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高2024级高二年级上学期质量监测
物理试题参考答案
一、单项选择题
1.B 2.A 3.B 4.D 5.D 6.C 7.D
二、多项选择题
8.AD 9.BD 10.AD
三、非选择题
11.
（1）C（2）B
（3）示例：原线圈匝数标注值与实际值不符（或副线圈存在漏磁、实验存在测量误差等）。
12.（1）R₁
（2）U =2.20V，I =0.44A；电压偏大
1 1
（3）R =5.0
x
13. (1) 解：等腰直角三角形，B到AC的距离为 ,......·················...2分
电势差: ....····························4分
(2) 粒子在垂直AC方向做匀速直线运动，沿AC方向做匀加速直线运动。·············5分
加速度 ····························7分
·运动时间 且 代入得 ··························10分
14. (1) 解： t=0时速度为0，无感应电流，安培力为0，由牛顿第二定律F=ma，········2分
得 ·····································4分
(2)v=2m/s 解析：匀速时受力平衡 ·························6分
代入数据得v=2m/s······································7分
(3)由能量守恒
总
，································9分
解得 ········································10分
电阻R与r串联，热量比
，故 ·································12分
15.(1) 解：粒子在磁场中做匀速圆周运动，半径r=d 由 得 ···············4分
(2) 第1区域：圆心角60°，时间 ..····························6分
·第1、2区域间：匀速直线运动，时间
·········································8分
学科网（北京）股份有限公司·第2区域：圆心角120°，时间
。·········································10分
·总时间 ···································12分
（3）第3个
速度改为2.5v时，半径 ··································13分
·第1区域：偏转后沿边界夹角60°，进入第2区域。························14分
·第2区域：偏转后速度方向与边界夹角60°，进入第3区域。····················15分
·第3区域：计算得粒子无法完全穿过，故最远到第3个磁场区域。··················16分
学科网（北京）股份有限公司
