電磁感應(yīng)綜合問題

1、2007、3E = NtE = BLvBta db c a db c 31R2R2R833R =R931E=RI I22B3 rv1R105上海上海)abR由由式解得式解得a a1010(0.6(0.60.250.250.8)m/s0.8)m/s2 2=4m/s=4m/s2 2 (2)(2)設(shè)金屬棒運動達到穩(wěn)定時，速度為設(shè)金屬棒運動達到穩(wěn)定時，速度為v v，所受安培力為，所受安培力為F F，棒在沿導(dǎo)軌方向受力平衡棒在沿導(dǎo)軌方向受力平衡 mgsinmgsinmgcosmgcosF F0 0 此時金屬棒克服安培力做功的功率等于電路中電阻此時金屬棒克服安培力做功的功率等于電路中電阻R R消耗消耗的電

2、功率的電功率 FvFvP P 由由、兩式解得兩式解得P8v = 10m /sF0. 210 0. 6 - 0. 250. 8(3)(3)設(shè)電路中電流為設(shè)電路中電流為I I，兩導(dǎo)軌間金屬棒的長為，兩導(dǎo)軌間金屬棒的長為l l，磁場的，磁場的磁感應(yīng)強度為磁感應(yīng)強度為B I=BB I=Bl lv v/R /R P PI I2 2R R 由由、兩式解得兩式解得8 20.410 1PRBTvl磁場方向垂直導(dǎo)軌平面向上磁場方向垂直導(dǎo)軌平面向上 專題聚焦專題聚焦04廣東廣東1MNPQ2v0 0E = Bl(v -v)1 12EI=2R +R 2BIm g3l 2Pm gv4220122 2m gPm gv(R

3、R )B l1MNPQ2v0 2F1210P = P -I (R +R )-m gv4 g20122 2m gPm gv(RR )5B l1MNPQ2v0/(3 )FE lBR總/ba = F /maa = F /(2m )abF = 2F0avv =30b2vv =3aa0-F t= 2m (v - v )bbF t= m v2evm g = mRev=gR22de11m v =m v + m g2R22dv=5gR2dv= v=5gR12F = 2F110-F t= 2m v - 2m v22F t= m v12F = 2F03 5gRv=20= 2m v = 3m5gRI I2 21 1

4、E E= = B B( (l l - -l l) )v vEIR電流沿順時針方向。兩金屬桿都要受到安培力作用，電流沿順時針方向。兩金屬桿都要受到安培力作用，作用于桿作用于桿x x1 1y y1 1的安培力為的安培力為11fBl I22fBl I12120Fm gm gff1221()()FmmgIB ll122221()()FmmgvRBll12()Pmmgv12122221()()()FmmgPR mmgBll2QI R21221()()FmmgQRB ll1221Fm gm gff/2112P = (f - f)v = F-(m+m)g v122221()()FmmgvRBll2/1221()()FmmgQPRB llBU = S= SKttqUF = Eq =dqUdqUa =- gdm2211 qUy =at =(- g)t22 dm22011() ()22qUlyatgdmv201222()v dUmdKgSr qlqUa =- g = 0dm22Um gdK=Sr q2202222()v dmgdmdKgr qr ql2022()v dmdKgr ql 即即 20022()v dmdBBgtr qlP1212BLvE=R +RR +RI IILIL2222P PA A1212B L vB L vF=B =F=B =R + RR + R2QQQQ2M mvG-