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共回答了17个问题采纳率:100% 举报
(1)5.23×10
-5 s(2)x=5m(3)当0<x′<3m时
![](https://img.yulucn.com/upload/0/cd/0cd8d236798c9f37c9a33424617c1245_thumb.jpg)
;当3m≤
![](https://img.yulucn.com/upload/2/73/273756ca672426903298b51af45a1765_thumb.jpg)
≤5m时
试题分析:⑴带电粒子在磁场中做匀速圆周运动,洛伦兹力提供向心力,根据牛顿第二定律有
![](https://img.yulucn.com/upload/7/7b/77b6d04d78a646f38b41142ae657cab9_thumb.jpg)
(1分)
代入数据得:
![](https://img.yulucn.com/upload/d/5b/d5bc69a1ce4d47a653082a97b7636d78_thumb.jpg)
(1分)
轨迹如图1交y轴于C点,过P点作v的垂线交y轴于O
1 点,
由几何关系得O
1 为粒子运动轨迹的圆心,且圆心角为60°。 (1分)
在磁场中运动时间
![](https://img.yulucn.com/upload/7/4a/74ae5d40e6a3e4f5b605dd787086fe66_thumb.jpg)
(1分)
代入数据得:t=5.23×10
-5 s(1分)
⑵带电粒子离开磁场垂直进入电场后做类平抛运动
方法 一:
粒子在电场中加速度
![](https://img.yulucn.com/upload/c/90/c904dc50729b61c2e4728d7b5984f115_thumb.jpg)
(1分)
运动时间
![](https://img.yulucn.com/upload/0/36/036698deeb4d8049ab61fb7f27b8b615_thumb.jpg)
(1分)
沿y方向分速度
![](https://img.yulucn.com/upload/3/96/396d30dedadaed48e4d77f807e625fd1_thumb.jpg)
(1分)
沿y方向位移
![](https://img.yulucn.com/upload/1/5f/15f0a0938f5122188a490b4109520563_thumb.jpg)
(1分)
粒子出电场后又经时间t
2 达x轴上Q点
故Q点的坐标为
![](https://img.yulucn.com/upload/b/84/b84c9290ad2f06bce49b836d9bea0aa4_thumb.jpg)
(1分)
方法二:
设带电粒子离开电场时的速度偏向角为θ,如图1,则:
![](https://img.yulucn.com/upload/9/1e/91ebd39f970fe47e17235abe75329cc7_thumb.jpg)
(2分)
设Q点的横坐标为x
则:
![](https://img.yulucn.com/upload/a/05/a05385fc988145d9b5e121ac423d5235_thumb.jpg)
(2分)
故x=5m。(1分)
⑶电场左边界的横坐标为x′。
当0<x′<3m时,如图2,设粒子离开电场时的速度偏向角为θ′,
则:
![](https://img.yulucn.com/upload/c/2f/c2fd63a78cdbeeb2a4d397baf550a2bf_thumb.jpg)
(1分)
又:
![](https://img.yulucn.com/upload/a/eb/aeb4e86fb3349603d580e31b75c2d92f_thumb.jpg)
(1分)
由上两式得:
![](https://img.yulucn.com/upload/0/cd/0cd8d236798c9f37c9a33424617c1245_thumb.jpg)
(1分)
当3m≤
![](https://img.yulucn.com/upload/2/73/273756ca672426903298b51af45a1765_thumb.jpg)
≤5m时,如图3,有
![](https://img.yulucn.com/upload/9/77/977e6916992863c72cf937c74878eaac_thumb.jpg)
(2分)
将y=1m及各数据代入上式得:
![](https://img.yulucn.com/upload/b/4d/b4da46e3f855ed9463e090c434d5964e_thumb.jpg)
(1分)
1年前
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