[题目]如图所示.两根相距为L＝1m的足够长的平行光滑金属导轨.位于水平的xOy平面内.一端接有阻值为R＝6Ω的电阻.在x>0的一侧存在垂直纸面向里的磁场.磁感应强度B只随t的增大而增大.且它们间的关系为B＝kt.其中k＝4T/s.一质量为m＝0．5kg的金属杆与金属导轨垂直.可在导轨上滑动.当t＝0时金属杆静止于x＝0处.有一大小可调节的外力题目和参考答案——青夏教育精英家教网—

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【题目】如图所示，两根相距为L＝1m的足够长的平行光滑金属导轨，位于水平的xOy平面内，一端接有阻值为R＝6Ω的电阻。在x>0的一侧存在垂直纸面向里的磁场，磁感应强度B只随t的增大而增大，且它们间的关系为B＝kt，其中k＝4T/s。一质量为m＝0．5kg的金属杆与金属导轨垂直，可在导轨上滑动，当t＝0时金属杆静止于x＝0处，有一大小可调节的外力F作用于金属杆，使金属杆以恒定加速度a＝2m/s2沿x轴正向做匀加速直线运动。除电阻R以外其余电阻都可以忽略不计。求：当t＝4s时施加于金属杆上的外力为多大。

【答案】513N

【解析】t＝4s时，金属杆移动的位移x＝at2＝×2×16m＝16m

此时的速度v＝at＝2×4m/s＝8m/s

切割产生的动生电动势E1＝BLv＝ktLv＝4×4×1×8V＝128V

磁场变化产生的感生电动势E2＝Lx＝4×1×16V＝64V

根据楞次定律和右手定则知，两个电动势的方向相同

则总电动势E＝E1＋E2＝192V

则感应电流的大小I＝＝32A

根据牛顿第二定律得，F－BIL＝ma

解得F＝BIL＋ma＝4×4×32×1＋0．5×2N＝513N

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