACCB222则ACBAACBACBCBAACBACB同理可证CBAACB，CBABAC111解（1）225XYZSYAXAAVVVV（2）113,2,1T，425XYZSAAAVVVV，164S2535V，42533535SXYZSAAAASVVVVVV112解电场线的切线方向为电场强度方向，则22D21DX42DY4ELXYXXYYCFVV即电场线方程为CYYXX422113解（1）3331111222222XYZRXXARYYARZZARVVV333XYZRXXARYYARZZAVVV3331111222222XYZRXXARYYARZZARVV333XYZRXXARYYARZZAVVV则有11RR（2）RRFRF，RRFRF又RR所以有RFRF114解设XZYX22F则2222XYZXYZAXAXAFVVV2,1,244YZAAF115解两个曲面的夹角实为它们的梯度的夹角的较小的一个1222XYZXAYAZAFVVV424XYZAAAVVV222XYZXAYAAFVVV42XYZAAAVVV1212COSFFFFQ16448COS621321Q08ARCCOS,090321QQ部分共同作用，使点（0，0，1）处BV方向为YAV5500225557051DD2121Y549102SSYYSYYJJBYAYAYJARCTGYAATMMMVVVVV216解如图（B）所示，在闭合回路ABCD上应用安培环路定律DXXLHLHLHLJLVV2SXJH磁场强度在导体板的上面是X的负方向XAV，在导体板下面是X的JAYXANANABCDB正方向XAV，NAV如图（B）所示，故12NHJAVVV217解设中间的无穷大平面在XOY面上，最上面的导体面在空间所产生的电场强度为61106120110,1110,1ZZEAZMMEAZMMEE时661000013111010244ZZEAAEEEEVVV01ZMM同理有222220004SINDDDBRKDRARRRQQJVV22RKDBARVV220解利用电场高斯定律0DDSSSESSERVV其中1230000DDDDSSSSESESESESEEEEVVVVVVVV其中1SV为高斯圆柱面的侧面，2SV和3SV分别为高斯圆柱面的上底面和下底面。根据题意，长直圆筒的轴向电场0ZE及周向电场0EJ，只存在径向电场RE，且径向电场和上下底面平行，于是100DDSSESESEEVVVV式中DDDRRRSRZAEEAJVVVV所以1200000DDD2LRRRRSESERZAARLEEEJEVVVV而200DDD2LSSSSSAZALRRJR所以0932211101001051036177/SRREACMREM221解圆筒内0RA022022201122220000222000DD22DDLN0442DLN22VVVRARAVVVVVRESVRELALAEARERERRAAACRAAARRCRARERERREFRRREEERRFEEVVVVVVVV222解圆弧上电流产生的场为1BV1000122D444RZZLIIILARBAARRRMMMVVVVVAB上电流产生的场为2BV120,2AA100AABCD10CM2M5CMXY02120SINSIN44ZZIBARIARMAAMVVVBC上产生的场为3BVABR10334120221201SINSIN45102445105104ZZZIBARIARIARMAAMMVVVVCD上产生的场为4BV044ZIBARMVV1234201202273125104125104510510210100201274000810310ZZBBBBBIARRRIATMMVVVVVVV223解在不完整的圆环上1001211D424RZLIILABAAAMMVVVV在一根长引线上12SIN125,212122RAAA02120SINSIN451SIN124SIN12ZZIBARIAAMAAMVVV同理另一根长引线上32BBVV所以1230051SIN1112212SIN12479ZZBBBBIAAAAMMVVVVVV1479ZHAAVV224解10RCM2100DDDRVLHLJRRJVVV20510005050512200DD80022224001RRRRRRHERRREEHEARRJJPJVV10RCM012200DDDVLHLJRRJVVV0050052005005228000122800212221400RHEEEEHARJJVV225解根据题意，取柱坐标系。设内导体的电流为I，由于电流分布是均匀的且具有轴对称性，它所产生的磁场也应该是轴对称的，即BV的大小只与半径R有关，与J无关。RA区域在该区域内均匀分布着电流密度为21JIA的电流，如取半径为R的圆环为积分回路，根据安培环路定律21100DDDRLHLJRRJVV得到2122IRHRAJ因而122IRHAJ则122IRHAAJVVARB区域同理取半径为R的圆为积分回路，则有220DHRIJJ所以22IHARJVVBRC区域在该区域中均匀分布着电流密度为222IJCB的反向电流。同样，取半径为R的积分回路，于是有223200DDDRBHRIJRRJJJ由此可得223222ICRHACBRJVVRC区域在半径为R的积分回路中电流的代数和为零，则40HV226解DJTDBVVV00MM080081102SIN21310042SIN631021DYYJBTXATXAMMMVVVV227解（1）由EDHBEVVVVV利用EV相等求A。TBEVV81100SIN1000RZRARAAAETAZARRZREJJJVVVVV888100100SIN10DCOS1010ABATAZTATAZARRJJVVV又TDHVV286000111SIN101000RZRZARAAAHBTAZARRRRBJJMMJMVVVVVVRAAZTARTDVV862010SIN101M又RAAZTREVV810COS100880010010SIN10RDETAZATTREEVVV比较式和式，得268001010010AEM8001103AME（2）JJAZTRAAZTRABVMVV3110COS31010COS101008688（3）RDAAZTRTDJVVV86010SIN10911M2188001DD167COS10COS103CDRIJRZATTPJVVM228解（1）VDRV而2422324505901901RRRRRRREV又20450VRREAR时DDVSVDSVRVV2222000004450SINDDDARERARRREEQJQVV254360RERAAVV5290REAARVVRA（3）AAR0EV0EV第三章31解因为导体表面的电场处处与导体表面垂直，切向电场为0；所以400290310/0NXYZTEEAAAVMEVVVVVV由边界条件得SNNDDR21222992014002903101051610/36SECMREV0TEV32解560COS10/CXJETAVMSSVVV0550556010SIN1079610SIN10/DRRXXDEJTTTATAVMEEEEVVVVV51096760S613310/SMS33解5,0,51SRRMMZCMJNHAHJRRVV5,0,50ZRMMZMHAJV2COS2R85,0,50COS410RMMZMZTJZAV38042COS4PT810ZAV（2）1100RZZRARAARHHHARARRZRRZRHJJJJJVVVVVVVV4RSIN2ZCOS4P810RTAVTDHVVQ4DSIN2DHTRVVP4COSZP810RTAVDT881SIN2SIN41010RZTARV（此处不考虑静态场，故省去常数项）80502SIN2SIN410RRDEZTAREEVVV123SNNDDRQ因此理想导体内0D818838781SIN2SIN410101SIN90SIN410201010510SIN410SNDDZTRTTRO8001,02,0258844SIN2COS410400COS410400COS410DRRDRDJZTATRTAJTARRRRRR34解299/109SIN106MAATEJYCVVSTDJDVV6920025610SIN910RYDETAC/MEEEVVV则699069225610910COS91011910COS910/DYDJTTTAAMEVVV35解CJESVV，TDJDVV1OCEJ71075V936111010103636DOOOOJEEEEWPPV131557361064410CDJJVV2OCEJ41002V93110108036DOJEV22108321080362PDCJJVV3OCEJ1610V931101025336DOJEV109361044710253CDJJVV36解（1）已知DDJTVV则D2COS5DSIN5DXXZDJTTZATTZAWWWVV0091SIN5/21810SIN5XRXDETZAVMTZAWEEEWWWVVVV（2）BETVVBHRVVMM0105COS5200XYZXYYXAAAEEATZAXYZZEWWEVVVVVV所以20055COS5DSIN522YYBTZATTZAWWWEWEVVV20001SIN52YRBHTZAWMMWEMVVV（3）2005COS52YDXXHJHATZAZWWMEVVVV620011521023510/2RADSWMEW37解设外导体为1，内导体为3，其间介质为2，有22/20MNCDSNR012TTEE由DDSSSDSSRVV得228025RDAMMRMMR时02MF由HMVF可知MF只与J相关01222JFMR21CCMJF0J时0MF故02CPJ2时IMFP21ICJPF2IM02AVAV只有ZA方向分量，只和R相关01RARRRZ21LNCRCAZZACRCALN21由A得连续性，从21RRR时RIRAABZPM20VVJPARIHVV2ILDHVVIRHP2JPARIHVV2由此可见。两种方法求得的HV、BV相同46解XYZABQQCDRBCRACR0,02,0203,02,0其中080402020222ACRZYACAAR2222，40402020222BCR，ZYBCAAR1010310106102797913ZYAADV47解1）23222DRDQSPR6232226103644040302401018P2）201021EEDDDVVVVVEEBCBCACACRRQRRQ442020PEPE1利用镜像法QQ0X的空间中222222221211211211214YXYXYXYXQOPEF在导体表面只存在法向场,所以XEFF23222322232223221221221221224YXXYXXYXXYXXQOPEQXOSEER23222322232223221221221221224YXXYXXYXXYXXQSPR230,0,052PRQSQP5522SHPR的0,02322320,01414HQHQHSPPR48解NMGG33106198910220216HQFPE要使得FG则MGHQ20216PECQ832291095106191021036116PP49解利用镜像法可知IIIOOMMMM“图一图二如图一所示上半空间中122IIHAARRJJPPV如图二所示下半空间22IIHARJPV故在上半空间中10101TNBHHMMVVVNOTARIRIARIRI2COSCOS22SINSIN20000000PJMMMMMJPMPJMMMMMJPMNOTARRIARRICOSCOS2SINSIN20000000JMMMMMJPMJMMMMMJPM在直角坐标系下找出X,Y,Z与,JJRR的关系，图一图二222YHXR，222YHXRRHXJSINSINRHXJ,NXTYAAAA即为即为RYJCOS，RYJCOS下半空间中NTBHB222VVVMNTARIARIICOS12SIN20000JMMMMPMJPMMMMMNTARIARICOSSIN000JPMMMMMMPJ222YHXRRHXJSINRYJCOS410解1）0Z时QQ2121EEEEQRREEEEEE0000QRREE11201044RQRQPEPEF2222220111141HZYXHZYXRREEPEXRRAHZYXXHZYXXE114123222232220EEPEFV232222322212HZYXHZAHZYXYYREZRRAHZYXHZ1123222EE2）0K时SINCOSJJKBKANNF由00F得0NA0FA得2APPANKNK3,2,1NSINJAPJNBNF02222RNRRRRRRAP为欧拉方程解为APAPNNNNRDRCR由00FJFR00R所以0NDJAPFAPNRCNNNSIN1UNACNNNJAPAPSIN1两边同乘以JAPMSIN从0到J积分，利用三角00SINCOS2MNMNMNAPPJAAA112NNNUCPN121211SINNNNUNRNPAPFJPA419已知0YBUF，其余五个面电位皆为零，求腔内电位分布。解电位满足三维拉普拉斯方程及边界条件2222220XYZFFF00,0,00,0,0,00,0,0,00,0,00,0,0,0,0,0XAYBZCXYBZCZXAYBZCXAYBYXAZCUYBXAZCFFFFFF由边界条件可知F在X方向，Z方向至少有两个零点，又因为对XXYYZZF分离变量后得2222221110DXDYDZXDXYDYZDZ所以SIN,SINMMNNXXAKXZZCKZ而对应的MNMNYYBSHKY由0,1,2,3,MMXAKMAP由0,1,2,3,NNZCKNCP又222222211,SINSINMNMNMNMNNMKKKMNKACMNMNXYZDXZSHYACACPPPPPPFQ022011SINSINYBMNNMUMNMNUDXZSHBACACFPPPPQ设22MNMNMNEDSHBACPP则011SINSINMNNMMNUEXZACPP方程两边同时乘以SINSINSTXZACPP并对,XZ取如下积分0000011SINSINSINSINSINSINCACAMNNMSTUXZDXDZACMNSTEXZXZDXDZACACPPPPPP由三角函数正交性可得右边2200SINSIN4CASTSTSTACEXZDXDZEACPP左边024,1,3,5,1,3,5,ACUSTSTP所以0216,1,3,5,1,3,5,STUESTSTP所以1,SINMMMMMMXXBCHKXBCHXBMMXYCYCHXBBPPPF420YEXA解选柱坐标系，空腔内02F且F和Z无关011222JFFRRRRR令JFFRR带入方式01222FFJRRRRRRR令KRRRRRRFF221J0222KRDRDRRRDRDR022FFKDDJ因为2PJJFF且F关于X轴对称，2NK所以PJNANCOSFN0123NNNNRDRCRR0COSNNNNNINNRDRCJFN0120R时，0INF所以0ND1COSNNNINNRCJF同理柱外电位0COSNNNNNOUTNRFREJFN0，1，2由RJJCOS0REOUT得11COSCOSNNNOUTNRFREJJF再利用AR时边界条件OUTINJJ及RROUTINJEJE0得JFJNAFAENACNNNNNNCOSCOSCOS101JEJEJENANFENANCNNNNNNCOSCOSCOS110110比较1N时0012ECEEE20001AEFEEEE1N时0NC0NF故JJEEEJEEEFAREAEERINSIN2COS200001V421解设22,XYUWZUXYJVXY则,12UVXYVXYCUVYX等位线方程可写为12XYC复电位可写为22222,WZXYJXYKK为常数422解1选对数函数LNWZKZC采用柱坐标，令JZREJ，则12LNUKRCVKCJ选2VKCJ为电位函数，则由边界条件可得127010FJ21270EAARRJJFFJUV3981011270101121036DEARARJJEPUVUV4无切向场811801019000001112101121051610SNSSDRQDVDRDZCRRR习题五51一圆柱形铝管，外半径为32MM，管壁厚6MM，求单位长度的电阻。解铝管的内半径为26MM，设流过铝管的电流为I，则6622222110348102632PPPIIRRIJ6/10348IEJSPS133495103486103486SPSPSIIIDER（W）52一长为LM的导体，电阻率为S,导体各处的横截面相似，一端的面积为AM2,另一端面的面积为KAM2，求两端面间的电阻。52解（此题应为导体各处的横截面相似且呈线性关系）。设流入导体的电流为I，则设任一截面面积为ZS，由KALSAS0得ALKAAB1则AAZLKZS1SEZSIJSZSIEKAKILAZALKDZIDZEULN111010SSAKKLIURS1LN53解BLUA本题所求电感为跨接在内外导体间的RAREEVRARLIJ2PVEJVVSRALRIE2SPVABLIDRLRILDEUBABALN22PSSPVVABUIGLN2PS54解R2R1球冠面积PQQPJQQ20220COS12SINRDDRSRAREEVRRARIASIJCOS122QPVSSJEVV21COS122RRDRRILDEUSQPVV2112COS12RRRRIURQPS55解设电容器板内的D为0D，则1RE2RE3RE方法一（1）NNDD10SNNSDDSDQ11101111RNNDDDEUEEFDSUQCR93911011096710131036130PEE（2）同理FC9210315（3）同理FC9310376FCCCC9321101221111方法二由介质边界条件NNNNDDDD0321121320321DDDDDNNNDZEDZEDZELDEUVVDZDDZDDDDRNDRN1212102001EEEEDZDDDDDRN3122303EE3213210RRRONDDDDEEEESDDSDQN01UQC56解设内导体单位长度带电量为Q，EV、DV只有R方向分量，电荷将均匀分布在导体表面上，QSDDSDD2211VVVVQRERE121102QEQPE在介质与空气的分界面上TTEE21且没有J方向分量，即21EEERQE12110EQQPELN2110ABQDREUBAEQQPELN/2110ABUQCEQQPE57MNACHHCR1R2LRLRP设电轴的位置偏离轴心CMMA8568MMH538M点N点的电位相等120LN2RRLPERFCACAHLM2LN20PERFCACAHLN2LN20PERF由此可得出CACAHCACAH22所以C满足0222AHCC可求出00030JPMARIB210V对于载流环所在平面JADZDRADZDXSDYV1212122112IIMMYYDYDZYISDBS20311012122PMYVV3LN10PMI3LN012121PMYIM510R1R2R3D1ND2N300052105020220434PEEPEPQDRRQDRRQU020PEUQC511ABVR21LNCRCRAEFAR时0UFBR时0F得ABBAAUCLN01EBABBAAUBACLNLN02EE所以BABBAAUBARABBAAURALNLNLNLN00EEEEF解RARDDVDDDNN213221003RRRRDRDDRDLDEUEEVV24RDSDDQPVV24RQDP解1）EERFRA2因为F仅为R的函数所以0JF0ZFEFRARRRR12）RARABBAAUAE1LN0EEFV1）内ABABAAUAEDARNSLN011EER外ABBBAAUAEDBRNSLN012EER4）000010LN22UABABAAUAAUAUQCSEPPR512由6261105104100PSPSEEIPSPS2185410E所以21121111154400SSSPSRESJI21222254500SSSSJI513解解ZAREEVEEETT21EJ11SEJ22S21121111111RREEDSJDLEIURPSPS2122221222221RRRREEIURPSPS212222112211RRERESJSJIPSPS（A）因为/MRKS，可见电场只与R有关。RAREEV设从内球壳流进外球壳的电流为I，则RRCAKRMRIARIJE4422PSPSVVLN141420BKMAAKMBMIRDKRMRIRDEUBABAPPVVBKMAAKMBMUILN40PLN410BKMAAKMBMIURP（B）LN44202BKMAAKMBKAMAMUKAMAIEDARNARNSEPPEER内LN20BKMAAKMBKBMBMUDBRNSER外CRAKRMRBKMAAKMBMUDVV201LNE201LN1RKMBKMAAKMBMKURRDRDVERVDLN4402BKMAAKMBAKMAMUAQSPEPR内内LN4402BKMAAKMBBKMBMUBQSPEPR外外ERRCABKMAAKMBRMUARIJLN4202PVFLN400BKMAAKMBMURUIP514解ABINDRRHINDSRINDSRINDSBBASSSLN2121220000PMPMPMPMYABHNINLLN202PMY515解一般情况下认为AL1，AL2ZANIB1101M2211021ANNIPMY221012121ANNIMPMY516解若AB为一对传输线，CD为一对传输线，A对CDJPMARIBA202V1LN220022LDDIDRLRISDBACADDDSAAADACPMPMYVV同理BDBCBDDILN202PMYIMBA2221YY517传输线的内自感/48200MHLIPMPMJPMARIB20DX处的磁感应强度2200ZZAXDIAXIBPMPMVAADIADXDYBADAZLN0PMYVAADILLN00PMYAADLLLILN4000PMPM518解直导线在螺线管内产生的磁场为20TIRBRPMMABHTIRHDRTIRDSTIRRBARRLN222000PMMPMMPMMY10751015LN10200036102500LN29290HABNHTINMRPPPMMY161解EV矢量为Y方向，电磁波沿Z方向传播，2106COS73728222ZTZEYPPP2106COS73710682822ZTTEYPPP又P2KQ,MEW22K,8106PW22222222282221062TETEKTEZEYYYYMEWPP2106COS7378ZTEYPP符合均匀平面波的一维波动方程，所以它属于均匀平面波。62解10328HZFPWP2K；MK12PL；SMKVP/1038W；MUEH10/773/EHVV波沿Z轴传播；由右手螺旋法则，H在X方向上振动。63解1HZVF8810924610103L2910032/1FTS331061022PLPK4122377/800/HEHVVA/M方向为YA64解由EUR和HUUR的关系可知YMXMAAZTHAAZTHHSINSINWWV2YMXMAAZTEAAZTESIN/SIN/00WHWHHESVVVZMMZMMAAZTEAZTEAAZTEAZTESIN/SINSIN/SIN00WHWWHWZMAAZTEV022/SIN2HW65解HZUF980001052120103L5001050111HEH又RRRRUUUEPEEH120001QPE120500RRUQ（1）在均匀媒质中有11VVPRRUCFEL129811081052103LEFCURR（2）由式（1）、（2）得991RU131RE66解MVAAAEZYX31024V1）333310782433318610210104ZYXZYXAAAAAAHEVV322231078243310782433ZYXAAAAZYXAAA8902703702）34210JKRXYZEAAAEV6183JKRXYZHAAAEV311RE3324781022AVXYZSEHAAAVVV33）HEURRVVQEPH120152RE67解1）不失一般性，可假设两圆极化波左旋101YXJKZAJAEEEV右旋202YXJKZAJAEEEV合成波21EEEVVVYJKZXJKZAEJEEAEEE20102010YJKZJXJKZAEEEEAEEE220102010PYXEEVVYXEEVV2PJJYXXEV与YEV振幅不等，相位相差2P为一个椭圆极化波故椭圆极化波可分解为一个左旋圆极化波和一个右旋圆极化波。2）不失一般性，设两个旋向相反的椭圆极化波JKZYMXMEAJEAEE221VJKZYMXMEAJEAEE22V故1EV为左旋圆极化2EV为右旋圆极化21EEVVJKZYMXMEAJEAEYXEEVVYXEEVV，2PJJYX，故合成波为一右旋圆极化波68解（1）线性极化；（2）916,244PPPPFFXY所以为左旋椭圆极化波。（正椭圆）。（3）ZXAYTAYTE50SIN450COS3VZXAYTAYT250COS450COS3P4220PPJD,43所以是右旋椭圆极化波。（正椭圆）。69解设一圆极化波00COSSINXYEETKZAETKZAWWV00COSSINYXEEHTKZATKZAWWHHV000020COSSINCOSSINXYYXZSEHEEETKZAETKZATKZATKZAEAWWWWHHHVVV所以SV是与距离、时间无关的常数。610解因为两个波传播方向相同，频率相同，电场方向垂直，有2P的相位差，且振幅不同，所以合成应为椭圆极化波。或这样证明COS3TEXWSIN22COS2TTEYWPW13322YXEE可见为一正椭圆又因为Y方向超前X方向，所以为左旋椭圆极化波。611解YXAZTEAZTEECOS4COS300BWBWVXXAZTEAZTECOS21COS2700BWBWYYAZTEAZTECOS21COS2700BWBW2121EEEE1E与1E构成一振幅为7/2的右旋圆极化波1E与2E构成一振幅为1/2的右旋圆极化波5612解由1001800WES，故海水为高损耗媒质。（1）522410410502276PPPWMSBA52BPLM66102551050LFVM/S61025BWPVM/S（2）已知4410PPPSWMHJJEEYXAXTEHVO355210COS100526526PPP610042MMHEH105210COS100426526ZXAXTEEOVPPPX1带入HV,EV表达式可求出该处的场（3）吸收功率10SSXAVXAVPVV吸XXAVAEHES481RE2154PVVVXAP1047112吸613解1）因为MS/10857S61022PPWF100EWS此时黄铜认为是良导体6WSEJ44441062621073JEEJJPPSWMEMH1212HHHHG4422371037722122371037722JJM510662MSWD2122HHHT44222371022022371037722JJ2）10036008010361102496PPEWS错误，应等于900故海水可视为良导体WSEJ447644114104102PPPPPSWMHJJJEEE1212HHHHG1439947037722224113772222411JEJJM4502MSWD（应为025M）2122HHHTJEJJ79000750377222241122224112614解WESWES海水为良导体M811524104102222274PPPWMSPBPL472104104040/22DBMWMSPPA1047441404104102PPPPSWMHJJEE2）当310100F时100WES海水为良导体同理M5LMDB/261A4440PHJE3）当6101F时100WES海水为良导体M581LMDB/973A441PHJE4）当61010F时90WES海水为一般媒质2211125021112642MESAWWEMESBWWE2050MPLBJ0781J3158931240I444E2MSHEWE5）当610100F时9WES海水为一般媒质同理3760/DBMA0042BM150B馈入天线的平均功率。解A按题意2PQ，利用镜像法，可按半波天线计算上半球区辐射场10160060ISINCOS2COSR2IE00QQPPHQ67A2I0B4L单极子天线辐射电阻为半波振子的一半，即W536RR28130WRI21PR20AV若利用坡印廷矢量，在PQ0半球内积分，所得结果一样。例18图为一半波天线，平行于地面，在离地面4HL处架设，求E平面的方向图。解远区电场为2COS2EE1YQJQPYCOSSIN2KH1E为JKR01ESINCOS2COSR60IJEQQP于是方向性函数为JQQQPQCOSKHSINSINSINCOS2COS,FD因系镜像问题，有效区为PQ0，E平面方向图如图所示。4习题如果振荡频率分别为50HZ和50MHZ，问在电偶极子一个波长远处，辐射场与感应场的振幅之比应为多大（4839）在自由空间中电流元，令MV24LI20PLH，L为振子长度，求；）1LR；）5LR29时，JQHE的值。（WO23J0367E，O0018J0376E）一中心馈电短振子天线位于自由空间，4ML，100ML，40AIO电流振幅沿天线作线性变化（图），求远区点OO55,75,5KMP处的电场QE及总的平均辐射功率。（MV1MJ29，253W）发射天线长度为20M，工作频率为1MHZF，该天线可视为电偶极子天线，天线电流振幅为A52IMO，求辐射功率AVP和辐射电阻RR。（W1001112，W513）一天线位于原点，周围媒质为空气，已知远区场MVESINR100ER2J2LPQQ，求辐射功率AVP。（9W88）6设半波振子电流振幅为1A，求距天线轴1609M处的QE）0Q；）2PQ。（；MMV29EJ371011JL）