高二物理上学期期末试题答案.pdf

高二物理答案 第 1页（共 3页） 2019－2020 学年度第一学期第二学段模块检测 高二物理答案及评分标准 一、单项选择题：本大题共 8 小题，每小题 3 分，共 24 分。 1——5：B、C、C、D、A；6——8：B、C、D 二、多项选择题：本大题共 4 小题，每小题 4 分，共 16 分，选不全得 2 分，有选错得 0 分。 9．ABD、10．AD、11．BD、12．AD 三、非选择题。 13．（1） C （3 分）（2） A （3 分） 14．（1）见右图（2 分） （2） 右（2 分） （3） A 、 断开开关。（每空 2 分，共 4 分） 15．（9 分） 解:（1）由几何关系可知，电子在磁场中的圆心角为θ＝240°····························· （1 分） 轨道半径为 0 3 2sin 60 3 dr d  ·································································（2 分） 根据牛顿第二定律： 2vevB m r  ································································· （1 分） 解得： 3 3 edBv m  ····················································································（1 分） （2）圆心角对于的弧长为 4 3 rs r   ·······················································（2 分） 粒子穿过磁场的时间 4 3 s rt v v   ································································ （1 分） 联立解得： 4 3 mt eB  ················································································· （1 分） 16. （9 分） 解：（1）导体棒切割磁感线运动，由于切割长度、磁感应强度均不变化，所以感应电动势 与切割速度 v 成正比，所以瞬时值表达式为 tTBLve m 2cos ···················································································（2 分）高二物理答案 第 2页（共 3页） 代入数据可得 te 10cos4.0 ···················································································· （2 分） （2）由瞬时值表达式可以看出，回路中的交流按余弦规律变化，所以产生的焦耳热为 2 mEE 有 ·························································································· （1 分） tR EQ m  总 1) 2 ( 2 ················································································（2 分） 代入数据可得： JQ 4 ················································································· （2 分） 17．（12 分） 解：（1）棒最大速度时 mBlvE  ··································································（2 分） 解得： smvm /3 ······················································································ （2 分） 对棒： mmvtBIl  ·····················································································（2 分） tIq  ···················································································（1 分） 对棒和电源 QmvqE m  2 2 1 ·······································································（2 分） QQ 3 2棒 ··············································································（1 分） 解得： JQ 5.1棒 ···························································································（2 分） 18．(16 分) 解：(1) 由题条件可判断粒子做圆周运动半径为 R＝d············ (1 分) 粒子在磁场中：qv0B＝mv20 R ······································(2 分) 解得 B＝mv0 qd ···························································(1 分) (2) 粒子运动轨迹如图示 粒子在磁场中运动时间： Tt 3 231  ································ (1 分) 2 24 TmrqvB  ·························································(1 分) qB mvr 0 =d······························································· (1 分)高二物理答案 第 3页（共 3页） 粒子在无场区运动时间： 0 2 323 v dt  ························································· 粒子再次回到 P 点时间：t＝t1＋t2····················································· (1 分) 解得： 0 364 v ddt   ···············································（1 分) (3) 粒子运动轨迹如图示 粒子速度变为 3v0，则在磁场中运动半径为 R′＝3d ··········································(1 分) 由 P 点出发后第一个圆弧的弧长： dds  5326 5 1  ··································· (1 分) 无磁场区圆的直径长度： dds 32322  1 粒子以 3v0 沿 y 轴正向经过 P 粒子运动路程 s＝k(6s1＋6s2)＝(30πd＋12 3d)k，其中 k＝1、2、3、 ···················(2 分) 2 粒子以 3v0 大小沿－y 方向经过 P s′＝3s1＋2s2＋k(6s1＋6s2 )，其中 k＝0、1、2、3、…········································ (1 分) 代入得 s′＝15πd＋4 3d＋k(30πd＋12 3d)，其中 k＝0、1、2、3、… ················(1 分)