大学物理作业答案

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大学物理作业答案

作业11.1B定义.1.2D质点带正电且沿曲线作加速运动,必有沿法向和运动方向的场强分量,故只有D满足条件。1.3对称:;NMdxxOX场强最大处:1.4坐标系建立如图:MN上长为dx的电荷受力,高斯定理:;;方向沿x轴正向。otherstaffoftheCentre.Duringthewar,ZhuwastransferredbacktoJiangxi,andDirectorofthenewOfficeinJingdezhen,JiangxiCommitteeSecretary.Startingin1939servedasrecorderoftheWestNorthOrganization,SecretaryoftheSpecialCommitteeAfterthevictoryofthelongMarch,hehasbeentheNorthwestOfficeoftheFederationofStateenterprisesMinister,ShenmufuguSARmissions,DirectorofNingxiaCountypartyCommitteeSecretaryandrecorderoftheCountypartyCommitteeSecretary,Ministersand41\n1.5如果点电荷是孤立的,则半径为R的球面上,场强大小一定处处相等。如果点电荷周围还有其它的带电体,则球面上的场强应是各场强的叠加,可能不处处相等。1.6设棒上电荷线密度为,则:,圆心处场强可以看成是半径为R,电荷线密度为的均匀带电园环在圆心处产生的场强与放在空隙处长为,电荷线密度为的均匀带电棒载圆心产生的场强的叠加。即:;(方向从圆心指向空隙处)。1.7按题给坐标,由于电荷呈线分布,则线密度为对上下段任意分割取电荷元dq,经分析可知上下两部分dq在圆心处产生关于y轴对称,由对称性可知:,otherstaffoftheCentre.Duringthewar,ZhuwastransferredbacktoJiangxi,andDirectorofthenewOfficeinJingdezhen,JiangxiCommitteeSecretary.Startingin1939servedasrecorderoftheWestNorthOrganization,SecretaryoftheSpecialCommitteeAfterthevictoryofthelongMarch,hehasbeentheNorthwestOfficeoftheFederationofStateenterprisesMinister,ShenmufuguSARmissions,DirectorofNingxiaCountypartyCommitteeSecretaryandrecorderoftheCountypartyCommitteeSecretary,Ministersand41\n方向沿y轴负方向。1.8按题给坐标,O点的场强可以看作是两根半无限长带电直线、半圆形带电细线在O点产生场强的叠加。即:(半无限长导线),(半圆)1.9失去自由电子。理论上说质量有所减少。但测量很困难。作业22.1B2.2B2.3C2.4(此时可视为点电荷)otherstaffoftheCentre.Duringthewar,ZhuwastransferredbacktoJiangxi,andDirectorofthenewOfficeinJingdezhen,JiangxiCommitteeSecretary.Startingin1939servedasrecorderoftheWestNorthOrganization,SecretaryoftheSpecialCommitteeAfterthevictoryofthelongMarch,hehasbeentheNorthwestOfficeoftheFederationofStateenterprisesMinister,ShenmufuguSARmissions,DirectorofNingxiaCountypartyCommitteeSecretaryandrecorderoftheCountypartyCommitteeSecretary,Ministersand41\n2.5()2.6[解]受力分析如图所示,小球在重力mg,绳中张力T及静电的共同作用下而处于受力平衡状态。其中为无限大均匀带电平面(电荷面密度为)产生的均匀电场。,的方向如图所示。于是有:(c/m2)2.7[解]由于电荷、电场分布具有球对称性,可利用高斯定理求场强。取同心的半径为r的球面为高斯面(图中虚线)如图所示。则通过高斯面的电通量为当<时,高斯面内包围的电荷代数和为所以,(〈)当>>时,otherstaffoftheCentre.Duringthewar,ZhuwastransferredbacktoJiangxi,andDirectorofthenewOfficeinJingdezhen,JiangxiCommitteeSecretary.Startingin1939servedasrecorderoftheWestNorthOrganization,SecretaryoftheSpecialCommitteeAfterthevictoryofthelongMarch,hehasbeentheNorthwestOfficeoftheFederationofStateenterprisesMinister,ShenmufuguSARmissions,DirectorofNingxiaCountypartyCommitteeSecretaryandrecorderoftheCountypartyCommitteeSecretary,Ministersand41\n所以或(>>)当>时,所以(>)2.7[解]由于电荷、电场分布具有轴对称性,可利用高斯定理求场强。取长为半径为r的同轴闭合圆柱面为高斯面,则通过高斯柱面的电通量为=当<时,∴当>>时,当>时,∴作业33.1C3.2CotherstaffoftheCentre.Duringthewar,ZhuwastransferredbacktoJiangxi,andDirectorofthenewOfficeinJingdezhen,JiangxiCommitteeSecretary.Startingin1939servedasrecorderoftheWestNorthOrganization,SecretaryoftheSpecialCommitteeAfterthevictoryofthelongMarch,hehasbeentheNorthwestOfficeoftheFederationofStateenterprisesMinister,ShenmufuguSARmissions,DirectorofNingxiaCountypartyCommitteeSecretaryandrecorderoftheCountypartyCommitteeSecretary,Ministersand41\n3.3C[解]由于球对称性,由高斯定理求得场强分布(>>)又因为外球壳接地,且地为电势零点,所以处为电势零点,有所以3.43.5[解]由于球对称性,由高斯定理求得场强分布<>∴∵d<∴>otherstaffoftheCentre.Duringthewar,ZhuwastransferredbacktoJiangxi,andDirectorofthenewOfficeinJingdezhen,JiangxiCommitteeSecretary.Startingin1939servedasrecorderoftheWestNorthOrganization,SecretaryoftheSpecialCommitteeAfterthevictoryofthelongMarch,hehasbeentheNorthwestOfficeoftheFederationofStateenterprisesMinister,ShenmufuguSARmissions,DirectorofNingxiaCountypartyCommitteeSecretaryandrecorderoftheCountypartyCommitteeSecretary,Ministersand41\n3.6[解](1)把圆盘无限分割成许多圆环,其中任一圆环半径为,宽为,此圆环在点产生的电势为由电势叠加原理,有(2)(3),3.7[解]经分析电荷分布呈线分布,无限分割带电圆弧为许多电荷元,有,其中任一电荷元可看成点电荷,它在点产生的场强为,方向如图,电势,以轴为对称轴,选另一电荷元与对称,则有和,由于对称性,otherstaffoftheCentre.Duringthewar,ZhuwastransferredbacktoJiangxi,andDirectorofthenewOfficeinJingdezhen,JiangxiCommitteeSecretary.Startingin1939servedasrecorderoftheWestNorthOrganization,SecretaryoftheSpecialCommitteeAfterthevictoryofthelongMarch,hehasbeentheNorthwestOfficeoftheFederationofStateenterprisesMinister,ShenmufuguSARmissions,DirectorofNingxiaCountypartyCommitteeSecretaryandrecorderoftheCountypartyCommitteeSecretary,Ministersand41\n点总的场强和电势为所有点电荷在该点产生的场强和电势的叠加。3.8静电场是保守力场,静电场是有势场;静电场也是无旋场。3.9不一定(详见课堂举例)。只有在电势不变的空间中场强才处处为零。3.10不一定(详见课堂举例)。作业44.1(C)可先假设内球带电,先不管正负,然后根据电势叠加原理及大地电势为零来计算内球的电势,就可做出正确判断。类似的接地问题都可以按类似的思路求解。4.2(B)4.3(B)4.4(B)重新达到一个新的静电平衡,电荷重新分配,整个导体是等势体(等势体)otherstaffoftheCentre.Duringthewar,ZhuwastransferredbacktoJiangxi,andDirectorofthenewOfficeinJingdezhen,JiangxiCommitteeSecretary.Startingin1939servedasrecorderoftheWestNorthOrganization,SecretaryoftheSpecialCommitteeAfterthevictoryofthelongMarch,hehasbeentheNorthwestOfficeoftheFederationofStateenterprisesMinister,ShenmufuguSARmissions,DirectorofNingxiaCountypartyCommitteeSecretaryandrecorderoftheCountypartyCommitteeSecretary,Ministersand41\n4.54.6Q内=-QQ外=Q或4.7(1)一定相等.是等势体.(2)不一定.4.8解:由题意和场强叠加原理,两导线间,距导线为x点的场强为(注:答案中黑体表示矢量)dx12-由高斯定理,有则两导线间的电势差为故单位长度的电容为4.9解:由高斯定理,有(注:答案中黑体表示矢量)1234S3S1S2EE或其中otherstaffoftheCentre.Duringthewar,ZhuwastransferredbacktoJiangxi,andDirectorofthenewOfficeinJingdezhen,JiangxiCommitteeSecretary.Startingin1939servedasrecorderoftheWestNorthOrganization,SecretaryoftheSpecialCommitteeAfterthevictoryofthelongMarch,hehasbeentheNorthwestOfficeoftheFederationofStateenterprisesMinister,ShenmufuguSARmissions,DirectorofNingxiaCountypartyCommitteeSecretaryandrecorderoftheCountypartyCommitteeSecretary,Ministersand41\n为单位矢量(要有必要的步骤,要求同作业2)=0(静电感应:要作简单的说明)4.10证明:(1)做出如图所示的高斯面S1,由于导体内部场强为零,侧面法线方向与场强方向垂直,故由高斯定理有所以,S1面内电荷数为零,即。(2)做出如图所示的高斯面S2,由于,又E左=E右=E,故由高斯定理可得,有。再做高斯面S3,可知此时有。两式联立,即可得证。(3)(注:答案中黑体表示矢量)4.11(1)方向为垂直导体面向外;(2)没有变化;(3)内部场强不变.otherstaffoftheCentre.Duringthewar,ZhuwastransferredbacktoJiangxi,andDirectorofthenewOfficeinJingdezhen,JiangxiCommitteeSecretary.Startingin1939servedasrecorderoftheWestNorthOrganization,SecretaryoftheSpecialCommitteeAfterthevictoryofthelongMarch,hehasbeentheNorthwestOfficeoftheFederationofStateenterprisesMinister,ShenmufuguSARmissions,DirectorofNingxiaCountypartyCommitteeSecretaryandrecorderoftheCountypartyCommitteeSecretary,Ministersand41\n作业55.1(C)5.2(A)5.3(B)5.4,(R1R3区域,同上一样,可求得:作业77.1B[解]7.2A[解]7.37.4,7.5铜的电阻率与铝的电阻率之比7.6[解](1)(2)otherstaffoftheCentre.Duringthewar,ZhuwastransferredbacktoJiangxi,andDirectorofthenewOfficeinJingdezhen,JiangxiCommitteeSecretary.Startingin1939servedasrecorderoftheWestNorthOrganization,SecretaryoftheSpecialCommitteeAfterthevictoryofthelongMarch,hehasbeentheNorthwestOfficeoftheFederationofStateenterprisesMinister,ShenmufuguSARmissions,DirectorofNingxiaCountypartyCommitteeSecretaryandrecorderoftheCountypartyCommitteeSecretary,Ministersand41\n7.7[解](1)这里的球壳层的横截面可认为相同,球壳层的漏电阻:(2)(3)(4)a1O1a2O2作业8稳恒磁场解答8.1.(B):8VVa1423,8.2.(A)B=0.8´10-4TotherstaffoftheCentre.Duringthewar,ZhuwastransferredbacktoJiangxi,andDirectorofthenewOfficeinJingdezhen,JiangxiCommitteeSecretary.Startingin1939servedasrecorderoftheWestNorthOrganization,SecretaryoftheSpecialCommitteeAfterthevictoryofthelongMarch,hehasbeentheNorthwestOfficeoftheFederationofStateenterprisesMinister,ShenmufuguSARmissions,DirectorofNingxiaCountypartyCommitteeSecretaryandrecorderoftheCountypartyCommitteeSecretary,Ministersand41\nTIabdec120°OR8.3.,方向垂直纸面向里。8.4.;a。(n的方向为垂直纸面向里,与电流之间满足右手螺旋关系)8.5.与电源相连的两根导线的电流方向相反,扭在一起可以使磁场尽可能相互抵消,以免产生磁干扰。8.6.(批改此题时要注意方向和数量级)由毕奥——萨伐尔定律可知,在本题中的电流元在各点处产生的磁感应强度表达式为,所以有①a②a③a④aotherstaffoftheCentre.Duringthewar,ZhuwastransferredbacktoJiangxi,andDirectorofthenewOfficeinJingdezhen,JiangxiCommitteeSecretary.Startingin1939servedasrecorderoftheWestNorthOrganization,SecretaryoftheSpecialCommitteeAfterthevictoryofthelongMarch,hehasbeentheNorthwestOfficeoftheFederationofStateenterprisesMinister,ShenmufuguSARmissions,DirectorofNingxiaCountypartyCommitteeSecretaryandrecorderoftheCountypartyCommitteeSecretary,Ministersand41\n⑤a8.7.(批改此题时要注意单位和数量级)电子绕核高速运动,相当于一个载流圆线圈,其运流电流为该圆线圈在其中心处产生的磁感应强度和对应的磁矩的大小分别为或B=12.5T8.8.用场强叠加原理求。半圆柱面上的面电流密度为,建立如图所示坐标系。在半圆柱面上取一平行于轴线的窄长条dl,其上载有电流,(此处)将其视为线电流,它在轴线上产生的磁感应强度为dB,其大小为方向如图,由对称性分析可知,otherstaffoftheCentre.Duringthewar,ZhuwastransferredbacktoJiangxi,andDirectorofthenewOfficeinJingdezhen,JiangxiCommitteeSecretary.Startingin1939servedasrecorderoftheWestNorthOrganization,SecretaryoftheSpecialCommitteeAfterthevictoryofthelongMarch,hehasbeentheNorthwestOfficeoftheFederationofStateenterprisesMinister,ShenmufuguSARmissions,DirectorofNingxiaCountypartyCommitteeSecretaryandrecorderoftheCountypartyCommitteeSecretary,Ministersand41\n方向与轴线垂直。作业9稳恒磁场解答9.1.(B)F不变,B增大F不变:磁感应线为闭合曲线(磁场的高斯定理);B增大:长直载流导线的磁场距离成反比。9-2.(A)反比于B,正比于v2。因为电子在磁场中运动的轨道半径为,通过其所围的面积内的磁通量为9.3.:①根据安培环路定理;②真实螺线管。9.4.;方向;;方向由对称性知道二者磁场通过矩形的磁通量相等,正向一致。_9.5.因为磁场的分布呈轴对称性,所以以轴线上一点为中心,以任意半径r分别在管内、管壁内、管外作三条安培环路,有otherstaffoftheCentre.Duringthewar,ZhuwastransferredbacktoJiangxi,andDirectorofthenewOfficeinJingdezhen,JiangxiCommitteeSecretary.Startingin1939servedasrecorderoftheWestNorthOrganization,SecretaryoftheSpecialCommitteeAfterthevictoryofthelongMarch,hehasbeentheNorthwestOfficeoftheFederationofStateenterprisesMinister,ShenmufuguSARmissions,DirectorofNingxiaCountypartyCommitteeSecretaryandrecorderoftheCountypartyCommitteeSecretary,Ministersand41\n当时,当时,当时,所以根据安培环路定律有由此得空间各点的磁场分布为9.6.解:(1)设线圈总匝数为N,以环心为中心,r为半径作一圆形安培环路(),由电流分布可知,环路上各点B大小相等,方向沿环路切向,根据安培环路定理可得得(2)在任意半径r处取长为h宽为dr的小截面,其磁通量为_otherstaffoftheCentre.Duringthewar,ZhuwastransferredbacktoJiangxi,andDirectorofthenewOfficeinJingdezhen,JiangxiCommitteeSecretary.Startingin1939servedasrecorderoftheWestNorthOrganization,SecretaryoftheSpecialCommitteeAfterthevictoryofthelongMarch,hehasbeentheNorthwestOfficeoftheFederationofStateenterprisesMinister,ShenmufuguSARmissions,DirectorofNingxiaCountypartyCommitteeSecretaryandrecorderoftheCountypartyCommitteeSecretary,Ministersand41\nabdcB1B2B¢1B¢29.7.用安培环路定理,可以证明图中B1=B2;用高斯定理,可以证明图中B¢1=B¢2。_命题得证具体证明如下:在该磁场中作任意尺寸的环路abcd,要求ab、cd与所在处的磁感应线平行,bc、da与所在处的磁感应线垂直,如图所示。则有因为空间中无电流选一个如图所示的高斯面,使其包围一条磁感应线,且对称轴与该磁感应线平行,则由磁场中的高斯定理可得所以命题得证。作业10稳恒磁场10-1.(C)N型半导体是电子导电,电子向b端漂移otherstaffoftheCentre.Duringthewar,ZhuwastransferredbacktoJiangxi,andDirectorofthenewOfficeinJingdezhen,JiangxiCommitteeSecretary.Startingin1939servedasrecorderoftheWestNorthOrganization,SecretaryoftheSpecialCommitteeAfterthevictoryofthelongMarch,hehasbeentheNorthwestOfficeoftheFederationofStateenterprisesMinister,ShenmufuguSARmissions,DirectorofNingxiaCountypartyCommitteeSecretaryandrecorderoftheCountypartyCommitteeSecretary,Ministersand41\n(霍尔效应)10-2.(D)竖直向上10-3.(B)10-4.电子的速率方向10-5.向三角形中心运动。因为同向电流导线相吸引。10-6.解:(1)当金属中自由电子所受磁场的洛仑兹力与电场力平衡时:,(2)由ab两端的霍尔电势差:得电子浓度为:(3)电子向b端偏转,所以a点电势高,b点电势低(4)P型半导体是空穴导电,即正电子导电,所以与上相反,b点电势高。10-7.解:(1)线圈所受的力矩大小:(2)力矩做的功:这里:;所以:otherstaffoftheCentre.Duringthewar,ZhuwastransferredbacktoJiangxi,andDirectorofthenewOfficeinJingdezhen,JiangxiCommitteeSecretary.Startingin1939servedasrecorderoftheWestNorthOrganization,SecretaryoftheSpecialCommitteeAfterthevictoryofthelongMarch,hehasbeentheNorthwestOfficeoftheFederationofStateenterprisesMinister,ShenmufuguSARmissions,DirectorofNingxiaCountypartyCommitteeSecretaryandrecorderoftheCountypartyCommitteeSecretary,Ministersand41\n10-8.不能。因为带电粒子在磁场中所受洛仑兹力始终与粒子运动方向垂直,它只能改变粒子的运动方向,不能改变速度的大小,所以不能用该力来增大粒子的动能。10-9.来自于磁场的能量。作业11稳恒磁场11-1.D.抗磁质:11-2.C.的环流只与L内所包围的传导电流有关。11-3.铁磁质;顺磁质;抗磁质。11-4.剩磁;矫顽力。11-5.解:(1)以环心为中心,r为半径作一圆形安培环路,由有介质的安培环路定理可知对管内,此时无磁介质,则:(2)管内充满磁介质时,(3)磁介质内由导线中电流产生的:由磁化电流产生的:otherstaffoftheCentre.Duringthewar,ZhuwastransferredbacktoJiangxi,andDirectorofthenewOfficeinJingdezhen,JiangxiCommitteeSecretary.Startingin1939servedasrecorderoftheWestNorthOrganization,SecretaryoftheSpecialCommitteeAfterthevictoryofthelongMarch,hehasbeentheNorthwestOfficeoftheFederationofStateenterprisesMinister,ShenmufuguSARmissions,DirectorofNingxiaCountypartyCommitteeSecretaryandrecorderoftheCountypartyCommitteeSecretary,Ministersand41\n11-6.解:在导体内部和外部分别以轴线为中心,做以r为任意半径的圆环为安培环路,则由安培环路定理可得根据B和H的关系式得11-7.不能。介质中的安培环路定理说明定理的左端,即的环流只与传导电流有关,与分子电流无关;但并不可以说只与传导电流有关,与分子电流无关或与磁介质无关。这里的环流和是两个不同的概念。在一般情况下otherstaffoftheCentre.Duringthewar,ZhuwastransferredbacktoJiangxi,andDirectorofthenewOfficeinJingdezhen,JiangxiCommitteeSecretary.Startingin1939servedasrecorderoftheWestNorthOrganization,SecretaryoftheSpecialCommitteeAfterthevictoryofthelongMarch,hehasbeentheNorthwestOfficeoftheFederationofStateenterprisesMinister,ShenmufuguSARmissions,DirectorofNingxiaCountypartyCommitteeSecretaryandrecorderoftheCountypartyCommitteeSecretary,Ministersand41\n是与分子电流有关的,因为介质磁化产生的分子电流也激发磁场,这可以从公式或中看到。作业12电磁感应(一)12-1.(D)12-3.(B)12-3.012-4.解::在棒上距长直导线为r处取一个线元,方向由a到b,,方向由a到b,所以金属棒中感应电动势为:a端电势高12-5.解:线框的上下两条边不切割磁力线,所以不产生感应电动势,只有左右两条边切割磁力线产生感应电动势,在d=10cm时,设左边处的磁感应强度为,右边处为,则此时线框中的磁感应电动势为:方向:顺时针方向12-6.解:otherstaffoftheCentre.Duringthewar,ZhuwastransferredbacktoJiangxi,andDirectorofthenewOfficeinJingdezhen,JiangxiCommitteeSecretary.Startingin1939servedasrecorderoftheWestNorthOrganization,SecretaryoftheSpecialCommitteeAfterthevictoryofthelongMarch,hehasbeentheNorthwestOfficeoftheFederationofStateenterprisesMinister,ShenmufuguSARmissions,DirectorofNingxiaCountypartyCommitteeSecretaryandrecorderoftheCountypartyCommitteeSecretary,Ministersand41\n对产生电动势起作用的是垂直于速度的磁场分量V12-7.答:由法拉第电磁感应定律可知,由于在两个环的尺寸相同,通过两个环的磁通量对时间的变化率也相同,所以两环所在处产生的感应电场和感应电动势的大小均相同,但由于木环中没有自由电子,所以在木环中没有感应电流产生,但在铜环中会有感应电流产生,因为铜是导体。作业13电磁感应(二)13-1.(D)(由即可得到)13-2.(A)(参看例题)13-3.解:穿过一匝线圈的磁通量为:由法拉第电磁感应定律可知,感应电动势为:otherstaffoftheCentre.Duringthewar,ZhuwastransferredbacktoJiangxi,andDirectorofthenewOfficeinJingdezhen,JiangxiCommitteeSecretary.Startingin1939servedasrecorderoftheWestNorthOrganization,SecretaryoftheSpecialCommitteeAfterthevictoryofthelongMarch,hehasbeentheNorthwestOfficeoftheFederationofStateenterprisesMinister,ShenmufuguSARmissions,DirectorofNingxiaCountypartyCommitteeSecretaryandrecorderoftheCountypartyCommitteeSecretary,Ministersand41\n所以,时的感应电动势的大小为V,方向为顺时针方向。13-1.解:线框abcda所围面积总只有abefa一部分有磁通量,此面积,感应电动势负号表示其方向由d到c。13-2.解:穿过导体杆L与导轨形成的线圈的磁通量:导体杆中的电动势为:6.块状金属导体处于变化的磁场中或者相对磁场运动时,在导体中都可能产生涡电流。所以,对于均匀磁场,当块状金属导体一部分进入或离开均匀磁场且切割磁力线运动时,或在均匀磁场中旋转有磁通量改变时,就能产生涡电流。otherstaffoftheCentre.Duringthewar,ZhuwastransferredbacktoJiangxi,andDirectorofthenewOfficeinJingdezhen,JiangxiCommitteeSecretary.Startingin1939servedasrecorderoftheWestNorthOrganization,SecretaryoftheSpecialCommitteeAfterthevictoryofthelongMarch,hehasbeentheNorthwestOfficeoftheFederationofStateenterprisesMinister,ShenmufuguSARmissions,DirectorofNingxiaCountypartyCommitteeSecretaryandrecorderoftheCountypartyCommitteeSecretary,Ministersand41\n7.是为了减小涡流损耗。使涡流受绝缘的限制,只能在薄片范围内流动,增大了电阻,减小了涡流,使损耗降低。作业14电磁感应(三)14-1.答案C。解:对于长直螺线管自感系数14-2.答案B。解:自感系数只与线圈的几何形状,大小及周围磁介质分布有关,与电流无关。14-3.解:磁能密度,若按质能关系其中c为真空光速。14-4解:(1)此题可以直接利用书上的结论求,,也可以按照定义(2)14-5.解:(1)螺线管2激发的磁场为通过线圈1的磁通量为其中(因为)otherstaffoftheCentre.Duringthewar,ZhuwastransferredbacktoJiangxi,andDirectorofthenewOfficeinJingdezhen,JiangxiCommitteeSecretary.Startingin1939servedasrecorderoftheWestNorthOrganization,SecretaryoftheSpecialCommitteeAfterthevictoryofthelongMarch,hehasbeentheNorthwestOfficeoftheFederationofStateenterprisesMinister,ShenmufuguSARmissions,DirectorofNingxiaCountypartyCommitteeSecretaryandrecorderoftheCountypartyCommitteeSecretary,Ministersand41\n(2)14-6.解:磁场被限制在同轴电缆内,由安培环路定理可得因而磁场能量也只储存在电缆内。由磁场分布的特点可知,在距轴线为r处磁场的大小相同,因而这些位置的磁能密度相同。在圆柱体内、外磁场的分布不同,因而磁能密度的表达式不同,我们分别求出长为l的电缆中在及区域的磁能,两者之和即为长为l的电缆中的总磁能。在()范围内,取长度为l半径为r和的同轴圆柱筒作为体积元dV,(),此体积元内的磁能密度为otherstaffoftheCentre.Duringthewar,ZhuwastransferredbacktoJiangxi,andDirectorofthenewOfficeinJingdezhen,JiangxiCommitteeSecretary.Startingin1939servedasrecorderoftheWestNorthOrganization,SecretaryoftheSpecialCommitteeAfterthevictoryofthelongMarch,hehasbeentheNorthwestOfficeoftheFederationofStateenterprisesMinister,ShenmufuguSARmissions,DirectorofNingxiaCountypartyCommitteeSecretaryandrecorderoftheCountypartyCommitteeSecretary,Ministersand41\n该体积元内储存的能量为则长为l的一段电缆中,在区域内储存的磁场能量为,同样在()范围内,取长度为l的一段柱壳(),其中储存的磁能为因此,长度为l的一段电缆中储存的总磁能为单位长度内储存的磁能为由自感磁能与自感系数的关系得14-7.这是由于电路突然断电,使得回路中产生了强烈的自感应现象,出现了一个很大的自感电动势,产生很强的电场,击穿空气。otherstaffoftheCentre.Duringthewar,ZhuwastransferredbacktoJiangxi,andDirectorofthenewOfficeinJingdezhen,JiangxiCommitteeSecretary.Startingin1939servedasrecorderoftheWestNorthOrganization,SecretaryoftheSpecialCommitteeAfterthevictoryofthelongMarch,hehasbeentheNorthwestOfficeoftheFederationofStateenterprisesMinister,ShenmufuguSARmissions,DirectorofNingxiaCountypartyCommitteeSecretaryandrecorderoftheCountypartyCommitteeSecretary,Ministersand41\n14-8当两个线圈平面相互平行放置,并且它们的中心在同一条直线上时,其互感最大;当一个线圈平面平行于两圆心连线,另一个线圈平面垂直于两圆心连线时,其互感为零。大学物理作业15电磁场理论15-1.答案D。15-2.答案D。解:,而可以是常数。15-3.答案B。解:电磁波的表达式得知,电磁波的传播方向沿y轴负方向,与电场、磁场方向满足如下关系15-4.解:相同;;垂直;垂直;横波。15-5.解:(1)极板上电量Q=It,电荷面密度,极板间电场强度,则电场的变化率otherstaffoftheCentre.Duringthewar,ZhuwastransferredbacktoJiangxi,andDirectorofthenewOfficeinJingdezhen,JiangxiCommitteeSecretary.Startingin1939servedasrecorderoftheWestNorthOrganization,SecretaryoftheSpecialCommitteeAfterthevictoryofthelongMarch,hehasbeentheNorthwestOfficeoftheFederationofStateenterprisesMinister,ShenmufuguSARmissions,DirectorofNingxiaCountypartyCommitteeSecretaryandrecorderoftheCountypartyCommitteeSecretary,Ministersand41\n(2)极板间的电位移为,则位移电流密度(3)极板间的位移电流,(4)由安培环路定理,15-6.解:极板间的位移电流密度以r为半径绕极板中心作圆形安培环路,由安培环路定理:解出yzx15-7.解:根据电磁波的特点可以判断,电场强度的振动方向一定沿着y轴的正方向,在真空中,有,所以作业16otherstaffoftheCentre.Duringthewar,ZhuwastransferredbacktoJiangxi,andDirectorofthenewOfficeinJingdezhen,JiangxiCommitteeSecretary.Startingin1939servedasrecorderoftheWestNorthOrganization,SecretaryoftheSpecialCommitteeAfterthevictoryofthelongMarch,hehasbeentheNorthwestOfficeoftheFederationofStateenterprisesMinister,ShenmufuguSARmissions,DirectorofNingxiaCountypartyCommitteeSecretaryandrecorderoftheCountypartyCommitteeSecretary,Ministersand41\n1.在地面参考系测得一星球离地球5光年,宇航员欲将此距离缩为3光年,他乘的飞船相对地球的速度应是[]A.B.C.D.答:[C]解:这里,要求宇航员的钟走3光年,是原时:;地面上的时钟测量,宇航员走5光年,是测时:。因此由,得到,。注意:地面上仍然认为宇航员走了5光年。2.火箭的固有长度为,其相对地面以作匀速直线运动。若火箭上尾部一射击口向火箭首部靶子以速度发射一子弹,则在火箭上测得子弹从出射到击中靶的时间间隔为[]。A.B.C.D.答:[B]解:事件发生在火箭上,与地面无关。当然,地面上测量这一时间间隔是不同的。3.在惯性系中轴上相距处有两只同步钟和,在相对系沿轴以速运动的惯性系中也有一只同样的钟。若轴平行,当相遇时,恰好两钟读数都为零,则当与相遇时系中钟的读数为,系中钟的读数为。答:,解:如图,在系测量,和的距离为,钟otherstaffoftheCentre.Duringthewar,ZhuwastransferredbacktoJiangxi,andDirectorofthenewOfficeinJingdezhen,JiangxiCommitteeSecretary.Startingin1939servedasrecorderoftheWestNorthOrganization,SecretaryoftheSpecialCommitteeAfterthevictoryofthelongMarch,hehasbeentheNorthwestOfficeoftheFederationofStateenterprisesMinister,ShenmufuguSARmissions,DirectorofNingxiaCountypartyCommitteeSecretaryandrecorderoftheCountypartyCommitteeSecretary,Ministersand41\n正在以速度从向运动,钟从到达所用的时间为这也就是钟的读数。由于和在系中是静止的,所以,系中测量,和的距离是原长;在系看来,和以速度运动,和的距离是测量长度,因此由于在系看来,以速度运动,运动距离所用时间为这就是钟的读数。可见,钟与otherstaffoftheCentre.Duringthewar,ZhuwastransferredbacktoJiangxi,andDirectorofthenewOfficeinJingdezhen,JiangxiCommitteeSecretary.Startingin1939servedasrecorderoftheWestNorthOrganization,SecretaryoftheSpecialCommitteeAfterthevictoryofthelongMarch,hehasbeentheNorthwestOfficeoftheFederationofStateenterprisesMinister,ShenmufuguSARmissions,DirectorofNingxiaCountypartyCommitteeSecretaryandrecorderoftheCountypartyCommitteeSecretary,Ministersand41\n钟相遇时,确实是:钟读数小、钟读数大,即似乎确实能分辨出来“钟慢、钟快”。钟相对于系运动,钟确实应该慢;而在系看来,钟也是运动的,也经该慢。这似乎出现了矛盾。如图,认为:钟与钟相遇时,钟与钟根本没有校准,钟的指针比钟提前。(或者,从经典物理大致考虑:信号的传播是需要时间的,钟指针指向“0”这一信号传到时,将钟调到“0”,此时,钟已经过“0”了,即钟比钟提前了)。如图,认为:在对“相遇到钟与钟相遇所用时间的测量”中,和钟的测量结果是一样的,都比钟测量的结果短,即和钟都慢;只不过是在相遇时,钟的指针“提前”了,从而在钟与钟相遇时,钟的“读数”比钟的“读数”大。可见,上面的结果,并不违反相对论,反而正是相对论的必然结果。4.根据狭义相对论的原理,时间和空间的测量值都是,它们与观测者的密切相关。答:相对的,相对运动状态。5.、系是坐标轴相互平行的两个惯性系,系相对与沿轴正方向匀速运动。一刚性尺静止于系中,且与轴成角,而在系中测得该尺与轴成角,试求:、系的相对运动速度。解:如图,在系中测量,,所以在系中测量,,所以由洛伦兹变换,得到,otherstaffoftheCentre.Duringthewar,ZhuwastransferredbacktoJiangxi,andDirectorofthenewOfficeinJingdezhen,JiangxiCommitteeSecretary.Startingin1939servedasrecorderoftheWestNorthOrganization,SecretaryoftheSpecialCommitteeAfterthevictoryofthelongMarch,hehasbeentheNorthwestOfficeoftheFederationofStateenterprisesMinister,ShenmufuguSARmissions,DirectorofNingxiaCountypartyCommitteeSecretaryandrecorderoftheCountypartyCommitteeSecretary,Ministersand41\n6.一匀质矩形薄板,静止时边长分别为和,质量,试计算在相对薄板沿一边长以速运动的惯性系中测得板的面密度。解:在相对于板运动的参照系中,长度收缩,同时质量增大。质量为:;长度为:,质量密度为7.列车和隧道静止时长度相等,当列车以的高速通过隧道时,分别在地面和列车上测量,列车长度与隧道长度的关系如何?若地面观测者发现当列车完全进入隧道时,隧道是的进、出口处同时发生了雷击,未击中列车,按相对论的理论,列车上的旅客会测得列车遭雷击了吗?为什么?解:(1)由于隧道相对于地面是静止的,而列车是运动的,所以,地面测量隧道的长度是原长,地面测量列车的长度是测长,即地面测量:隧道长,列车长地面测量隧道长与列车长的关系为:由于列车相对于列车是静止的,而隧道是运动的,所以,列车测量列车的长度是原长,列车测量隧道的长度是测长,即列车测量:列车长,隧道长地面测量隧道长与列车长的关系为:otherstaffoftheCentre.Duringthewar,ZhuwastransferredbacktoJiangxi,andDirectorofthenewOfficeinJingdezhen,JiangxiCommitteeSecretary.Startingin1939servedasrecorderoftheWestNorthOrganization,SecretaryoftheSpecialCommitteeAfterthevictoryofthelongMarch,hehasbeentheNorthwestOfficeoftheFederationofStateenterprisesMinister,ShenmufuguSARmissions,DirectorofNingxiaCountypartyCommitteeSecretaryandrecorderoftheCountypartyCommitteeSecretary,Ministersand41\n(2)地面测得雷击时刻火车完全位于隧道内,没有遭雷击。列车上的测量同样得出列车没有遭雷击。设列车头到达隧道出口为事件,闪电到达隧道出口为事件;列车尾到达隧道进口为事件,闪电到达隧道进口为事件。在地面上测量,事件与事件是同时同地发生的两个事件,在任何惯性系中测量都是同时发生的,因此在列车上测量,事件与事件是同时同地发生的两个事件,即在列车上测量,列车头与闪电同时到达隧道出口,闪电没有击中列车头;在地面上测量,事件与事件是同时同地发生的两个事件,在任何惯性系中测量都是同时发生的,因此在列车上测量,事件与事件是同时同地发生的两个事件,即在列车上测量,列车尾与闪电同时到达隧道进口,闪电没有击中列车尾。事实上,“列车头到达隧道出口的事件”与“列车尾到达隧道进口的事件”,是在地面这一惯性系中不同地点同时发生的两个事件,在列车这一惯性系中测量就不可能是同时的;“闪电到达隧道出口的事件”与“闪电到达隧道进口的事件”,是在地面这一惯性系中不同地点同时发生的两个事件,在列车这一惯性系中测量就不可能是同时的。设“闪电到达隧道出口的事件”在地面测量,在列车上测量;“闪电到达隧道进口的事件”在地面测量,在列车上测量。由于,则可见,出口处雷击先发生,此时列车头部未出隧道;入口处雷击后发生,此时列车尾部进入隧道。otherstaffoftheCentre.Duringthewar,ZhuwastransferredbacktoJiangxi,andDirectorofthenewOfficeinJingdezhen,JiangxiCommitteeSecretary.Startingin1939servedasrecorderoftheWestNorthOrganization,SecretaryoftheSpecialCommitteeAfterthevictoryofthelongMarch,hehasbeentheNorthwestOfficeoftheFederationofStateenterprisesMinister,ShenmufuguSARmissions,DirectorofNingxiaCountypartyCommitteeSecretaryandrecorderoftheCountypartyCommitteeSecretary,Ministersand41\n作业171.实验室测得粒子的总能量是其静止能量的K倍,则其相对实验室的运动速度为[]A.B.C.D.答:[C]解:,,,2.把一静止质量为的粒子,由静止加速到,所需作的功为[]A.B.C.D.答:[D]解:3.观测者乙以的速率相对观测者甲运动,若甲携带质量为的物体,则(1)乙测得物体的质量为:;(2)甲测得物体的总能量为:;(3)乙测得物体的总能量为:。答:;;。解:,(1)物体相对于乙在运动,测得的是运动质量otherstaffoftheCentre.Duringthewar,ZhuwastransferredbacktoJiangxi,andDirectorofthenewOfficeinJingdezhen,JiangxiCommitteeSecretary.Startingin1939servedasrecorderoftheWestNorthOrganization,SecretaryoftheSpecialCommitteeAfterthevictoryofthelongMarch,hehasbeentheNorthwestOfficeoftheFederationofStateenterprisesMinister,ShenmufuguSARmissions,DirectorofNingxiaCountypartyCommitteeSecretaryandrecorderoftheCountypartyCommitteeSecretary,Ministersand41\n(2)物体相对于甲静止,测得的是静止能量(3)物体相对于乙运动,测得的是运动的总能量4.电子静止质量,当它以的速度运动时,按相对论理论,其总能量为,动能为,按经典理论,其动能为。答:;;。解:(1)(2)动能等于总能与静止能量之差(3)按经典物理,电子动能为5.子的静止质量是电子质量的207倍,在其自身参照系中平均寿命。若在实验室参照系中得其平均寿命。试问:实验室测得其质量是电子静止质量的多少倍?解:依题,由子的静止寿命(原时)与运动寿命(测时),可以求得子的运动速度otherstaffoftheCentre.Duringthewar,ZhuwastransferredbacktoJiangxi,andDirectorofthenewOfficeinJingdezhen,JiangxiCommitteeSecretary.Startingin1939servedasrecorderoftheWestNorthOrganization,SecretaryoftheSpecialCommitteeAfterthevictoryofthelongMarch,hehasbeentheNorthwestOfficeoftheFederationofStateenterprisesMinister,ShenmufuguSARmissions,DirectorofNingxiaCountypartyCommitteeSecretaryandrecorderoftheCountypartyCommitteeSecretary,Ministersand41\n则子的运动质量为子的静止质量与电子的静止质量之比为则子的运动质量与电子的运动质量之比为6.粒子的动能等于其静止能量的一半,求其运动速度。解:;;7.已知系相对系以的速度沿X轴正向运动。一静止质量为的粒子也沿X轴运动,在系中测得粒子速率。求:(1)相对系,粒子的动能;(2)相对系,粒子的速度;(3)在系中测,粒子的总能量。解:(1);;(2);(3)otherstaffoftheCentre.Duringthewar,ZhuwastransferredbacktoJiangxi,andDirectorofthenewOfficeinJingdezhen,JiangxiCommitteeSecretary.Startingin1939servedasrecorderoftheWestNorthOrganization,SecretaryoftheSpecialCommitteeAfterthevictoryofthelongMarch,hehasbeentheNorthwestOfficeoftheFederationofStateenterprisesMinister,ShenmufuguSARmissions,DirectorofNingxiaCountypartyCommitteeSecretaryandrecorderoftheCountypartyCommitteeSecretary,Ministersand41\n8.根据相对论的理论,实物粒子在介质中的运动速度是否有可能大于光在该介质中的传播速度。讨论:相对论只给出真空中的光速是物体的极限速度,光在介质中的速度小于光速,则实物粒子在介质中的速度有可能大于光在介质中的传播速度。9.如果、是惯性系中互为因果关系的两个事件(是的原因,先于发生),问:是否能找到一个惯性系,在该系中测得先于发生,出现时间顺序颠倒的现象?答:不能。解:(数学讨论略)相对论理论也不可能不遵守客观事实。因果律不能颠倒,但没有因果联系的两个事件,在不同的惯性系中,发生的先后顺序有可能颠倒。otherstaffoftheCentre.Duringthewar,ZhuwastransferredbacktoJiangxi,andDirectorofthenewOfficeinJingdezhen,JiangxiCommitteeSecretary.Startingin1939servedasrecorderoftheWestNorthOrganization,SecretaryoftheSpecialCommitteeAfterthevictoryofthelongMarch,hehasbeentheNorthwestOfficeoftheFederationofStateenterprisesMinister,ShenmufuguSARmissions,DirectorofNingxiaCountypartyCommitteeSecretaryandrecorderoftheCountypartyCommitteeSecretary,Ministersand41
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