首页 大学物理作业答案

大学物理作业答案

举报
开通vip

大学物理作业答案大学物理作业答案 第一章作业解答 1-3 一质点在平面上运动,运动方程为 xOy 12=3+5, =+3-4. yxttt2 式中以 s计,,以m计((1)以时间为变量,写出质点位置矢量的表示式;(2)求出=1 s 时刻和yxtttt ,2s 时刻的位置矢量,计算这1秒内质点的位移;(3)计算,0 s时刻到,4s时刻内的平均速度;(4)tt求出质点速度矢量表示式,计算,4 s 时质点的速度;(5)计算,0s 到,4s 内质点的平均加速度;ttt(6)求出质点加速度矢量的表示式,计算,4s 时质点的加速度(请把位...

大学物理作业答案
大学物理作业答案 第一章作业解答 1-3 一质点在平面上运动,运动方程为 xOy 12=3+5, =+3-4. yxttt2 式中以 s计,,以m计((1)以时间为变量,写出质点位置矢量的表示式;(2)求出=1 s 时刻和yxtttt ,2s 时刻的位置矢量,计算这1秒内质点的位移;(3)计算,0 s时刻到,4s时刻内的平均速度;(4)tt求出质点速度矢量表示式,计算,4 s 时质点的速度;(5)计算,0s 到,4s 内质点的平均加速度;ttt(6)求出质点加速度矢量的表示式,计算,4s 时质点的加速度(请把位置矢量、位移、平均速度、瞬t 时速度、平均加速度、瞬时加速度都表示成直角坐标系中的矢量式)( ,,1,2解:(1) r,(3t,5)i,(t,3t,4)j m2 t,1t,2(2)将,代入上式即有 ,,, r,8i,0.5jm1 ,,, r,11j,4jm2 ,,,,, ,r,r,r,3j,4.5jm21 ,,,,,,(3)? r,5j,4j,r,17i,16j04 ,,,,,,,r,r,r12i,20j,,140v,,,,3i,5jm,s? ,t4,04 ,,,dr,,1v,,3i,(t,3)jm,s(4) dt ,,,,1m,s则 v,3i,7j4 ,,,,,,(5)? v,3i,3j,v,3i,7j 04 ,,,,v,v,v4,,240a,,,,1jm,s ,t44 ,,dv,,2a,,1jm,s(6) dt y这说明该点只有方向的加速度,且为恒量。 2,2m,sx1-5 质点沿x轴运动,其加速度和位置的关系为 a,2+6,a的单位为,x的单位为 m. 质点 ,1m,sx在,0处,速度为10,试求质点在任何坐标处的速度值( dvdvdxdva,,,v解: ? dtdxdtdx 2分离变量: ,d,,adx,(2,6x)dx两边积分得 123v,2x,2x,c 2 quickly. (23) Gao Changhai night rescued compatriots in January 1943, the Japanese control after Su Jiahu along the highway, to intensify guerrilla "raids". One day, the Japanese army from Wuzhen area caught 53 "Shina" is held in spring qiaotu farm 2 x,0c,50由题知,时,,? v,100 3,1? v,2x,x,25m,s ,21-6 已知一质点作直线运动,其加速度为 ,4+3 ,开始运动时,,5 m, =0,求该m,saxvt质点在,10s 时的速度和位置( t dv 解:? a,,4,3tdt分离变量,得 dv,(4,3t)dt 32v,4t,t,c积分,得 12 t,0由题知,, ,? v,0c,001 32故 v,4t,t 2 dx32v,,4t,t又因为 dt2 32dx,(4t,t)dt分离变量, 2 123x,2t,t,c积分得 22 t,0由题知 , ,? x,5c,502 123x,2t,t,5故 2所以时 t,10s 321,v,4,10,,10,190m,s102 123x,2,10,,10,5,705m102 3,,t1-7 一质点沿半径为1 m 的圆周运动,运动方程为 =2+3,式中以弧度计,以秒计,求:(1) t ,2 s 时,质点的切向和法向加速度;(2)当加速度的方向和半径成45?角时,其角位移是多少? t dd,,2,,9t,,,18t 解: ,, dtdt ,2 (1)t,2s时, a,R,,1,18,2,36m,s , 222,2 a,R,,1,(9,2),1296m,sn ο45(2)当加速度方向与半径成角时,有 quickly. (23) Gao Changhai night rescued compatriots in January 1943, the Japanese control after Su Jiahu along the highway, to intensify guerrilla "raids". One day, the Japanese army from Wuzhen area caught 53 "Shina" is held in spring qiaotu farm 大学物理作业答案 3 a, tan45:,,1an 2即 R,,R, 22亦即 (9t),18t 23则解得 t,9 于是角位移为 23 ,,2,3t,2,3,,2.67rad9 第二章作业解答 2-9 一质量为的质点在平面上运动,其位置矢量为 mxOy ,,, r,acos,ti,bsin,tj ,求质点的动量及,0 到t, 时间内质点所受的合力的冲量和质点动量的改变量( t2, 解: 质点的动量为 ,,,, p,mv,m,(,asin,ti,bcos,tj) ,t,0t,将和分别代入上式,得 2, ,,,,, , p,m,bjp,,m,ai12 则动量的增量亦即质点所受外力的冲量为 ,,,,,, I,,p,p,p,,m,(ai,bj)21 ,,,,,,,,F2-12 设((1) 当一质点从原点运动到时,求所作的功((2)F,7i,6jNr,,3i,4j,16km合 如果质点到处时需0.6s,试求平均功率((3)如果质点的质量为1kg,试求动能的变化( r ,解: (1)由题知,为恒力, F合 ,,,,,,,? A,F,r,(7i,6j),(,3i,4j,16k)合 ,,21,24,,45J A45P,,,75w(2) ,t0.6 ,E,A,,45J(3)由动能定理, k ,OO2-26 固定在一起的两个同轴均匀圆柱体可绕其光滑的水平对称轴转动(设大小圆柱体的半径分别 RMmmmmm为和,质量分别为和(绕在两柱体上的细绳分别与物体和相连,和则挂在圆柱r1212 RMmmm体的两侧,如题2-26图所示(设,0.20m, ,0.10m,,4 kg,,10 kg,,,2 kg,r12quickly. (23) Gao Changhai night rescued compatriots in January 1943, the Japanese control after Su Jiahu along the highway, to intensify guerrilla "raids". One day, the Japanese army from Wuzhen area caught 53 "Shina" is held in spring qiaotu farm 4 h且开始时,离地均为,2m(求: mm12 (1)柱体转动时的角加速度; (2)两侧细绳的张力( 解: 设,和β分别为,和柱体的加速度及角加速度,方向如图(如图b)( aamm1212 题2-26(a)图 题2-26(b)图 (1) ,和柱体的运动方程如下: mm12 ? T,mg,ma2222 ? mg,T,ma1111 ,,TR,Tr,I, ? 12 ,,式中 T,T,T,T,a,r,,a,R,112221 1122I,MR,mr而 22 由上式求得 Rm,rm12,g,22I,mR,mr12 0.2,2,0.1,2,,9.8 112222,10,0.20,,4,0.10,2,0.20,2,0.1022 ,2,6.13rad,s (2)由?式 NT,mr,,mg,2,0.10,6.13,2,9.8,20.8 222 由?式 NT,mg,mR,,2,9.8,2,0.2.,6.13,17.1 111 M2-27 计算题2-27图所示系统中物体的加速度(设滑轮为质量均匀分布的圆柱体,其质量为,半径为 mm,在绳与轮缘的摩擦力作用下旋转,忽略桌面与物体间的摩擦,设,50 kg,,200 kg,M,15 r12quickly. (23) Gao Changhai night rescued compatriots in January 1943, the Japanese control after Su Jiahu along the highway, to intensify guerrilla "raids". One day, the Japanese army from Wuzhen area caught 53 "Shina" is held in spring qiaotu farm 大学物理作业答案 5 kg, ,0.1 m r 解: 分别以,滑轮为研究对象,受力图如图(b)所示(对,运用牛顿定律,有 mmmm1212 ? mg,T,ma222 ? T,ma11对滑轮运用转动定律,有 12 ? Tr,Tr,(Mr),212又, ? a,r, 联立以上4个方程,得 mg200,9.8,22a,,,7.6m,s M15m,m,5,200,1222 题2-27(a)图 题2-27(b)图 题2-28图 lO2-28 如题2-28图所示,一匀质细杆质量为,长为,可绕过一端的水平轴自由转动,杆于水平位m 置由静止开始摆下(求: (1)初始时刻的角加速度; ,(2)杆转过角时的角速度. 解: (1)由转动定律,有 112mg,(ml), 23 3g,,? 2l(2)由机械能守恒定律,有 l1122mgsin,,(ml), 223quickly. (23) Gao Changhai night rescued compatriots in January 1943, the Japanese control after Su Jiahu along the highway, to intensify guerrilla "raids". One day, the Japanese army from Wuzhen area caught 53 "Shina" is held in spring qiaotu farm 6 g3sin,? ,,l 第四章作业解答 ,4-3 如题4-3图所示,物体的质量为,放在光滑斜面上,斜面与水平面的夹角为,弹簧的倔强系m k数为,滑轮的转动惯量为,半径为(先把物体托住,使弹簧维持原长,然 后由静止释放,试证IR 明物体作简谐振动,并求振动周期( 题4-3图 解:分别以物体和滑轮为对象,其受力如题4-3图(b)所示,以重物在斜面上静平衡时位置为坐标原m 点,沿斜面向下为轴正向,则当重物偏离原点的坐标为时,有 xx 2dxmgsin,T,m, ? 12dt ? TR,TR,I,12 2dx,R, ? T,k(x,x)202dt 式中,为静平衡时弹簧之伸长量,联立以上三式,有 x,mgsin,/k0 2dIx()mR,,,kxR 2dRt 2kR2,,令 2mR,I 则有 2dx2,,x,0 2dt 故知该系统是作简谐振动,其振动周期为 22mR,Im,IR,2/T,,,2,(2,) 2KkR, 2,,3x,0.1cos(8,)(SI)4-4 质量为的小球与轻弹簧组成的系统,按,的规律作谐振动,10,10kg3求: (1)振动的周期、振幅和初位相及速度与加速度的最大值; (2)最大的回复力、振动能量、平均动能和平均势能,在哪些位置上动能与势能相等? quickly. (23) Gao Changhai night rescued compatriots in January 1943, the Japanese control after Su Jiahu along the highway, to intensify guerrilla "raids". One day, the Japanese army from Wuzhen area caught 53 "Shina" is held in spring qiaotu farm 大学物理作业答案 7 (3)与两个时刻的位相差; t,5st,1s21 解:(1)设谐振动的标准方程为,则知: x,Acos(,t,,)0 ,21 A,0.1m,,,8,,?T,,s,,,2,/304, ,1,1,2.51又 v,,A,0.8,m,sm,sm 2,2 a,,A,63.2m,sm (2) F,a,0.63Nmm 12,2E,mv,3.16,10J m2 1,2E,E,E,1.58,10J pk2 当时,有, E,EE,2Epkp 11122kx,,(kA)即 222 22 ? x,,A,,m220 (3) ,,,,(t,t),8,(5,1),32,21 AT4-5 一个沿轴作简谐振动的弹簧振子,振幅为,周期为,其振动方程用余弦函数表示(如果x t,0时质点的状态分别是: (1)x,,A; 0 (2)过平衡位置向正向运动; Ax,(3)过处向负向运动; 2 Ax,,(4)过处向正向运动( 2 试求出相应的初位相,并写出振动方程( ,x,Acos,00解:因为 ,v,Asin,,,00, 将以上初值条件代入上式,使两式同时成立之值即为该条件下的初位相(故有 ,2,,,x,Acos(t,,) 1T ,323,,,x,Acos(t,,) 22T2 2,,,,x,Acos(t,), 33T3quickly. (23) Gao Changhai night rescued compatriots in January 1943, the Japanese control after Su Jiahu along the highway, to intensify guerrilla "raids". One day, the Japanese army from Wuzhen area caught 53 "Shina" is held in spring qiaotu farm 8 ,,525 ,,x,Acos(t,,)44T4 ,324cm4.0st,04-6 一质量为的物体作谐振动,振幅为,周期为,当时位移为10,10kg ,24cm(求: t,0.5s(1)时,物体所在的位置及此时所受力的大小和方向; x,12cm(2)由起始位置运动到处所需的最短时间; x,12cm(3)在处物体的总能量( ,2解:由题已知 A,24,10m,T,4.0s ,2,1? ,,,0.5,rad,sT t,0又,时, x,,A,?,,000 故振动方程为 ,2 x,24,10cos(0.5,t)m t,0.5s (1)将代入得 ,2 x,24,10cos(0.5,t)m,0.17m0.5 2,,,,,Fmamx ,,32,3,,10,10,(),0.17,,4.2,10N2 方向指向坐标原点,即沿轴负向( x t,0(2)由题知,时,, ,,00 A,,0,x,,且v,故,时 ,t,t0t23 ,2,,,t,,/,s? 323, (3)由于谐振动中能量守恒,故在任一位置处或任一时刻的系统的总能量均为 11222,,,EkAmA22 1,,322 ,,10,10(),(0.24)22 ,4,7.1,10J 4-8 图为两个谐振动的曲线,试分别写出其谐振动方程( x,t 题4-8图 quickly. (23) Gao Changhai night rescued compatriots in January 1943, the Japanese control after Su Jiahu along the highway, to intensify guerrilla "raids". One day, the Japanese army from Wuzhen area caught 53 "Shina" is held in spring qiaotu farm 大学物理作业答案 9 3t,0解:由题4-8图(a),?时, x,0,v,0,?,,,,又,A,10cm,T,2s0002 ,2,1即 ,,,,rad,sT 3故 x,0.1cos(,t,,)ma2 5A,t,0,0,由题4-8图(b)?时, x,v,?,,00023 ,0,0,2时, x,v,?,,t,0,,11112 551又 ,,,,,,,,132 5? ,,,6 55,故 x,0.1cos(t,)m ,b63 第五章作业解答 Bt,CxCAAB5-8 已知波源在原点的一列平面简谐波,波动方程为=cos(),其中,, 为正值y恒量(求: (1)波的振幅、波速、频率、周期与波长; l(2)写出传播方向上距离波源为处一点的振动方程; d(3)任一时刻,在波的传播方向上相距为的两点的位相差( 解: (1)已知平面简谐波的波动方程 x,0 () y,Acos(Bt,Cx)将上式与波动方程的标准形式 x,,,cos(22)y,At, ,比较,可知: B,,A波振幅为,频率, 2, B2,u,,,,,波长,波速, ,CC 12,T,,波动周期( B, x,l(2)将代入波动方程即可得到该点的振动方程 y,Acos(Bt,Cl) (3)因任一时刻同一波线上两点之间的位相差为 t ,2,,,(x,x) 21, 2,,x,x,d,将,及代入上式,即得 21C ,,,Cd( quickly. (23) Gao Changhai night rescued compatriots in January 1943, the Japanese control after Su Jiahu along the highway, to intensify guerrilla "raids". One day, the Japanese army from Wuzhen area caught 53 "Shina" is held in spring qiaotu farm 10 ,t,4,x5-9 沿绳子传播的平面简谐波的波动方程为=0.05cos(10),式中,以米计,以秒计(求: yyxt (1)波的波速、频率和波长; (2)绳子上各质点振动时的最大速度和最大加速度; (3)求=0.2m 处质点在=1s时的位相,它是原点在哪一时刻的位相?这一位相所代表的运动状态在xt =1.25s时刻到达哪一点? t 解: (1)将题给方程与标准式 ,2,, y,Acos(2t,x), ,1,1A,0.05,,5,,0.5u,,,,2.5相比,得振幅,频率,波长,波速( m,ssmm(2)绳上各点的最大振速,最大加速度分别为 ,1 m,sv,,A,10,,0.05,0.5,max ,2222 m,sa,,A,(10,),0.05,5,max x,0.2(3) m处的振动比原点落后的时间为 x0.2,,0.08 su2.5 x,0.2t,1x,0故,时的位相就是原点(),在时的位相, t,1,0.08,0.92ssm0即 π( ,,9.2 t,1.25设这一位相所代表的运动状态在s时刻到达点,则 x x,x,u(t,t),0.2,2.5(1.25,1.0),0.825m11 A5-14 如题5-14图所示,有一平面简谐波在空间传播,已知P点的振动方程为= cos(,t,,)( y0P (1)分别就图中给出的两种坐标写出其波动方程; bP(2)写出距点距离为的点的振动方程( Q 解: (1)如题5-14图(a),则波动方程为 lxy,Acos[,(t,,),,] 0uu如图(b),则波动方程为 题5-14图 xy,Acos[,(t,),,] 0u Q(2) 如题5-14图(a),则点的振动方程为 quickly. (23) Gao Changhai night rescued compatriots in January 1943, the Japanese control after Su Jiahu along the highway, to intensify guerrilla "raids". One day, the Japanese army from Wuzhen area caught 53 "Shina" is held in spring qiaotu farm 大学物理作业答案 11 b A,Acos[,(t,),,]Q0u 如题5-14图(b),则点的振动方程为 Q b A,Acos[,(t,),,]Q0u 5-19 如题5-19图所示,设点发出的平面横波沿方向传播,它在点的振动方程为BBPB ,3CCPC;点发出的平面横波沿方向传播,它在点的振动方程为y,2,10cos2,t1 -1,3CP,本题中以m计,以s计(设BP,0.4m,,0.5 m,波速=0.2m?s,yuy,2,10cos(2,t,,)t2 求: (1)两波传到P点时的位相差; (2)当这两列波的振动方向相同时,P处合振动的振幅; *(3)当这两列波的振动方向互相垂直时,P处合振动的振幅( ,2,,, 解: (1) ,,(,),(CP,BP) 21, ,,,(CP,BP) ,u 2,,,(0.5,0.4),0 ,0.2 题5-19图 P(2)点是相长干涉,且振动方向相同,所以 ,3 A,A,A,4,10mP12 0(3)若两振动方向垂直,又两分振动位相差为,这时合振动轨迹是通过?,?象限的直线,所以合振 幅为 22,3,3A,A,A,2A,22,10,2.83,10 m121 BBPB5-19 如题5-19图所示,设点发出的平面横波沿方向传播,它在点的振动方程为 ,3CCPC;点发出的平面横波沿方向传播,它在点的振动方程为y,2,10cos2,t1 ,3-1CPBPy,本题中以m计,以s计(设,0.4m,,0.5 m,波速u=0.2m?s,y,2,10cos(2,t,,)t2 求: (1)两波传到P点时的位相差; P(2)当这两列波的振动方向相同时,处合振动的振幅; P*(3)当这两列波的振动方向互相垂直时,处合振动的振幅( ,2,,,,,(,),(CP,BP) 解: (1) 21, ,,,(CP,BP), u quickly. (23) Gao Changhai night rescued compatriots in January 1943, the Japanese control after Su Jiahu along the highway, to intensify guerrilla "raids". One day, the Japanese army from Wuzhen area caught 53 "Shina" is held in spring qiaotu farm 12 2, ,,(0.5,0.4),0,0.2 题5-19图 (2)点是相长干涉,且振动方向相同,所以 P ,3 A,A,A,4,10mP12 0(3)若两振动方向垂直,又两分振动位相差为,这时合振动轨迹是通过?,?象限的直线,所以合振 幅为 22,3,3A,A,A,2A,22,10,2.83,10 m121 第八章作业解答 l,8-2 两小球的质量都是,都用长为的细绳挂在同一点,它们带有相同电量,静止时两线夹角为2 ,m 如题8-2图所示(设小球的半径和线的质量都可以忽略不计,求每个小球所带的电量( 解: 如图示 ,Tcosmg,,,21q ,,TsinF,,e2,4π,(2lsin,)0, 解得 q,2lsin,4,,mgtan,0 -9-1l,8-6 长=15.0cm 的直导线AB上均匀地分布着线密度=5.0x10C?m 的正电荷(试求:(1)在导线 P的延长线上与导线B端相距=5.0cm处点的场强;(2)在导线的垂直平分线上与导线中点相距a1 =5.0cm 处点的场强( dQ2 解: 如图所示 dxP(1)在带电直线上取线元,其上电量在点产生场强为 dq 1dx,dE, P24π(a,x),0 l,dx2E,dE, PPl2,,,,4π(a,x)02 ,11,, []ll,4π0a,a, 22 ,l, 22π,(4a,l)0 ,9,1l,15a,12.5,,5.0,10C,mcmcm用,, 代入得 quickly. (23) Gao Changhai night rescued compatriots in January 1943, the Japanese control after Su Jiahu along the highway, to intensify guerrilla "raids". One day, the Japanese army from Wuzhen area caught 53 "Shina" is held in spring qiaotu farm 大学物理作业答案 13 ,12方向水平向右 N,CE,6.74,10P 1dx,(2)同理 方向如图所示 dE,Q224πx,d,02 ,dE,0由于对称性,即只有分量, EyQxQ,l 1dxd,2? dE,Qy22224πx,dx,d,022 ldx,d22 EE,,dl,3,lQyQy,2224π,20(x,d)2 ,l ,222πl,4d,02 ,9,1l,15以, ,代入得 ,,5.0,10C,cmcmd,5cm2 2,1E,E,14.96,10N,C,方向沿轴正向 yQQy ,5-3108-10 均匀带电球壳内半径6cm,外半径10cm,电荷体密度为2×C?m求距球心5cm,8cm ,12cm 各 点的场强( ,,qq,,2 高斯定理解:, E,dS,E4πr,,s,,00 ,r,5q,0E,0当时,, cm, 4π33r,8,pq,r)时, cm(r,内3 4π32,rr,,,内4,13E,3.48,10N,C,? , 方向沿半径向外( 24π,r0 4π33r,12q,,r)cm时, (r,,内外3 4π33,r,r,,外内4,13 E,,4.10,10N,C? 沿半径向外. 2,4πr0 ABABRqq8-16 如图所示,在,两点处放有电量分别为+,-的点电荷,间距离为2,现将另一正试 OCq验点电荷从点经过半圆弧移到点,求移动过程中电场力作的功( 0 解: 如图示 quickly. (23) Gao Changhai night rescued compatriots in January 1943, the Japanese control after Su Jiahu along the highway, to intensify guerrilla "raids". One day, the Japanese army from Wuzhen area caught 53 "Shina" is held in spring qiaotu farm 14 1qq (,),0U,ORR4π,0 qqq1 (,),,U,C3RR6π,R04π,0 qqo? Aq(UU),,,0OC6π,R0 ,8-17 如图所示的绝缘细线上均匀分布着线密度为的正电荷,两直导线的长度和半圆环的半径都等于OR(试求环中心点处的场强和电势( CDOdl,Rd,解: (1)由于电荷均匀分布与对称性,和段电荷在点产生的场强互相抵消,取 AB ,轴场强沿yOO则产生点如图,由于对称性,点dEdq,,Rd, 负方向 ,,,Rd2 E,dE,cos,y,,,2,4πR,02 ,,,sinsin(,),,, ,224π,R0 ,, ,2π,R0 OAB(2) 电荷在点产生电势,以 U,0, A2R,,,xxddU ,,,ln21,,BR,x,x,4π4π4π000 ,CD同理产生 U,ln22,4π0 ,,πR半圆环产生 U,, 34π,4,R00 ,,ln2U,U,U,U,,? 123O2π,4,00 第九章作业解答 :CDOABRBC9-7 如题9-7图所示,、为长直导线,为圆心在点的一段圆弧形导线,其半径为(若OI通以电流,求点的磁感应强度( :OCDABBC解:如题9-7图所示,点磁场由、、三部分电流产生(其中 ,AB 产生 B,01 quickly. (23) Gao Changhai night rescued compatriots in January 1943, the Japanese control after Su Jiahu along the highway, to intensify guerrilla "raids". One day, the Japanese army from Wuzhen area caught 53 "Shina" is held in spring qiaotu farm 大学物理作业答案 15 ,I0CD 产生,方向垂直向里 ,B212R II,,3::00CD 段产生 ,方向向里 ,B,(sin90,sin60),(1,)3RR2,24,2 I,3,0?,方向向里( ,B,B,B,B,(1,,)01232R26, 9-8 在真空中,有两根互相平行的无限长直导线和,相距0.1m,通有方向相反的电流,LL21 =20A,=10A,如题9-8图所示(,两点与导线在同一平面内(这两点与导线的距离均为ABIIL122 5.0cm(试求A,B两点处的磁感应强度,以及磁感应强度为零的点的位置( 题9-8图 ,8图所示,方向垂直纸面向里 解:如题9-BA ,,II,40102T ,,,1.2,10BA,,2(0.1,0.05)2,0.05 , B,0(2)设在外侧距离为处 LLr22 ,,II02则 ,,0,,2(r,0.1)2r r,0.1解得 m d9-12 两平行长直导线相距=40cm,每根导线载有电流==20A,如题9-12图所示(求: II12 A(1)两导线所在平面内与该两导线等距的一点处的磁感应强度; lrr(2)通过图中斜线所示面积的磁通量((==10cm, =25cm)( 31 ,,II,50102B,,,4,10, 解:(1) T 方向纸面向外 Add2,()2,()22 dS,ldr(2)取面元 quickly. (23) Gao Changhai night rescued compatriots in January 1943, the Japanese control after Su Jiahu along the highway, to intensify guerrilla "raids". One day, the Japanese army from Wuzhen area caught 53 "Shina" is held in spring qiaotu farm 16 ,I,rr,,,1,IIlIlIl121,601010211Wb ,,[,],ln3,ln,ln3,2.2,10ldr,r12,2,(,)2,2,3,rdr CDEF9-20 如题9-20图所示,在长直导线内通以电流=20A,在矩形线圈中通有电流=10 A,ABII12 CDbd与线圈共面,且,都与平行(已知=9.0cm,=20.0cm,=1.0 cm,求: ABEFABa(1)导线的磁场对矩形线圈每边所作用的力; AB (2)矩形线圈所受合力和合力矩( ,CD 解:(1)方向垂直向左,大小 FCD ,I,401NF,Ib,8.0,10 CD2,2d ,FE同理方向垂直向右,大小 FFE ,I,501N F,Ib,8.0,10FE2,2(d,a),CF方向垂直向上,大小为 FCF d,aIIIIda,,,5,012012FrN,d,ln,9.2,10 CF,drd2,2,,ED方向垂直向下,大小为 FED ,5N F,F,9.2,10EDCF ,,,,, (2)合力方向向左,大小为 F,F,F,F,FCDFECFED ,4NF,7.2,10 ,,,合力矩 MPB,,m ? 线圈与导线共面 ,,? PB//m ,M,0( 第十章作业解答 10-5如题10-5所示,在两平行载流的无限长直导线的平面内有一矩形线圈(两导线中的电流方向相反、 dI大小相等,且电流以的变化率增大,求: dt (1)任一时刻线圈内所通过的磁通量; (2)线圈中的感应电动势( 解: 以向外磁通为正则 (1) quickly. (23) Gao Changhai night rescued compatriots in January 1943, the Japanese control after Su Jiahu along the highway, to intensify guerrilla "raids". One day, the Japanese army from Wuzhen area caught 53 "Shina" is held in spring qiaotu farm 大学物理作业答案 17 b,,adaIIIlbada,,,,,000lrlr dd[lnln],,,,,m,,bdrrbd2π2π2π lddabadI,,,,0[lnln](2) ,,,,,dt2πdbdt b10-7 如题10-7图所示,长直导线通以电流=5A,在其右方放一长方形线圈,两者共面(线圈长=0.06m,I -1d宽=0.04m,线圈以速度=0.03m?s 垂直于直线平移远离(求:=0.05m时线圈中感应电动势的大av 小和方向( 题10-7图 ,CDv解: AB、运动速度方向与磁力线平行,不产生感应电动势( DA产生电动势 ,,A,I,0,,,,,, (vB)dlvBbvb1,D,2dBC产生电动势 ,,CI,,0 ,(v,B),dl,,vb,2,B2π(a,d)?回路中总感应电动势 Ibv11,,80V ,,,(,),1.6,10 ,,,122πdd,a方向沿顺时针( 长直导线所在平面2b10-11 如题10-11图所示,长度为的金属杆位于两无限,v的正中间,并以速度平行于两直导线运动(两直导线通以大小相等、方向相 I反的电流,两导线相距2(试求:金属杆两端的电势差及其方向( a dr解:在金属杆上取距左边直导线为,则 r ,,Ba,b,,Iv,Iva,b11,00,,v,B,l,,,r,()d()dln AB,,Aa,b,ra,r,a,b22 ,,0? AB B,A?实际上感应电动势方向从,即从图中从右向左, ,Iva,b0U,ln? AB,a,b quickly. (23) Gao Changhai night rescued compatriots in January 1943, the Japanese control after Su Jiahu along the highway, to intensify guerrilla "raids". One day, the Japanese army from Wuzhen area caught 53 "Shina" is held in spring qiaotu farm
本文档为【大学物理作业答案】,请使用软件OFFICE或WPS软件打开。作品中的文字与图均可以修改和编辑, 图片更改请在作品中右键图片并更换,文字修改请直接点击文字进行修改,也可以新增和删除文档中的内容。
该文档来自用户分享,如有侵权行为请发邮件ishare@vip.sina.com联系网站客服,我们会及时删除。
[版权声明] 本站所有资料为用户分享产生,若发现您的权利被侵害,请联系客服邮件isharekefu@iask.cn,我们尽快处理。
本作品所展示的图片、画像、字体、音乐的版权可能需版权方额外授权,请谨慎使用。
网站提供的党政主题相关内容(国旗、国徽、党徽..)目的在于配合国家政策宣传,仅限个人学习分享使用,禁止用于任何广告和商用目的。
下载需要: 免费 已有0 人下载
最新资料
资料动态
专题动态
is_215732
暂无简介~
格式:doc
大小:122KB
软件:Word
页数:24
分类:生活休闲
上传时间:2017-10-13
浏览量:8