電磁感應(yīng)綜合應(yīng)用之--導(dǎo)軌類問題

1、-電磁感應(yīng)電磁感應(yīng)中的導(dǎo)軌類問題中的導(dǎo)軌類問題1.練習(xí)練習(xí) 如圖所示，兩光滑平行金屬導(dǎo)軌間距為L，直導(dǎo)線MN垂直跨在導(dǎo)軌上，且與導(dǎo)軌接觸良好，整個(gè)裝置處于垂直于紙面向里的勻強(qiáng)磁場中，磁感應(yīng)強(qiáng)度為B.電容器的電容為C，除電阻R外，導(dǎo)軌和導(dǎo)線的電阻均不計(jì)現(xiàn)給導(dǎo)線MN一初速度，使導(dǎo)線MN向右運(yùn)動，當(dāng)電路穩(wěn)定后，MN以速度v向右做勻速運(yùn)動時(shí)()A電容器兩端的電壓為零B電阻兩端的電壓為BLvC電容器所帶電荷量為CBLvD為保持MN勻速運(yùn)動，需對其施加的拉力大小為 v v0 0 2 2BB l vFBIlRr2 2()BFB l vamm Rr v v0 0 2 2()BFB l vamm Rr20102

2、mvQ00BIltmv 0mvqBlBlsqnRrRrRrQRQr F F BFFmgam BFBIlBlvBlRr2 2B l vRrv2 2()FB l vgmm Rr F F mFmgam2 2()()mFmgRrvB l BFFmgam 2 20()FB l vgmm Rr BlsqnRrRr F F 0mFtBLqmgtmv212EmFsQmgSmvBFFmgam2()mmmBLvFvmgvRr2 20()FB l vgmm Rr FB F F F N M 0QCEmQCUCBlv0mQQQCECBlvmmvBIltBl Q 2 2mBlCEvmB l C2 2mmBlCEImvmB

3、 l C安222 21()22()mm BlCEWmvmB l C安 v v0 0 CBlvUIR v v0 0 qCUUBlv0mvmvBIltBlq 2 20B l Cvvm v v0 0 1 1 2 2 21211212BlvBlvBl(vv )IRRRR012mBlvIRR v v0 0 1 1 2 2 2 22112BB l (vv )FBIlRR v v0 0 1 1 2 2 2 012m v(mm )v共21222011m v(mm )vQ22共1122QRQR v v0 0 1 1 2 2 v v2 2 1 1 2 2 v v1 1 212211RRvBlvBlI1 122Bl

4、 vBl v121122FBIllFBIll112212IFlIFl11 01 1Imvmv2220Im v1 122Bl vBl v21 2221 22 110mlvvmlm l1 2 1221 22 120ml lvvmlm l1122QRQR2221 01 122111222mvmvm vQ2202Bl qm v1221BlvBlvIRRBFBIl11BFam22BFFam12F(mm )a1BFm aBFBIl2112Bl(vv )IRR121212 212(RR )m FvvB l (mm )111vvat 222vva t 11122212Bl (vat )Bl (va t )IR

5、R1 1221 12212B(l vl v )B(l al a )tRR1 122l al a111BFFam222BFam1122BBFlFl2121221221Bl mFFl ml m1 122l al a111BFFam222BFam1122BBFlFl1 22221221l laFl ml m221221221laFl ml m1 222221221Bl l mFFl ml m12221221l mFIl ml mB222814LBRFtavvtavvcdab2211,RtaavvBLREI5)4()4(51212111amBIL 222amBILFmFa982mFa921BLFI9。BFmgam BFBIl(EEBlRr反）(BElvBlRr）(B()ElvBlgm Rr）= mEIRr,mmFBI lmmFmgammin,mEBlvIRrlrRBlvEBmminminmgFBIl22)(lBrRmgBlEvm minmin()2minmIEIEIRrmgv反0mBLqmgtmv212EmqEQmgSmvBFmgam(B()ElvBlgm Rr）=BlsqnRrRr B