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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 :When rod having length d is moving with velocity v perpendicular to MAGNETIC field EMF is induced between two end of rod. Induced emf `epsilon`= Bvd When switch is on at t = 0 time, current from inductor start increasing Suppose current in inductor is I at t = t time, ACCORDING to Kirchhoff.s second law, `epsilon=V_R+V_L` `Bvd=IR+L (dI)/(dt)` `therefore (dI)/(dt) +R/L I=(Bvd)/L` Solution of this linear differential equation, `I=(Bvd//L)/(R//L)+A "exp" (-R/L t)` `I=(Bvd)/R +Ae^(-R/L t)`...(1) Where A is constant that can be found by INITIAL condition , when t=0, I=0 `therefore 0=(Bvd)/R+A(1)` `A=(Bvd)/R` From equation (1), `I=(Bvd)/R-(Bvd)/R e^(-R/L t)` `I=(Bvd)/R(1-e^(-R/Lt))` which is required equation of current. |
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