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时间:2018-07-19
《电磁场与电磁波基础教程习题解-》由会员上传分享,免费在线阅读,更多相关内容在教育资源-天天文库。
1、电磁场与电磁波基础教程习题解-1《电磁场与电磁波基础教程》(符果行编著)习题解答第1章1.1解:(1)A?C?5???2??22Ax?Ay?Az222?1?2???3??2214,B???4?2?1?17,29;?114(2)aA?AA?ax?ay2?az3?,aB?117?-a4?az?,aC?y129?ax5?az2?;(3)A+B?ax?ay?2?4??az??3?1??ax?ay2?az2?A?B?1?(?2)2?(?2)2=3;(4)A?B??ax?ay2?az3????ay4?az???1
2、1;(5)A?B??ax?ay2?az3????ay4?az???ax10?ay?az4;(6)?A?B??C???ax10?ay?az4???ax5?az2???42;(7)?A?B??C???ax10?ay?az4???ax5?az2??ax2?ay40?a25。301.2解:cos??AA?BAB?121078,??68.56?;121078121078?1.37;?3.21在B上的投影AB?Acos??14?在A上的投影BA?Bcos??67?B。1.3解:A?B??4???2???6??4?
3、???2??8??0?正交?。1.4解:ax?ax?1,ay?ay?1,az?az?1;ax?ay?ay?az?az?ay?0;ax?ax?ay?ay?az?az?0;ax?ay?az;ay?az?ax,az?ax?ay。1.5解:(1)a??a??1,a??a??1,az?az?1;a??a??0,a??az?0,az?a??0;a??a??0,a??a??0,az?az?0;a??a??az,a??az?a?,az?a??a?。1(2)ar?ar?1,a??a??1,a??a??1;ar?a??0
4、,a??a??0,a??ar?0;ar?a??0,a??a??0,a??ar?0;ar?a??a?,a??a??ar,a??ar?a?。1.6解:???ax???x???ay???y?az???z?axy?az?2xy?z22??az3yz2在点(2,-1,1)处30-1,1??2,?ax?ay3?az3;???l????al????AA??ax?ay3?az3??13?ax2?ay2?az???213。。1.7解:???ax1.8解:??r?????x?ay??y???y?az????z?axy?a
5、yz?az,??221??1,,?ax4?ay?az?x?x???y???z?z??1?1?1?3。1?1.9解:对r?a???azz取散度,??r???r?1?r?r2??????????z?z?3,对r?arr取散度,?r2看出对同一位置矢量r?r??3,取散度不论选取什么坐标系都应得同一值,坐标系的选取只是表示形式不同而已。1.10解:??B?1??c?-1??c?1??c??0,??B=a?a?????????0,由亥姆霍兹?z????????z?????????定理判定这是载流源在无源区(G
6、?J?0)产生的无散场。1.11解:??E?a???c?1??c?1??a?0,由亥姆霍兹定理判定?????0,??E?z?z????????????这是电荷源在无源区?g?q?0?产生的无旋场;将??E?0与恒等式??(?u)?0对比,可知E与±?u等效,令标量位??u得E????。301.12解:F满足无旋场的条件为??F?0,在直角坐标系中表示为axayaz??x??y??z?03y?azbx?2z??cy?z?解得a=0,b=3和c=2。21.13解:??F???x?x22?y2????y?2
7、xy??0,??F??ax??z?2xy??ay??z?x?y2???22??a2xy?x?y???z?????az4y?x?y??由亥姆霍兹定理判定知,这是属于第三类的无散有旋场。1.14解:取F?arCr2:??F?1??2C?11??c?1??c??r?2??0,??F=a??2??a??2??02r?r?r?rsin????r?r???r?,属于第一类的无散无旋场,由无旋性可以引入标量位的梯度来表示;取F?ar:??F?rc1??2c?c11??c?1??c?r??,??F?a?a?r????
8、???022r?r?r?rrsin????r?r???r?,属于第二类的有散无旋场,由无旋性可以引入标量位的梯度来表示;30取F?a?cr:??F?1??c????0,rsin????r?1??F?ar??c1??c?1??c?c??sin???a??r???a??r???ar2cot?rsin????rr?r?r?r?r?r??r,属于第三类的无散有旋场。3第2章2.1解:q3受到q1和q2的作用力应当等值反向,所以q3应位于q1和q2的连线上某点处。
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