如图所示,两根不计电阻的金属导线MN与PQ放在水平面内,MN是直导线,PQ的PQ1段是直导线,Q1Q2段是弧形导线,Q2Q3段是直导线,MN、PQ1、Q2Q3相互平行,M、P间接入一个阻值R=0
◎ 题目
如图所示,两根不计电阻的金属导线MN与PQ放在水平面内,MN是直导线,PQ的PQ1段是直导线,Q1Q2段是弧形导线,Q2Q3段是直导线,MN、PQ1、Q2Q3相互平行,M、P间接入一个阻值R=0.25Ω的电阻。一根质量为1.0 kg不计电阻的金属棒AB能在MN、PQ上无摩擦地滑动,金属棒始终垂直于MN,整个装置处于磁感应强度B=0.5T的匀强磁场中,磁场方向竖直向下。金属棒处于位置(I)时,给金属棒一个向右的速度v1=4 m/s,同时方向水平向右的外力F1 ="3" N作用在金属棒上使金属棒向右做匀减速直线运动;当金属棒运动到位置(Ⅱ)时,外力方向不变,大小变为F2,金属棒向右做匀速直线运动,经过时间t ="2" s到达位置(Ⅲ)。金属棒在位置(I)时,与MN、Q1Q2相接触于a、b两点,a、b的间距L1=1 m,金属棒在位置(Ⅱ)时,棒与MN、Q1Q2相接触于c、d两点。已知s1="7.5" m。求:(1)金属棒向右匀减速运动时的加速度大小? (2)c、d两点间的距离L2=? (3)外力F2的大小? (4)金属棒从位置(I)运动到位置(Ⅲ)的过程中,电阻R上放出的热量Q=? |
◎ 答案
(1)金属棒从位置(I)到位置(Ⅱ)的过程中,加速度不变,方向向左,设大小为a,在位置I时,a、b间的感应电动势为E1,感应电流为I1,受到的安培力为F安1,则 E1=BL1 v1,,F安1···································① F安1="4" N·································································② 根据牛顿第二定律得 F安1-F1 =ma·····························································③ a=" 1" m / s2·································································④ (2)设金属棒在位置(Ⅱ)时速度为v2,由运动学规律得 =-2a s1···························································⑤ v2=" 1" m / s·································································⑥ 由于在(I)和(II)之间做匀减速直线运动,即加速度大小保持不变,外力F1恒定,所以AB棒受到的安培力不变即F安1=F安2 ···························································⑦ m······················································⑧ (3)金属棒从位置(Ⅱ)到位置(Ⅲ)的过程中,做匀速直线运动,感应电动势大小与位置(Ⅱ)时的感应电动势大小相等,安培力与位置(Ⅱ)时的安培力大小相等,所以 F2= F安2="4" N······························································⑨ (4) 设位置(II)和(Ⅲ)之间的距离为s2,则 s2= v2t="2" m ································································⑩ 设从位置(I)到位置(Ⅱ)的过程中,外力做功为W1,从位置(Ⅱ)到位置(Ⅲ)的过程中,外力做功为W2,则 W1= F1 s1="22.5" J ···························································11 W2= F2 s2="8" J······························································12 根据能量守恒得W1+ W2·······························13· 解得Q =" 38" J ·····························································14 |
◎ 解析
略 |
◎ 知识点
专家分析,试题“如图所示,两根不计电阻的金属导线MN与PQ放在水平面内,MN是直导线,PQ的PQ1段是直导线,Q1Q2段是弧形导线,Q2Q3段是直导线,MN、PQ1、Q2Q3相互平行,M、P间接入一个阻值R=0…”主要考查了你对 【电磁感应现象的综合应用】 等知识点的理解和应用能力。关于这些知识点的“档案”,你可以点击相应的链接进行查看和学习。- 最新内容
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