电磁感应综合问题

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)设金属棒运动达到稳定时，速度为设金属棒运动达到稳定时，速度为v v，所受安培力为，所受安培力为F F，棒在沿导轨方向受力平衡棒在沿导轨方向受力平衡 mgsinmgsinmgcosmgcosF F0 0 此时金属棒克服安培力做功的功率等于电路中电阻此时金属棒克服安培力做功的功率等于电路中电阻R R消耗消耗的电

2、功率的电功率 FvFvP P 由由、两式解得两式解得P8v = 10m /sF0. 210 0. 6 - 0. 250. 8(3)(3)设电路中电流为设电路中电流为I I，两导轨间金属棒的长为，两导轨间金属棒的长为l l，磁场的，磁场的磁感应强度为磁感应强度为B I=BB I=Bl lv v/R /R P PI I2 2R R 由由、两式解得两式解得8 20.410 1PRBTvl磁场方向垂直导轨平面向上磁场方向垂直导轨平面向上 专题聚焦专题聚焦04广东广东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总/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电流沿顺时针方向。两金属杆都要受到安培力作用，电流沿顺时针方向。两金属杆都要受到安培力作用，作用于杆作用于杆x x1 1y y1 1的安培力为的安培力为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-