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1、2012届高考物理磁场专题复习教案22012高考复习 电学部分 精讲精练 磁场 4 带电粒子在复合场中的运动 【课标要求】 1(掌握带电粒子在复合场中运动规律。 2(掌握带电粒子在复合场中运动的分析方法。 【知识精要】 1(带电粒子在复合场中的直线运动的条件是:带电粒子所受的合外力为零，或者所受的合外力与速度方向在一条直线上。 2(带电粒子在复合场中的匀速圆周运动运动的条件是:带电粒子所受的恒力互相平衡，洛仑兹力提供向心力。 3(带电粒子在复合场中的变加速直线运动，往往根据能量关系加以解决。 【名师点拨】 例1:(2011银川模拟)如图所示，? ? ? ? ? ? ? ? ? ? ? ? ?
2、? ? ? ? ? ? ? 空间有一垂直纸面的磁感应强度为B? ? ? ? ? ? ? ? ? ? 0.5T的匀强磁场，一质量为0.20kg? ? ? ? ? ? ? ? ? ? F? ? ? ? ? ? ? ? ? ? 且足够长的绝缘木板静止在光滑水平面上，在木板左端无初速放置一质量为0.1kg、电荷量q=+0.2C的滑块，滑块与绝缘木板之间动摩擦因数为0.5，滑块受到的最大静摩擦力可认为等于滑动摩擦力。现对木板施加方向水平向左，大小为0.6N恒力，g2取10m/s.则( ) 2A(木板和滑块一直做加速度为2m/s的匀加速运动 B(滑块开始做匀加速直线运动，然后做加速度减小的加速运动，最后做
3、匀速直线运动 2C(最终木板做加速度为2 m/s的匀加速运动，滑块做速度为10m/s的匀速运动 2D(最终木板做加速度为3 m/s的匀加速运动，滑块做速度为10m/s的匀速运动 解析:刚开始，滑块和木板一起做匀加速直线运动，随着速度的不断增加，滑块受到的竖直向上的洛仑兹力不断增加，滑块所受的弹力减小，合力减小，滑块做变加速运动，一段时间后，滑块的重力和洛仑兹力相平衡，滑块做匀速直线运动，而木块作匀加速直线运动。2根据qvB=mg，得v=10m/s;根据F=Ma，得a=3m/s。 of work enthusiasm and forward-looking. The difficulties a
4、nd problems of individual cadres indifferent masses as the buck passing, long, make some simple complex problems. Some cadres general talk about pay, do not take the initiative to undertake for the bitter and tired of the work, the lack of courage to play a positive attitude. corrective measures: (L
5、ED Leadership: Luo Mingjun, rectification time: before September 25th, insist for a long time) 1, effectively solve the enterprise less, help is not enough. In order to turn style, solve problems, and do practical things, heart to heart as the core, in accordance with the provisions of division of L
6、abor Bureau, by the Bureau of Party members and cadres room composition the working group , to help enterprises solve problems, promote the construction of major projects; close ties with the masses, to ask for the people, ask for people to know the people, public opinion, the people, improve people
7、s livelihood. 点拨:洛仑兹力随速度的变化而变化，从而引起其它力发生变化，引起合力发生变化，引起运动性质发生变化。顺着这一思路可以迅速找到解决问题的金钥匙。 例2:如图所示，匀强电场方向水平向右，匀强磁场方向垂直于纸面向里，一质量为m，带电量为q的微粒以速度v与磁场方向垂直，与电场成45?角射入复合场中，恰能做匀速直线运动，求电场强度E的大小，磁感应强度B的大小. 解答:由带电粒子所受的洛伦兹力与v垂直，电场力方向与电场 线平行，微粒如图所示方向进入磁场中，如果只受到电场力与洛伦兹力作用，合力不可能为零，也就不可能做匀速直线运动.由此可知本题必须考虑到微粒所受的重力，才可能使微粒做
8、匀速直线运动。 假设粒子不带正电，则所受电场力方向水平向左，洛伦兹力方向斜向右下方与v垂直，同学们可以从力的平衡条件判断出这样的粒子不可能做匀速直线 运动，所以粒子应带正电荷，受力情况如图所示，根据合外力为 零可得 mg=qvBsin45? ? qE=qvBcos45? ? 2mg由?式可得B=;E=mg/q. qv点评:此类问题明确带电粒子在复合场中做匀速直线运动，首先根据假设法判断粒子是否受到重力作用，进而根据平衡条件判定粒子的电性，最后根据平衡方程解出未知量。 例3:(2011沈阳模拟)如图所示，质量为m，带电量为球+q的小环沿着穿B E 过它的竖直棒下落，棒与环孔间的动摩擦因数为.匀强
9、电场水平向右，场强为E，匀强磁场垂直于纸面向外，磁感应强度为B.在小环下落，求: (1)小环的速度为多大时，它的加速度最大, (2)小环运动的最大速度可达到多少, 解析:(1)小球由静止开始下落 故小球的最大加速度为a=g; mmgqE，,mg(qvBqE),小球的加速度为0时，有 ，解得:v, mmqB,点拨:粒子在复合场中的动态分析，由速度变化入手，进行受力分析，找出合力变化和速度变化的关系，判定粒子加速度和速度最大的条件。 of work enthusiasm and forward-looking. The difficulties and problems of individual
10、 cadres indifferent masses as the buck passing, long, make some simple complex problems. Some cadres general talk about pay, do not take the initiative to undertake for the bitter and tired of the work, the lack of courage to play a positive attitude. corrective measures: (LED Leadership: Luo Mingju
11、n, rectification time: before September 25th, insist for a long time) 1, effectively solve the enterprise less, help is not enough. In order to turn style, solve problems, and do practical things, heart to heart as the core, in accordance with the provisions of division of Labor Bureau, by the Burea
12、u of Party members and cadres room composition the working group , to help enterprises solve problems, promote the construction of major projects; close ties with the masses, to ask for the people, ask for people to know the people, public opinion, the people, improve peoples livelihood. 例4:如图所示K与虚线
13、MN之间是加速电场。虚线MN与PQ之间是匀强电场，虚线PQ与荧光屏之间是匀强磁场，且MN、PQ与荧光屏三者互相平行。电场和磁场的方向如图所示。图中A点与O点的连线垂直于荧光屏。一带正电的粒子从A点离开加速电场，速度方向垂直于偏转电场方向射入偏转电场，在离开偏转电场后进入匀强磁场，最后恰好垂直地打在图中的荧光屏上。已知电场和磁场区域在竖直方向足够长，加速电场电压与偏转电场的场强关系为U=Ed/2，式中的d是偏转电场的宽度且为已知量，磁场的磁感应强度B与偏转电场的电场强度E和带电粒子离开加速电场的速度v关系符合表达式v=E/B，如图所示，00试求: (1)磁场的宽度L为多少, P M d L (2
14、)带电粒子最后在电场和磁场中总的偏转距离是多K B A v0 O 少, U 解析:?其轨迹如图所示。偏转角为: E N Q vqEdqEdyP tan,1M d L 2vmvvmvK 0000B A v0 O 即带电粒子在电场中的偏转角=45。 y U E 带电粒子离开偏转电场速度为2分 在磁2vv0 N v Q 场中偏转的半径为 22mv2mvmv00 ，由图可知，磁场宽度L=Rsin=d R,2dqBqE/vqE0?带电粒子在偏转电场中距离为，在磁场中偏转距离为,y,0.5d1,y,0.414d2点评:带电粒子在复合场中的运动，必须搞清粒子在每个过程中的受力特点和速度的关系，从而搞清粒子在
15、不同过程的运动性质，为选择合理的解题方法提供科学的依据。 例5:如图甲所示，两个共轴的圆筒形金属电极，外电极接地.其上均匀分布着平行于轴线的四条狭缝a、b、c和d，外筒的半径为r.在外筒之外的足够大区域中有平行于轴线方向0的均匀磁场，磁感应强度的大小为B.在两极间加上电压，使两圆筒之间的区域内有沿半径向外的电场.一质量为m、带电量为+q的粒子，从紧靠内筒且正对狭缝a的s点出发，初速为零，如果该粒子经过一段时间的运动之后恰好又回到出发点s，则两极之间的电压U应是of work enthusiasm and forward-looking. The difficulties and problem
16、s of individual cadres indifferent masses as the buck passing, long, make some simple complex problems. Some cadres general talk about pay, do not take the initiative to undertake for the bitter and tired of the work, the lack of courage to play a positive attitude. corrective measures: (LED Leaders
17、hip: Luo Mingjun, rectification time: before September 25th, insist for a long time) 1, effectively solve the enterprise less, help is not enough. In order to turn style, solve problems, and do practical things, heart to heart as the core, in accordance with the provisions of division of Labor Burea
18、u, by the Bureau of Party members and cadres room composition the working group , to help enterprises solve problems, promote the construction of major projects; close ties with the masses, to ask for the people, ask for people to know the people, public opinion, the people, improve peoples liveliho
19、od. 多少,(不计重力，整个装置在真空) 解答:粒子从s点出发后，经两圆筒之间的电场加速，从a孔 o a 1+q S b 垂直筒壁射入磁场.设射入磁场时的速度为v，有关系式: o d 12c ? qU,mv2乙 进入磁场后，受洛伦兹力作用作匀速圆周运动，有关系式: 2v qvBm ? ,R根据题意，要求粒子在最后又回到出发点s，必须使粒子能从 狭缝b(或d)沿径向飞入，在两筒壁间先作减速运动，然后又反向 加速从原狭缝飞出，依次循环，如图乙所示.因此要求粒子在磁场 中作圆周运动的半径应满足条件: R=r ? 022qrB0联立?、?、?式可得两极之间的电压为 U,2m点评:此类问题需要明确带电
20、粒子的运动过程，科学大胆想象粒子的运动轨迹，结合圆周运动解题的要领:找轨道圆心，定轨道半径，洛仑兹力提供向心力，建立方程组解决相关问题。 例6:(2011重庆模拟)在竖直平面内有一圆形绝缘轨道，半径R=1m，匀强磁场垂直于,3,2轨道平面向里，一质量为m=110kg，带电量为q = -310C的小球，可在内壁滑动(现在最低点处给小球一个水平初速度v，使小球在竖直平面内0-1 v/m?s 逆时针做圆周运动，图甲是小球在竖直平面内做圆周运动的甲 速率v随时间变化的情况，图乙是小球所受轨道的弹力F随2 时间变化的情况，小球一直沿圆形轨道运动.结合图象所给数0 t/s 2据，g取10m/s(求: 8.
21、0F/N 乙 -2 10(1)磁感应强度的大小( (2)小球从开始运动至图甲中速度为t/s 0 2m/s的过程中，摩擦力对小球做的功( v 0解析:(1)从甲图可知，小球第二次过最高点时，速度大小为2m/s，2而由乙图可知，此时轨道与球间弹力为零， 代入数据，得B=0.1T ？mg,qvB,mv/Rof work enthusiasm and forward-looking. The difficulties and problems of individual cadres indifferent masses as the buck passing, long, make some sim
22、ple complex problems. Some cadres general talk about pay, do not take the initiative to undertake for the bitter and tired of the work, the lack of courage to play a positive attitude. corrective measures: (LED Leadership: Luo Mingjun, rectification time: before September 25th, insist for a long tim
23、e) 1, effectively solve the enterprise less, help is not enough. In order to turn style, solve problems, and do practical things, heart to heart as the core, in accordance with the provisions of division of Labor Bureau, by the Bureau of Party members and cadres room composition the working group ,
24、to help enterprises solve problems, promote the construction of major projects; close ties with the masses, to ask for the people, ask for people to know the people, public opinion, the people, improve peoples livelihood. ,2(2)从乙图可知，小球第一次过最低点时，轨道与球面之间的弹力为F=8.010N，2根据牛顿第二定律， 代入数据，得v=7m/s. F,mg,qvB,mv
25、/R000以上过程，由于洛仑兹力不做功,由动能定理可得: 2 -32-mg2R+W = mv/2 - mv/2代入数据得: W=-2.510J f0f 点拨:实物和图形相结合是高考中的热点问题，解题时，结合受力分析和图形判定粒子在圆弧的最低点所受弹力最大，在圆弧的最高点所受弹力最小，因为摩擦力不断变化，所以本题适宜用动能定理和向心力的有关知识加以解决。 例7:如图中甲所示，真空中两水平放置的平行金属板C、D，上面分别开有正对的小孔O和O，金属板C、D接在正弦交流电源上，C、D两板间的电压U随时间t变化的图线如图12CD-25中乙所示。t=0时刻开始，从C板小孔O处连续不断飘入质量为m=3.21
26、0kg、电荷量1-19q=1.610C的带正电的粒子(设飘入速度很小，可视为零)。在D板外侧有以MN为边界的匀强磁场，MN与D金属板相距d=10cm，匀强磁场的大小为B=0.1T，方向如图中所示，粒子的重力及粒子间相互作用力不计，平行金属板C、D之间的距离足够小，粒子在两板间的运动时间可忽略不计。求: U/V M N CD50 B 0.020.030.04 0 O2 0.01 t/s C U,CD D-50 O1 乙甲 (1)带电粒子经小孔O进入磁场后，能飞出磁场边界MN的最小速度为多大, 2(2)从0到0.04s末时间内哪些时刻飘入小孔O的粒子能穿过电场并飞出磁场边界MN, 1(3)磁场边界
27、MN有粒子射出的长度范围。(保留一位有效数字) 解析:(1)设粒子飞出磁场边界MN的最小速度为v，粒子在磁场中做匀速圆周运动，02根据洛伦兹力提供向心力知: qvB=mv/R 000粒子恰好飞出磁场，则有:R=d 03所以最小速度 v=qBd/m=510m/s 02)由于C、D两板间距离足够小，带电粒子在电场中运动时间可忽略不计，故在粒(子通过电场过程中，两极板间电压可视为不变，设恰能飞出磁场边界MN的粒子在电场中of work enthusiasm and forward-looking. The difficulties and problems of individual cadres
28、indifferent masses as the buck passing, long, make some simple complex problems. Some cadres general talk about pay, do not take the initiative to undertake for the bitter and tired of the work, the lack of courage to play a positive attitude. corrective measures: (LED Leadership: Luo Mingjun, recti
29、fication time: before September 25th, insist for a long time) 1, effectively solve the enterprise less, help is not enough. In order to turn style, solve problems, and do practical things, heart to heart as the core, in accordance with the provisions of division of Labor Bureau, by the Bureau of Par
30、ty members and cadres room composition the working group , to help enterprises solve problems, promote the construction of major projects; close ties with the masses, to ask for the people, ask for people to know the people, public opinion, the people, improve peoples livelihood. 运动时CD板对应的电压为U，则根据动能
31、定理知: 022qU=mv/2 得:U=mv/2q=25V 0000根据图像可知:U=50sin50t，25V(或-25V)电压对应的时间分别为:1/300s和1/60s(或CD7/300s和11/300s)，所以粒子在0到0.04s内飞出磁场边界的时间为1/300s1/60s(或7/300s11/300s) 。 (3)设粒子在磁场中运动的最大速度为v，对应的运动半径为R，则有: mm22qU=mv/2 qvB=mv/R mmmmm粒子飞出磁场边界时相对小孔向左偏移的最小距离为:21/21/22x=R-(R-d)=0.1(2-1)m?0.04m mm磁场边界MN有粒子射出的长度范围为:?x=d
32、-x=0.06m 。 点拨:遇到复杂问题，首先把物理过程搞清楚，把大化小，逐个突破，遇到临界问题可以利用极限分析的方法迅速找到临界条件。 【及时反馈】 1(质量为m、带电量为q的小球，从倾角为的光滑绝缘斜面上由静止下滑，整个斜面置于方向水平向外的匀强磁场中，其磁感强度为B，如图所示。若带电小球下滑后某时刻对斜面的作用力恰好为零，下面说法中正确的是( ) A(小球带正电 B(小球在斜面上运动时做匀加速直线运动 C(小球在斜面上运动时做加速度增大，而速度也增大的变加速直线运动 D(则小球在斜面上下滑过程中，当小球对斜面压力为零时的速率为mgcos,Bq 2(如下图所示，匀强磁场中有一个开口向上的绝
33、缘半球，将带有正电荷的小球从半球左边最高处静止释放，物块沿半球内壁只能滑到C点处;若将该物块自半球右边最高点静止释放，能滑到位置的为 ( ) A(与C等高的D点处 B(比D高的某处 C(比D低的某处 D(上述情况都有可能 3(在匀强磁场中有一带电粒子做匀速圆周运动，当它运动到M点，突然与一不带of work enthusiasm and forward-looking. The difficulties and problems of individual cadres indifferent masses as the buck passing, long, make some simple
34、 complex problems. Some cadres general talk about pay, do not take the initiative to undertake for the bitter and tired of the work, the lack of courage to play a positive attitude. corrective measures: (LED Leadership: Luo Mingjun, rectification time: before September 25th, insist for a long time)
35、1, effectively solve the enterprise less, help is not enough. In order to turn style, solve problems, and do practical things, heart to heart as the core, in accordance with the provisions of division of Labor Bureau, by the Bureau of Party members and cadres room composition the working group , to
36、help enterprises solve problems, promote the construction of major projects; close ties with the masses, to ask for the people, ask for people to know the people, public opinion, the people, improve peoples livelihood. 电的静止粒子碰撞合为一体，碰撞后的运动轨迹应是图中的哪一个,(实线为原轨迹，虚线为碰后轨迹，且不计粒子的重力)( ) 4(一带电微粒M在相互垂直的匀强电场、匀强磁
37、场中作匀速圆周运动，匀强电场竖直向上，匀强磁场水平且垂直纸面向里，如图所示，下列说法正确的是 E ( ) M A(沿垂直纸面方向向里看，微粒M的绕行方向为逆时针方向 B B(运动过程中外力对微粒作功的代数和为零，故机械能守恒 C(在微粒旋转一周的时间内重力作功为零 D(沿垂直纸面方向向里看，微粒M的绕行方向既可以是顺时针也可以是逆时针方向图 5(如图所示，长L的丝线的一端固定，另一端拴一带正电的小球，小球质量为m，带电量为q，使丝线与竖直方向成角由静止释放小球，小球运动的空间有方向垂直纸面向里的匀强磁场，磁感应强度为B。求小球每次运动到最低点时，线的拉力。 6(2011西安模拟)如图所示，坐标
38、系xoy位于竖直平面内，所在空间有沿水平方向垂直于纸面向里的匀强磁场，磁感应强度大小为B，在x,0的空间内还有沿x轴负方向的匀强电场，场强大小为E。一个带电油滴经图中x轴上的M点，沿着直线MP做匀速运动，0图中=30，经过P点后油滴进入x,0的区域，要使油滴在x,0的区域内做匀速圆周运动，需要在该区域内加一个匀强电场。若带电油滴沿弧PN做匀速圆周运动，并垂直于x轴通过轴上的N点。已知重力加速度为g。 of work enthusiasm and forward-looking. The difficulties and problems of individual cadres indiffe
39、rent masses as the buck passing, long, make some simple complex problems. Some cadres general talk about pay, do not take the initiative to undertake for the bitter and tired of the work, the lack of courage to play a positive attitude. corrective measures: (LED Leadership: Luo Mingjun, rectificatio
40、n time: before September 25th, insist for a long time) 1, effectively solve the enterprise less, help is not enough. In order to turn style, solve problems, and do practical things, heart to heart as the core, in accordance with the provisions of division of Labor Bureau, by the Bureau of Party memb
41、ers and cadres room composition the working group , to help enterprises solve problems, promote the construction of major projects; close ties with the masses, to ask for the people, ask for people to know the people, public opinion, the people, improve peoples livelihood. y 10.三角函数的应用B 在ABC中，C为直角，A
42、、B、C所对的边分别为a、b、c，则有(1)判断油滴的带电性质; M a x N O (2)求油滴运动的速率; E (3)求在x,0的区域内所加电场(1)二次函数yax2的图象：是一条顶点在原点且关于y轴对称的抛物线。是二次函数的特例，此时常数b=c=0.P 的场强; (4)求油滴从M点出发运动到N点所用的时间。 7(2011深圳模拟)如图所示的区域中，左边为垂直纸面向里的匀强磁场，磁感应强度为B，右边是一个电场强度大小未知的匀强电场，其方向平行于OC且垂直于磁场方向(一个质量为m，电荷量为,q的带电粒子从P孔以初速度v沿垂直于磁场方向进入匀强磁场0中，初速度方向与边界线的夹角=60?，粒子恰
43、好从C孔垂直于OC射入匀强电场，最后打在Q点，已知OQ=2OC，不计粒子的重力，求: 增减性：若a0，当x时，y随x的增大而增大。(1)粒子从P运动到Q所用的时间t( 三三角函数的计算(2)电场强度E的大小( 2、第三单元“生活中的数”。通过数铅笔等活动，经历从具体情境中抽象出数的模型的过程，会数，会读，会写100以内的数，在具体情境中把握数的相对大小关系，能够运用数进行表达和交流，体会数与日常生活的密切联系。(3)粒子到达Q点的动能E( kQ一、指导思想：1(BD 2(C 3(A 4( AC 5( mg(3,2cos,)，qB2gL(1,cos,)最值：若a0，则当x=时，；若a0，则当x=
44、时，(23，9)E,2E3E(?带正电 ?6 ? 向上 ? B3gB(6)直角三角形的外接圆半径Bv,(2，9)m027(? ? ? mv033qB2、加强基础知识的教学，使学生切实掌握好这些基础知识。特别是加强计算教学。计算是本册教材的重点，一方面引导学生探索并理解基本的计算方法，另一方面也通过相应的练习，帮助学生形成必要的计算技能，同时注意教材之间的衔接，对内容进行有机的整合，提高解决实际问题的能力。of work enthusiasm and forward-looking. The difficulties and problems of individual cadres indif
45、ferent masses as the buck passing, long, make some simple complex problems. Some cadres general talk about pay, do not take the initiative to undertake for the bitter and tired of the work, the lack of courage to play a positive attitude. corrective measures: (LED Leadership: Luo Mingjun, rectificat
46、ion time: before September 25th, insist for a long time) 1, effectively solve the enterprise less, help is not enough. In order to turn style, solve problems, and do practical things, heart to heart as the core, in accordance with the provisions of division of Labor Bureau, by the Bureau of Party members and cadres room composition the working group , to help enterprises solve problems, promote the construction of major projects; close ties with the masses, to ask for the people, ask for people to know the people, public opinion, the people, improve peoples livelihood.