4.关于磁感线，下列说法正确的是
2023~2024学年第一学期高二年级期末学业诊断
A.磁感线是不闭合的有向曲线，从N极出发到S极终止
B.磁感线是为了研究磁场根据实验事实而假想出来的
物理试卷
C.磁感线是带电粒子在磁场中运动的轨迹
D.没有磁感线的地方就没有磁场存在
(考试时间：上午10：30一12：00)】
5.下列说法正确的是
说明：本试卷为闭卷笔答，答题时间90分钟，满分100分。
A.安培力的方向一定与磁场的方向垂直
题号
二
三
四
总分
B.通电导线在磁场中一定受到安培力的作用
得分
C.判定通电导线在磁场中受力的方向用右手定则
D.安培力是洛伦兹力的宏观表现，洛伦兹力不做功，安培力也不做功
一、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
6.如图所示，两个同心放置的共面金属圆环α和b,一条形磁铁穿过圆心且与环面垂直，关于
相应位置。
穿过两环的磁通量Φ和Φ，下列说法正确的是
题号
1
2
3
4
5
6
7
8
9
10
A.a、b环内磁通量由下向上，中>中。
答案
B.a、b环内磁通量由上向下，Φ<Φ。
C.a、b环内磁通量由下向上，中<中
1.关于电源电动势，下列说法正确的是
A.电源电动势是反映电源把电能转化为其他形式能量本领大小的物理量
D.a、b环内磁通量由上向下，中>中。
B.电动势E=”与电压U=”中的W都是指电场力做的功
7.R,和R是两个材料相同、厚度相同、表面为正方形的电阻，R的尺寸小于R,的尺寸。现给
R,、R加以相同的电压U,下列选项正确的是
C.外电路断路时，路端电压的大小等于电源电动势的大小
A.R、R的电阻R>R
D.电动势越大表明电源储存的电能越多
B.通过R、R的电流I>I2
2.下列说法正确的是
C.R、R的电功率P=P
电流方向
A.丹麦物理学家法拉第通过实验发现了电流的磁效应
B.两条通电导线平行且电流方向相同时会相互排斥
D.R、R,在相同时间内产生的焦耳热Q,C.电流周围的磁场方向是用右手螺旋定则来判断的
8.单根长直导线周围磁感应强度的大小B=k(k为常数，为导线中电流的大小，为到导线的
D.安培认为磁体的磁场是由静止的电荷产生的
距离)。等边三角形PMN的三个顶点处，三根完全相同的长直导线
3.关于电流做功，下列说法正确的是
通有等大、垂直纸面向外的电流1。每根导线在中心0处产生的磁
A.电流做功的实质是导体中的恒定电场对自由电荷的静电力在做功
B.R是电动机的内阻，通过电动机的电流与其两端的电压满足U=R
感应强度大小均为B。,若将P处导线的电流变为2/，中心O处磁感
C.焦耳定律O=PR是由焦耳通过公式推导得到的
应强度的大小变为
M
D.W=U只适用于纯电阻电路计算电流做的功
A.V5B
B.Bo
C.2Bo
D.3Bp
料
高二物理第1页（共8页）
高二物理第2页（共8页）
的2023～2024学年第一学期高二年级期末学业诊断
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 C C A B A A C B D C
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 AD AD CD BCD BD
三、实验题：共 14分。
16.（6分）
（1）R2（2分） （2）3.8 （2分） （3） 8.2 （2分）
17.（8分）
（1）红（1分） （2） 1（1分）
（3）30（2分） （4）140 （2分） （5）131（2分）
四、计算题：共 41分。
18.（10分）
（1）根据左手定则，可判断磁场方向垂直纸面向里····························（3分）
（2）根据题意，由几何关系得
= ······················································································（2分）
带电粒子在磁场中做匀速圆周运动
= ·················································································（3分）
= ·····················································································（2分）
{#{QQABKQSAggCIAABAAAhCQw1oCEGQkBAAAKoGBFAAsAIAiBNABAA=}#}
19.（10分）
（1）小球在到达 A点之前做匀速直线运动
° = °··································································（1分）
=
E = 100V/m ················································································（1分）
电场方向变化后，小球所受重力与电场力平衡，小球在洛伦兹力作用下
做匀速圆周运动
= ··············································································（1分）
由几何关系得
= ··················································································· （1分）
= ····················································································（1分）
B=100T ····················································································（1分）
（2）小球做匀速直线运动的时间
= = ··············································································（1分）
小球做匀速圆周运动的周期
= ······················································································（1分）
小球做匀速圆周运动转过的圆心角
=
小球做匀速圆周运动的时间
=
= ·················································································（1分）
= + = (
+ )
t = （ + ）s ········································································（1分）
{#{QQABKQSAggCIAABAAAhCQw1oCEGQkBAAAKoGBFAAsAIAiBNABAA=}#}
20.（10分）
（1）带电小球水平方向只受电场力做初速度为零的匀加速直线运动、竖直方向
只受重力做竖直上抛运动，从 A到M的时间与从M到 B的时间相等
x 1= 2 t ···················································································（1分）
x1+x =
2 （2t）2······································································（1分）
x1：x2= 1：3················································································（1分）
（2）合运动与分运动具有等时性，小球所受的电场力为 qE、重力为 mg
小球从 A到M
qEx1 - mgh = 0·············································································（1分）
x = 1 ················································································· （1分）
h = ·················································································· （1分）
求得 qE=mg···········································································································（1分）
由图可知 tan θ＝1 ······································································ （1分）
θ= ，sin θ ＝
小球从 A运动到 B的过程中，速度与合外力 G′垂直时可取最小值·······（1分）
E 1kmin ＝ m(v0sin θ)2 ＝ 1012 J························································（1分）
2
{#{QQABKQSAggCIAABAAAhCQw1oCEGQkBAAAKoGBFAAsAIAiBNABAA=}#}
21.（11分）
（1）粒子进入磁场后运动轨迹如图所示，带电粒子从 O点运动到MN边界
= 0··········································································（1分）
带电粒子在磁场中做匀速圆周运动，半径为 R
= ················································································（1分）
由几何关系得
= ·················································································· （1分）
= ····················································································（1分）
= ····················································································（1分）
= ···················································································（1分）
（2）带电粒子在电场中加速时间为
= = ····································································· （1分）
两个磁场 B大小相等，周期相同 =
带电粒子在MN、PQ之间做匀速圆周运动时，运动时间为
= ·······················································································（1分）
带电粒子在 PQ下方磁场运动，运动时间为
=
····················································································（1分）
带电粒子第一次回到释放点 O的时间为
= ( + ) + ········································································（1分）
根据周期性，粒子回到 O点 总 = ( + )（ 取正整数）······（1分）
{#{QQABKQSAggCIAABAAAhCQw1oCEGQkBAAAKoGBFAAsAIAiBNABAA=}#}
