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Find the current in the sliding rod AB (resistance = R) for the arrangement shown in figure. vecB is constant and is out of the paper. Parallel wires have no resistance, v is constant. Switch S is closed at time t = 0. |
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Answer» Solution :A rod of length d is moving with velocity v perpendicular to magnetic field B which produces emf between TWO ends of rod induced emf `epsilon=BVD` Now when SWITCH is on at t = 0, capacitor start charging, suppose charge on capacitor at t time is Q(t). Now according to Kirchhoff.s second law, `epsilon=V_R+V_C` `Bvd=IR+Q/C` `THEREFORE R=(dQ)/(dt)+Q/C=Bvd` `therefore (dQ)/(dt)+1/(RC)Q=(Bvd)/R` Now solution of this linear different equation, `Q=(Bvd//R)/(1//RC)+A "exp"(-1/"RC"t)` `Q=BvdC+Ae^(-t//RC)`...(1) where A is constant that can be found by initial CONDITION. At t=0, Q=0 `therefore ` 0=BvdC+A(1) `therefore` A=-BvdC From equation (1) `Q=BvdC-BvdC e^(-(t/(RC)))` `Q=BvdC-(1-e^(-(t/(RC)))` Induced charge `I=(dQ)/(dt)=BvdC [ 0-("-1"/"RC")e^(-(t/"RC"))]` `I=(BvdC)/(RC)e^(-(t)/(RC))` `I=(Bvd)/R e^(-(t/"RC"))` which is required equation of current. |
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