如图所示,光滑的平行导轨倾角为θ,处在磁感应强度为B的匀强磁场

如图所示,光滑的平行导轨倾角为θ,处在磁感应强度为B的匀强磁场中,导轨中接入电动势为E、内阻为r的直流电源.电路中有一阻值为R的电阻,其余电阻不计,将质量为m、长度为L的导体棒由静止释放,求导体棒在释放瞬间的加速度的大小.

 

答案

a=gsinθ-.

解析:受力分析如图所示,导体棒受重力mg、支持力FN和安培力F,由牛顿第二定律:

mgsinθ-Fcosθ=ma   ①

F=BIL     ②

I=     ③

由①②③式可得

a=gsinθ-.

 

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