大学物理第8章习题答案.ppt

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1、IIddddr解:B顺时针方向选择顺时针方向为正方向.8-7Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.8-12P解:Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.abc解:v方向ab方向cbEvaluationonly.

2、CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.8-13解电动势的方向:由B指向A,A端电势较高。BEvaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.8-16BOR证明：Evaluationonly.CreatedwithAspose.S

3、lidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.8-18解：连接oa和ob与ab构成回路oab.××××××××××××××××××××××××××××××××labLBO方向abEvaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.解：××××××××××××××××××××××××××××××××

4、labLBO方向abdlEKhEvaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.8-19解：螺绕环中磁场分布hR1R2R1R2rdrEvaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.解:µ1S1µ2S2N8-20Evaluation

5、only.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.解8-21Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.AA'BB'1)连接A和A'8-22Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5Clie

6、ntProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.AA'BB'2)连接A'和BEvaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.ABRBNBSNA8-23解：设线圈B中通有电流Ｉ，在小线圈A处产生的磁场：方向与线圈B中的电流同向．Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5Clien

7、tProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.解:8-26(1)Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.(2)Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.解：由安培环路

8、定律可得,在导线内部与导体中心轴线相距为r处的磁感强度的大小：8-27Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011A