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【題目】如圖所示，兩根相距為L＝1m的足夠長的平行光滑金屬導軌，位于水平的xOy平面內，一端接有阻值為R＝6Ω的電阻。在x>0的一側存在垂直紙面向里的磁場，磁感應強度B只隨t的增大而增大，且它們間的關系為B＝kt，其中k＝4T/s。一質量為m＝0．5kg的金屬桿與金屬導軌垂直，可在導軌上滑動，當t＝0時金屬桿靜止于x＝0處，有一大小可調節(jié)的外力F作用于金屬桿，使金屬桿以恒定加速度a＝2m/s2沿x軸正向做勻加速直線運動。除電阻R以外其余電阻都可以忽略不計。求：當t＝4s時施加于金屬桿上的外力為多大。

【答案】513N

【解析】t＝4s時，金屬桿移動的位移x＝at2＝×2×16m＝16m

此時的速度v＝at＝2×4m/s＝8m/s

切割產生的動生電動勢E1＝BLv＝ktLv＝4×4×1×8V＝128V

磁場變化產生的感生電動勢E2＝Lx＝4×1×16V＝64V

根據楞次定律和右手定則知，兩個電動勢的方向相同

則總電動勢E＝E1＋E2＝192V

則感應電流的大小I＝＝32A

根據牛頓第二定律得，F－BIL＝ma

解得F＝BIL＋ma＝4×4×32×1＋0．5×2N＝513N

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