大学物理复习资料桂林电子科技大学(4)

16. 1.11×105 V 3

-分 A端 2分 17. 解：由题意，大线圈中的电流Ｉ在小线圈回路处产生的磁场可视为均匀的．

?0IR22?IR2B??4?(R2?x2)3/22(R2?x2)3/2 3分 ?0故穿过小回路的磁通量为

22???0IR22??rRI0??B?S??r?223/22(R?x)2x3 2分 由于小线圈的运动，小线圈中的感应电动势为

d?3?0?r2IR2dx3?0?r2R2I?i???v44dtdt2x2x 2分 当x =NR时，小线圈回路中的感应电动势为

242??3??rIv/(2NR) 1分 0 i I (t) a y ?i d y x (t) ?v 解：线框内既有感生又有动生电动势．设顺时针绕向为??i的正方向．由??i = ?d??/dt出发，先求任意时刻t的??(t)

???(t)??B?dS18.

?I(t)??0x(t)dy2?ya 2分 ?a?b?0I(t)x(t)lna 2分 2?a?b再求??(t)对t的导数：

d?(t)?0a?bdIdx?(ln)(x?I)2?bdtdt dt??0I0e??tv(1??t)lna?ba (x?vt) 2????d??0vI0e??t(?t?1)lna?bdt2?a 4分 ∴ ??i

?i方向：??t <1时，逆时针；??t >1时，顺时针． 2分 19. 解： S?12a23/2?3a2/4 ??BScos?t， ??2?n/60 ∴ ?OO???(d?/dt)?BS?sin?t?(2?BSn/60)sin(2?nt/60)

?(3?na2B/120)sin(2?nt/60) 20. 解：由静电学计算： r?0代表r方向单位矢量 E??q(t)r? 4???200rr

U?q(t)(1?1)?q(t)(R2?R1) 4??0?rR1R24??0?rR1R2 E?

?UR1R2?RR?∴ r2(Rr0?122U0sin?t?r02?R1) r(R2?R1)J???D? ?E???0?rR1R2?位移电流密度为 ?t??0?r?tr2(RU0?co?st?r02?R1)

???4??0?rR1R2U过球面的总位移电流 I??J?dS?J?4?r2RR0?co?st2?1

21. 解：长直带电线运动相当于电流I?v(t)??． 正方形线圈内的磁通量可如下求出

dΦ??02??Ia?xadx Φ??a0 2?Iadx?x?02?Ia?ln2

??0a

2分 3分 5分 4分 3分 2

分 2分 2分 ?0dv(t)?0adIdΦ??aln2?i???ln22?dtdt2?dt 2分 i(t)?

?iR??02?R?adv(t)ln2dt 2分 22. 解：(1)由于线框垂直下落，线框所包围面积内的磁通量无变化，故感应电流

Ii = 0 2分 (2) 设dc边长为l′，则由图可见

l?= L + 2Lcos60°= 2L

取d→c的方向为dc边内感应电动势的正向，则

l?c?c?0I??dr?dc??(v?B)?dl??vBdl??2gH?2?(r?l)?dd ?0Il??l?0Il?2Ll 3分

?dc?0，说明cd段内电动势的方向由d→c 2分

?2?2gHlnl?2?2gHln2?dc由于回路内无电流 Vcd?Uc?Ud??因为c点电势最高，d点电势最低，故：Vcd为电势最高处与电势最低处之间的 ??0I2gHln2L?ll 2分 电势差． 1分 s 1? v//?vco?23. 解： v??vsin

分 x2???????????????????????i

(?i指向以A到B为正) 3分 s s x1?a?vtco?式中： x2?a?l?vtco??Ia?l?vtco?s?i??0vsin?ln2?a?vtco?s 2分 A端的电势高． 2分 ??d?i???vsin?dx2?xx1?0I 24. 解∶(1) 由

mdvvBl?mg?BIlI?dtR 3，分 dvB2l2?g?vmR 得 dtdv??dt22?Blv00g?mR积分 RmgB2l2v?22?exp(?t)mR 4分 Bl得

vt其中

Lre3ult$?ltrch^lingnp2052^pict鸤*Lpacpsop0?x635etavile8_45010009000003080100_00240150000000

00050000000902000000000400000002010100050000000102ffffff00040000082e011<0085000000310201000000050000000b0200800000050000000c024002400712

2000026060f001a00ffffffff000010000002c0ffffffb7ffffvf_0070002f70q00000`00800_26060f000c004d617068547970610000501015000000fB0220ff00000000000090010100

0000002001054694d6573204e657'20526f6d616e000083040040002d0100p008000000320af400930601000000780015400000fb0280fe0_00000000009001010000000402007054696d6573204u6u7720526f6d616e000083P60000002d01010044000000f00100000x000000320qa001f8020100

000680015000000fb0280fe0002000000009001_00000000402001054696d6573204e6577205?6f6d616e0000?3040000002d01000004000000f001010008000p00320ai001c10501000000650008000000320aa001ac03010_0020290009040000320aa001340004_000006578702810000000fb0280fe0000000000009001000000020002001053796d636f6c000204

00_000d01010004000?00f001000008000000220aa0019204010000003d000a000_0026060&010a00ffffffff01000000040010p00q0

fb021000070000000010bc_2000p008>010_0222537973746?6d0086040004802d01000004000000f0010100?70000000000Lrtlch|zchè\\`f1#]hich￡??￥?lch

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trchttlch|kerning0_7 ×_1?′ó?òòch

hco\\6336317_disrs~± Laf13 \\ltrch?s0 :?2^sv 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

二、静电场习题 1. 如图所示，两个同心球壳．内球壳半径为R1，均匀带有电荷Q；外球壳半径为R2，壳的厚度忽略，原先不带电，但与地相连接．设地为电势零点，则在两球之间、距离球心为r的P点处电场强度的大小与电势分别为： QR1R2 rOPQQ24??r0 (A) E＝，U＝4??0r．

Q?11?Q????24??0?R1r?4??r?． 0 (B) E＝，U＝Q?11?Q??r?R??24??4??r02??． 0 (C) E＝，U＝Q (D) E＝0，U＝4??0R2． ［ ］ Q2 Q1 R2 P r O R1 如图所示，两个同心的均匀带电球面，内球面半径为R1、带电荷Q1，外球面半2.

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