3.科考队逃人某一磁矿区域后，发现指南针静止时，N极指问为北偏东30°
30
2022~2023学年第二学期高二年级期中质量监测
如图所示，设该位置地磁场磁感应强度的水平分量为5，磁矿所产生的
磁感应强度水平分量最小位为
物理试卷
A号
3.B
D.V3B
4.质量为m的通电细杆置于倾角为0的光滑斜面上，杆巾电流方向垂直于纸面向内。电流恒
(考试时间：上午8：009：30
定，余属杆长度不变，现加一匀强磁场，使杆能在斜所上静止。所加磁场磁感应强度最小
说明：本试卷为闭卷笔答，答题时间90分钟，满分100分。
的，其方向为
⑧
四
好
A.水平向右
B.竖直向下
分
C.沿斜面向下
D.垂直斜面向上
得分
5.如阁所示，甲为某无线门铃按辄，其原理如图乙所示，按下门铃按钮，磁铁靠近螺线管；松开
门铃按红，磁铁远离螺线督，何归原位置。下列说法正确的是
螺线管
磁佚
第1卷（选样题共45分）
-按下按机
松开按钮
、单项选择题：本题共10小题，每小题3分，共30分。请将正确选项前字母标号填入下表内
相应位置。
题号
1
2
3
4
5
89
10
.按柱按卸不动，门铃可以持纹响
答案
B.松开按钮过，螺线管A端电势较高
G.按下安钮过程，通过门铃的电流方向是A到B
1.东汉时期思想家王充在《论衡》书中有关于“可南之杓，投之于地，其柢（即勺柄）指南”的记
D.无论快速按下按钮，还是缓憾按下按钮，螺线管产生的感应电流大小都相同
载，如图所示。司南是用天然磁铁矿石琢成一个勺形的东叫，放在一
平
6如图所示，粗细均匀的同种导线制成的单匝正方形线框αcd齿定丁匀强磁场中，线框平面
个刻着方位的光滑盘上。下列说法正确的是
磁场方向垂直，线框顶点a、d与直流电源相接，若山边受到的安培力大×××
A.“柢”相当于磁体的N极
小为F,则整个线框受到的安培力的大小为
B.司南在转动过程中产生感生电场
A.F
B.2
C.地球的磁场是人为想象出来的
C.3
D.4F
+2
D.司南的勺柄能够指南是中司南的磁场决定的
7.威尔逊云室是能城示带电粒子运动径迹的实验装苴，是研究微观粒子的重要器材：某次实
2.有关电和磁现象，下列说法正确的是
验时，在云室巾爪上匀强磁场，电子垂直磁场方向射人云室的运动轨迹如图所示，由于电子
运动过程中受到阻力的作用，因此其动能逐渐减小，下列说法正确的是
A.磁场垂直纸面问外，电子从A点运动到B点
B.磁场垂直纸向外，电子从B点运动到A点
C.磁场垂直纸而向甲，电子从A点运动到乃点
D.磁场垂直纸面向里，电子从B点运动到A点
甲
乙
丙
人
8.一台质谱仪的工作原理如图所示，电荷量和质量均相同的粒子飘人电压为的加速比场，
人.图甲中若该元心件州金属材料制作，则通入示的电流时，上表面电势比下表面电势低
其初速度几乎为零，这些粒了经加速后通过狭缝0沿着与磁场垂
B.图乙中用外力顺时针（从左边看）转动铜盘，电路中会产生感应电流，通过R的电流白上
直的万向逃人匀强磁场，粒子好能在底片上的点。已知放
而下
置底片的区域M=L,且OM=,若想要粒子始终能打在底片MW
G.图丙中使用电磁炉热食物时可以使用陶瓷锅
上，则加速电场的电压最大为
D.图门]中探雷器工作时，线圈中要有恒定的电流
A.v。
B.2Ug
C.4tp
D.8Ua
高二物理第1页〔共8页)
高一物理第2贞（共8）2022～2023学年第二学期高二年级期中质量监测
物理参考答案及评分建议
一、单项选择题：本题包含 10小题，每小题 3分，共 30分。
题号 1 2 3 4 5 6 7 8 9 10
选项 B A A D B D D C B D
二、多项选择题：本题包含 5小题，每小题 3分，共 15分。
题号 11 12 13 14 15
选项 BD BD AC ACD BD
三、实验题：共 14分。
16.（6分）
（1）T2＝4π2LC （2分）
（2）
（2分）
（3）28.2mH （27.3mH～29.0mH） （2分）
17.（8分）
（1）红 （4分） （2）400（4分）
四、计算题：共 41分。
18.（8分）
(1)根据几何关系可知离子在磁场中的运动半径为
L
R-Rcos60°= ············································································ （1分）
2
R=L
qU= mv02···················································································· （1分）
mv2
根据 qv0B= 0 ········································································· （1分）
L
可得： = ···········································································（1分）
T
(2)t= ·······················································································（1分）
6
2 L
T= ·····················································································（1分）
v0
2
可得：t = ··············································································（2分）
6
19.（10分）
（1）由图乙可知 时刻 ··········································（1分）
根据法拉第电磁感应定律得 ····································（1分）
代入数据解得： ································································· （1分）
（2） 内根据闭合电路欧姆定律可以得到： ········（1分）
根据焦耳定律可以得到
上产生的焦耳热为： ······························（1分）
内，同理可以得到： ，
根据焦耳定律可以得到
上产生的焦耳热为： ······························（1分）
所以 ····················································· （1分）
（3） 内，根据法拉第电磁感应定律得
根据闭合电路欧姆定律可以得到：
电荷量 ··············································································（1分）
代入数据可以解得： C················································（2分）
20A.（11分）
（1）粒子一定做匀速直线运动························································（1分）
q v0 B=qE····················································································（2分）
E=B v0························································································（1分）
方向与 x 轴负向夹角为 60°斜向左下················································ （1分）
（2）粒子做类平抛运动，等效高度 y′，等效射程 x′
tg60° = ' ···················································································（1分）
'
y′ = ···················································································（1分）
x′ = v0t ·······················································································（1分）
t = ···················································································（1分）
y′ = x′ = ································································（1分）
=
B 点坐标（ ， ） ·······················································（1分）
20B.（11分）
（1）做匀速直线运动····································································（2分）
Eminq= mgsin30°，E
min= ···························································（2分）
方向与 y轴正方向夹角为 60°斜向左上 ·············································（1分）
（2）小球做类平抛运动，等效高度 y′，等效射程 x′
tg60° = ' y′ = 洛 F 洛 = mgcos30° x′ = v0 t ·························· （2分） '
t = ·······················································································（1分）
y′= x′= ····································································（2分）
=
B点坐标（ ， ）······························································ （1分）
21A.（12分）
解析：（1）由机械能守恒： = ··········································（1分）
= ····················································································（1分）
= ·························································································（1分）
= ········································································（1分）
得 = ·····································································（1分）
（2） = ··············································································（1分）
= ·······················································································（1分）
= ·····················································································（1分）
= ·················································································（1分）
得 = ···················································································（1分）
（3 ）由能量守恒： + = ·······························（1分）
得 = ( + ) ···································································· （1分）
21B.（12分）
解析：（1）线框在上升阶段离开磁场到最大高度的过程中，
由动能定理： + = ··········································（2分）
+
解得： = ··································································· （1分）
（2）由题意知，线框在上升阶段刚进入磁场的速度为 2v。
= ·················································································· （1分）
= ·························································································（1分）
+ + = ····································································（2分）
+
= + + 解得 ························································（2分）
（3）由能量守恒： ( ) = + + ····················（2分）
解得 = （ + ） ······················································· （1分）
