Question

矩形線圈abcd的長(zhǎng)ab=20 cm,寬bc=10 cm,匝數(shù)n=200,線圈總電阻R=5 Ω,整個(gè)線圈位于垂直于線圈平面的勻強(qiáng)磁場(chǎng)內(nèi),并保持靜止.

圖11

(1)若勻強(qiáng)磁場(chǎng)的磁感應(yīng)強(qiáng)度B隨時(shí)間的變化如圖11甲所示,求線圈的感應(yīng)電動(dòng)勢(shì)E及t=0.30 s時(shí)線圈ab邊所受的安培力;

(2)若勻強(qiáng)磁場(chǎng)的磁感應(yīng)強(qiáng)度B隨時(shí)間按正弦變化的規(guī)律如圖11乙所示,線圈1 min產(chǎn)生多少熱量?

Answer 1

解析：(1)E=nS=200××0.2×0.1 V=2 V                          ①

I==0.4 A                                                                  ②

0.30 s時(shí)磁感應(yīng)強(qiáng)度B=0.20 T

③所以此時(shí)ab邊所受安培力F=BIl_ab=0.20×0.4×0.2 N=0.016 N.                       ④

(2)由E_m=nBSω=nΦ_mω=200×0.2×0.2×0.1×V=80 π V⑤所以E=π V⑥Q=×60 J=3.79×10⁵ J.                                      ⑦

答案：（1）E=2 V  F=0.016 N（2）3.79×10⁵ J