第三章 习 题
1.
B???A
?zI?0x在直角坐标中，Bz??Ay?x??Ax ?y
在柱坐标中，B和H只有eθ分量。在分界面防
因此，依题意设：A?xB0ey，或A??yB0ex ?结果都为B?B0ez?B0
???xB0ey???yB0ex?????xB0ey????yB0ex??B
0?B0?03.
z?I0?y
依?H?dl?I，在柱坐标中得
H?H??I2?r 或H?I2?reθ
因此
???0I2?reθ,z?0B??
??I?2?reθ,z?0?0，z?0M???mH??z?0
?????0?1?I2?reθ,JM???M??m??H???0,z?0 ????0?1?I??r?ez,z?0或
JM???M??m??HI???0,其余
M????0?1?I,z?0,r?04.
??2(x?0)处，B的法向(eθ方向)连续。因此，
设
B?BBI1?2?Creθ，得
?x?0H???CIeθ,??0r
???CI?reθ,x?0依?H?dl?I，得
???CI?CI?????r0r?r??I，因此 ?C???01???
0?B???0I????reθ
0J??M??????B?H?M1?1?0??0 ???J??M?B?M2?2?????????H20??????0?1???H2??????????01I?
0?1??????0??reθ???????0??01I??????r?ez0?0??JM?JM1?JM2????0???I??r?ez，或 I???0M????I，沿z轴 ??B?1?????B???1?B??????Bz??z 依题意，B?为?、z的函数，因此
1??????B???2Cz?0 ?B??C?2z?c?z?
B??C?z?c?z??
依轴对称性有，B?与?无关
??H?11????B?????1?Bz??B???e???????z?????
??B???Bz????z??????eθ???1???????B???1?B????????e?z??依??H?0得
?B??z??Bz???0 C??1dc?z??dz?C??0，因此， c?z??c0为常数
绕z轴作一小柱面，柱中心在z?0处。设柱半径为??，高为?z。柱面的磁通量为
?B?dS????C??z?c?z??????????2????z?
B?C?z??z?2?????2?B?Cz220?0????????0时
??B?lim?B?dS????0,?z??0?????2?z
?lim2c0????0,?z??0????2
若??B?0，则c?z??c0?0，因此
B??C?z 7.
z AA1A12A2
?2??1??r?r??r????r???1?2?r2??2??2??z2 ?2A???0J。依题意A?A?r?ez。 在柱内
1d?rdr??rdA1?dr?????dA1120J，rdr??2?0Jr?C1dA1dr??12??C10Jrr A?11?4?0Jr2?C1ln?r??C2
B1???A?A1C?1???re?????1θ?2?0Jr?1r??eθ r?0 处无线电流，A值有限
因此，C?0，A??1114?20Jr?C2
在柱外
1d?dA2?dA2rdr??rdr???0，rdr?C3 A2?C3ln?r??C4
BA22???A2???C?reθ??3reθ r?a分界面处，H的切向连续。因此
1Ja??C3， C122?a3??2?Ja
5.
?????1r2?C?r?a??4?02?A???J,
??2?????a2ln?r??C?4?J,r?a??A在r?a分界面处连续，若设其值为零，
则：C1?a222?4?0a，C4?2lna ?1?a2?r2J,r?aA???0???4
??a2?2ln?a????r??J,r?a9.
zH=HP(R,?,?)0ez ?1区?02区
无传导电流时，??H?0。引人磁标势?m
H????m。依??B??0???H?M??0，得
??H????M???????B???H????BB?0?????????0?????????11??????????B?00 因此，??H??????m????2?m?0
以下在上图所示的球坐标中求解?2?m?0 边界条件，?mR????H0Rcos?
1区，R为零时解有限，因此
???bl?m1???alRl?Rl?1??Pl?cos??l?0??
??alRlPl?cos??l?02区，依边界条件?mR????H0Rcos?，
得 ???dl?m2???cllR?Rl?1??Pl?cos??l?0??
??H??dl0Rcos?l?0Rl?1Pl?cos??以下依边界条件确定系数al、cl和dl ⑴在R?R0处，?m连续，因此
??aRll0Pl?cos?????H0R0cos??l?0??d
lRl?1Pl?cos??l?00a1R0??Hd10R0?R2 ① 0l?1
时，aldllR0?Rl?1 ② 0⑵在R?R0处，B法向连续，因此
???m1R?R????m2?R0 0?RR?R0???lal?1lR0Pl?cos???l?0???d
l0????H0cos????l?1?Rl?2Pl?cos????l?00??由上式得
d0?0
?a?2d1?1???0??H0?R3? ?0? ③ ?l?2时，?lal?1?dllR0???0?l?1Rl?2 ④0由①、③两式得
a3?0H0????31????2?，d?0?R0H010??2?0由②、④两式得
l?2，al?dl?0。因此
?0H0m1??3???2?Rcos?
0?????30?R0H0cosm2??H0Rcos?????2?2 0RB1??H1?????m1????????m11??m11????Rem1?R?R??eθ?Rsin???eφ????????3?0H0co3?0H0??
???2??seR?sin?eθ0??2?0???3?? 0H03??0H0??2?ez?0??2?0(上式中：ez?cos?eR?sin?eθ)
B2??0H2???0??m2????????30?R0H0cos??0??H0cos??2??2?3?e?0R??R???30???H???0?R0H0sin??0sin????2?R3??e0?θ??0H0ez??????0?R30H0??31?0??2?0?R3cos?eR?R3ez????H??30???0?R0?3?H0?R?RH0?00???2???R5?R3?0?m?B1?1????M1dV?????????H1?0??dV??B1B1???? ?????????0???dV?0??0???B1dV??
??0?????4?R30433??0H00B1??R00??03??2?0?4????0??2?R30H00zH=H0ez?m2 ?m1?m3?
无传导电流时，??H?0。引人磁标势?m，
H????m，??H??????m????2?m?0 以下求解?2?m?0 依边条件
??lm1??alRPl?cos??
l?0???cl?m2???blRl?l?Rl?1??Pl?cos?? 0????m3??H0Rcos??dlRl?1Pl?cos?? l?0 以下依边界条件确定系数al、bl、cl和dl ⑴在边界R?R1和R?R2处，磁标势?m连续
??al??lR1Pl?cos??????blcl?lR1?Rl?1??Pl?cos??
l?0l?0?1??????blRl2?cl?Rl?1???Pl?cos???l?0?2?
?H0R2cos???dlRl?1Pl?cos??l?02因此
allcllR1?blR1?Rl?1 ① 1b1R2?c1R2??Hd10R2?R2 ② 22l?1
时，bl?cllR2Rl?1?dll?1 ③ 2R2⑵在边界R?R1和R?R2处，B法向连续
???m1??0?R??m2?RRR 1?RR?1???m2????m3?RR?R0，得 2?RR?R2 10.
??l?10?lalR1Pl?cos???l?0?????lbRl?1c
l??l1??l?1?Rl?2??Pl?cos??l?0?1??????lbRl?1??l?1cl?l2?Rl?2?l?0??2?P?l?cos??????
0???H?0cos????l?1?dlP?cos???l?0Rl?2l?2??因此
?l?1???0lalR1??lbl?1??l?1?cl?lR1l?2?? ④ ?R1??bcd1?2?13???1R0H0?2?03 ⑤ 2R2l?1，时
????lbl?1lR2??l?1?cl?Rl?2?d????l?0?l?1?Rl?2⑥ 2?2l?1时，由①、③、④和⑥得
al?bl?cl?dl?0
l?1
时
由由①、②、④和⑤得以下四方程
a1R1?b1R1?c1R2 1b1R2?c1R2??Hd10R2?2 2R2?a?c1?01????b1?2R3???? 1???c?b1??1?2R3??????0?d??H1??0?23?2? 2R?求解得
a?9??30R2H01??2R3231????0??R2?2???0????2?0?因此
?9??0R2H0Rcos?m1??3?2R331????20??R2?2???0????2?0?B1???0??m1???????m11??m11??m1?0???ReR?R??eθ?Rsin???eφ???9??230R2H0?cos?eR?sin?eθ??2R3231????0??R2?2???0????2?0?9??23?0R2?2R3????23H010??R2?2???0????2?0?书中答案经通分整理即为上式。 11.
zP(R,?,?) M0?'1区2区
无传导电流时，??H?0。引人磁标势?m
H????m。在1区，依
??B?????H??0M0?????H??0??M0????H?0，得??H?0
因此，?2?m1?0。另有?2?m2?0 在1区(R?R0)，R为零时解有限，因此
??m1??alRlPl?cos??
l?0在2区(R?R0)，R??时?m2有限，因此
?m2???dll?0Rl?1Pl?cos?? 以下依边界条件确定系数al和dl
⑴在边界R?R0处，磁标势?m1连续
??al?dllR0Pl?cos???l?0?l?0Rl?1Pl?cos?? 0