Question

重為G=0.1N的金屬棒ab，放在光滑的平行金屬導(dǎo)軌上，如圖所示，軌道間距為L(zhǎng)=0.5m，所在平面與水平面的夾角為30°，勻強(qiáng)磁場(chǎng)垂直于軌道平面向上，電源電動(dòng)勢(shì)E=3V，金屬棒電阻R=6Ω，其余電阻不計(jì)，若金屬棒恰好靜止，求：                                                                                              

（1）金屬棒所受安培力的方向；                                                                                        

（2）磁感應(yīng)強(qiáng)度的大小．                                                                                                   

                                                                                                  

                                                                                                                                      

Answer 1

（1）根據(jù)左手定則，磁感線穿過(guò)手心，四指指向電流方向，大拇指指向?yàn)榘才嗔Ψ较�，由此可知安培力方向沿�?dǎo)軌平面向上．

答：金屬棒所受安培力的方向沿斜面向上．

（2）回路中的電流I==0.5A         

對(duì)金屬棒受力分析如圖所示，根據(jù)平衡條件，有

F=Gsin30°                          

又 F=BIL                         

由以上三式，代入數(shù)據(jù)，解得：B=0.2T．

答：磁感應(yīng)強(qiáng)度的大小為：B=0.2T．