大连市 2023～2024 学年度第一学期期末考试
高二物理参考答案
一、选择题：本题共 10 小题，共 46分。在每小题给出的四个选项中，第 1～7 题只有
一项符合题目要求，每个小题 4分；第 8～10题有多项符合题目要求，每小题 6分，全
部选对的得 6分，选对但不全的得 3分，有错选或不答的得 0分。
1 2 3 4 5 6 7 8 9 10
B C D A D A C AD BC BCD
二、实验题（本题共 2小题，共 14 分）
15
11．（1）1.875或 (2分)（2）AC (2分) （2）左(1分)、是(1分)
8
12．（1）如下图(共 3分，画出 V2给 1分，画出 R1给 1分，电路 1分) （3）0.34 (1分)
（4）1.41—1.43(1分)、0.837—0.915 (1分) （4）偏小(1分)、 偏小(1分)
2 （拟合图像参考图）
三、计算题（本题共 3小题，共 40 分。其中 13 题 10 分、14 题 14 分、15 题 16 分。解
答应写出必要的文字说明、方程式和重要的演算步骤，只写出最后答案的不能得分，有
数值计算的题答案中必须明确写出数值和单位。）
13．(10分)
（1）从左向右或从 p向 q ·········································································(1分)
（2）线圈产生电动势的最大值 Em=nBSω ······················································(1分)
e=Emcosωt ·····························································································(1分)
e=nBSωcosωt ························································································· (1分)
Em
（3）线圈产生电动势的有效值 E ·······················································(1分)
2
E
电路中电流的有效值 I ····································································(1分)
R r
电路中产生的焦耳热Q I 2 (R r)T ·····························································(1分)
T 2 ································································································(1分)
由题意，线圈匀速转动，则外力做的功等于电路中产生的焦耳热 W=Q ················(1分)
高二物理答案 第 1页 共 4页
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n2B2S 2
外力驱动线圈转动一周所做的功W ·············································(1分)
R r
14．(14分)
（1）在匀强电场中，粒子做类平抛运动，轨迹如图中 AO段所示
x方向由运动学公式 2L v0t ·····································································(1分)
1
y 2方向由运动学公式 L at ···································································(1分)
2
F
由牛顿第二定律得 a 电 F电 qE ····················································(共 1分)m
E mv
2
联立解得 0 ··················································································(1分)
2qL
（2）设粒子在 O点的速度 v与 x轴负方向夹角为α
粒子在 O点沿 y轴负方向分速度的大小 vy at ··············································(1分)
v
v v20 v
2
y tan
y ·····························································(共 1分)
v0
解得 v 2v0 45
粒子在第三象限做匀速圆周运动的轨迹为 OP段，半径为 R
v2
由牛顿第二定律得 qBv m ··································································(1分)
R
R 2mv解得 0
qB
由几何关系得 OP 2Rsin 45 ··································································(1分)
OP 2mv 0
qB
2mv
故 P点坐标为（0、 0 ）·····································································(1分)
qB
（3）解法一：粒子在第四象限做匀速圆周运动的轨迹为 PQ段，半径为 R1，转过的圆
心角为β，通过的弧长为 l
由几何关系得 R1=2R ················································································(1分)
3 ·······························································································(1分)
4
l 2 R1 ························································································(1分)2
l
粒子在第四象限运动的时间 t ······························································ (1分)
v
3 m
联立解得 t ···················································································(1分)
2qB
高二物理答案 第 2页 共 4页
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解法二：粒子在第四象限做匀速圆周运动的轨迹为 PQ段，半径为 R1，转过的圆心角为
β，由几何关系得 R1=2R ············································································ (1分)
3 ·······························································································(1分)
4
v2
由牛顿第二定律得 qB1v m ·································································(1分)R
v2 B
联立 qBv m 可得B1 R 2
2 R
粒子在第四象限运动的周期T 1 ························································(1分)
v
解得T 4 m
qB
t T= 3 m联立解得粒子在第四象限运动的时间 ····································(1分)
2 2qB
15．(16分)
（1）M刚进入磁场时
M产生的电动势 E 2BLv0 ······································································(1分)
E
回路中的电流 I0 ·············································································(1分)2R
M所受的安培力 F 2BI0L ·····································································(1分)安
对M由牛顿第二定律得 F 2ma ·····························································(1分)
安
B2L2v
联立解得M刚进入磁场时的加速度 a 0 ·············································(1分)
mR
（2）M进入磁场后在安培力作用下做减速运动，N做加速运动，直到两金属杆产生的
电动势等大、反向，回路中的电流为零，最终两金属杆都做匀速直线运动达到稳定状态。
设两金属杆的速度大小分别为 vM 、vN则有
高二物理答案 第 3页 共 4页
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2BLvM BLvN ·····················································································(1分)
可得 vN 2vM
从M刚进入磁场到两金属杆达到稳定状态的过程：
取M初速度方向为正方向
对M由动量定理得 2BILt 2mvM 2mv0 ·················································(1分)
对 N由动量定理得 BILt mvN 0 ·····························································(1分)
联立以上两式得 vM vN v0 ·····································································(1分)
v 2v
解得 vM 0 ，v 03 N 3
Q 1 2mv2 (1 2mv2 1 2两金属杆产生的总的焦耳热 0 M mvN ) ·····················(2分)2 2 2
2 2
解得Q mv
3 0
Q 1 2
所以此过程中M中产生的焦耳热QM mv0 ·········································(1分)2 3
（3）从M刚进入磁场到两金属杆距离最近的过程：
由（2）可知两杆的速度大小始终满足 vM vN v0
当两金属杆速度相等时距离最近，即 vM1 vN1 ··············································(1分)
v
解得 v v 0M1 N1 2
对 N由动量定理得 BI1Lt1 mvN1 0 ·························································(1分)
q I1t1 ·································································································(1分)
mv
联立解得 q 0 ··················································································(1分)
2BL
高二物理答案 第 4页 共 4页
{#{QQABYYaEogAIAAJAABhCUQUoCACQkACCACoOxAAMMAAAgBNABCA=}#}大连市2023~2024学年度第二学期期末考机
高二物理
2.答选择题时，选出每小题答案后，用铅笔把答题卡对应题目的答案标号涂黑。加
1、答卷前，考生务必将自己的姓名、准考证号填写在答题卡上：
注意事项：
需改动，用橡皮擦干净后，再选涂其他答案标号。答非选择题时，将答案写在答题卡上。
写在本试题卷上无效。
一、选择题（本题共10小题，第1一7题只有一项符合题目要求，每小题4分。第8一10
3.考试结束后，将本试卷和答题卡一并交回。
第I卷选择题（共46分）
厂题有多项符合题目要求，每小题6分，全部选对的得6分，选对但不全的得3分，有选
错的得0分。共46分。)
干件海人向面平点热
1.关于下列四幅图的说法中正确的是
接高频交
流电源
接姿流电
图(d)
A.如图()所示，真空治炼炉的炉外线圈通入高频交流电时，线圈中会产生大量热量，
图b)
图(c)
图(a
从而治炼金属
B.如图(b)所示，回旋加速器是利用磁场控制轨道，使带电粒子“转圈圈”，利用电场
进行加速的仪器
C.如图（ ）所示，运输时要把毫安表的正、负接线柱用导线连在一起，这是为了保护
电表指针，利用了电磁驱动原理
D.如图()所示，摇动手柄使蹄形磁铁转动，则铝框会和磁铁同向转动，且和磁铁转
得一样快
2.在赤道表面某位置小磁针静止时N极指向北偏东30，经研究是因为小磁针正下方有
一条通电直导线沿南北方向放置，己知该位置地磁场的磁感应强度大小为B(不考虑地
磁偏角的影响)则该通电直导线在小磁针处产生的磁场的磁感应强度大小为
品容由.A
A.√5B
B.2B
0
不大确装面中密的走光中圈题
3
:武卫中的微两霸抛司宝管
高二物理试卷第1页（共8页）
页8共)项》原
高有装g物活年中所原人e子我之
a经发我月未o
一C为偏转分高器，慰感应强度为：现有一质量为m电荷量为g(90)的粒子（不计
力)，经加速后，该粒子恰能沿直线通过速度选择器，粒子进入分离器后做匀速圆周
动，最后打到照相底片D上，下列说法正确的是
A.速度选择器中的磁场方向垂直纸面向外
平水济航写边元泡(d)图时，A
棉盟分谢言()图态锡学盖红
B.粒子经加速器加速后的速度大小为2，
qU,
司止轴大单微的O
m
H五数1前由兰6a2@
C.速度选择器两板间电压为B,d
qU
限青的卧个两的结中闻空客示图低
长4圳贸其，页除古
2mU
D。粒子在B,磁场中做匀速圆周运动的半径为
B
g
4.如图所示，
实线表示在竖直平面内的电场线，
电场线与水平方向成a角，水平方向的
匀强磁场与电场正交。有一带电液滴沿虚线AB斜向上做直线运动，AB与水平方向成B
角，且a心B,下列说法中正确的是
经的同的计变同W圆
A,液滴一定做匀速直线运动
B.
液滴一定带负电
C.
电场力可能做负功
D.
液滴有可能做匀变速直线运动
高二物理试卷第2页（共8页）
