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走向高考8-1

走向高考8-1
走向高考8-1

第八章第一讲

一、选择题

1.在磁场中某区域的磁感线,如图所示,则 ( )

A .a 、b 两处的磁感应强度的大小不等B

a >B b

B .a 、b 两处的磁感应强度的大小不等,B a

C .同一通电导线放在a 处受力一定比放在b 处受力大

D .同一通电导线放在a 处受力一定比放在b 处受力小

[答案] B

[解析] 磁感线的疏密程度表示B 的大小,但安培力的大小 除跟该处的B 的大小和I 、L 有关外,还跟导线放置的方向与B 的方向的夹角有关,故C 、D 错误;由a 、b 两处磁感线的疏密程度可判断出B b >B a ,所以B 正确.

2.(2010·海门测试)下列四幅图中涉及经典的物理实验研究,其中说法正确的是( )

[答案] BCD

[解析] 通电导线的周围有磁场,处在磁场中的小磁针要受到磁场力作用,若A 图中导线有电流,小磁针要发生偏转,图景与实际不符,A 项错误;B 图中开关闭合,线圈中磁通量变化,产生感应电流,导致电流表的指针偏转;C 图中电子在磁场中受到磁场力作用,用左手定则判断电子束将向下偏转;D 图中汽车速度达到第一宇宙速度时,汽车环绕地球近地做圆周运动,处在车内的物体处于完全失重状态.

3.在磁场中的同一位置,先后引入长度相等的直导线a 和b ,a 、b 导线的方向均与磁场方向垂直,但两导线中的电流不同,因此所受的力也不一样.下列几幅图象表现的是导线所受的安培力F 与通过导线的电流I 的关系.a 、b 各自有一组F 、I 的数据,在图象中各描出一个点.在下图中,请判断描绘正确的是 ( )

[答案] BC [解析] 由题意“同一位置同一磁场”,“导线长度相同”可知F I

斜率相等;电流为零时不受力,所以A 、D 选项错误.

4.一根电缆埋藏在一堵南北走向的墙里,在墙的西侧处,当放一指南针时,其指向刚

好比原来旋转180°,由此可以断定,这根电缆中电流的方向为() A.可能是向北B.可能是竖直向下

C.可能是向南D.可能是竖直向上

[答案] D

[解析]在地磁场作用下,小磁针静止时N指向北方,现改变为N极指向南方,故应有竖直向上的通电电流,即选D.

5.如图所示,两个完全相同的线圈套在一水平光滑绝缘圆柱上,但能自由移动,若两线圈内通过大小不等的同向电流,则它们的运动情况是()

A.都绕圆柱转动

B.以不等的加速度相向运动

C.以相等的加速度相向运动

D.以相等的加速度相背运动

[答案] C

[解析]同向环形电流间相互吸引,虽然两电流大小不等,但据牛顿第三定律知两线圈间相互作用力必大小相等,所以选C项.

6.(2009·全国卷Ⅰ)如图,一段导线abcd位于磁感线强度大小为B的匀强磁场中,且与磁场方向(垂直于纸面向里)垂直.线段ab、bc和cd的长度均为L,且∠abc=∠bcd=135°.流经导线的电流为I.方向如图中箭头所示.导线段abcd所受到的磁场的作用力的合力()

A.方向沿纸面向上,大小为(2+1)ILB

B.方向沿纸面向上,大小为(2-1)ILB

C.方向沿纸面向下,大小为(2+1)ILB

D.方向沿纸面向下,大小为(2-1)ILB

[答案] A

[解析]该导线可以用a和d之间的直导线长为(2+1)L来等效代替.根据F=BIL,可知大小为(2+1)BIL,方向由左手定则知:沿纸面向上.故A项正确.7.(2009·合肥一中二模)在匀强磁场中有一用相同材料制成的导体框abc,b为半圆弧的顶点.磁场方向垂直于导体框平面向里,在ac两端接一直流电源,如图所示,则()

A.导体框abc所受安培力的合力为零

B.导体框abc所受安培力的合力垂直于ac向上

C.导体框abc所受安培力的合力垂直于ac向下

D.导体框abc的圆弧段所受安培力为零

[答案] B

[解析] 由左手定则可知,导体框abc 所受安培力的合力垂直于ac 向上.

8.(2009·广东模拟)如图所示,水平放置的扁平条形磁铁,在磁铁的左端正上方有一金属线框,线框平面与磁铁垂直,当线框从左端正上方沿水平方向平移到右端正上方的过程中,穿过它的磁通量的变化情况是 ( )

A .先减小后增大

B .始终减小

C .始终增大

D .先增大后减小

[答案] D

[解析] 线框在磁铁两端的正上方时穿过该线框的磁通量最少,在磁铁中央时穿过该线框的磁通量最多,所以该过程中的磁通量先增大后减小,故D 对.

二、非选择题

9.某兴趣小组在研究长直导线周围的磁场时,为增大电流,用多根导线捆在一起代替

______T .(要求估算出比例系数,用等式表示)

[答案] 2×10-7I r

[解析] 分析表格中数据可得,B 与电流I 成正比,与测试点到直导线的距离r 成反比,

即B =k ·I r ,取表格中的第一单元格进行计算可得k ≈2×10-7,即B =2×10-7×I r

T. 10.如图所示,在倾角为37°的光滑斜面上有一根长为0.4m ,质量为6×10-2kg 的通电

直导线,电流强度I =1A ,方向垂直于纸面向外,导线用平行斜面的轻绳拴住不动,整个装置放在磁感应强度每秒增加0.4T ,方向竖直向上的磁场中.设t =0时,B =0,则需要多长时间,斜面对导线的支持力为零?(g 取10m/s 2)

[答案] 5s

[解析] 斜面对导线的支持力为零时导线的受力如图所示.

由平衡条件 F T cos37°=F ①

F T sin37°=mg ②

由①②解得:F =mg tan37°

代入数值得:F =0.8N

由F =BIL 得:

B =F IL =0.81×0.4

T =2T. B 与t 的变化关系为B =0.4t .

所以:t =5s

11.如图所示,质量为0.05kg ,长l =0.1m 的铜棒,用长度也为l 的

两根轻软导线水平悬挂在竖直向上的匀强磁场中,磁感应强度为B =0.5T.

不通电时,轻线在竖直方向,通入恒定电流后,棒向外偏转的最大角度θ

=37°,求此棒中恒定电流多大?(不考虑棒摆动过程中产生的感应电流,

g 取10N/kg)

同学甲的解法如下:对铜棒受力分析如图所示:

当最大偏转角θ=37°时,棒受力平衡,有:

F T cos θ=mg ,F T sin θ=F 安=BIl

得I =mg tan θBl =0.05×10×340.5×0.1

A =7.5A 同学乙的解法如下:

F 安做功:W F =Fx 1=BIl sin37°×l sin37°=BI (l sin37°)2

重力做功:

W G =-mgx 2=-mgl (1-cos37°)

由动能定理得:W F +W G =0

代入数据解得:I =509

A ≈5.56A 请你对甲、乙两同学的解法作出评价:若你对两者都不支持,则给出你认为正确的解答.

[答案] 评价见解析 3.33A

[解析] 甲同学的错误原因:认为物体速度为零时,一定处于平衡位置,或者认为偏角最大时为平衡位置.

乙同学的错误原因:将安培力表达式误写为

F 安=BIl sin37°,应为:F 安=BIl .

正确的解法如下:铜棒向外偏转过程中

F 安做功:W F =Fx 1=BIl ×l sin37°

重力做功:

W G =-mgx 2=-mgl (1-cos37°)

由动能定理得:W F +W G =0

代入数据解得:I =103

A ≈0.33A. 12.如图所示,PQ 和MN 为水平平行放置的金属导轨,相距L =1m.PM 间接有一个电动势为E =6V ,内阻r =1Ω的电源和一只滑动变阻器.导体棒ab 跨放在导轨上,棒的质量为m =0.2kg ,棒的中点用细绳经定滑轮与物体相连,物体的质量M =0.3kg.棒与导轨的动摩擦因数为μ=0.5,匀强磁场的磁感应强度

B =2T ,方向竖直向下,求为使物体保持静止,滑动变阻器连入电路的阻值为多大?设最大静摩擦力与滑动摩擦力相等,导轨与棒的电阻不计.(g 取10m/s 2)

[答案] 2Ω≤R ≤5Ω

[解析] 导体棒受到的最大静摩擦力为

F f =μF N =μmg =0.5×0.2×10N =1N

绳对导体棒的拉力F 拉=Mg =0.3×10N =3N

导体棒将要向左滑动时

BI max L =F f +F 拉,I max =2A

由闭合电路欧姆定律I max =E R min +r =6R min +1

得R max =2Ω

导体棒将要向右滑动时F f +BI min L =F 拉,I min =1A

由闭合电路欧姆定律I min =E R max +r =6R max +1

得R max =5Ω

滑动变阻器连入电路的阻值为2Ω≤R ≤5Ω.

13.(2009·黄冈一模)如图所示是一种电磁泵,泵体是一个长方体,端面是一个边长为σ的正方形,ab 长为l ,上下两面接在电源上,电压为U (内阻不计).磁感应强度为B 的磁场指向cdfe 面,液体电阻率为ρ,密度为D (液体原来不导电,在泵头通入导电剂后才导电).求:

(1)最大抽液高度;

(2)每秒钟抽液的质量.

[答案] (1)UlB /(σρDg ) (2)UσD B

[解析] (1)泵体内液体的电阻为R =ρσσl =ρl

. 通过泵体的电流为I =U R =Ul ρ

. 安培力F =BIσ=UlσB ρ,安培力产生的压强p =UlB σρ

. 对液体来说,p =Dgh ,则有UlB σρ

=Dgh . 所以h =UlB /(σρDg ).

(2)在阻力不计的情况下,有mgh =U 2R t , 所以m t =U 2ghR =UσD B

.

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