Q.1} Two parallel long smooth conducting rails separated by a distance ‘l ’ are connected by a movable conducting connector of mass m. Terminals of the rails are connected by the resistor R and a capacitor C (other end). A uniform magnetic field B perpendicular to the plane of rails is switched on (in to the paper). The connector is dragged by a force F. find the speed of the connector as function of time if the force is applied at t = 0. also find the terminal velocity of the connector. (Ans: v = FR/B2l2 (1- e^[- B2l2t / R(m+ B2l2 C]) )Q.2) A conducting rod of length l, resistance R & mass M is moved with a constant velocity on the two parallel rails whose one ends are at infinity and other is connected with capacitor C. The magnetic field B varies w.r.t time t as B=5t. at t=0, the area of the loop containing capacitor and the rod is zero and the capacitor is uncharged. The rod starts moving at t=0 on the fixed smooth conducting rails which have negligible resistance. Find:-a) The current in circuit as function of time t. (Ans: I = 10Clv {1- e^ -t/RC} )b) If the above system is kept in vertical plane such that the rod can move vertical plane such that rod moves vertically downward due to gravity and other parts are kept fixed & B= constant = B0, then find the maximum current in the circuit.Please sir, kindly reply these solutions as early as possible

aku -- kumar , 14 Years ago

Grade 12