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A pair of parallel horizontal conducting rails of negligible resistance, shorted at one end is fixed on a table. The distance between the rails is L. A conducting massless rod of resistance R can slide on the rails frictionlessly. The rod is tied to a massless string which passes over a pulley fixed to the edge of the table. A mass m, tied to the other end of the string, hangs vertically. A constant magnetic field B exists perpendicular to the table. If the system is released from rest, calculate : i) the terminal velocity achieved by the rod. ii) the acceleration of the mass at the instant when the velocity of the rod, is half the terminal velocity. |
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Answer» Solution :i) Let the VELOCITY of ROD = V Intensity of magnetic field = B ` therefore ` emf induced in rod (e) = BLV current induced in rod (i) = `(BLv)/(R )` Force on the rod F ` = BIL = (B^2 vL^2)/( R)` Net force on the system = MG - T mg- T = ma but `T = F = (B^2 vL^2)/(R ) = ma` hence `mg- (B^2 v L^2)/( R) =ma` or`a= g - (B^2 vL^2)/(mR)`........(i) For rod to achieve terminal velocity `V_T, a = 0` ` therefore 0 = g - (B^2 v_T L^2)/(mR)` or Terminal velocity`(v_T) = (mg R)/(B^2 L^2)`.....(ii) (ii) Acceleration of mass when `v = (v_T)/(2)` or `v = (mgR)/(2B^2 L^2)`.Put this value of v in (i) ` therefore a = g- (B^2 L^2)/(mR) xx ((mgR)/(2B^2 L^2)) " or " a = g - g/2` or`a = g/2`.......(iii) |
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