如图所示,足够长的光滑平行导轨MN、PQ倾斜放置,两导轨间距离为L=

如图所示,足够长的光滑平行导轨MN、PQ倾斜放置,两导轨间距离为L=1.0 m,导轨平面与水平面间的夹角为30°,磁感应强度为B的匀强磁场垂直于导轨平面向上,导轨的M、P两端连接阻值为R=3.0 Ω的电阻,金属棒ab垂直于导轨放置并用细线通过光滑定滑轮与重物相连,金属棒ab的质量m=0.20 kg,电阻r=0.50 Ω,重物的质量M=0.60 kg,如果将金属棒和重物由静止释放,金属棒沿斜面上滑的距离与时间的关系如下表所示,不计导轨电阻,取g=10 m/s2。求:

(1)所加磁场的磁感应强度B为多大?

(2)电阻R在0.6 s内产生的热量为多少?

时间t/s

0

0.1

0.2

0.3

0.4

0.5

0.6

上滑距离(m)

0

0.05

0.15

0.35

0.70

1.05

1.40

答案

解:(1)由表中数据可以看出最终ab棒将匀速运动,匀速运动速度:v=

=3.5 m/s    ①

棒受力如图,由平衡条件可得:

T=F+mgsin30°                                                                ②

T=Mg                                                                       ③

F=

                                                                 

联立解得:B=

T。                                                        ⑤

(2)在0.6 s内s=1.40 m,由能的转化与守恒得

Mgs=mgssin30°+Q+

(M+m)v2                                                  ⑥

又因为:QR=

                                                           ⑦

联立解得:QR=1.8 J。                                                          ⑧


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